Selecta

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schinzel_vol_1_titelei

5.3.2007

18:32 Uhr

Seite 1

Heritage of European Mathematics

Advisory Board Michèle Audin, Strasbourg Ciro Ciliberto, Roma Ildar A. Ibragimov, St. Petersburg Wladyslaw Narkiewicz, Wroclaw Peter M. Neumann, Oxford Samuel J. Patterson, Göttingen


5.3.2007

18:33 Uhr

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Andrzej Schinzel in Cambridge (England), 1961


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Andrzej Schinzel

Selecta Volume I Diophantine Problems and Polynomials

Edited by Henryk Iwaniec Wladyslaw Narkiewicz Jerzy Urbanowicz


5.3.2007

18:33 Uhr

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Author: Andrzej Schinzel Institute of Mathematics Polish Academy of Sciences ul. Śniadeckich 8, skr. poczt. 21 00-956 Warszawa 10 Poland

Editors: Henryk Iwaniec Department of Mathematics Rutgers University New Brunswick, NJ 08903 U.S.A. [email protected]

Władysław Narkiewicz Institute of Mathematics University of Wrocław pl. Grunwaldzki 2/4 50-384 Wrocław Poland [email protected]

Jerzy Urbanowicz Institute of Mathematics Polish Academy of Sciences ul. Śniadeckich 8, skr. poczt. 21 00-956 Warszawa 10 Poland [email protected]

2000 Mathematics Subject Classification: 11, 12

ISBN 978-3-03719-038-8 (Set Vol I & Vol II) The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.helveticat.ch. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2007 European Mathematical Society Contact address: European Mathematical Society Publishing House Seminar for Applied Mathematics ETH-Zentrum FLI C4 CH-8092 Zürich Switzerland Phone: + 41 (0)44 632 34 36 Email: [email protected] Homepage: www.ems-ph.org Printed in Germany 987654321

Preface Andrzej Bobola Maria Schinzel, born on April 5, 1937 at Sandomierz (Poland), is well known for his original results in various areas of number theory appearing in over 200 research papers, of which the first thirty were published while he was an undergraduate at the Warsaw University. Working under the guidance of Wacław Sierpiński, he became interested in elementary number theory, and the subjects of his early papers range from properties of arithmetical functions, like Euler’s ϕ-function or the number of divisors, to Diophantine equations. Paul Erd˝os, a big champion of elementary number theory, wrote in his letter to Sierpiński of October 23, 1960—when Andrzej was studying at the University of Cambridge under the supervision of Harold Davenport—“Schinzel’s completion of my proof is much simpler than anything I had in mind”. Many mathematicians cooperating with Andrzej Schinzel could repeat the words of Erd˝os. Since completing his studies at Warsaw University in 1958, Andrzej Schinzel has been employed by the Institute of Mathematics of the Polish Academy of Sciences, where he obtained his Ph.D. in 1960. On his return from a Rockefeller Foundation Fellowship at the University of Cambridge and the University of Uppsala (where he studied under Trygve Nagell) he completed his habilitation in 1962. In 1967 he was promoted to Associate Professor, and in 1974 to Full Professor. In 1979 he was elected to Corresponding Member of the Polish Academy of Sciences and in 1994 to Full Member. Andrzej Schinzel’s very first paper (1 )—published at the age of 17—is a postscript to a result of H.-E. Richert who proved a general theorem about partitions of integers into distinct summands from a given set which implied in particular that every integer > 33 is a sum of distinct triangular numbers. Schinzel observed that every integer > 51 is a sum of at most four distinct triangular numbers. The favorite subject of the early research of Schinzel, Euler’s totient function, is considered here in five papers; the earliest one, published in 1954, is not included(2 ). In 1958 a joint work J1 with Sierpiński appeared, analyzing various consequences of the conjecture stating that if f1 , . . . , fs ∈ Z[x] are irreducible polynomials having positive leading coefficients and there is no natural number > 1 that is a divisor of each of the numbers f1 (n) · · · fs (n) for n being an integer then for infinitely many natural n the values f1 (n), . . . , fs (n) are primes. This celebrated conjecture—with many unexpected consequences—is called “Schinzel’s Hypothesis H”. Schinzel’s doctoral thesis B1 dealt with the period of a class of continued fractions and was related (see B2) to a question concerning pseudo-elliptic integrals, considered already (1 )

Sur la décomposition des nombres naturels en sommes de nombres triangulaires distincts, Bull. Acad. Polon. Sci. Cl. III 2 (1954), 409–410. (2 ) Sur quelques propriétés des fonctions ϕ(n) et σ (n), Bull. Acad. Polon. Sci. Cl. III 2 (1954), 463–466 (with W. Sierpiński).

vi

Preface

by N. H. Abel in the very first volume of Crelle’s Journal. In his habilitation thesis— consisting of four papers I1, I2, I3 and a paper not included (3 )—Schinzel generalized a classical theorem of Zsigmondy of 1892 (often called the Birkhoff–Vandiver theorem) on primitive divisors. The central theme of Schinzel’s work is arithmetical and algebraic properties of polynomials in one or several variables, in particular questions of irreducibility and zeros of polynomials. To this topic he devoted about one-third of his papers and two books (4 ). In the books Schinzel presented several classical results and included many extensions, improvements and generalizations of his own. Undoubtedly Schinzel and his beloved journal Acta Arithmetica influenced many mathematicians, stimulating their thinking and mathematical careers. Andrzej Schinzel has since 1969 been the editor of this first international journal devoted exclusively to number theory, being a successor of his teacher W. Sierpiński. Among the other editors of Acta Arithmetica during these years were/are J. W. S. Cassels, H. Davenport, P. Erd˝os, V. Jarník, J. Kaczorowski, Yu. V. Linnik, L. J. Mordell, W. M. Schmidt, V. G. Sprindzhuk, R. Tijdeman and P. Turán. These people and the other outstanding mathematicians from the advisory board of Acta Arithmetica have determined the line of the journal. Andrzej Schinzel’s work has been influential in the development of many areas of mathematics, and his 70th birthday gives us an opportunity to honor his accomplishments by putting together his most important papers. This selection of Schinzel’s papers—published during more than five decades—is divided into two volumes containing 100 articles. We have asked some outstanding mathematicians for commentaries to the selected papers. Also included is a list of unsolved problems and unproved conjectures proposed by Schinzel in the years 1956–2006, arranged chronologically. The first volume covers six themes: A. Diophantine equations and integral forms (with commentaries by Robert Tijdeman) B. Continued fractions (with commentaries by Eugène Dubois) C. Algebraic number theory (with commentaries by David W. Boyd and Donald J. Lewis) D. Polynomials in one variable (with commentaries by Michael Filaseta) E. Polynomials in several variables (with commentaries by Umberto Zannier) F. Hilbert’s Irreducibility Theorem (with commentaries by Umberto Zannier) The second volume contains papers covering seven themes: G. H. I. J. K. L. M.

Arithmetic functions (with commentaries by Kevin Ford) Divisibility and congruences (with commentaries by Hendrik W. Lenstra, Jr.) Primitive divisors (with commentaries by Cameron L. Stewart) Prime numbers (with commentaries by Jerzy Kaczorowski) Analytic number theory (with commentaries by Jerzy Kaczorowski) Geometry of numbers (with commentaries by Wolfgang M. Schmidt) Other papers (with commentaries by Stanisław Kwapień and Endre Szemerédi)

(3 )

The intrinsic divisors of Lehmer numbers in the case of negative discriminant, Ark. Mat. 4 (1962), 413–416. Selected Topics on Polynomials, XXII+250 pp., University of Michigan Press, Ann Arbor 1982, and Polynomials with Special Regard to Reducibility, X+558 pp., Encyclopaedia of Mathematics and its Applications 77, Cambridge Univ. Press, Cambridge 2000.

(4 )

Preface

vii

Many people helped with the editing of the volumes. First of all, we gratefully thank the authors of the commentaries we had the pleasure to work with. Second, our special thanks go to Stanisław Janeczko, Director of the Institute of Mathematics, Polish Academy of Sciences, for his support, and to Manfred Karbe, Publishing Director of the European Mathematical Society Publishing House, for his invaluable assistance during the work on the Selecta. Third, we wish to thank Jerzy Browkin for reading the papers and some corrections, and Jan K. Kowalski for retyping the papers and offering valuable suggestions for improving the presentation of the material. Finally, we wish to express our gratitude to the staff of the European Mathematical Society Publishing House, especially to Irene Zimmermann, for the very pleasant cooperation. We have decided to unify some notations, using the “blackboard bold” type for most common sets; so C, R, Q, Qp and Fp always stand for the complex, real, rational, p-adic and finite fields respectively; Z, Zp for the rings of integers and p-adic integers; and N, N0 for the sets of positive and nonnegative integers. The greatest common divisor of integers a1 , a2 , . . . , an is denoted by (a1 , a2 , . . . , an ). If (a1 , a2 , . . . , an ) = 1, the integers are called relatively prime, and if (ai , aj ) = 1 for any 1 i = j n, the integers are called coprime. As usual, for x ∈ R set [x] = max{n ∈ Z : n x}, x = min{n ∈ Z : x n} and x = min{x − [x], x − x}. We denote by {x} the fractional part of x. Lines where minor corrections of the original text have been made are marked “c” in the left margin. March 2007

Henryk Iwaniec Władysław Narkiewicz Jerzy Urbanowicz

Contents

Volume 1 A.

A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 A12 A13 A14 A15 A16

Diophantine equations and integral forms Commentary on A: Diophantine equations and integral forms by R. Tijdeman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sur les nombres de Mersenne qui sont triangulaires avec Georges Browkin . . . . . . . . . . . . . . . . . . . . . . . . . . Sur quelques propriétés des nombres 3/n et 4/n, où n est un nombre impair Sur l’existence d’un cercle passant par un nombre donné de points aux coordonnées entières . . . . . . . . . . . . . . . . . . . . . . . . . . . Sur les sommes de trois carrés . . . . . . . . . . . . . . . . . . . . . . . . On the Diophantine equation nk=1 Ak xkϑk = 0 . . . . . . . . . . . . . . . Polynomials of certain special types with H. Davenport and D. J. Lewis . . . . . . . . . . . . . . . . . . . . An improvement of Runge’s theorem on Diophantine equations . . . . . . On the equation y m = P (x) with R. Tijdeman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Zeta functions and the equivalence of integral forms with R. Perlis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Quadratic Diophantine equations with parameters with D. J. Lewis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Selmer’s conjecture and families of elliptic curves with J. W. S. Cassels . . . . . . . . . . . . . . . . . . . . . . . . . . . Families of curves having each an integer point . . . . . . . . . . . . . . . Hasse’s principle for systems of ternary quadratic forms and for one biquadratic form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On Runge’s theorem about Diophantine equations with A. Grytczuk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On sums of three unit fractions with polynomial denominators . . . . . . . On equations y 2 = x n + k in a finite field with M. Skałba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 3 11 13 17 18 22 27 36 41 47 54 62 67 87 93 116 124

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B.

Continued fractions Commentary on B: Continued fractions by Eugène Dubois . . . . . . . . . . . . . . . . . . . . . . . . . B1 On some problems of the arithmetical theory of continued fractions . B2 On some problems of the arithmetical theory of continued fractions II B3 On two conjectures of P. Chowla and S. Chowla concerning continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

127 . . . . . . . . .

129 131 149

. . .

161

C. Algebraic number theory Commentary on C: Algebraic numbers by David W. Boyd and D. J. Lewis . . . . . . . . . . . . . . . . . . . . C1 A refinement of two theorems of Kronecker with H. Zassenhaus . . . . . . . . . . . . . . . . . . . . . . . . . . . . C2 On a theorem of Bauer and some of its applications . . . . . . . . . . . . C3 An extension of the theorem of Bauer and polynomials of certain special types with D. J. Lewis and H. Zassenhaus . . . . . . . . . . . . . . . . . . . C4 On sums of roots of unity. (Solution of two problems of R. M. Robinson) . C5 On a theorem of Bauer and some of its applications II . . . . . . . . . . . C6 On the product of the conjugates outside the unit circle of an algebraic number C7 On linear dependence of roots . . . . . . . . . . . . . . . . . . . . . . . . C8 On Sylow 2-subgroups of K2 OF for quadratic number fields F with J. Browkin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C9 A class of algebraic numbers . . . . . . . . . . . . . . . . . . . . . . . . C10 On values of the Mahler measure in a quadratic field (solution of a problem of Dixon and Dubickas) . . . . . . . . . . . . . . . . . . . . . . . . . D.

Polynomials in one variable Commentary on D: Polynomials in one variable by Michael Filaseta . . . . . . . . . . . . . . . . . . . . . . . . . . D1 Solution d’un problème de K. Zarankiewicz sur les suites de puissances consécutives de nombres irrationnels . . . . . . . . . . . . . . . . D2 On the reducibility of polynomials and in particular of trinomials . . . D3 Reducibility of polynomials and covering systems of congruences . . . D4 Reducibility of lacunary polynomials I . . . . . . . . . . . . . . . . . D5 Reducibility of lacunary polynomials II . . . . . . . . . . . . . . . . . D6 A note on the paper “Reducibility of lacunary polynomials I” with J. Wójcik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D7 Reducibility of lacunary polynomials III . . . . . . . . . . . . . . . . D8 Reducibility of lacunary polynomials IV . . . . . . . . . . . . . . . . D9 On the number of terms of a power of a polynomial . . . . . . . . . . D10 On reducible trinomials . . . . . . . . . . . . . . . . . . . . . . . . .

167 169 175 179 190 197 210 221 238 253 264 272 281

. .

283

. . . . .

. . . . .

295 301 333 344 381

. . . . .

. . . . .

403 409 447 450 466

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D11 On a conjecture of Posner and Rumsey with K. Gy˝ory . . . . . . . . . . . . . . . . . . . . . . . . . D12 Reducibility of lacunary polynomials XII . . . . . . . . . . . . D13 On reducible trinomials II . . . . . . . . . . . . . . . . . . . . D14 On reducible trinomials III . . . . . . . . . . . . . . . . . . . . D15 On the greatest common divisor of two univariate polynomials I D16 On the greatest common divisor of two univariate polynomials II D17 On the reduced length of a polynomial with real coefficients . . E.

E1 E2 E3 E4 E5 E6 E7 E8 E9 E10

. . . . .

. . . . . . . .

Polynomials in several variables Commentary on E: Polynomials in several variables by Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . . Some unsolved problems on polynomials . . . . . . . . . . . . . . Reducibility of polynomials in several variables . . . . . . . . . . Reducibility of polynomials of the form f (x) − g(y) . . . . . . . Reducibility of quadrinomials with M. Fried . . . . . . . . . . . . . . . . . . . . . . . . . . . A general irreducibility criterion . . . . . . . . . . . . . . . . . . . Some arithmetic properties of polynomials in several variables with H. L. Montgomery . . . . . . . . . . . . . . . . . . . . . . On difference polynomials and hereditarily irreducible polynomials with L. A. Rubel and H. Tverberg . . . . . . . . . . . . . . . . On a decomposition of polynomials in several variables . . . . . . On weak automorphs of binary forms over an arbitrary field . . . . Reducibility of symmetric polynomials . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

549 563 580 605 632 646 658 693

. . . .

. . . .

695 703 709 715

. . . . . . . .

720 739

. . . .

747

. . . .

755 760 779 828

Hilbert’s Irreducibility Theorem Commentary on F: Hilbert’s Irreducibility Theorem by Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . . . F1 On Hilbert’s Irreducibility Theorem . . . . . . . . . . . . . . . . . . F2 A class of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . F3 The least admissible value of the parameter in Hilbert’s Irreducibility Theorem with Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

F.

835 . . . . . . . . .

837 839 846

. . .

849

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Contents

Volume 2 G. Arithmetic functions Commentary on G: Arithmetic functions by Kevin Ford . . . . . . . . . . . . . . . . . . . . . . G1 On functions ϕ(n) and σ (n) . . . . . . . . . . . . . . . . G2 Sur l’équation ϕ(x) = m . . . . . . . . . . . . . . . . . . G3 Sur un problème concernant la fonction ϕ(n) . . . . . . . G4 Distributions of the values of some arithmetical functions with P. Erd˝os . . . . . . . . . . . . . . . . . . . . . . G5 On the functions ϕ(n) and σ (n) with A. Makowski . . . . . . . . . . . . . . . . . . . . G6 On integers not of the form n − ϕ(n) with J. Browkin . . . . . . . . . . . . . . . . . . . . .

859 . . . .

. . . .

. . . .

. . . .

861 866 871 875

. . . . . . . . .

877

. . . . . . . . .

890

. . . . . . . . .

895

H. Divisibility and congruences Commentary on H: Divisibility and congruences by H. W. Lenstra jr. . . . . . . . . . . . . . . . . . . . . . . H1 Sur un problème de P. Erd˝os . . . . . . . . . . . . . . . . . . . H2 On the congruence a x ≡ b (mod p) . . . . . . . . . . . . . . . H3 On the composite integers of the form c(ak + b)! ± 1 . . . . . H4 On power residues and exponential congruences . . . . . . . . H5 Abelian binomials, power residues and exponential congruences H6 An extension of Wilson’s theorem with G. Baron . . . . . . . . . . . . . . . . . . . . . . . . . H7 Systems of exponential congruences . . . . . . . . . . . . . . H8 On a problem in elementary number theory with J. Wójcik . . . . . . . . . . . . . . . . . . . . . . . . . H9 On exponential congruences . . . . . . . . . . . . . . . . . . . H10 Une caractérisation arithmétique de suites récurrentes linéaires avec Daniel Barsky et Jean-Paul Bézivin . . . . . . . . . . . H11 On power residues with M. Skałba . . . . . . . . . . . . . . . . . . . . . . . . I.

I1 I2 I3 I4 I5

. . . .

. . . .

. . . .

. . . .

. . . .

899 . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

901 903 909 912 915 939

. . . . . . . . . . . .

971 975

. . . . . . . . . . . .

987 996

. . . . . . 1001 . . . . . . 1012

Primitive divisors Commentary on I: Primitive divisors by C. L. Stewart . . . . . . . . . . . . . . . . . . . . . . . . . . . . On primitive prime factors of a n − bn . . . . . . . . . . . . . . . . . . On primitive prime factors of Lehmer numbers I . . . . . . . . . . . . On primitive prime factors of Lehmer numbers II . . . . . . . . . . . . On primitive prime factors of Lehmer numbers III . . . . . . . . . . . Primitive divisors of the expression An − B n in algebraic number fields

1031 . . . . . .

. . . . . .

1033 1036 1046 1059 1066 1090

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Contents

I6 J.

J1 J2 J3 J4 J5

An extension of the theorem on primitive divisors in algebraic number fields 1098 Prime numbers Commentary on J: Prime numbers by Jerzy Kaczorowski . . . . . . . . . . . . . . . . . . . . . . . . . . Sur certaines hypothèses concernant les nombres premiers with W. Sierpiński . . . . . . . . . . . . . . . . . . . . . . . . . . . . Remarks on the paper “Sur certaines hypothèses concernant les nombres premiers” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A remark on a paper of Bateman and Horn . . . . . . . . . . . . . . . . On two theorems of Gelfond and some of their applications Section 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On the relation between two conjectures on polynomials . . . . . . . . .

K. Analytic number theory Commentary on K: Analytic number theory by Jerzy Kaczorowski . . . . . . . . . . . . . . . . . . . . . . K1 On Siegel’s zero with D. M. Goldfeld . . . . . . . . . . . . . . . . . . . . . . K2 Multiplicative properties of the partition function with E. Wirsing . . . . . . . . . . . . . . . . . . . . . . . . . K3 On an analytic problem considered by Sierpiński and Ramanujan K4 Class numbers and short sums of Kronecker symbols with J. Urbanowicz and P. Van Wamelen . . . . . . . . . . . . L.

L1 L2 L3 L4

Geometry of numbers Commentary on L: Geometry of numbers by Wolfgang M. Schmidt . . . . . . . . . . . . . . . . . . . A decomposition of integer vectors II with S. Chaładus . . . . . . . . . . . . . . . . . . . . . . . A decomposition of integer vectors IV . . . . . . . . . . . . . A property of polynomials with an application to Siegel’s lemma On vectors whose span contains a given linear subspace with I. Aliev and W. M. Schmidt . . . . . . . . . . . . . . .

1103 . 1105 . 1113 . 1134 . 1142 . 1145 . 1154 1193

. . . . . 1195 . . . . . 1199 . . . . . 1211 . . . . . 1217 . . . . . 1224 1245

. . . . . . 1247 . . . . . . 1249 . . . . . . 1259 . . . . . 1274 . . . . . . 1288

M. Other papers 1303 Commentary on M: Other papers by Stanisław Kwapień . . . . . . . . . . . . . . . . . . . . . . . . . . 1305 The influence of the Davenport–Schinzel paper in discrete and computational geometry by Endre Szemerédi . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1311 M1 Sur l’équation fonctionnelle f [x + y · f (x)] = f (x) · f (y) avec S. Gołab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1314

xiv M2 M3 M4 M5

Contents

A combinatorial problem connected with differential equations with H. Davenport . . . . . . . . . . . . . . . . . . . . . . An analogue of Harnack’s inequality for discrete superharmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . An inequality for determinants with real entries . . . . . . . . . Comparison of L1 - and L∞ -norms of squares of polynomials with W. M. Schmidt . . . . . . . . . . . . . . . . . . . . . .

. . . . . . 1327 . . . . . . 1338 . . . . . . 1347 . . . . . . 1350

Unsolved problems and unproved conjectures Unsolved problems and unproved conjectures proposed by Andrzej Schinzel in the years 1956–2006 arranged chronologically Publication list of Andrzej Schinzel

1365 . . . . 1367 1375

Part A Diophantine equations and integral forms

Andrzej Schinzel Selecta

Commentary on A: Diophantine equations and integral forms by R. Tijdeman

A1. In 1960 the authors published a paper [12] in which they remark that the equation investigated in their 1956 paper is equivalent with the equation 2n − 7 = y 2 . Already in 1915 Ramanujan claimed that the latter equation in positive integers n, x has only solutions for n = 3, 4, 5, 7, 15. This result follows immediately from the paper by Browkin and Schinzel and had been confirmed earlier by Nagell [34], however by a different method. For a survey of the further developments up to 1966 see Hasse [27]. In the 1960 paper the authors used the method from their 1956 paper to prove results of the following type: if D ≡ 0, 4, 7 (mod 8), then the equation (1)

2n − D = y 2

has one solution at the most. If any solution exists, then n 2. They conjectured that if D > 0 and D = 7, 23, 2k − 1 (k ∈ N) then (1) has at most one solution. The conjecture was established by Beukers [6]. By use of√hypergeometric functions he first gave a good lower bound for the approximation to 2 by rationals whose denominators are a power of two. From this Beukers obtained an explicit upper bound n < 435 + 10(log |D|/ log 2) for any solution (y, n) of (1) and subsequently he proved the Browkin–Schinzel conjecture. He also dealt with the case D < 0 and showed that there are at most four solutions. M. Le [31] sharpened this result by proving among other results that if D is not of the form 22m − 3 · 2m+1 + 1, then there are at most three solutions. In a second paper [7] Beukers extended his investigations to the equation y 2 −D = pn , where D is a positive integer and p is an odd prime not dividing D. He showed that there are at most four solutions and gave a family of such equations having three solutions. That there are at most three solutions was proved much later by Bauer and Bennett [3]. For related later work, see e.g. Bender and Herzberg [4], Le [32], Yuan [62] and Bugeaud and Shorey [14]. Many papers have been written on individual Diophantine equation of Ramanujan– Nagell type (1). The theory on linear forms in logarithms of algebraic numbers made it possible to treat whole classes of such equations at the same time. By first applying this theory, next the basis reduction algorithm of Lenstra, Lenstra and Lovász, and finally the Fincke–Pohst algorithm for finding short lattice vectors, de Weger [60] extended the

4

A. Diophantine equations and integral forms

theory into two directions. In the first place he studied for fixed integer k and primes p1 , . . . , ps the generalized Ramanujan–Nagell equation y 2 + k = p1z1 · · · pszs in y ∈ N, 0 z1 , . . . , zs ∈ Z. As an illustration of his method he computed all 16 nonnegative numbers y such that y 2 + 7 has no prime divisors larger than 20 explicitly, the largest being y = 273. Secondly he considered the equation x + z = y 2 in integers x, y, z with x > z, y > 0, x and |z| composed of fixed primes. As an illustration he computed all 388 solutions where x and |z| are composed of the primes 2, 3, 5 and 7, the largest solution being 199290375 − 686 = 141172 . For further developments into this direction see e.g. Smart [48] and Wildanger [61]. A2. Suryanarayana and Rao [50] gave another solution to the problem of writing 3/(2n + 1) for any integer n > 1 as the sum of reciprocals of three distinct odd positive integers. They also showed that 2/(2n + 1) can be expressed as the sum of two such reciprocals if and only if 2n + 1 is not a prime ≡ 3 (mod 4). The result of paper A2 has been recently rediscovered by Hagedorn [26]. See also Sierpiński [44]. A3. The result of this paper is presented in the book by Honsberger [29]. It has been generalized by Kulikowski [30] to the corresponding problem in higher dimensions. A4. Schinzel’s main result in this paper was later applied by Schinzel and Sierpiński [42] to prove that every sufficiently large integer is a sum of four powers a b with integers a > 1, b > 1. The hypothesis that the restriction n < 101200 in Corollaire 3 can be dropped was earlier made in E. Grosswald, A. Calloway, and J. Calloway [24] who proved that there exists a finite set S of numbers with the following property: if n > 0 is not divisible by 4, not ≡ −1 (mod 8) and not in S, then n is the sum of three strictly positive squares. See further Mordell [33] and, for an application of such results to the non-relativistic quantum statistical mechanics of an ideal gas at low temperature, Baltes, Draxl and Hilf [2]. A5. An immediate consequence of Corollary 2 is that the Diophantine equation x 2 + y m = z2n has infinitely many primitive solutions if (m, n) = 1. This observation was made by Schinzel after D. W. Boyd had proved this fact in another way. See Boyd [10]. For later work on equations x l + y m = zn see Darmon and Granville [19] who showed that there are only finitely many integer solutions in x, y, z with gcd(x, y, z) = 1 provided that 1/ l + 1/m + 1/n < 1 by using descent arguments and Faltings’ theorem, remarked that there are no solutions at all if 1/ l + 1/m + 1/n = 1, and for each case with 1/ l + 1/m + 1/n > 1 wrote down a parametrization that gives rise to infinitely many non-zero coprime integer solutions (when one plugs in integers for the parameters). For further work see Beukers [8] and Edwards [22]. A6. Theorem 1 of this paper was generalized by Schinzel in F1 to the case of polynomials in several variables. An immediate consequence of Theorem 1 of the above paper is that if f ∈ Q[x] is an integer-valued polynomial such that every arithmetic progression contains an integer a for which f (a) is a kth power, then f = g k , where g ∈ Q[x] is integer-valued. Berstel [5] extended this under suitable conditions to rational functions.

Diophantine equations and integral forms

5

Perelli and Zannier [35] gave the following related version. Let f (x) be a polynomial with integral coefficients. Assume that every arithmetic progression contains an integer x such that f (x) = A(x)y k(x) , where A(x), y are integers, k(x) k0 , and the prime divisors of A(x) belong to a finite set S. Then f (x) = A(P (x))k identically, where A is an integral constant, P (x) is a polynomial with integral coefficients and k k0 . S. Chowla [16] proved the following variant of the Corollary to Theorem 2 in paper A6: If g(x) is a sum of two squares for every sufficiently large integer x, then there exist two polynomials P1 (x) and P2 (x) with integer coefficients such that g(x) = P1 (x)2 + P2 (x)2 for all x. In a subsequent paper [20] the three authors extended the Corollary to Theorem 2 as follows. Let F (x, y, t) be any polynomial with integral coefficients which is of degree at most two in x and y. Suppose that every arithmetic progression contains an integer t such that the equation F (x, y, t) = 0 is soluble in rationals x, y. Then there exist rational functions x(t) and y(t) with rational coefficients such that F (x(t), y(t), t) = 0 identically in t. Far reaching generalizations of results in the paper by Davenport, Lewis and Schinzel can be found in Theorems 51 and 57, and Corollary 4 in Section 5.6 of Schinzel [41]. A7. A refinement of Schinzel’s result was given by Ayad [1] who gave explicit forms of the Puiseux expansions corresponding to the places at infinity of the associated function field. The “only if” part of the following statement for n = 2 is a direct consequence of Schinzel’s above paper. Let C/Q be an irreducible affine curve of geometric genus 0 and fix an embedding of C into An so that the ideal of C is generated by polynomials with integer coefficients. Further, let C∞ be the set of points “at infinity” on the projective closure C of C in Pn , and let ∞ be the points “at infinity” on the desingularization of C. Assume that C(Z) contains at least one nonsingular point. Then C(Z) is infinite if and only if one of the following two conditions is satisfied: (a) ∞ consists of a single point; or (b) ∞ consists of exactly two points which are conjugate over a real quadratic field. The general statement was proved by Silverman [46]. The “if” part was proved by Poulakis [38] who corrected an error of Silverman who formulated the conditions in terms of C∞ , instead of in terms of ∞ . Counterexamples for Silverman’s statement involving ∞ can be found in the review MR 2001h:11080. For more information on Runge’s method see A14 and the added notes. A8. The result in this paper was generalized by Shorey, van der Poorten, Tijdeman, and Schinzel in [43] as follows. Let S be the set of all non-zero integers composed of primes from some fixed finite set. Let f ∈ Q[x, y] be a binary form with f (1, 0) = 0 such that among the linear factors in the factorization of f at least two are distinct. Let d be a positive integer. Then the equation wzq = f (x, y) in integers q, w, x, y, z with w ∈ S, y ∈ S, (x, y) = d, |z| > 1 implies q < C where C is an effectively computable constant depending only on f, d and S. As an application they proved the following generalization of a result of Mahler about the greatest prime factor of ax n + by q tending to ∞ if max(|x|, |y|) → ∞ with (x, y) = 1. Let n ∈ Z, n > 1. Let S be as above. The equation ux n + vy q = w in non-zero integers q > 1, u ∈ S, v ∈ S, w ∈ S, x ∈ Z, y ∈ Z with |x| > 1, |y| > 1, (ux, w) = 1 and nq 6 has only finitely many solutions.

6


Quantitative versions of the Schinzel–Tijdeman theorem were given by Turk [53]. He proved for example: Let a ∈ Z be non-zero and let F ∈ Z[X] have at least two distinct zeros, degree n and height H . Suppose F (x) = ay m for some x, y ∈ Z with |y| > 1. Then Cn5 (log(3H ))2 (log 3|a|)(log log 3|a|)2 , m < exp log(n log 3H ) where C is an absolute constant. Later Pink [36] gave new explicit lower bounds for the difference |F (x) − by m | where F (X) is an integer polynomial of degree n 2 and x, y, b, m are non-zero integers with m 2, |y| 2 and F (x) = by m . Various results on the representation of powers by polynomials with coefficients from an algebraic number field were derived by Trelina [52]. Gy˝ory, Pink and Pintér [25] have given effectively computable upper bounds for n in the equations f (x) = wy n and F (x, z) = wy n , where f is a polynomial, F a binary form, both with discriminants belonging to some set S of S-integers, x, y, w, z, n are unknown integers with z, w ∈ S, y ∈ / S and n 3. The upper bounds only depend on the product p1 · · · ps and on the degrees of f (X) and of F (X, Y ). Quantitative versions of the Schinzel–Tijdeman approach have been used to solve Diophantine equations. For example Herrmann, Járási and Pethö [28] completed the study of the Diophantine equation x n = Dy 2 + 1 in variables (n, x, y) for 1 D 100. Using a combination of elementary techniques, and a clever criterion due to Nagell, J. H. E. Cohn [17] had determined all solutions except for the six cases (n, D) ∈ {(3, 31), (5, 31), (3, 38), (3, 61), (5, 71), (7, 71)}. The remaining cases were treated by using the methods on linear forms in logarithms. Bugeaud, Mignotte and Siksek [13] studied the equations Fn = q k y p and Ln = q k y p , where Fn and Ln denote the nth Fibonacci and Lucas numbers, respectively, and k > 0 and p, q are primes. They have announced to have proved that the only Fibonacci numbers which are powers y n with integers y > 1, n > 1 are 8 and 144. Other examples where the Schinzel–Tijdeman theorem has been applied are Voorhoeve, Gy˝ory and Tijdeman [58], Brindza [11] for the equation 1k + 2k + . . . + (x − 1)k + R(x) = by z , and Dilcher [21], Urbanowicz [54], [55], [56] for some generalizations of the equation, and Bilu, Kulkarni and Sury [9] for the equation x(x + 1) · · · (x + (m − 1)) + r = y n . A9. Call two number fields arithmetically equivalent if they have the same zeta function. Define the normset of a number field K as the set of elements of Z that are norms of integers of K. Coykendall [18] has given an example fields that have √ of arithmetically equivalent √ different normsets. The fields are K = Q( 8 −15) and L = Q( 8 −240). B. de Smit and R. Perlis [49] have shown that these fields are arithmetically equivalent and in fact have the same class number. The paper A9 shows that there are non-isomorphic fields which are arithmetically equivalent and have class number 1, so consequently have the same normset. Thus the normset cannot, in general, distinguish between non-isomorphic arithmetically equivalent fields. A10. Generalizations of Theorem 2 to polynomials with coefficients in a number field K can be found as Theorems 53 and 55 in Schinzel [41].


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A11. Chapter 5 of Schinzel’s book [41] is mainly devoted to the study of the following question. Let K be a number field. Assume that F ∈ C[x, t] has a zero in K s for a sufficiently large set t ∗ ∈ Zr . Does it follow that F viewed as a polynomial in x has a zero in K(x)s ? Schinzel proves that for s = 1 the answer is yes. By an example he shows that the answer is in general no for s 3. The problem is open for s = 2. The example in the above paper shows that the answer is no provided that Selmer’s conjecture is correct. The paper induced J. Silverman [45] to conjecture that almost all ranks of the curves from a certain family he studied are at most 1. A12. Generalizations of results from this paper to the case t ∈ Zr for any positive integer r have been given in Schinzel [40]. Further generalizations of Theorems 1 and 2 where coefficients are taken from a number field can be found as Theorems 58 and 59 in Schinzel’s book [41]. A13. Theorem 1 of this paper has been stated as Exercise 11.4 in Cassels’ book [15] with the following hint: the intersection of Fj (x, y, z) = 0 is either (i) a conic, (ii) a line or (iii) a set of n 4 points conjugate over k. A14. Later bounds for solutions of Diophantine equations which do not satisfy Runge’s condition have been given by Walsh [59] and in the special case that the irreducible polynomial is of the form F (x) − G(y) where F and G are monic polynomials with rational coefficients and gcd(deg F, deg G) > 1 by Tengely [51]. In a slightly different setting various estimates were obtained by D. Poulakis, partly in collaboration with others. For example, in [37] he obtained the following sharp bound. Let K be a number field and denote by OK its ring of integers. Let F (X, Y ) be an irreducible polynomial in K[X, Y ] \ K[Y ] of total degree N and of degree n > 0 in Y . Suppose K is totally real, and FN (1, Y ) is a polynomial of degree n without real roots, where FN (X, Y ) denotes the homogeneous part of degree N of F (X, Y ). Denote by H the (multiplicative) absolute height. If F (x, y) = 0 with (x, y) ∈ OK × K, then N 3 3 H (F )2 N . max{H (x), H (y)} < 2N N In [39] Poulakis improved upon Walsh’s estimates especially with respect to the dependence on gcd(x, y) in the following special case: F (X, Y ) is an irreducible, integer polynomial of degree 2 such that the curve C defined by the equation F (X, Y ) = 0 has infinitely many integer points and the point (0, 0) is simple on C whereas d 1 is a real number. A15. A general conjecture is that for all positive integers m, k with m > k 3 the equations (2)

1 1 1 m = + + ... + n x1 x2 xk

is solvable in positive integers x1 , x2 , . . . , xk for all but finitely many values of n. For (m, n) = (4, 3) Erd˝os and Straus even conjectured that there are no exceptions at all, for m > k = 3 it is Schinzel’s conjecture, both cited in the above paper. The conjecture for k > 3 follows trivially from the conjecture for k = 3. The conjecture is still wide open. Some authors have derived upper bounds for the number of exceptions Em,k (N ) = |{n N : equation (2) has no solution}|.

8


Vaughan [57] proved that Em,3 (N ) bm N exp(−cm (log N )2/3 ) where bm and cm are positive constants. Elsholtz [23] improved upon earlier results of C. Viola and Z. Shen by generalizing Vaughan’s result to Em,k (N ) cm,k N exp(−bm,k (log N )ek ) where bm,k and cm,k are positive constants and ek = 1 − 1/(2k−1 − 1) for all m, k specified above. A16. This paper provides an effective method for finding points over a finite field on an elliptic curve of the form E : Y 2 = X3 + B. In a subsequent paper Skałba [47] has given a deterministic polynomial time algorithm for finding points over a finite field on an elliptic curve E : Y 2 = X3 + AX + B provided that A = 0.

References [1] M. Ayad, Sur le théorème de Runge. Acta Arith. 58 (1991), 203–209. [2] H. P. Baltes, P. K. J. Draxl, E. R. Hilf, Quadratsummen und gewisse Randwertprobleme der mathematischen Physik. J. Reine Angew. Math. 268/269 (1974), 410–417. [3] M. Bauer, M. A. Bennett, Applications of the hypergeometric method to the generalized Ramanujan–Nagell equation. Ramanujan J. 6 (2002), 209–270. [4] E. A. Bender, N. P. Herzberg, Some Diophantine equations related to the quadratic form ax 2 + by 2 . In: Studies in Algebra and Number Theory, Academic Press, New York 1979, 219–272. [5] J. Berstel, Sur des fractions rationnelles particulières. In: Sém. M. P. Schützenberger, A. Lentin et M. Nivat, 1969/70: Problèmes Mathématiques de la Théorie des Automates, Exp. 2. [6] F. Beukers, On the generalized Ramanujan–Nagell equation I. Acta Arith. 38 (1980/81), 389–410. [7] −−, On the generalized Ramanujan–Nagell equation II. Acta Arith. 39 (1981), 113–123. [8] −−, The Diophantine equation Ax p + By q = Czr . Duke Math. J. 91 (1998), 61–88. [9] Yu. F. Bilu, M. Kulkarni, B. Sury, The Diophantine equation x(x+1) · · · (x+(m−1))+r = y n . Acta Arith. 113 (2004), 303–308. [10] D. W. Boyd, The Diophantine equation x 2 + y m = z2n . Amer. Math. Monthly 95 (1988), 544–547; Addendum, ibid. 97 (1990), 411–412. [11] B. Brindza, On some generalizations of the Diophantine equation 1k + 2k + . . . + x k = y z . Acta Arith. 44 (1984), 99–107. [12] J. Browkin, A. Schinzel, On the equation 2n − D = y 2 . Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 8 (1960), 311–318. [13] Y. Bugeaud, M. Mignotte, S. Siksek, Sur les nombres de Fibonacci de la forme q k y p . C. R. Math. Acad. Sci. Paris 339 (2004), 327–330. [14] Y. Bugeaud, T. N. Shorey, On the number of solutions of the generalized Ramanujan–Nagell equation. J. Reine Angew. Math. 539 (2001), 55–74. [15] J. W. S. Cassels, Local Fields. Cambridge Univ. Press, Cambridge 1986. [16] S. Chowla, Some problems of elementary number theory. J. Reine Angew. Math. 222 (1966), 71–74. [17] J. H. E. Cohn, The Diophantine equation x n = Dy 2 + 1. Acta Arith. 106 (2003), 73–83. [18] J. Coykendall, A remark on arithmetic equivalence and the normset. Acta Arith. 92 (2000), 105–108.


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[19] H. Darmon, A. Granville, On the equations zm = F (x, y) and Ax p + By q = Czr . Bull. London Math. Soc. 27 (1995), 513–543. [20] H. Davenport, D. J. Lewis, A. Schinzel, Quadratic Diophantine equations with a parameter. Acta Arith. 11 (1966), 353–358. [21] K. Dilcher, On a Diophantine equation involving quadratic characters. Compositio Math. 57 (1986), 383–403. [22] J. Edwards, A complete solution to X 2 + Y 3 + Z 5 = 0. J. Reine Angew. Math. 571 (2004), 213–236. [23] C. Elsholtz, Sums of k unit fractions. Trans. Amer. Math. Soc. 353 (2001), 3209–3227. [24] E. Grosswald, A. Calloway, J. Calloway, The representation of integers by three positive squares. Proc. Amer. Math. Soc. 10 (1959), 451–455. [25] K. Gy˝ory, I. Pink, A. Pintér, Power values of polynomials and binomial Thue–Mahler equations. Publ. Math. Debrecen 65 (2004), 341–362. [26] Th. R. Hagedorn, A proof of a conjecture on Egyptian fractions. Amer. Math. Monthly 107 (2000), 62–63. [27] H. Hasse, Über eine diophantische Gleichung von Ramanujan–Nagell und ihre Verallgemeinerung. Nagoya Math. J. 27 (1966), 77–102. [28] E. Herrmann, I. Járási, A. Pethö, Note on “The Diophantine equation x n = Dy 2 +1” (Acta Arith. 106 (2003), 73–83) by J. H. E. Cohn. Acta Arith. 113 (2004), 69–76. [29] R. Honsberger, Mathematical Gems from Elementary Combinatorics, Number Theory, and Geometry. The Mathematical Association of America, Buffalo 1973. [30] Th. Kulikowski, Sur l’existence d’une sphère passant par un nombre donné de points aux coordonnées entières. Enseignement Math. (2) 5 (1959), 89–90. [31] M. Le, On the number of solutions of the generalized Ramanujan–Nagell equation x 2 − D = 2n+2 . Acta Arith. 60 (1991), 149–167. [32] M. Le, A note on the number of solutions of the generalized Ramanujan–Nagell equation. J. Number Th. 62 (1997), 100–106. [33] L. J. Mordell, The representation of integers by three positive squares. Michigan Math. J. 7 (1960), 289–290. [34] T. Nagell, The Diophantine equation x 2 + 7 = 2n . Norsk Mat. Tidskr. 30 (1948), 62–64; English transl.: Ark. Mat. 4 (1961), 185–187. [35] A. Perelli, U. Zannier, An arithmetic property of polynomials. Boll. Un. Mat. Ital. A (5) 17 (1980), 199–202. [36] I. Pink, On the differences between polynomial values and perfect powers. Publ. Math. Debrecen 63 (2003), 461–472. [37] D. Poulakis, Polynomial bounds for the solutions of a class of Diophantine equations. J. Number Theory 66 (1997), 271–281. [38] −−, Affine curves with infinitely many integral points. Proc. Amer. Math. Soc. 131 (2003), 1357–1359. [39] −−, Integer points on rational curves with fixed gcd. Publ. Math. Debrecen 64 (2004), 369–379. [40] A. Schinzel, An application of Hilbert’s irreducibility theorem to Diophantine equations. Acta Arith. 41 (1982), 203–211.

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[41] A. Schinzel, Polynomials with Special Regard to Reducibility. Encyclopaedia of Mathematics and its Applications 77, Cambridge Univ. Press, Cambridge 2000. [42] A. Schinzel, W. Sierpiński, Sur les puissances propres. Bull. Soc. Roy. Sci. Liège 34 (1965), 550–554. [43] T. N. Shorey, A. J. van der Poorten, R. Tijdeman, A. Schinzel, Applications of the Gel fond– Baker method to Diophantine equations. In: Transcendence Theory: Advances and Applications (ed. A. Baker and D. Masser), Academic Press, London 1977, 59–77. [44] W. Sierpiński, O rozkładach liczb wymiernych na ułamki proste (On Decompositions of Rational Numbers into Unit Fractions). PWN, Warszawa 1957 (Polish). [45] J. H. Silverman, Divisibility of the specialization map for families of elliptic curves. Amer. J. Math. 107 (1985), 555–565. [46] −−, On the distribution of integer points on curves of genus zero. Theoret. Comput. Sci. 235 (2000), 163–170. [47] M. Skałba, Points on elliptic curves over finite fields. Acta Arith. 117 (2005), 293–301. [48] N. P. Smart, Determining the small solutions to S-unit equations. Math. Comp. 68 (1999), 1687–1699. [49] B. de Smit, R. Perlis, Zeta functions do not determine class numbers. Bull. Amer. Math. Soc. (N.S.) 31 (1994), 213–215. [50] D. Suryanarayana, N. V. Rao, On a paper of André Schinzel. J. Indian Math. Soc. (N.S.) 29 (1965), 165–167. [51] Sz. Tengely, On the Diophantine equation F (x) = G(y). Acta Arith. 110 (2003), 185–200. [52] L. A. Trelina, Representation of powers by polynomials in algebraic number fields. Dokl. Akad. Nauk BSSR 29 (1985), no. 1, 5–8 (Russian). [53] J. Turk, On the difference between perfect powers. Acta Arith. 45 (1986), 289–307. [54] J. Urbanowicz, On the equation f (1)1k + f (2)2k + . . . + f (x)x k + R(x) = by z . Acta Arith. 51 (1988), 349–368. [55] −−, On Diophantine equations involving sums of powers with quadratic characters as coefficients I. Compositio Math. 92 (1994), 249–271. [56] −−, On Diophantine equations involving sums of powers with quadratic characters as coefficients II. Compositio Math. 102 (1996), 125–140. [57] R. C. Vaughan, On a problem of Erd˝os, Straus and Schinzel. Mathematika 17 (1970), 193–198. [58] M.Voorhoeve, K. Gy˝ory, R. Tijdeman, On the Diophantine equation 1k +2k +. . .+x k +R(x) = y z . Acta Math. 143 (1979), 1–8; Corr. ibid. 159 (1987), 151. [59] P. G. Walsh, A quantitative version of Runge’s theorem for Diophantine equations. Acta Arith. 62 (1992), 157–172; Corr. ibid. 73 (1995), 397–398. [60] B. M. M. de Weger, Algorithms for Diophantine equations. CWI Tract 65, Stichting Mathematisch Centrum, Amsterdam 1989. [61] K. Wildanger, Über das Lösen von Einheiten- und Indexformgleichungen in algebraischen Zahlkörpern. J. Number Theory 82 (2000), 188–224. [62] P. Z. Yuan, On the number of solutions of x 2 − D = p n . Sichuan Daxue Xuebao 35 (1998), 311–316.

Originally published in Comptes rendus des séances de l’Académie des Sciences 242, 1780–1781


Sur les nombres de Mersenne qui sont triangulaires* avec Georges Browkin

Démonstration que l’équation 2x − 1 = (1/2)y(y + 1) en entiers positifs x et y n’a que les solutions (x, y) = (1, 1), (2, 2), (4, 5) et (12, 90).

Théorème. Les nombres de Mersenne Mn = 2n − 1 (où n = 1, 2, . . . ) qui sont en même temps triangulaires, tk = (1/2)k(k + 1), sont seulement les nombres M1 = t1 , M2 = t2 , M4 = t5 et M12 = t90 . Démonstration. L’équation 2n − 1 = (1/2)k(k + 1) équivaut à 1 2 7 1 2 7 n+1 k + + = + , c 2 4 2 4 donc à 1 1 √ 1 1 √ 1 1 √ n+1 1 1 √ n+1 (1) k+ + −7 k + − −7 = −7 −7 . + − 2 2 2 2 2 2 2 2 √ √ et, en tant que Les nombres (1/2) ± (1/2) −7 sont des entiers du corps Q( −7) √ −7). facteurs du nombre premier rationnel 2, ils sont premiers dans le corps Q( √ √ Posons α = k + (1/2) + (1/2) −7 et β = k + (1/2) − (1/2) −7. On a αk − βk = k k √ c −7 et αk βk = 2n+1 . Le plus grand diviseur commun des nombres αk et βk [dans le corps √ √ Q( −7)] divise donc les normes des nombres −7 et 2n+1 , c’est-à-dire les nombres 7 et 2n+1 , donc aussi le plus √ grand diviseur commun de ces derniers, c’est-à-dire le nombre 1. Or, dans le corps Q( −7) les seuls diviseurs du nombre 1 sont les nombres 1 et √−1. On en conclut que les nombres αk et βk sont premiers entre eux [dans le corps Q( −7)] et √ évidemment distincts des unités de ce corps. Or, puisque dans le corps Q( −7) a lieu le théorème sur l’unicité du développement en facteurs premiers, on en déduit de (1) que αk = ±α0n+1

βk = ±β0n+1 ou bien αk = ±β0n+1 et βk = ±α0n+1 , √ et, comme αk − βk = −7, on trouve √ √ −7 = ±α0n+1 ∓ β0n+1 ou bien −7 = ±β0n+1 ∓ α0n+1 *

et

Note transmise par M. Wacław Sierpiński, séance du 27 février 1956.

12


√ et en posant uk = (1/ −7)(α0k − β0k ) pour k = 1, 2, . . . , on trouve |un+1 | = 1.

(2)

On a u1 = u2 = 1 et, comme l’a remarqué M. Antoine Wakulicz, uk+1 = uk − 2uk−1 pour k = 2, 3, . . . . Les nombres uk (k = 1, 2, . . . ) sont donc des entiers rationnels. Pour k et l naturels le nombre k(l−j ) ukl = α0 β0k uk l−1

j =1

√ est entier dans le corps Q( −7), donc, comme nombre réel, est un entier rationnel. On a donc (3)

uk | ukl

pour k et l naturels.

On démontre sans peine que les restes mod 64 des nombres uk (k = 1, 2, . . . ) forment une suite infinie périodique dont la période, précédée par la suite 1, 1, −1, −3, −1 est formée de 16 termes : 5, 7, −3, −17, −11, 23, −19, −1, −27, −25, 29, 15, 21, −9, 13, 31. Le nombre ±1 pour k > 5 figure dans cette suite périodique seulement pour k ≡ 13 (mod 16) et alors on a uk ≡ −1 (mod 64). Il résulte donc de (2) que n = 0, 1, 2, 4 ou bien (4)

n + 1 ≡ 13 (mod 16)

et

un+1 = −1.

La suite vk = u16k+13 (k = 0, 1, 2, . . . ) donne mod 17 les restes qui forment une suite périodique dont la période (commençant au premier terme de la suite) a 9 termes : −1, −5, 0, 5, 1, 2, −4, 4, −2. Le nombre −1 figure dans cette suite seulement pour k ≡ 0 (mod 9), d’où l’on déduit que n + 1 ≡ 13 (mod 9 · 16), donc que n + 1 ≡ 1 (mod 3). La suite wk = u3k+1 (k = 0, 1, 2, . . . ) donne mod 79, les restes formant une suite périodique dont la période (commençant dès le premier terme de la suite) a 13 termes 1, −3, 7, −11, −1, 14, 17, −39, −20, 17, −4, −37, −20 et le nombre −1 figure dans cette suite seulement pour k ≡ 4 (mod 13), ce qui donne n + 1 = 3k + 1 ≡ 13 (mod 39), d’où n + 1 = 13(3s + 1) et, d’après (3) u3s+1 | un+1 = −1,

d’où

|u3s+1 | = 1.

S’il était 3s = 0, on aurait n = 12. Pour 3s = 0, d’après 3s = 1, 2, 4, on conclut comme plus haut que 3s + 1 = 13(3t + 1) (où s et t sont des entiers 0), n + 1 = 169(3t + 1) et, d’après (3) u169 | un+1 = −1,

d’où

ce qui contredit à (4), puisque 169 ≡ 13 (mod 16). Notre théorème se trouve ainsi démontré.

u169 = ±1,

Originally published in Mathesis 65:4-5-6 (1956), 219–222


Sur quelques propriétés des nombres 3/n et 4/n, où n est un nombre impair

En rapport avec un théorème connu des Égyptiens d’après lequel tout nombre rationnel positif est une somme d’un nombre fini de fractions primaires distinctes (c’est-à-dire de nombres 1/k, où k est un entier positif), E. P. Starke a posé en 1952 la question de savoir si tout nombre rationnel positif au dénominateur impair est une somme d’un nombre fini de fractions primaires distinctes aux dénominateurs impairs (1 ). Deux années plus tard ce problème a été résolu positivement, entre autres par R. Breusch (2 ). Or, les démonstrations qu’on a données ne contiennent aucun renseignement sur le nombre des termes nécessaires pour représenter un nombre rationnel donné comme somme de fractions primaires distinctes aux dénominateurs impairs. W. Sierpiński a examiné les nombres rationnels positifs aux numérateurs 1, 2 et 3. Il a démontré que aucun nombre 1/n, où n est un nombre impair n’est une somme de deux fractions primaires aux dénominateurs impairs, et que pour tout nombre impair n > 1 le nombre 1/n est une somme de trois fractions primaires distinctes aux dénominateurs impairs. Il a aussi démontré que pour n impair > 1 le nombre 2/n est une somme de 4 fractions primaires distinctes aux dénominateurs impairs, et que pour aucun nombre impair n le nombre 2/n n’est une somme de trois fractions primaires aux dénominateurs impairs. Or pour que le nombre 2/n, où n est un nombre impair, soit une somme de deux fractions primaires distinctes aux dénominateurs impairs, il faut et il suffit que le nombre n soit > 1 et qu’il ne soit pas un nombre premier de la forme 4k + 3 (3 ). Quant aux nombres 3/n, W. Sierpiński a exprimé l’hypothèse que pour n impair > 3 ils sont des sommes de trois fractions primaires distinctes aux dénominateurs impairs. Je démontrerai ici cette hypothèse.

1. Théorème. n étant un nombre impair > 3, le nombre 3/n est une somme de trois fractions primaires distinctes aux dénominateurs impairs. Si le nombre 3/n est une somme de trois fractions primaires distinctes aux dénominateurs impairs 3 1 1 1 = + + , n x y z (1 ) (2 ) (3 )

American Mathematical Monthly 59 (1952), p. 640, problem 4512. Ibidem, 61 (1954), p. 200–201 ; voir aussi [2], p. 785. Les démonstrations de ces théorèmes paraîtront dans le livre [1] de W. Sierpiński, Théorie des nombres, vol. II (en polonais), préparé pour l’impression.

14


tout nombre 3/kn, où k = 1, 3, 5, . . . jouit de la même propriété puisqu’on a alors 3 1 1 1 = + + . kn kx ky kz 3 1 1 1 Vu encore que = + + , il en résulte que pour démontrer notre théorème, 9 5 9 45 il suffit de le démontrer pour les nombres 3/n, où n est un nombre premier > 3. De tels nombres sont de la forme 6k ± 1, où k est un nombre naturel. Si n = 6k + 1, la démonstration résulte tout de suite de l’identité suivante, trouvée par W. Sierpiński : c

(1)

1 1 1 3 = + + . 6k + 1 2k + 1 (2k + 1)(4k + 1) (4k + 1)(6k + 1)

Si n = 6k − 1, on a n + 1 = 6k et n + 1 = 3s · 2t, où s et t sont des entiers positifs et t n’est pas divisible par 3, donc t = 3u + 1 ou bien t = 3u + 2, où u est un entier 0. On a donc n + 1 = 3s (6u + 2) ou bien n + 1 = 3s (6u + 4). Si n + 1 = 3s (6u + 2), on a, comme on le vérifie sans peine (2)

1 1 1 3 = s + + s . n 3 (2u + 1) (2u + 1)n 3 (2u + 1)n

Tous les dénominateurs à droite sont impairs et distincts, puisque, d’après n > 3 on a n = 3s (6u + 2) − 1 3s · 2 − 1 > 3s . Si n + 1 = 3s (6u + 4), alors, comme on le vérifie aisément, en posant m = s−1 3 (6u + 4) + 2u + 1, on trouve 1 1 1 3 = + + n m 3s m 3s mn et, comme 3s > 1 et n > 3, tous les dénominateurs à droite sont distincts. Le théorème est ainsi démontré. Notre démonstration donne en même temps le moyen de trouver pour chaque nombre impair n > 3 donné une décomposition désirée. Par exemple, pour n = 191, on a n + 1 = 192 = 3(6 · 10 + 4), d’où la décomposition

(3)

3 1 1 1 = + + . 191 85 3 · 85 3 · 85 · 191 Pour n = 863, on a n + 1 = 864 = 32 (6 · 5 + 2), d’où la décomposition 3 1 1 1 = + + . 863 27 · 11 11 · 863 27 · 11 · 863 Il est à remarquer qu’il peut exister d’autres décompositions que celles qui sont déterminées par notre démonstration. Par exemple, pour n = 359, on a n+1 = 360 = 32 (6·6+4) 3 1 1 1 et notre démonstration donne la décomposition = + 2 + 2 , 359 133 3 · 133 3 · 133 · 359 3 1 1 1 mais on a aussi la décomposition = 2+ + . 359 11 3 · 11 · 359 3 · 112 · 359 Vu que pour tout nombre impair n > 1 le nombre 1/n est une somme de trois fractions primaires aux dénominateurs impairs (évidemment > n), en partant du théorème et en appliquant cette décomposition au plus petit terme de la décomposition du nombre 3/n en

A2. Propriétés des nombres 3/n et 4/n

15

somme de fractions primaires distinctes aux dénominateurs impairs successivement s − 1 fois (où s est un nombre naturel), nous obtenons le corollaire : s étant un nombre naturel quelconque et n un nombre naturel > 3, le nombre 3/n est une somme de 2s + 1 fractions primaires distinctes aux dénominateurs impairs. D’autre part, il est clair que pour n impair le nombre 3/n n’est pas une somme d’un nombre pair de fractions primaires aux dénominateurs impairs (puisque la somme de ces dernières est un nombre rationnel au numérateur pair et dénominateur impair, donc un nombre distinct de 3/n). 2. Théorème. n étant un nombre naturel tel que n > 1 et n = 5, le nombre 4/n est une somme de quatre fractions primaires distinctes aux dénominateurs impairs. Soit n un nombre impair > 1 et n = 5. Si n = 3 on a 4 1 1 1 1 = + + + , 3 1 5 9 45 d’où il résulte tout de suite que le théorème est vrai pour tout nombre impair n divisible par 3. Il nous reste donc les nombres de la forme 6k ± 1, où k = 1, 2, . . . . Si n = 6k + 1, on a 4 1 3 = + n 6k + 1 6k + 1 et la vérité du théorème pour le nombre n résulte de la formule (1) et de la remarque que, pour k naturel, on a 2k + 1 < 6k + 1 < (2k + 1)(4k + 1) < (4k + 1)(6k + 1). Si n = 6k − 1, alors, comme dans la démonstration du premier théorème, nous distinguerons les cas n + 1 = 3s (6u + 2) et n + 1 = 3s (6u + 4), où s = 1, 2, . . . et u = 0, 1, 2, . . . . Si n+ 1 = 3s (6u + 2) et u = 0, donc si n = 3s · 2 − 1, alors pour s pair on a n = 8k +1, où k est un nombre naturel et la vérité du théorème pour le nombre n résulte de l’identité de W. Sierpiński 4 1 1 = + 8k + 1 2k + 1 (2k + 1)(4k + 1) +

1 1 + , (2k + 1)(8k + 1) (2k + 1)(4k + 1)(8k + 1)

et pour s impair on a, comme on le démontre sans peine, n = 16k + 5, où, vu que n = 5 et que n n’est pas divisible par 3, k est un entier > 1, et la vérité du théorème pour le nombre n résulte de l’identité 4 1 1 1 1 = + + + 16k + 5 3(2k + 1) 16k + 5 3(16k + 5) (2k + 1)(16k + 5)

16


et de la remarque que pour k > 1 on a 3(2k + 1) < 16k + 5 < 3(16k + 5) < (2k + 1)(16k + 5). Si n + 1 = + 2) et u > 0, alors, vu que 4/n = 1/n + 3/n, la vérité du théorème pour le nombre n résulte de la formule (2) et de la remarque que les nombres n, 3s (2u + 1), (2u + 1)n et 3s (2u + 1)n sont tous distincts. Enfin, si n + 1 = 3s (6u + 4), la vérité du théorème pour le nombre n résulte des formules 4/n = 1/n + 3/n et (3) [où m = 3s−1 (6u + 4) + 2u + 1] et de la remarque que les nombres n, m, 3s m et 3s mn sont tous distincts. Le théorème se trouve ainsi démontré. Or, en ce qui concerne le nombre n = 5, W. Sierpiński a démontré que le nombre 4/5 n’est pas une somme de moins de six fractions primaires aux dénominateurs impairs et qu’on a 3s (6u

4 1 1 1 1 1 1 = + + + + + . 5 3 5 7 15 21 105 On peut démontrer sans peine que pour que le nombre 4/n, où n est un nombre impair, soit une somme de deux fractions primaires aux dénominateurs impairs, il faut et il suffit que n ait un diviseur naturel de la forme 8k + 3.

Ouvrages cités [1] W. Sierpiński, Teoria liczb, cze´ sc´ 2. Monografie Matematyczne 38, PWN, Warszawa 1959. [2] B. M. Stewart, Sums of distinct divisors. Amer. J. Math. 76 (1954), 779–785.


Originally published in L’Enseignement mathématique IV:1 (1958), 71–72

Sur l’existence d’un cercle passant par un nombre donné de points aux coordonnées entières

Le but de cette Note est de démontrer ce Théorème. Quel que soit le nombre naturel n, il existe dans le plan un cercle dont la circonférence contient précisément n points aux coordonnées entières. (Ce théorème a été mentionnée dans l’article de M. W. Sierpiński Sur quelques problèmes concernant les points aux coordonnées entières paru dans ce fascicule, page 25.) Pour n impair, n = 2k + 1, où k est un entier 0, le cercle au centre Démonstration. 1 , 0 et au rayon 5k /3 satisfait à notre théorème. 3 En effet, d’après un théorème connu sur le nombre de décomposition en deux carrés, l’équation x 2 + y 2 = 52k a 4(2k + 1) solutions en nombres entiers x et y. Comme 52k ≡ 1 (mod 3) pour k = 0, 1, 2, . . . , on démontre sans peine que dans chaque telle solution, un et un seul des nombres x et y est divisible par 3. Les solutions se divisent donc en 2k + 1 quadruples disjoints : (x, y), (x, −y), (y, x), (−y, x), où x est un entier divisible par 3 et y un entier qui n’est pas divisible par 3. Dans chaque tel quadruple, une et une seule solution satisfait à la condition que le premier terme de la paire ≡ −1 (mod 3) et le second ≡ 0 (mod 3). Il existe donc précisément 2k + 1 = n solutions en nombres entiers z et t de 2 k 2 l’équation (3z − 1)2 + (3t)2 = 52k , c’est-à-dire de l’équation z − 13 + t 2 = 53 . Cela prouve qu’il existe précisément n points aux coordonnées entières sur le cercle déterminé par cette équation, ce qui démontre notre théorème pour n impair. Pour n pair, n = 2k, où k est un nombre naturel, le cercle au centre 21 , 0 et au rayon 5(k−1)/2 /2 satisfait à notre théorème. En effet, d’après le théorème mentionné plus haut, l’équation x 2 + y 2 = 5k−1 a précisément 4k solutions en nombres entiers x et y. Or, des nombres x, y, un et seul un est pair et ainsi toutes les solutions se divisent en 2k paires disjointes (x, y) et (y, x) qui ne diffèrent entre elles que par l’ordre de leurs termes. Dans chaque telle paire précisément une solution satisfait à la condition que le premier élément est impair et le second pair. Il existe donc précisément 2k = n solutions en nombres entiers z, t de l’équation (2z − 1)2 + (2t)2 = 5k−1 , c’est-à-dire à l’équation (k−1)/2 2 2 z − 21 + t 2 = 5 2 , ce qui démontre notre théorème pour n pair. Notre théorème se trouve ainsi démontré.


Originally published in Bulletin de l’Academie Polonaise des Sciences Série des sci. math., astr. et phys. VII (1959), 307–310

Sur les sommes de trois carrés Présénte par W. Sierpiński le 18 mars 1959

Une de mes communications antérieures [1] contient la remarque que l’hypothèse de G. Pall ([2]) soutenant que tout nombre naturel de la forme 2(8n + 1) > 2 est une somme de trois carrés > 0 est en défaut pour n = 8. Or, le problème suivant s’impose : quels sont les nombres naturels qui sont sommes de trois carrés positifs ? Le but de la présente communication est de démontrer le théorème qui suit. Théorème 1. Pour que le nombre n admette une décomposition n = x 2 + y 2 + z2 , où x, y, z > 0 et (x, y, z) = 1, il faut et il suffit que n ait à la fois deux propriétés suivantes : (i) n ≡ 0, 4, 7 (mod 8); (ii) n a un diviseur naturel de la forme 4k − 1 ou bien n n’est pas un “numerus idoneus”. Les “numeri idonei” sont — comme on sait — des nombres naturels D vérifiant l’équation p(−4D) = 1, où p(d) est un nombre des classes de formes binaires au discriminant d de genre principal. Une définition arithmétique de “numeri idonei” a été donnée par Euler (cf. [3], vol. 1, p. 361 ; pour les corrections voir [4]). Démonstration. Le nombre N3 (n) des réprésentations propres du nombre n > 3 par la forme x 2 + y 2 + z2 est donné par la formule (cf. [3], vol. 2, p. 265) ⎧ ⎪ ⎨12h(−4n) pour n ≡ 1, 2, 5, 6 (mod 8) (1) N3 (n) = 24h(−4n) pour n ≡ 3 (mod 8) ⎪ ⎩ 0 pour n ≡ 0, 4, 7 (mod 8), où h(d) est le nombre des classes de formes binaires au discriminant d. Mais, d’après le théorème sur la duplication, on a : ⎧ λ−2 pour d ≡ 4 (mod 16) ⎪ ⎨2 (2) h(d) = p(d) · 2λ pour d ≡ 0 (mod 32) ⎪ ⎩ λ−1 2 pour d’autres cas, où λ est le nombre des facteurs premiers du nombre d.

A4. Sur les sommes de trois carrés

19

Pour n > 3, on obtient de formules (1) et (2) ⎧ μ+2 p(−4n) pour n ≡ 1, 2, 5, 6 (mod 8) ⎪ ⎨3 · 2 (3) N3 (n) = 3 · 2μ+2 p(−n) pour n ≡ 3 (mod 8) ⎪ ⎩ 0 pour n ≡ 0, 4, 7 (mod 8), où μ est le nombre des facteurs premiers impairs du nombre n. D’autre part, le nombre des décompositions du nombre n en sommes de deux carrés rélativement premiers est donné par l’égalité 0 lorsque 4 | n ou 4k − 1 | n pour k = 1, 2, . . . (4) N2 (n) = μ+2 2 en cas contraire. c Or, si n > 1, le nombre des réprésentations de n dans la forme x 2 + y 2 + z2 , où 3 (x, y, z) = 1 et xyz = 0, est égale à 2 N2 (n) = 3N2 (n). La condition nécessaire et suffisante pour que le nombre n > 1 soit une somme de trois carrés positifs, rélativement premiers est donc l’inégalité N3 (n) > 3N2 (n). En vertu de (3) et de (4) cette inégalité exprime que n ≡ 0, 4, 7 (mod 8) et que N2 (n) = 0, où bien p(−4n) > 1. Vu (4) et la définition des “numeri idonei”, la démonstration est achevée. Corollaire 1. Si n ≡ 0, 4, 7 (mod 8), n = 25 et n a un diviseur quadratique > 1, n est une somme de trois carrés positifs, rélativement premiers. C’est une conséquence immédiate du Théorème 1 et du théorème de Grube ([4]), d’après lequel les nombres 9, 18, 25, 45, 72 sont les seuls “numeri idonei” ayant un diviseur quadratique impair > 1. Remarque. Corollaire 1 affirme davantage que le théorème de Pall ([2]), à savoir que le réprésentation par la forme x 2 + y 2 + z2 est propre. Corollaire 2. Tout nombre naturel n ≡ 0, 4, 7 (mod 8) suffisamment grand est une somme de trois carrés positifs rélativement premiers. En effet, il résulte directement du Théorème 1 et du théorème de Chowla [5] * que limd→∞ p(−d) = ∞. Corollaire 3. Les seuls nombres naturels n < 101200 tels, que n ≡ 0, 4, 7 (mod 8) et qui ne sont pas sommes de trois carrés > 0, sont les suivants 1, 2, 5, 10, 13, 25, 37, 58, 85, 130. La démonstration consiste à appliquer le Théorème 1 et les résultats de Cunningham et Cullen [7] sur les “numeri idonei” < 101200. L’autre hypothèse de Pall [2], à savoir que tout nombre naturel = 1, 25 de la forme 8k + 1 est somme de trois carrés > 0, résulte directement de l’hypothèse suivante : *

La démonstration de Chowla n’est pas effective (cf. [6]).

20 c


Hypothèse 1. Dans le Corollaire 3 la condition n < 101200 peut être supprimée. Théorème 2. Pour qu’un nombre naturel n se décompose en une somme de trois carrés distincts et rélativement premiers, il faut et il suffit que n ait à la fois la propriété (i) et la suivante : (iii) ou bien n a un diviseur premier ≡ 5, 7 (mod 8), ou bien n ≡ 1, 2, 6 (mod 8) et p(−4n) > 1, ou bien n ≡ 3 (mod 8) et p(−n) > 1. La démonstration du Théorème 2 est tout à fait analogue à celle du Théorème 1. Elle est basée sur la formule pour le nombre des réprésentations de n dans la forme x 2 + 2y 2 . Corollaire 4. Tout nombre n ≡ 0, 4, 7 (mod 8) suffisament grand est une somme de trois carrés distincts et rélativement premiers. C’est une conséquence du Théorème 2 et de celui de Chowla [5].

c

Corollaire 5. Les seuls nombres naturels n ≡ 0, 4, 7 (mod 8) qui ne sont pas des sommes de trois carrés distincts sont les suivants : 1, 2, 6, 9, 18, 22, 33, 57, 102, 177 (pour n ≡ 1, 2, 5, 6 (mod 8) et n < 101200) ; 3, 11, 19, 27, 43, 51, 67, 99, 123, 163, 187, 267, 627 (pour n ≡ 3 (mod 8) et n < 23000). La démonstration consiste à appliquer le Théorème 2 et les résultats de Cunningham et Cullen [7] de même que ceux de S. B. Townes (cf. [8] p. 89). Corollaire 6. Les seuls nombres naturels n < 2875 qui ne sont pas des sommes de trois nombres triangulaires distincts sont les suivants : 1, 2, 3, 5, 6, 8, 12, 15, 20, 23, 33, 78. C’est une conséquence de la partie du Corollaire 5 concernant les nombres n ≡ 3 (mod 8). Si l’on admet l’hypothèse suivante : Hypothèse 2. Les inégalités n < 101200 et n < 23000 peuvent être supprimées dans le Corollaire 5 et l’inégalité n < 2875 peut l’être dans le Corollaire 6, le Corollaire 6 entraîne un théorème plus précis que celui établi dans ma communication précedente [1].

A4. Sur les sommes de trois carrés

21

Ouvrages cités [1] A. Schinzel, Sur la décomposition des nombres naturels en sommes de nombres triangulaires distincts. Bull. Acad. Polon. Sci. Cl. III 2 (1954), 409–410. [2] G. Pall, On sums of squares. Amer. Math. Monthly 40 (1933), 10–18. [3] L. E. Dickson, History of the Theory of Numbers. Chelsea, New York 1952. [4] F. Grube, Über einige Euler’sche Sätze aus der Theorie der quadratischen Formen. Zeitschrift für Math. u. Phys. 19 (1874), 492–519. [5] S. Chowla, An extension of Heilbronn’s Class-Number Theorem. Q. J. Math. 5 (1934), 304–307. [6] C. L. Siegel, Über die Classenzahl quadratischer Zahlkörper. Acta Arith. 1 (1935), 83–86. [7] A. Cunningham, H. Cullen, Report of the British Association 1901, p. 552. [8] L. E. Dickson, Introduction to the Theory of Numbers. Chicago 1936.

Originally published in Polish in Roczniki Polskiego Towarzystwa Matematycznego Seria I: Prace Matematyczne IV (1960), 45–51


On the Diophantine equation

n

A k xk ϑ k = 0 *

k=1

The equations ϕ=

(1)

n k=1 n

(1 )

Ak xkϑk = 0

(Ak , ϑk , mk are non-zero integers),

Ak Xkmk = 0

k=1

will be called equivalent by a birational G transformation, if there exists a mutually rational transformation in the sense of Georgiev [2] which transforms the function ϕ into the function F =

n

Xrμr

r=1

n

Ak Xkmk

(μr an integer).

k=1

It is easy to prove that the above relation of equivalence is reflexive, symmetric and transitive. It is also clear that if all solutions of equation (1) in non-zero rationals are known, then substituting them into the formulas of the appropriate birational G transformation we will obtain all solutions in non-zero rationals of an arbitrary equation equivalent to (1). Georgiev proved ([2], p. 216) that for a birational G transformation which takes the function ϕ to F to exist, it is necessary and sufficient that the following condition be fulfilled: (2)

the numbers λr,p = μr /ϑp (r = p) and λr,r = (μr + mr )/ϑr be integers,

and also that (3)

1+

n n n μk mi = ± ϑi . mk k=1

*

i=1

i=1

Presented at the meeting of the Wrocław Branch of the Polish Mathematical Society on 3 June 1955. Corrigendum: Prace Mat. 44 (2004), 293–294.

23

A5. On the Diophantine equation

The relevant transformation is then given by the formulas: xp =

(4)

n

λ

(1 p n).

Xr r,p

r=1 c

This has been deduced by Georgiev via his Theorems 3 and 10 from his Theorem 2, which needs a modification (allowing a permutation of terms in the considered sum). When this modification is made the condition (2) is replaced by (2a)

there exists a permutation σ of {1, 2, . . . , n} such that Aσ (k) = Ak for all k and the numbers λr,p = μσ (r) /ϑp (r = p) and λr,r = μσ (r) + mσ (r) /ϑr are integers

and formula (4) by xp =

(4a)

n

λ

Xσ r,p (r)

(1 p n).

r=1

The conditions (2) and (4) correspond to the case σ (r) = r. We shall prove Theorem 1. Equation (1) is equivalent by a birational G transformation (4) to the equation n

(5)

Ak Xkmk = 0, where mk = (ϑk , [ϑ1 , . . . , ϑk−1 , ϑk+1 , . . . , ϑn ]).

k=1

Moreover m

p

xp

(6)

xsϑs

=

Xp p Xsms

(1 p, s n).

Proof. Let n

ϑi = ϑ,

[ϑ1 , . . . , ϑk−1 , ϑk+1 , . . . , ϑn ] = θk ,

i=1

n

mi = m.

i=1

Suppose that the prime number q is a factor of ϑk with exponent ek . Without loss of generality we can assume that ek ek+1 for k = 1, . . . , n − 1. Therefore q is a factor of θk with exponent en for k n − 1, and en−1 for k = n. The number q is a factor of (ϑk , θk ) with exponent ek for k n − 1 and en−1 for k = n. The number q is a factor of θn m/mn with the same exponent as that of m. Because q was arbitrary and the formula below is symmetric, it means that g.c.d. (θk m/mk ) | m. k=1,...,n

But m | ϑ − m and therefore the equation (7)

m+

n k=1

has a solution.

ξk θk m/mk = ϑ

24


Let μk = ξk θk correspond to some specific solution {ξk }k=1,...,n . We will show that for j = k we have m j ϑk | θj mk .

(8) We distinguish two cases: 1◦

k n − 1. In this case mj | θj , and q is a factor of mk with exponent ek .

2◦

k = n. In this case q is a factor of θj with exponent en , of mj with exponent ej , therefore of the divisor with exponent ej + en , and of the dividend with exponent en + en−1 .

In both cases q is a factor of the dividend with exponent not smaller than it is a factor of the divisor, which is sufficient to establish formula (8). Multiplying (8) by m/mj mk we have ϑk m/mk | θj m/mj

(j = k)

and, because of (7), ϑk m/mk | m + μk m/mk , and therefore ϑk | μk + mk (1 k n) and the numbers λr,r = (μr + mr )/ϑr are integers. Since the numbers λr,p =

θr μr = ξr ϑp ϑp

(r = p),

are integers because of the way θr is defined, condition (2) holds. From formula (7) we have n μk m = ϑ, 1+ mk k=1

so also condition (3) holds, which finishes the proof of the first part of the theorem. By formulas (4) and (2) we have n ϑ m ϑp λr,p n Xp p xp p ϑp λr,p −ϑs λr,s r=1 Xr = = Xr = ms ϑs λr,s n Xs xsϑs r=1 r=1 Xr which finishes the proof of the second part of the theorem. Theorem 2. The equations n

(1)

Ak xkϑk = 0

k=1

and (9)

n

η

Ak yk k = 0

(ηk non-zero integers)

k=1

are equivalent by a birational G transformation of the form (10)

xp =

n r=1

k

yr r,p

(1 p n)

25

A5. On the Diophantine equation

if and only if, for k = 1, . . . , n c

mk = lσ (k) ,

(11)

where mk = (ϑk , [ϑ1 , . . . , ϑk−1 , ϑk+1 , . . . , ϑn ]), lk = (ηk , [η1 , . . . , ηk−1 , ηk+1 , . . . , ηn ]) and σ (k) is a permutation of {1, 2, . . . , n} such that Aσ (k) = Ak for all k. Then we have η

ϑ

xp p

(12)

xsϑs

=

(p) yσ σ(p)

η

(s) yσ σ(s)

.

Proof. By Theorem 1, equation (1) is equivalent to n

(5)

Ak Xkmk = 0,

k=1

and equation (9) is equivalent to the equation n

(13)

Ak Yklk = 0.

k=1

By reflexivity, symmetry and transitivity of the equivalence considered, the sufficiency of condition (11) is obvious. By formula (6), formula (12) is also obvious. To show the necessity of condition (11), suppose that equations (1) and (9) are equivalent. It then follows that equations (1) and (13) are equivalent, and (5) and (9) are equivalent too. c Replacing in condition (2a) mr by lr and μr by λr we have ϑp | λσ (r) (r = p), hence θr | λσ (r) and mr | λσ (r) and ϑr | λσ (r) + lσ (r) , thus mr | λσ (r) + lσ (r) . Therefore, mr | lσ (r) . From the equivalence of (5) and (9) it follows similarly that lr | mτ (r) , where τ is a permutation of {1, 2, . . . , k} such that Aτ (r) = Ar for all r. Thus mr | mτ (σ (r)) for all r, which gives mτ (σ (r)) = mr and lσ (r) = mr . Theorem 2 immediately implies Theorem 12 from the quoted paper of Georgiev [2]. By formula (12) we also have:

c

Corollary 1. If equations (1) and (9) are equivalent by a birational G transformation ϑ (10), Ak are distinct, and xs and ys are greater than zero, then xp p > xsϑs if and only if ηp ηs yp > ys . We assign the solutions {xk }k=1,2,...,n , {xk }k=1,2,...,n of equation (1) (respectively {yk }k=1,2,...,n , {yk }k=1,2,...,n of equation (9)) to the same class, if and only if ϑ η η x1 ϑ1 y 11 x pp y pp resp. (1 p n). = ϑ η = η1 ϑ y1 ypp x1 1 xp p Writing this condition in the form ϑ

x pp /x 1ϑ1 ϑ

xp p /x1ϑ1

=1

η

resp.

y pp /y 11 η

η

η

ypp /y1 1

=1

26


by formula (12) we have Corollary 2. Under a birational G transformation of the form (10), which gives the equivalence of (1) and (9), every class of solutions of equation (9) maps to a class of solutions of equation (1). Now suppose that in equation (1) we have (ϑn , ϑ1 · · · · · ϑn−1 ) = 1 (n > 1). Therefore mn = (ϑn , [ϑ1 , . . . , ϑn−1 ]) = 1, and by Theorem 1 equation (1) is equivalent by a birational G transformation to the equation n−1

(14)

Ak Xkmk + An Xn = 0,

k=1

all of whose rational solutions are given by the formulas: Xk = tk

(1 k n − 1),

Xn =

−

n−1 k=1

Ak tkmk

An

,

where tk (1 k n − 1) are rational parameters. The case n = 2 gives one class of solutions of equation (14), the case n = 3, m1 = m2 = 1 was considered by L. Tchacaloff and C. Karanicoloff [3], the case n 3 and mk = g (1 k n − 1) was considered by N. M. Basu [1], the general case was considered by T. Vijayaraghavan [4]. For n 3 equation (14) has infinitely many classes of rational solutions. Therefore from Corollary 2 and the observation that every class of rational solutions of equation (1) contains integer solutions, it follows that Corollary 3. If n 3 and if (ϑn , ϑ1 · · · · · ϑn−1 ) = 1, then equation (1) has infinitely many classes of integer solutions.

References [1] N. M. Basu, On a Diophantine equation. Bull. Calcutta Math. Soc. 32 (1940), 15–20. [2] G. Georgiev, O rozwiazaniu w liczbach wymiernych pewnych równań diofantycznych (On the solution in rational numbers of certain Diophantic equations). Prace Mat. 1 (1955), 201–238 (Polish). [3] L. Tchacaloff et C. Karanicoloff, Résolution de l’équation Ax m + By n = zp en nombres rationnels. C. R. Acad. Sci. Paris 210 (1940), 281–283. [4] T. The general rational solution of some Diophantine equations of the form Vijayaraghavan, nr k+1 r=1 Ar xr = 0. Proc. Indian Acad. Sci., Sect. A. 12 (1940), 284–289.

Originally published in Acta Arithmetica IX (1964), 107–116


Polynomials of certain special types with H. Davenport (Cambridge) and D. J. Lewis* (Ann Arbor)

1. Let f (x) be a polynomial with integral coefficients. It is well known that if f (x) is k a k-th power for every positive integer x, then f (x) = g(x) identically, where g(x) has integral coefficients. For proofs and references, see Pólya and Szegö [8], Section VIII, Problems 114 and 190; also Fried and Surányi [2]. In this connection, we shall prove the following general theorem: Theorem 1. Let f (x, y) be a polynomial with integral coefficients. Suppose that every arithmetical progression contains some integer x such that the equation f (x, y) = 0 has an integral solution in y. Then there exists a polynomial g(x) with rational coefficients such that (1) f x, g(x) = 0 identically. Corollary. Let k > 1 be an integer and let f (x) be a polynomial with integral coefficients. Suppose that every arithmetical progression contains some integer x such that f (x) is a k k-th power. Then f (x) = g(x) identically, where g(x) is a polynomial with integral coefficients. Professor LeVeque raised the question (in conversation) whether, if f (x) is representable as a sum of two squares for every positive integer x, or for every sufficiently large integer x, then f (x) is identically a sum of two squares. We shall prove that this is true, and we shall deduce it from the following general theorem. Theorem 2. Let K be any normal algebraic number field of degree n, with integral basis ω1 , ω2 , . . . , ωn , and let N (u1 , u2 , . . . , un ) = norm(u1 ω1 + u2 ω2 + . . . + un ωn ) *

This author was partially supported by a grant from the National Science Foundation.

28


denote the norm-form corresponding to K. Let f (x) be a polynomial with rational coefficients, and suppose that every arithmetical progression contains an integer x such that f (x) = N (u1 , u2 , . . . , un ) for some rational numbers u1 , u2 , . . . , un . Suppose further that either K is cyclic or the multiplicity of every zero of f (x) is relatively prime to n. Then f (x) = N u1 (x), u2 (x), . . . , un (x) identically, where u1 (x), u2 (x), . . . , un (x) are polynomials with rational coefficients. We observe that the hypotheses on K are always satisfied if K is normal and of prime degree n. The two alternatives in the hypothesis—one relating to K and the other to f (x) —are appropriate conditions to impose, in the sense that if both are violated, the conclusion may not hold. This is shown by the example (see §6) f (x) = x 2 ,

K = Q(e2πi/8 ),

where Q denotes the rational number field. The property of f (x) postulated in the theorem implies the solubility of the congruence f (x) ≡ N (u1 , . . . , un ) (mod m) in u1 , . . . , un for every integer x and every positive integer m. The congruence is to be understood in the multiplicative sense; see Hasse [3], 25, footnote *. If K is cyclic, then by a theorem of Hasse [4] this implies the apparently stronger statement that for every x we have f (x) = N (v1 , . . . , vn ) for some rational v1 , . . . , vn . Thus when K is cyclic, we have three apparently different conditions on f (x) which are in reality equivalent. Corollary to Theorem 2. Let f (x) be a polynomial with integral coefficients, and suppose that every arithmetical progression contains an integer x such that f (x) is a sum of two squares. Then f (x) = u21 (x) + u22 (x) identically, where u1 (x) and u2 (x) are polynomials with integral coefficients. In the particular case of Theorem 2, namely the case K = Q(i), which is needed for this Corollary, our method of proof has much in common with that used by Lubelski [7] in his investigation of the primes p for which f (x) ≡ 0 (mod p) is soluble. It will be seen that in the conclusion of the Corollary, it is asserted that u1 (x), u2 (x) have integral coefficients. In the more general Theorem 2, if it is postulated that f (x) has integral coefficients and that u1 , . . . , un are integers, it is not in general possible to draw the conclusion with u1 (x), . . . , un (x) having integral coefficients. This is illustrated by

A6. Polynomials of certain special types

the example (see §6) f (x) = 2x 2 (x + 1)2 + 3x(x + 1) + 4,

29

√ K = Q( −23).

However, it is possible to draw the conclusion stated above if the highest coefficient in f (x) is 1. This can be proved by first comparing the highest terms on both sides, and then appealing to Gauss’s lemma. There are other problems, of the same general character as those considered in this paper, which we are quite unable to attack. The simplest of them is that in which f (x) is representable as a sum of two integral cubes for every sufficiently large integer x.

2. Proof of Theorem 1. We note first that the hypothesis implies that every arithmetical progression contains infinitely many integers x such that the equation has an integral solution in y. For if d is the common difference of the progression, and x0 is one integer with the property, there exists an integer xn ≡ x0 + d n (mod d n+1 ) for n = 1, 2, . . . which has the property, and the integers xn are all distinct. We factorize f (x, y) into a product of powers of polynomials which are irreducible over the rational field Q; by Gauss’s lemma we can take these polynomials to have integral coefficients. We can omit any factor f0 (x, y) for which the equation f0 (x, y) = 0 has only finitely many integral solutions, since its omission will not invalidate the hypothesis. We can also omit any factor which does not contain y. Hence we can take (2)

f (x, y) = f1 (x, y)f2 (x, y) · · · fk (x, y),

where f1 (x, y), . . . , fk (x, y) are irreducible over Q and are such that each of the equations fj (x, y) = 0 has infinitely many integral solutions. It follows from Hilbert’s Irreducibility Theorem (Hilbert [5], p. 275; for references to later work, see Lang [6], pp. 163–164) that there exists an integer x0 such that all the polynomials fj (x0 , y), considered as polynomials in y, are irreducible over Q and are of the same degree in y as fj (x, y). Suppose first that all these degrees are greater than 1, and let nj denote the degree of fj (x0 , y) in y. Let η be a root of fj (x0 , η) = 0, and consider the prime ideal factorization of a rational prime p in Q(η) and in its least normal extension Q∗ (η). Let dr denote the density (in the Dirichlet series sense) of those primes which have exactly r prime ideal factors of the first degree in Q(η). Then (Hasse [3], p. 129) n r=0

dr = 1,

n

rdr = 1.

r=0

To prove that d0 > 0, it will suffice to prove that d1 < 1. Now any large prime p which has just one prime ideal factor of degree 1 in Q(η) will have some prime ideal factor of degree greater than 1 in Q(η), and so also in Q∗ (η). Since Q∗ (η) is normal, all prime ideal factors of p in Q∗ (η) will be of degree greater than 1, and the density of such p is exactly 1 − 1/n∗j , where n∗j denotes the degree of Q∗ (η) (Hasse [3], pp. 138–139). Hence

30


d1 1 − 1/n∗j , whence the result. In particular, there are infinitely many primes which have no prime ideal factor of the first degree in Q(η). By a well-known principle of Dedekind, if qj is such a prime (and is sufficiently large) we have (3)

fj (x0 , y) ≡ 0 (mod qj )

for all integers y. There is such a prime qj for each j . On the other hand, the hypothesis of the theorem implies that the arithmetical progression x ≡ x0 (mod q1 q2 · · · qk ) contains an integer x such that f (x, y) = 0 for some integer y. But then fj (x, y) = 0 for some j , whence fj (x0 , y) ≡ fj (x, y) ≡ 0 (mod qj ), contrary to (3). It follows that there is some j for which fj (x, y) is linear in y, say fj (x, y) = yA(x) − B(x), where A(x), B(x) are relatively prime polynomials with integral coefficients. There exist polynomials A1 (x), B1 (x) with integral coefficients such that A(x)A1 (x) + B(x)B1 (x) = c identically, where c is a non-zero constant. If x is an integer for which there is an integer y satisfying fj (x, y) = 0, then A(x) must divide c, and since this happens for infinitely many x, it follows that A(x) is a constant. Hence fj x, g(x) = 0 identically, where g(x) is the polynomial B(x)/A. This proves Theorem 1.

k The deduction of the Corollary is immediate, since we get f (x) = g(x) , where g(x) has rational coefficients, and then it follows from Gauss’s lemma that g(x) has integral coefficients.

3. Lemma 1. Suppose that the hypotheses of Theorem 2 hold. Let e e e (4) f (x) = c f1 (x) 1 f2 (x) 2 · · · fm (x) m , where c = 0 is a rational number and f1 (x), f2 (x), . . . , fm (x) are distinct primitive polynomials with integral coefficients, each irreducible over Q, and where e1 , e2 , . . . , em are positive integers. For any j , let q be a sufficiently large prime for which the congruence (5)

fj (x) ≡ 0 (mod q)

is soluble. If (ej , n) = 1 then q factorizes completely in K into prime ideals of the first degree. If K is cyclic then q factorizes completely into prime ideals of the first degree in the unique subfield Kj of K of degree n/(ej , n).


31

Proof. Put F (x) = f1 (x)f2 (x) · · · fm (x). Since the discriminant of F (x) is not zero, there exist polynomials ϕ(x), ψ(x) with integral coefficients such that F (x)ϕ(x) + F (x)ψ(x) = D

(6)

identically, where D is a non-zero integer. Let q be a large prime for which the congruence (5) is soluble, and let x0 be a solution. By (6) we have F (x0 ) ≡ 0 (mod q), whence F (x0 + q) ≡ F (x0 ) (mod q 2 ). By choice of x1 as either x0 or x0 + q, we can ensure that fj (x1 ) ≡ 0 (mod q),

F (x1 ) ≡ 0 (mod q 2 ),

whence fj (x1 ) ≡ 0 (mod q 2 )

and fi (x1 ) ≡ 0 (mod q) for i = j.

By the hypothesis of Theorem 2, there exists x2 ≡ x1 (mod q 2 ) such that f (x2 ) = N (u1 , u2 , . . . , un )

(7)

for some rational u1 , u2 , . . . , un . From the preceding congruences we have fj (x2 ) ≡ 0 (mod q),

fj (x2 ) ≡ 0 (mod q 2 ),

fi (x2 ) ≡ 0 (mod q)

for i = j.

Hence f (x2 ) ≡ 0 (mod q ej ),

(8)

f (x2 ) ≡ 0 (mod q ej +1 ).

Let the prime ideal factorization of q in K be (9)

q = q 1 q 2 · · · ql ;

the factors are distinct since q is supposed to be sufficiently large. We note that l divides n because K is a normal field, and that Nqi = q n/ l .

(10)

Write the prime ideal factorization of u1 ω1 + . . . + un ωn in K in the form u1 ω1 + . . . + un ωn = qα1 1 · · · qαl l ab−1 , where a, b are ideals in K which are relatively prime to q. Then (11)

N (u1 ω1 + . . . + un ωn ) = ±q n(α1 +...+αl )/ l N a(N b)−1 ,

and Na, N b are relatively prime to q. It follows from (7), (8), (11) that n(α1 + . . . + αl )/ l = ej ,

32


whence n divides l. (ej , n) If (ej , n) = 1 we get that n divides l, whence l = n and it follows from (9) and (10) that q factorizes completely in K into prime ideal factors of the first degree. Now suppose that K is cyclic(1 ). The Galois group of K is a cyclic group G of order n; it has a unique subgroup H of order n/ l, and each qi is invariant under the automorphisms of H . The subgroup H determines a subfield L of K, of degree l, and H is the Galois group of K relative to L. A prime ideal factor of q in L cannot split further in K, since any such factors would be derived from one another by the automorphisms of H and so would not be distinct. Hence the factorization of q in L is also of the form (9). Comparison of norms shows that the qi , considered as prime ideals in L, are of the first degree. The unique subfield Kj of K, of degree n/(ej , n), is a subfield of L, and therefore q also factorizes completely in Kj , the number of prime ideal factors being equal to the degree of Kj and each being of the first degree. This completes the proof of Lemma 1. Lemma 2. Let G(x) be a polynomial with integral coefficients, irreducible over Q, and let G(θ ) = 0. Let J be any subfield of Q(θ ). Then (12) G(x) = aNJ H (x) identically, where H (x) is a polynomial over J , and NJ denotes the norm from J to Q, extended in the obvious way to apply to J [x], and a is rational. Proof. Let ω be a generating element of J and let ω(1) = ω, . . . , ω(m) be the conjugates of ω, where m is the degree of J . Since J is contained in Q(θ ), we have ω = g(θ), where g is a polynomial with rational coefficients. Thus G(x) has a zero in common with the polynomial (13)

m

g(x) − ω(j ) ,

j =1

which has rational coefficients, and since G(x) is irreducible, it must divide this polynomial. The factors of (13) are relatively prime in pairs, since their differences are non-zero constants. Hence the polynomials H (j ) (x) = G(x), g(x) − ω(j ) are relatively prime in pairs, and since each of them divides G(x), their product must (1 )

In dealing with this case we do not need to exclude the possibility that (ej , n) = 1.


33

divide G(x). Thus G(x) = A(x)

m

H (j ) (x) = A(x)NJ H (1) (x) .

j =1

The norm on the right is a non-constant polynomial with rational coefficients, so it follows from the irreducibility of G(x) that A(x) is a constant. This proves the result. Lemma 3 (Bauer). Let J be a normal number field and let k be any number field. Suppose that every sufficiently large prime which has at least one prime ideal factor of the first degree in k also has at least one prime ideal factor of the first degree in J . Then J is contained in k. Proof. See Bauer [1] or Hasse [3], pp. 138 and 141.

4. Proof of Theorem 2. Let f (x) be the polynomial of the theorem, and fj (x) any one of its irreducible factors, as in (4). Let θ be any zero of fj (x) and q any large prime which has at least one prime ideal factor of the first degree in Q(θ ). Then by Dedekind’s theorem the congruence fj (x) ≡ 0 (mod q) is soluble. If (ej , n) = 1, it follows from Lemma 1 that q factorizes completely in the field K. By Lemma 3, with J = K and k = Q(θ ), this implies that K is contained in Q(θ ). It follows now from Lemma 2, with G(x) = fj (x), that fj (x) is expressible identically in the form fj (x) = aj NK Hj (x) , as in (12). Hence (14)

e e e fj (x) j = aj j NK Hj j (x) = bj NK Hj∗ (x) .

Now suppose that K is cyclic. It follows from Lemma 1 that q factorizes completely in the field Kj . By Lemma 3 with J = Kj and k = Q(θ ), this implies that Kj is contained in Q(θ ). It follows now from Lemma 2, with G(x) = fj (x), that fj (x) is expressible identically in the form fj (x) = aj NKj Hj (x) . Now

(e ,n) NK Hj (x) = NKj Hj (x) j ,

since the degree of K relative to Kj is (ej , n). Hence e e /(e ,n) e (15) fj (x) j = aj j NK Hj (x) j j = bj NK Hj∗ (x) .

34


The conclusions (14) and (15), reached on two alternative hypotheses, are the same. By (4) and the multiplicative property of the norm, we have f (x) = aNK h(x) , where h(x) is a polynomial over K. By the hypothesis of the theorem, taking x to be a suitable integer, we infer that a is the norm of an element α of K. Putting αh(x) = ω1 u1 (x) + . . . + ωn un (x), we obtain

f (x) = N u1 (x), . . . , un (x)

identically.

5. Proof of the Corollary to Theorem 2. It follows from the theorem, on taking K = Q(i), that f (x) = U12 (x) + U22 (x), where U1 , U2 are polynomials with rational coefficients. Let U1 (x) + iU2 (x) = αν(x), where ν(x) is a primitive polynomial whose coefficients are integers in Q(i) and α is an element of Q(i). Then f (x) = |α|2 ν(x)¯ν (x). Since ν(x) and ν¯ (x) are both primitive and f (x) has integral coefficients, it follows from Gauss’s lemma that |α|2 is an integer. But |α|2 is a sum of two rational squares, and so it must be a sum of two integral squares, i.e. |α|2 = |β|2 , where β is an integer in Q(i). Putting βν(x) = u1 (x) + iu2 (x), where u1 , u2 are polynomials with (rational) integral coefficients, we get f (x) = u21 (x) + u22 (x).

6. Two examples (1) Suppose that f (x) = x 2 ,

K = Q(e2πi/8 ).

We prove first that every square is expressible as a value of the norm form of K. This norm form is √ N u1 + iu2 + iu3 + i 3 u4 = (u21 − u23 + 2u2 u4 )2 + (u22 − u24 − 2u1 u3 )2 .


35

Plainly 22 is representable with u1 = u3 = 0, u2 = u4 = 1. Also if p is a prime and p ≡ 1 (mod 4) then p = a 2 + b2 and p2 is representable with u1 = u3 = 0, u2 = a, u4 = b. Finally, if p ≡ 3 (mod 4) then p is representable either as a 2 − 2b2 or as a 2 + 2b2 , and we take u1 = b, u3 = ±b, u2 = a, u4 = 0. On the other hand, x 2 is not representable in the form x 2 = N u1 (x), . . . , u4 (x) , where u1 (x), …, u4 (x) are polynomials with rational coefficients. For if the greatest degree of any of these polynomials is g 1, then the coefficient of x 4g on the right is N(c1 , . . . , c4 ), where c1 , . . . , c4 are rational numbers, not all zero, and this coefficient is not 0. In this example, K is normal but not cyclic, and the multiplicity of the zero of f (x) is not relatively prime to the degree (namely 4) of K. (2) Suppose that f (x) = 2x 2 (x + 1)2 + 3x(x + 1) + 4,

√ K = Q −23 .

Here the norm form of K is N (u1 , u2 ) = u21 + u1 u2 + 6u22 . For every integer x we have x(x + 1) = 2t, where t is an integer, and f (x) = 8t 2 + 6t + 4 = N (t + 2, t). On the other hand, if u1 (x), u2 (x) are polynomials in x with integral coefficients, the coefficient of the highest power of x in N u1 (x), u2 (x) is an integer of the form a 2 + ab + 6b2 , and cannot be 2, since the least positive integer other than 1 represented by this form is 6.

References [1] M. Bauer, Zur Theorie der algebraischen Zahlkörper. Math. Ann. 77 (1916), 353–356. [2] E. Fried, J. Surányi, Neuer Beweis eines zahlentheoretischen Satzes über Polynome. Mat. Lapok 11 (1960), 75–84 (Hungarian). [3] H. Hasse, Bericht über Klassenkörpertheorie II. Jahresber. Deutsch. Math.-Verein., suppl. vol. 6 (1930). [4] −−, Beweis eines Satzes und Widerlegung einer Vermutung über das allgemeine Normenrestsymbol. Göttinger Nachr., 1931, 64–69. [5] D. Hilbert, Über die Irreducibilität ganzer rationaler Functionen mit ganzzahligen Coefficienten. J. Reine Angew. Math. 110 (1892), 104–129; Ges. Abhandlungen II, 264–286. Chelsea, New York 1965. [6] S. Lang, Diophantine Geometry. Interscience, New York and London 1962. [7] S. Lubelski, Zur Reduzibilität von Polynomen in der Kongruenztheorie. Acta Arith. 1 (1936), 169–183, and 2 (1938), 242–261. [8] G. Pólya, G. Szegö, Aufgaben und Lehrsätze aus der Analysis, II. Springer, Berlin 1954.

Originally published in Commentarii. Pontificia Academia Scientiarum II:20 (1969), 1–9


An improvement of Runge’s theorem on Diophantine equations

Summarium. Auctor investigat quando aequatio cum duabus variabilibus infinitum solutionum integralium numerum habere possit.

I The first general result concerning the number of integer solutions of a Diophantine equation is due to Runge [3] and in its simplest form is as follows. (i) If f (x, y) is a polynomial with integer coefficients irreducible in the rational field and the equation f (x, y) = 0 has infinitely many integer solutions, then the highest homogeneous part of f (x, y) is up to a constant factor a power of an irreducible form. The more general formulation refers to the highest isobaric part of f (x, y). The final result permitting to decide whether any given equation f (x, y) = 0 has infinitely many integer solutions is due to Siegel [4] and is as follows. (ii) If f (x, y) = 0 has infinitely many integer solutions, then there exist rational functions R(t), S(t) not both constant such that (1) f R(t), S(t) = 0 identically in t and either (2)

R(t) =

A(t) , L(t)m

S(t) =

B(t) L(t)m

or C(t) D(t) , S(t) = , Q(t)m Q(t)m where A, B, C, D, L, Q are polynomials with integer coefficients, L is linear, Q irreducible indefinite quadratic.

(3)

R(t) =

The aim of this note is to deduce from the two above results the following improvement of the first one. Paper presented on October 17th, 1968 by Pontifical Academician H.E. Wacław Sierpiński.

A7. Improvement of Runge’s theorem

37

Theorem. If f (x, y) is a polynomial with integer coefficients irreducible in the rational field and the equation f (x, y) = 0 has infinitely many integer solutions then the highest homogeneous part of f (x, y) is up to a constant factor a power of a linear or irreducible indefinite quadratic form. Proof. Let f (x, y) have degree n and denote by fn (x, y) its highest homogeneous part. By (i) either fn (x, y) = ax n or fn (x, y) = by n or (4)

fn (x, y) = ax n + . . . + by n

(ab = 0).

It remains to consider the last case. By (ii) we have (1) where either 1.

R, S are polynomials not both constant or

2.

(2) holds with m > 0, (A, B, L) = 1 or

3.

(3) holds with m > 0, (C, D, Q) = 1.

In the case 1. it follows from (1) and (4) that R and S are of the same degree. Denoting this degree by d and the leading coefficients of R and S by r and s, respectively, we get fn (r, s) = lim t −nd f R(t), S(t) = 0. t=∞

Hence fn (x, y) is divisible by sx − ry and by (i) fn (x, y) = c(sx − ry)n . In the case 2. let t0 be the zero of L(t). Clearly A(t0 ) = 0 or B(t0 ) = 0. Multiplying (1) by L(t)mn and substituting afterwards t = t0 we obtain fn A(t0 ), B(t0 ) = 0. Hence fn (x, y) is divisible by B(t0 )x − A(t0 )y and by (i) n fn (x, y) = c B(t0 )x − A(t0 )y . In the case 3. let t1 , t2 be the zeros of Q(t). Clearly C(ti ) = 0 or D(ti ) = 0

(i = 1, 2).

Multiplying (1) by Q(t)mn and substituting afterwards t = ti we obtain fn C(ti ), D(ti ) = 0 (i = 1, 2). Hence fn (x, y) is divisible by D(ti )x − C(ti )y and by (4) D(ti ) = 0 (i = 1, 2). If C(t1 )D(t1 )−1 is rational then by (i) n fn (x, y) = c D(t1 )x − C(t1 )y . If C(t1 )D(t1 )−1 is irrational, the C(ti )D(ti )−1 are conjugate in a real quadratic field and by (1) n/2 . fn (x, y) = c D(t1 )x − C(t1 )y D(t2 )x − C(t2 )y

38


Corollary. If fn (x, y) is an irreducible form of degree n > 2 and g(x, y) is a polynomial with integer coefficients of degree m < n then the equation fn (x, y) = g(x, y) has only finitely many integer solutions. The corollary represents an improvement on the analogous results with Roth [2] deduced from his famous theorem; this had stronger hypothesis m < n − 2. I conclude by expressing my thanks to Professors H. Davenport and D. J. Lewis for their helpful suggestion and in particular for pointing out the corollary which they were c first to prove.

II In this second part I wish to extend the result of the first part so as to improve on Runge’s theorem in its full generality. Let f (x, y) be a polynomial with integer coefficients irreducible in the rational field and suppose that the equation f (x, y) = 0 has infinitely many integer solutions. Then according to Runge [3] (see [6], p. 89): (1) the highest terms in x and y occur in f separately as ax m , by n ; (2) each branch of the algebraic function y of x defined by f = 0 tends to infinity with x and is of order x m/n , every term cx ρ y σ in f has nρ + mσ mn; (3) the sum g(x, y) of the terms with nρ + mσ = mn is expressible as b y ν − d (β) x μ (β = 1, . . . , nν ), where

β

u−d

(β)

β

is a power of an irreducible polynomial.

Runge does not say explicitly that m n = = (m, n), ν μ c

but what he really proves is that g(x, y) is up to a constant factor a power of an irreducible polynomial (for another proof see Skolem [5]). Therefore, factorizing if necessary y ν − d (β) x μ we can conclude that λ (4) g(x, y) = bh x m/d , y n/d , d = (m, n), where h(u, v) is an irreducible form. We shall prove: Theorem. If f (x, y) is a polynomial with integer coefficients irreducible in the rational field, of degree m in x and n in y, and the equation f (x, y) = 0 has infinitely many integer solutions then (1) and (2) hold and the sum g(x, y) of all terms cx ρ y σ of f with λ nρ + mσ = mn is of the form bh x m/d , y n/d , where d = (m, n) and h is a linear or irreducible indefinite quadratic form.

A7. Improvement of Runge’s theorem

39

The proof is based on the theorem of Siegel [4] quoted in part I, it will be however a little simpler if we reformulate the said theorem, examining Siegel’s argument. Siegel proves that if f (x, y) = 0 has infinitely many integer solutions then the genus of f (x, y) = 0 is zero(∗ ). In this case (cf. Skolem [6], p. 102) there is a parametrization (5)

x=

ϕ(u, v) , χ (u, v)

y=

ψ(u, v) , χ (u, v)

where ϕ, ψ, χ are relatively prime forms of the same positive degree with rational coefficients and where the equation χ (u, v) = h has infinitely many integer solutions for some h = 0. Now, as proved by Maillet [1] (cf. [6], p. 100) the last condition implies that (6)

χ (u, v) = c1 (a1 + b1 v)l

or χ (u, v) = d1 (a2 u2 + b2 uv + c2 v 2 )l ,

where b22 − 4a2 c2 is positive and is not a perfect square. The latter case by the substitution u = t, v = 1 leads to a parametrization C(t) D(t) (7) x(t) = , y(t) = , f x(t), y(t) = 0, Q(t)α Q(t)β where C, D, Q are polynomials with rational coefficients, Q is irreducible indefinite quadratic, α 0, β 0 and x(t), y(t) are not both constant. Moreover, and this remark of Maillet seems to have been so far overlooked, the former case leads to the same parametrization (7) with α = β = 0. Indeed on substituting u = t, v = b1−1 (1 − a1 t) we get from (5) and (6) ϕ t, b1−1 (1 − a1 t) ψ t, b1−1 (1 − a1 t) x(t) = , y(t) = c1 c1 and the polynomials on the right hand side which are not both constant can be taken as C(t), B(t) in (7). Therefore, if f (x, y) = 0 has infinitely many integer solutions then (7) holds and either (8)

α = β = 0,

C, D are not both constant

α + β > 0,

(C, Qα ) = (D, Qβ ) = 1.

or (9)

Proof of the theorem. By Runge’s theorem we have (1), (2) and (4) and it remains to show that h is linear or indefinite quadratic. Set m/d = μ, n/d = ν. In the case (8) let γ , δ be the degrees of C, D respectively and c0 , d0 their leading coefficients. If t tends to infinity then x is of order t γ , y of order t δ and by (2) δ = γ m/n. Thus we get from (7) g(c0 , d0 ) = lim t −γ m f x(t), y(t) = 0, t=∞

(∗ ) The assumptions imply the absolute irreducibility of f , hence the genus is defined.

40


from (4)

μ h c0 , d0ν = 0 μ

and h(u, v) is divisible by d0ν u − c0 v. Since h is irreducible it must be linear. In the case (9) let t1 , t2 be the zeros of Q(t). If t tends to ti then x is of order (t − ti )−α c (possibly tends to 0 if α = C(ti ) = 0), y of order (t − ti )−β (possibly tends to 0 if β = D(ti ) = 0) and by (2) β = αm/n, C(ti ) = 0 = D(ti ) (i = 1, 2). Thus we get from (7) g C(ti ), D(ti ) = lim Q(t)αm f x(t), y(t) = 0, t=ti

from (4)

h C(ti )μ , D(ti )ν = 0

and h(u, v) is divisible by D(ti )ν u − C(ti )μ v (i = 1, 2). If C(t1 )−μ D(t1 )ν is rational h must be linear as before. If C(t1 )−μ D(t1 )ν is irrational then C(ti )−μ D(ti )ν are conjugate in a real quadratic field, h is divisible by D(t1 )ν u − C(t1 )μ v D(t2 )ν u − C(t2 )μ v and h is indefinite quadratic. This completes the proof. It should be noted that the above proof does not share an essential advantage of Runge’s proof, namely it does not permit to estimate the size of solutions of f (x, y) = 0 if the theorem implies the finiteness of their number. The reason for this defect is the noneffective character of Siegel’s theorem.

References [1] E. Maillet, Détermination des points entiers des courbes algébriques unicursales à coefficients entiers. C. R. Acad. Sci. Paris 168 (1919), 217–220. [2] K. F. Roth, Rational approximations to algebraic numbers. Mathematika 2 (1955), 1–20; corrigendum, ibid., 168. [3] C. Runge, Über ganzzahlige Lösungen von Gleichungen zwischen zwei Veränderlichen. J. Reine Angew. Math. 100 (1887), 425–435. [4] C. L. Siegel, Über einige Anwendungen diophantischer Approximationen. Abh. Preuss. Akad. Wiss. Phys.-math. Kl. Nr. 1 (1929); Ges. Abhandlungen I, Springer, Berlin 1966, 209–266. [5] Th. Skolem, Über ganzzahlige Lösungen einer Klasse unbestimmter Gleichungen. Norsk. Mat. Forenings Skrifter (I) Nr. 10 (1922). [6] −−, Diophantische Gleichungen. Berlin, 1938.

Originally published in Acta Arithmetica XXXI (1976), 199–204


On the equation y m = P (x) with R. Tijdeman (Leiden)

The aim of this paper is to prove the following Theorem. If a polynomial P (x) with rational coefficients has at least two distinct zeros then the equation (1)

y m = P (x),

x, y integers,

|y| > 1,

implies m < c(P ) where c(P ) is an effectively computable constant. For a fixed m the Diophantine equation (1) has been thoroughly investigated before (see [1] and [4]) and the known results together with the above theorem imply immediately Corollary 1. If a polynomial P (x) with rational coefficients has at least two simple zeros then the equation (1) has only finitely many integer solutions m, x, y with m > 2, |y| > 1 and these solutions can be found effectively. Corollary 2. If a polynomial P (x) with rational coefficients has at least three simple zeros then the equation (1) has only finitely many integer solutions m, x, y with m > 1, |y| > 1 and these solutions can be found effectively. A simple proof of the special case of Corollary 1 that P (x) has at least two simple rational zeros can be found in a survey paper by the second named author [6]. Corollary 2 is a step towards the following Conjecture. If a polynomial P (x) with rational coefficients has at least three simple zeros then the equation y 2 z3 = P (x) has only finitely many solutions in integers x, y, z with yz = 0. This conjecture lies rather deep, since it implies the existence of infinitely many primes p such that 2p−1 ≡ 1 (mod p2 ). The proof of the theorem is based on Baker’s work [2] and on two lemmata. We denote by x the distance from x to the nearest integer.

42


Lemma 1. For any complex numbers X, Y different from 0, a positive integer h and any choice of the roots X 1/ h , Y 1/ h we have |X 1/ h − Y 1/ h |

(2)

⎧ 1 1 ⎪ log |XY −1 | if |X| = |Y |, min 1, 1 − ⎨ 1/ h e h max |X|, |Y | · log XY −1 ⎪ ⎩4 if |X| = |Y |. h 2πi

Proof. We can assume without loss of generality that |X| 1 = Y 1/ h . If |X| > 1 we have

|X1/ h − 1| |X|1/ h − 1 = |X|1/ h 1 − |X|1/ h ,

and if |X| eh the inequality (2) follows immediately. To settle the case eh > |X| > 1 we verify by differentiation that the function f (t) = (1 − t −1 )/ log t is decreasing in the interval (1, e). Since f (e) = 1−e−1 , (2) follows on taking t = |X|1/ h . Suppose now that |X| = 1, X = cos ϕ + i sin ϕ,

ϕ = i −1 log X.

Then X1/ h = cos

ϕ + 2πj ϕ + 2πj + i sin h h

for some integer j

and

ϕ + 2πj |X1/ h − 1| = 2 sin . 2h However, sin ψ/ψ is decreasing on (0, π/2). Hence for all real ψ ψ |sin ψ| 2 π and ϕ + 2πj 4 log X |X1/ h − 1| 4 . 2πh h 2π i In the following lemma we denote the height of an algebraic number x by H (x).

Lemma 2. If γ1 , γ2 are algebraic integers of a field K of degree d then (3)

H (γ1 /γ2 ) 3d2d

σ

max |γ1σ |, |γ2σ | ,

A8. On the equation y m = P (x)

43

where σ runs through all the isomorphic injections of K into the complex field. Moreover, if K = K (the bar denoting complex conjugation) then 2 max |γ1σ |, |γ2σ | . H |γ1 /γ2 |2 3d2d σ

Proof. Clearly γ1 /γ2 satisfies the equation γ2σ x − γ1σ = 0. F (x) = σ

F (x) has rational integral coefficients, but it may be reducible. We have F (x) = NK/Q γ2 · f (x)r , where f is the minimal polynomial of γ1 /γ2 . By Gauss’s lemma F (x) = c · g(x)r , where c is an integer, g has integral coefficients and is irreducible as a constant multiple of f . By an inequality of Gel fond ([3], p. 139) we have H (F )

c

1 1 H (g)r H (g), 3dr 3d

unless H (g) = 1,

where H (P ) denotes the height of the polynomial P . On the other hand, max |γ1σ |, |γ2σ | . |γ1σ | + |γ2σ | 2d H (F ) σ

σ

This implies (3). Now if K = K we have |γi2 | = γi γ i ∈ K (i = 1, 2). Hence H |γ1 /γ2 |2 3d2d max |γ1σ γ σ1 |, |γ2σ γ σ2 | σ

3d2

d

= 3d2

d

max |γ1σ |, |γ2σ | · max |γ σ1 |, |γ σ2 |

σ

max

2 |γ1σ |, |γ2σ | .

σ

σ

Proof of the Theorem. Let K be the splitting field of P and let bP (x) = a

n

(x − αi )ri

(αi distinct, b integer)

i=1

have integral coefficients. It follows from (1) that (4)

n i=1

(ax − aαi )ri = ba N−1 y m ,

N=

n

ri ,

i=1

where the numbers aαi are algebraic integers. Since for integer x (ax − aαi , ax − aαj ) | (aαi − aαj ),

44


the highest common ideal divisor of any two factors on the left hand side of (4) is composed exclusively of prime ideals of K dividing Δ= (aαi − aαj ). 1i<j n

Hence, for each i n we have (ax − aαi )ri = dcm

(5)

for some ideals d and c such that d is composed exclusively of prime factors of abΔ and (c, abΔ) = 1. If p is a prime ideal and pt cm then clearly m | t and by (5) ri | t, thus m t r [m, ri ] | t. It follows that . Moreover d = di i and we get from (5) (m, ri ) ri m (6) (ax − aαi ) = di csi , s = . (m, [r1 , . . . , rn ]) Let p1 , . . . , pk be all prime ideal divisors of abΔ in K and let h be the class number of K. We have phj = (πj )

(1 j k),

chi = (γi )

(1 i n),

and by (6) for suitable integer exponents yij 0 (ax − aαi )h =

k

y πj ij γis .

j =1

If ε0 , ε1 , . . . , εr are a basis for the group of units in K we get (ax − aαi )h =

(7)

r

x

εqiq

k

y

πj ij γis

(1 i n),

j =1

q=0

where we can suppose without loss of generality that 0 xiq < s,

(8)

0 yij < s,

since any product r q=0

x εq q

k

y

πj j

with xq ≡ yj ≡ 0 (mod s), yj 0,

j =1

can be incorporated in γi . By our assumption n 2. We use (7) for i = 1, 2, denoting the right hand side of (7) by X and Y , respectively. If X = Y we have (ax − aα1 )h = (ax − aα2 )h and it follows, from α1 = α2 , that ax − aα1 = e2πig/ h (ax − aα2 ), 0 < g < h, and |x|

|α1 | + |α2 | . 2 sin(π/ h)

A8. On the equation y m = P (x)

45

Since |y| > 1, equation (1) gives m < c1 , where c1 as the subsequent constants c2 , c3 , . . . depends only on P and is effectively computable. log XY −1 If X = Y we have either |X| = |Y | or |X| = |Y | and = 0. In the former 2π i case we infer by (8) from Baker’s theorem [2] that log |XY −1 | > H |γ1 /γ2 |2 −c2 log s , in the latter case similarly log XY −1 −c log s , > H γ1 /γ2 3 2πi where in case H ( ) = 1, it should be replaced by 2. In virtue of Lemmata 1 and 2 we have in both cases |aα1 − aα2 | = |X1/ h − Y 1/ h | 1/ h −c log s max |γ1σ |, |γ2σ | 5 > c4 max |X|, |Y | >

c6−s

σ

s/ h

max |γ1 |, |γ2 |

−c log s max |γ1σ |, |γ2σ | 5

σ

for some constant c6 > 1. Applying any isomorphic injection τ of K into C to both sides of (7) and arguing as c before we get s/ h −c log s |aα1τ − aα2τ | > c6−s max |γ1τ |, |γ2τ | max |γ1σ |, |γ2σ | 5 . σ c

On taking the product over all injections τ we obtain s/ h−c5 d log s |NK/Q (aα1 − aα2 )| > c6−ds max |γ1σ |, |γ2σ | . σ

Since the left hand side is independent of s, this implies that either s c7 or max |γ1σ |, |γ2σ | < c62dh . σ

In the former case we have m c7 [r1 , . . . , rn ], in the latter case, by (7), (9)

k N (πj )y1j +y2j G s , NK/Q (ax − aα1 )h (ax − aα2 )h = ± j =1

where G = |N γ1 γ2 | < c64dh . The greatest prime factor of the right hand side of (9) is bounded by abΔc64dh . The left hand side of (9) is a polynomial in x with integer coefficients and at least two distinct zeros. It has been proved by the first named author, M. Keates, S.V. Kotov andV. G. Sprindzhuk (see [5]) that the greatest prime factor of such a polynomial exceeds c8 log log |x|. So we obtain |x| c9 and in view of (1) with |y| > 1, m c10 .

46


References [1] A. Baker, Bounds for the solutions of the hyperelliptic equation. Proc. Cambridge Philos. Soc. 65 (1969), 439–444. [2] −−, A sharpening of the bounds for linear forms in logarithms. Acta Arith. 21 (1972), 117–129. [3] A. O. Gel fond, Transcendental and Algebraic Numbers. Gosudarstv. Izdat. Tekhn.-Teor. Lit., Moscow 1952 (Russian). English translation: Dover Publ., New York 1960. [4] W. J. LeVeque, On the equation y m = f (x). Acta Arith. 9 (1964), 209–219. [5] V. G. Sprindzhuk, An effective analysis of the Thue and Thue–Mahler equations. In: Current Problems of Analytic Number Theory (Minsk 1972), Nauka i Tekhnika, Minsk 1974, 199–222 (Russian). [6] R. Tijdeman, Applications of the Gel fond–Baker method to rational number theory. In: Topics in Number Theory (Debrecen 1974), Colloq. Math. Soc. János Bolyai 13, North-Holland, Amsterdam 1976, 399–416.


Originally published in Journal für die reine und angewandte Mathematik 309 (1979), 176–182

Zeta functions and the equivalence of integral forms with R. Perlis* (Bonn)

1. Two algebraic number fields K, K are said to be arithmetically equivalent when their zeta functions ζK (s) and ζK (s) coincide. In this paper we give an example of two nonisomorphic arithmetically equivalent fields of class number one, and then show that the norm forms from the rings of integers of these fields provide a negative answer to an old conjecture concerning integral forms. We will consider the values of integral forms when the variables run through Z. The fixed divisor of a form g ∈ Z[X1 , . . . , Xn ] is the gcd of its values. Let w(R) be the number of integers in the interval (−R, R) which appear as values of g. Then g has density zero if the quotient w(R)/R approaches zero as a limit as R increases to infinity. Let X denote the n-tupel of variables (X1 , . . . , Xn ), let A be an n-by-n matrix, and let X · A be the linear change of variables defined by A. Chowla has made the following conjecture (see [2], and [4], p. 23): Let g and h be irreducible integral forms of degree d in n variables, each having density zero and fixed divisor one, and representing the same numbers when the variables run through Z. Then there is an integral matrix A of determinant ±1 for which g(X) = h(X · A). To support this conjecture, Chowla claimed it is true for binary quadratic forms, but this was due to an oversight. For example, g = X2 + XY + Y 2 and h = X2 + 3Y 2 satisfy all the hypotheses of the conjecture, but it is easy to check that any linear transformation taking h to g has determinant 1/2, and hence cannot be integral.

*

Supported by the Sonderforschungsbereich „Theoretische Mathematik” of Bonn University.

48


2. There are several ways in which this conjecture can be mended for binary quadratic forms: a)

One can additionally require that g and h have the same discriminant.

b)

One can additionally require that the integers represented properly by g (i.e., those values g(x1 , . . . , xn ) with (x1 , . . . , xn ) = 1) coincide with the integers represented properly by h.

c)

One can additionally require that g (and h) is integrally equivalent to any form f ∈ Z[X1 , . . . , Xn ] representing it integrally.

d)

One can weaken the conclusion to assert only that g and h are equivalent via a rational transformation of non-zero determinant.

With each of these alterations, the validity of Chowla’s conjecture for binary quadratic forms follows easily from the following reformulation of a theorem of Schering ([9]): Let g and h be two primitive binary quadratic forms with |disc. g| |disc. h|. Then g and h have the same sets of values if and only if either (i) g and h are integrally equivalent, or (ii) disc. g = D ≡ 5 (mod 8), the equation X2 − DY 2 = 4 has proper solutions, and h(X, Y ) is integrally equivalent to the form g(X, 2Y ). Let g and h be the norm forms from the rings of integers of the fields Q (−3)1/8 and Q (−48)1/8 . We will show below that these forms are a counterexample to the conjecture even when one adds all the extra hypotheses a), b), c), and weakens the conclusion to d). Despite this negative result, certain special versions of Chowla’s conjecture may be true. For binary forms of any degree the situation has been studied in [10], and results of a positive nature have been found. For more than two variables, the problem of relating the equivalence of forms to their values seems much more complicated; however, the following comment may be of some interest.Although discriminants have long been defined for forms of any degree and any number of variables (see for example [5], lecture 44, p. 159), their arithmetic meaning is far from clear in general: when the number of variables exceeds two, the discriminant is zero for any form that is not absolutely irreducible. In particular, it vanishes for such well-behaved forms as norm forms (the forms of our counterexample). This suggests that perhaps further investigations of Chowla’s conjecture should initially be restricted to nonsingular forms.

3. Let K = Q (−3)1/8 and K = Q (−48)1/8 . These are nonisomorphic fields of degree eight whose zeta functions coincide (see [7], p. 351). We will show in Sections 4 and 5 that both K and K have class number one. Using this result from below, we will now construct a counterexample to Chowla’s conjecture with all the variations a), b), c), and d) of Section 2.

A9. Equivalence of integral forms

49

Select a basis {θi } for the integers of K and {θi } for the integers of K and define (1)

g(X1 , . . . , X8 ) = normK/Q (X1 θ1 + . . . + X8 θ8 ) h(X1 , . . . , X8 ) = normK /Q (X1 θ1 + . . . + X8 θ8 ).

Then g and h are irreducible ([1], Th. 2, p. 80) integral forms of degree eight in eight variables, of density zero ([6]) and fixed divisor one. By the remark at the end of Section 2, they have the common discriminant zero. To check the validity of assumption b), we will characterize the sets of numbers properly represented by either form. Consider one of these forms, say g. Any value g(x1 , . . . , x8 ) is the norm of the algebraic integer xi θi . This norm is positive, since K is totally imaginary, and equals the norm of the ideal xi θi . Conversely, K has class number one, as will be shown below, so the norm of any ideal is the norm of an element, and this means that the values of g are precisely the norms of integral ideals. One once that x1 , . . . , x8 are relatively prime if and only if the corresponding sees at ideal xi θi is primitive, i.e., not divisible by any rational integer. Consider two ideals A and B whose norms are relatively prime. If their product AB is not primitive, then it is divisible by some prime number p and hence by every prime ideal factor of p. All of these latter divide exactly one of A or B, since (norm A, norm B) = 1, and therefore p divides either A or B. That is, the product of primitive ideals with relatively prime norms is again primitive. It is obvious that any factor of a primitive ideal is primitive, and this together with the previous fact means: g (and similarly, h) properly represents a product of relatively prime numbers if and only if it properly represents each factor. Thus, in order to know which values are properly represented by g, it suffices to know which prime powers pt are properly represented. Let p = i Piei (i = 1, . . . , s) be the factorization of a prime number p in K, with ramification indices ei and inertia degrees fi , and supposethat pt is the norm of an integral ideal A. Then this ideal necessarily has the form A = i Pivi (i = 1, . . . , s) with nonnegative v’s, and is primitive if and only if at least one of these vi is strictly less than the shows: pt is properly represented corresponding ramification index ei . Taking norms then by g if and only if the exponent t can be written as t = vi fi with nonnegative v’s, and vi < ei for at least one i. The analogous characterization holds for the prime powers properly represented by the form h, of course, with the ramification indices ei and the inertia degrees fi now computed in K instead of in K. As has already been mentioned, K and K have equal zeta functions. From this, it automatically follows that the prime ideal factors Pi of p in K can be paired with the factors Pi in K so that the corresponding inertia degrees are equal, fi = fi (i = 1, . . . , s) (see [7], Th. 1, p. 345). In general, ei and ei do not have to agree under this correspondence, but for the specific fields K, K in question, they do. The ramified primes are 2, 3 and 3, being an eighth power, obviously ramifies totally in K. It follows from the equality of the zeta functions that there is a unique prime of K lying over 3, and that it has inertia degree one, so 3 ramifies totally in K . We will show in the next section that (2) = P 4 in K and (2) = P 4 in K . With this, it follows that either both of g and h or neither of them properly represents a given pt , and hence they properly represent the same sets of values.

50


We now turn to the task of verifying assumption c). Suppose g(X) = f (X · A) with f ∈ Z[X1 , . . . , X8 ] and an integral matrix A of determinant d = 0. We will show that d = ±1. Set B = d · A−1 . Then B = (bij ) is an integral matrix of determinant d 7 , and g(X · B) = f (dX), which can be rewritten to give (2) where λi =

j

norm(X1 λ1 + . . . + X8 λ8 ) = d 8 · f (X) bij θj . If A is the integral ideal generated by the λ’s, then (2) implies

(3)

norm A d 8 .

(4)

(y1 , . . . , y8 ) = (x1 , . . . , x8 ) · A

Consider the lattice Λ = Zλi contained in A. Then B is the matrix of the transition map from the lattice O = Zθi of all integers in K to the lattice Λ, so the index of Λ in O is [O : Λ] = |det B| = |d|7 . On the other hand, Λ A so [O : Λ] d 8 , by (3), implying d = ±1. Suppose finally there were a rational matrix A giving a linear substitution

for which (5)

g(x1 , . . . , x8 ) = h(y1 , . . . , y8 ).

For xi ∈ Q let yj be defined by (4) and consider the map T : K → K taking xi θi to yj θj . This is an additive map which preserves norm. Hence for α ∈ K and for every natural number n (6) norm(n − α) = norm T (1) · n − T (α) , and this implies that the polynomials (7)

s(X) = norm(X − α) and

t (X) = norm T (1) · X − T (α)

coincide. If we select α to generate K over Q, then the roots of s(X) are the conjugates of α. Since T (α) · T (1)−1 lies in K and is a root of t (X) = s(X), it follows that K contains a conjugate of α. Since K and K have the same degree, this implies K ∼ = K , and this contradiction shows that g and h are not rationally equivalent.

4. It remains to show that the class numbers of K = Q (−3)1/8 and K = Q (−48)1/8 are one. We begin by proving the implication (8)

hK = 1 =⇒ hK = 1;

√ then we will show in the next section that hK = 1. So, assume hK = 1 and let L = K( 2). √ Then L is also K ( 2). We contend that each of the quadratic extensions L/K and L/K has exactly one ramified prime, which then necessarily ramifies totally. Clearly any ramified prime is finite and lies over 2. Consider first K. We check easily that (2) = P 4 in K where P = 1 + (−3)1/8 has norm 4. Since 2 is inert in the subfield √ Q( −3) of K, it follows that P is a prime ideal, and the only prime ideal of K lying

A9. Equivalence of integral forms

51

above 2. Hence P is the only prime of K that can possibly ramify in L. Since hK = 1, the abelian extension L/K must ramify, so P ramifies; i.e., 2 is the eighth power of a prime ideal in L. Intersecting this ideal with K shows that there is only one prime P of K lying above 2, and P ramifies in L. This proves the statement in italics above, and also shows that (2) = P 4 in K , settling the question of ramification indices left open in Section 3. Iwasawa has shown that the validity of the italic statement just proved suffices to conclude that (9)

2 | hK ⇐⇒ 2 | hL ⇐⇒ 2 | hK

(see [3]). Since hK = 1, it follows that hK is odd. But it has been proved in [8] that the class number quotient hK / hK = 2i for an integer i. The left side of this equality being odd, we have i = 0, which proves (8).

5. It is to be shown that hK = 1. Set θ = (−3)1/8 and let (x0 , x1 , . . . , x7 ) denote i xi θ i . By considering the tower of quadratic extensions of Q leading up to K, it is easy to verify that 1 a, b, c, d, e, f, g, h ∈ Z (10) a, b, c, d, a + 2e, b + 2f, c + 2g, d + 2h c 2 is the full ring of integers of K. From this, we calculate the discriminant of K to be DK = 216 · 37 . The ideal class group of K is generated by the prime √ ideals whose norm does not exceed the Minkowski bound MK = N ! · N −N · (4/π )t · |DK |, where t = 4 is the number of complex valuations of K and N = 8 is the field degree. We have MK < 76. The ramified primes are 2 and 3. We have already seen in Section 4 that each of these has exactly one prime factor in K and this factor is principal. Let P be a prime ideal of K whose norm pf does not exceed 76. Excluding the ramified primes, we have p 5 and hence f 2. If f = 2, then p is either 5 or 7. The irreducible polynomial of θ is F (X) = X8 + 3 and has the discriminant 224 · 37 . The only primes dividing this discriminant are the ramified primes 2 and 3, so there are no inessential discriminant divisors, and hence the splitting of an unramified prime number p is determined by the factorization of F (X) modulo p. Since F (X) is irreducible modulo 5, the prime number 5 is inert in K and hence has no prime factor of degree f = 2. Modulo 7, the polynomial factors as F (X) = (X + 2)(X − 2)(X2 + 4)(X2 − X + 4)(X2 + X + 4), so 7 splits into the product of five prime ideals in K, two of norm 7 (f = 1) and three of norm 49 (f = 2). We will show that each of these five ideals is principal. To this end, it suffices to exhibit five integers in K, two of norm 7 and three of norm 49, such that none is divisible by any of the others. These integers are given in Table 1 at the end of this paper. With this, we know that the ideal class group of K is generated by prime ideals of degree f = 1, lying over prime numbers p in the range 7 < p < 76. Now, a prime number p has a factor of degree f = 1 if and only if X8 + 3 has a root modulo p. In particular, −3 must be a square modulo p, and with the law of quadratic reciprocity it

52


follows that p ≡ 1 (mod 3). There are eight prime numbers satisfying this congruence and lying in our range; we divide them into three sets, as follows: A = {p with 2 (p − 1)} = {19, 31, 43, 67}; (11)

B = {p with 4 (p − 1)} = {13, 37, 61}; C = {p with 8 | (p − 1)} = {73}.

The multiplicative group of Z/pZ is cyclic of order p − 1. This implies that −3, being a quadratic residue, is automatically an eighth-power residue modulo each of the four primes in the set A, and in fact it implies that each of these primes has exactly two prime factors in K of degree f = 1. A moment’s thought shows that these two prime factors are conjugate via the automorphism of K sending θ to −θ , and thus one factor is principal if and only if the other one is. Hence, for p ∈ A, we will have proved that every prime factor of p with f = 1 is principal as soon as we have exhibited an integer of K with norm p. These are given in Table 2. Let p be a prime in the set B. Then −3 is an eighth power if and only if it is a fourth power modulo p. This shows that 13 has no factor of degree one in K, while each of 37 and 61 has precisely four prime factors of degree one. We will prove all these to be principal by exhibiting four non-associated integers of norm p, for p = 37 and 61. The only element of C is p = 73. One checks directly that −3 is not a fourth power and hence certainly not an eighth power modulo 73, implying that 73 has no prime factor of degree f = 1. Thus, the tables below prove that K, and consequently, as shown above in Section 4, also that K , has class number one. We conclude with a comment of the tables. Due to the simplicity of the defining polynomial X 8 + 3, it is more-or-less trivial to multiply two integers of K given in the form (10) and to express the result again in this form. Computing norms via the tower of quadratic subfields leading from K down to Q, it takes three such multiplications to check that an integer in these tables has the norm claimed for it. When this has been done, it remains to check that various entries are not divisible by others. This can be done efficiently as follows. Let I denote an integer with norm N . The entry T appearing below denotes the quotient N/I ; its value can be checked by computing T · I . Finally, to show that an integer J is not divisible by I , it suffices to multiply J · T and to find a single coordinate which is not divisible by N (up to a factor of 1/2, due to the form (10) of integers in K). Let I σ denote the integer obtained from I by changing the sign of the second, fourth, sixth, and eighth component; i.e., by replacing θ by −θ . Since the primes in these tables are unramified in K, one can show that when the norm of I is prime, then I cannot divide I σ . With these easy observations, it is possible to check the accurracy of the following tables, and hence that hK = hK = 1, in a short amount of time, by hand.

53

A9. Equivalence of integral forms Table 1 Integer I I1 = ( 1, 1, 1, 0, 0, 0, I1σ = ( 1, −1, 1, 0, 0, 0, I2 = (−1, 0, 1, 0, −1, 0, I3 = ( 1, −1, −1, 0, −1, 1, I3σ = ( 1, −1, −1, 0, −1, −1,

0, 0, 0, 0, 0,

N = norm I 0) 7 0) 7 0) 49 0) 49 0) 49

T = N/I (−2, −1, 3, −2, −1, 3, −2, −1) (−2, 1, 3, 2, −1, −3, −2, 1) (−7, 0, 14, 0, 21, 0, 7, 0) ( 7, 35, 14, 0, 21, 7, −7, 0) ( 7, −35, 14, 0, 21, −7, −7, 0)

Table 2 Prime p 19 31 43 67 37 37 37 37 61 61 61 61

Integer I of norm p = (2, 1, 1, 0, 0, −1, −1, −1) = (1, −1, −1, 1, 0, −1, 0, 0) = (1, 0, −1, 1, −1, 1, −1, 1) = (1, 0, 0, 0, −1, 1, 0, 0) I1 = (1, 1, 0, 1, 0, 0, 0, 0) I1σ = (1, −1, 0, −1, 0, 0, 0, 0) I2 = (1, 1, 1, 0, 0, 0, −1, −1) I2σ = (1, −1, 1, 0, 0, 0, −1, 1) J1 = (1, 1, 1, 1, 1, 0, 0, 0) J1σ = (1, −1, 1, −1, 1, 0, 0, 0) J2 = (1, −2, 23 , −2, 1, −1, 21 , 0)

T = N/I (−25, −1, 22, −9, −11, 14, −4, −7) (−35, 28, −10, 8, 6, −11, 15, −12) (−29, −39, −5, 17, 11, −3, −7, −2) ( 64, 42, 15, −9, −8, −22, −27, −24) ( 7, −16, −11, 4, 12, −1, −3, −9) ( 7, 16, −11, −4, 12, 1, −3, 9) ( 31, −40, 54, −47, 32, −21, 8, 4) ( 31, 40, 54, 47, 32, 21, 8, −4) ( −2, −7, 6, 21, −18, −2, −7, 6) ( −2, 7, 6, −21, −18, 2, −7, −6) −13 43 7 5 ( 59 8, −67 2 , 2 , 2 , 2 , 11, 2, 2)

−67 13 43 J2σ = (1, 2, 23 , 2, 1, 1, 21 , 0) ( 59 2 , −8, 2 , 2 , 2 , −11,

7 −5 2, 2 )

References [1] Z. I. Borevich, I. R. Shafarevich, Number Theory. Academic Press, New York 1966. [2] S. Chowla, Some problems of elementary number theory. J. Reine Angew. Math. 222 (1966), 71–74. [3] K. Iwasawa, A note on class numbers of algebraic number fields. Abh. Math. Sem. Univ. Hamburg 20 (1956), 257–258. [4] W. J. LeVeque, A brief survey of Diophantine equations. In: Studies in Number Theory, Math. Assoc. Amer., Buffalo 1969, 4–24. [5] E. Netto, Vorlesungen über Algebra, II. Leipzig 1900. [6] R. W. K. Odoni, On the norms of algebraic integers. Mathematika 22 (1975), 71–80. [7] R. Perlis, On the equation ζK (s) = ζK (s). J. Number Theory 9 (1977), 342–360. [8] R. Perlis, On the class numbers of arithmetically equivalent fields. J. Number Theory 10 (1978), 489–509. [9] E. Schering, Théorèmes relatifs aux formes binaires quadratiques qui représentent les mêmes nombres. J. Math. Pures Appl. (2) 4 (1859), 253–270. [10] A. Schinzel, On the relation between two conjectures on polynomials. Acta Arith. 38 (1980), 285–322; this collection: J5, 1154–1191.

Originally published in Acta Arithmetica XXXVII (1980), 133–141


Quadratic Diophantine equations with parameters with D. J. Lewis* (Ann Arbor)

To the memory of Paul Turán

1. In an earlier paper [3] written in collaboration with the late Harold Davenport we proved: Theorem A. Let a(t), b(t) be polynomials with integral coefficients. Suppose that every arithmetical progression contains an integer τ such that the equation a(τ )x 2 + b(τ )y 2 = z2 has a solution in integers x, y, z, not all 0. Then there exist polynomials x(t), y(t), z(t) in Z[t], not all identically 0, such that a(t)x(t)2 + b(t)y(t)2 ≡ z(t)2 identically in t. From this result we derived: Theorem B. Let F (x, y, t) be a polynomial with integral coefficients which is of degree at most 2 in x and y. Suppose that every arithmetical progression contains an integer τ such that the equation F (x, y, τ ) = 0 is soluble in rational numbers for x and y. Then there exist rational functions x(t), y(t) in Q(t) such that F x(t), y(t), t ≡ 0 identically in t. Earlier, one of us asked [6] whether a result similar to Theorem B holds if F (x, y, t) is replaced by any polynomial F (x, y, t1 , . . . , tr ) and the stronger assumption is made that for all integral r-tuples τ1 , . . . , τr , the equation F (x, y, τ1 , . . . , τr ) = 0 is soluble in the rational numbers for x and y. The stronger assumption is needed since the hypothesis analogous to the one of Theorem B involving arithmetical progressions is not sufficient already for F (x, y, t) = x 2 − y 3 − t. We shall show here that if F is of degree at most 2 in x and y a hypothesis analogous to the one of Theorem B suffices for any number of *

This paper was written while the authors were partially supported by an NSF grant.

A10. Quadratic Diophantine equations

55

parameters ti . We shall also indicate an equation of an elliptic curve over Q(t) for which the stronger assumption involving all integers t does not seem to suffice. As for allowing more variables, we note that in virtue of Gauss’s theorem, for every integer τ , the equation x 2 + y 2 + z2 = 28τ 2 + 1 is soluble in integers x, y, z, but there do not exist rational functions x(t), y(t), z(t) in Q(t) such that x(t)2 + y(t)2 + z(t)2 ≡ 28t 2 + 1 identically in t, since 28 is not the sum of three rational squares. A. Pfister has shown us a more refined example of the equation x 2 + y 2 + z2 = 5t 2 + 13 which for all rational values of t is soluble with x, y, z in Q, without being soluble with x, y, z in Q(t). We now turn to the crucial lemma from which the generalization of Theorems A and B in the case of several parameters will be deduced in §3.

2. Lemma 1. Let a(t1 , . . . , tr ), b(t1 , . . . , tr ), c(t1 , . . . , tr ) ≡ 0 be polynomials with integral coefficients. Suppose that for all r-tuples of integers τ1 , . . . , τr such that c(τ1 , . . . , τr ) = 0 the equation a(τ1 , . . . , τr )x 2 + b(τ1 , . . . , τr )y 2 = z2

(1)

has a solution in integers x, y, z not all 0. Then there exist polynomials x(t1 , . . . , tr ), y(t1 , . . . , tr ), z(t1 , . . . , tr ) with integral coefficients, not all identically 0, such that (2)

a(t1 , . . . , tr )x(t1 , . . . , tr )2 + b(t1 , . . . , tr )y(t1 , . . . , tr )2 ≡ z(t1 , . . . , tr )2

identically in t1 , . . . , tr . Proof. The proof is by induction on r. For r = 1 the result follows from Theorem A since clearly every arithmetical progression contains an integer τ for which c(τ ) = 0. Alternatively, with the stronger hypothesis of our lemma one can give a simpler direct proof for the case r = 1 following the arguments of Theorem A. Suppose the lemma is true for fewer than r parameters. We can obviously suppose that neither a(t1 , . . . , tr ) nor b(t1 , . . . , tr ) is identically 0, since otherwise the conclusion follows trivially. Denote the degree of a polynomial q in tr by |q|. We now proceed by induction on the degree of ab with respect to tr . If |a| + |b| = 0, the hypothesis of the lemma holds for c (t1 , . . . , tr−1 ) = c(t1 , . . . , tr−1 , τ ), where τ is an integer so chosen that c ≡ 0; and, hence, the lemma is true from our induction assumption. Suppose the result holds for all a, b, c satisfying |a| + |b| < n and c = 0 where n is some positive integer; we have to prove the result for polynomials a, b, c when |a| + |b| = n and c = 0. We can suppose, without loss of generality, that |a| |b|, and, so, in particular |a| > 0.

56


Suppose first that a(t1 , . . . , tr ) is not square free as a polynomial in tr , say a(t1 , . . . , tr ) = k(t1 , . . . , tr )2 a1 (t1 , . . . , tr ), where k has integral coefficients and |k| 1. The hypothesis of the lemma regarding a, b, c insures that this hypothesis also holds for the polynomials a1 (t1 , . . . , tr ), b(t1 , . . . , tr ) and c1 (t1 , . . . , tr ) = k(t1 , . . . , tr )c(t1 , . . . , tr ). Indeed, if τ1 , . . . , τr are integers such that c1 (τ1 , . . . , τr ) = 0, then the hypothesis for a, b, c asserts there are integers x, y, z, not all 0, satisfying (1). But then a1 (τ1 , . . . , τr )x 2 + b(τ1 , . . . , τr )y 2 = z2 has xk(τ1 , . . . , τr ), y, z as a nontrivial integral solution. Since |a1 | + |b| < |a| + |b| = n, the inductive hypothesis implies the existence of polynomials x1 (t1 , . . . , tr ), y1 (t1 , . . . , tr ), z1 (t1 , . . . , tr ) with integer coefficients and not all identically 0, such that a1 (t1 , . . . , tr )x1 (t1 , . . . , tr )2 + b(t1 , . . . , tr )y1 (t1 , . . . , tr )2 = z1 (t1 , . . . , tr )2 . On taking x(t1 , . . . , tr ) = x1 (t1 , . . . , tr ), y(t1 , . . . , tr ) = y1 (t1 , . . . , tr )k(t1 , . . . , tr ), z(t1 , . . . , tr ) = z1 (t1 , . . . , tr )k(t1 , . . . , tr ), we obtain an identical solution of (2). Hence we can suppose that a(t1 , . . . , tr ) is square free as a polynomial in tr and hence its discriminant D(t1 , . . . , tr−1 ) with respect to tr is not identically 0. Let a0 (t1 , . . . , tr−1 ), c0 (t1 , . . . , tr−1 ) be the leading coefficient of a and c with respect to tr ; taking c0 = c if |c| = 0. Let T be the set of points t = (t1 , . . . , tr−1 ) in (r − 1)-dimensional affine space defined by the inequality a0 (t1 , . . . , tr−1 )c0 (t1 , . . . , tr−1 )D(t1 , . . . , tr−1 ) = 0, and let T be the set of all integral (r − 1)-tuples τ = (τ1 , . . . , τr−1 ) in the set T . For every τ in T the polynomial cτ (tr ) = c(τ , tr ) ≡ 0. Our hypothesis on a, b, c asserts that for every integer τr such that cτ (τr ) = 0 the equation a(τ , τr )x 2 + b(τ , τr )y 2 = z2 is soluble nontrivially in integers x, y, z. Hence for each τ in T , by the case r = 1 of our theorem, there exist polynomials xτ (tr ), yτ (tr ), zτ (tr ) with integral coefficients, not all identically 0, such that a(τ , tr )xτ (tr )2 + b(τ , tr )yτ (tr )2 ≡ zτ (tr )2 identically in tr . We can suppose that xτ (tr ), yτ (tr ), zτ (tr ) = 1. Since a0 (τ )D(τ ) ≡ 0, a(τ , tr ) has no multiple factors, thus setting dτ (tr ) = a(τ , tr ), yτ (tr )

(3)

we get successively from (3): dτ (tr ) | zτ (tr )2 , dτ (tr ) | zτ (tr ), dτ (tr )2 | a(τ , tr )xτ (tr )2 , dτ (tr ) | xτ (tr ) and hence dτ (tr ) ≡ 1. Therefore, for τ in T we have z (t ) 2 τ r (4) b(τ , tr ) ≡ ≡ βτ (tr )2 mod a(τ , tr ), yτ (tr )

57

A10. Quadratic Diophantine equations c

where βτ is in Q[tr ] and |βτ | < |a| or βτ = 0. In order to exploit the congruence (4) we note that for all nonnegative integers h, |a|−1

trh ≡

αhl (t)trl mod a(t, tr ),

l=0

where αhl (t) are rational functions of t1 , . . . , tr−1 with powers of a0 (t) in the denominator. For τ in T we have a0 (τ ) = 0, hence αhl (t) are defined. Let |a|−1

βτ =

(5)

ξi ∈ Q.

ξi tri ,

i=0

From (4) we get for τ in T , b(τ , tr ) ≡

|a|−1

trl

l=0

|a|−1

ξi ξj αi+j,l (τ ) mod a(τ , tr ),

i,j =0

and if b(t, tr ) =

|a|

bi (t)tri ,

bi (t) in Z[t]

i=0

we get (6)

bl (τ ) + b|a| (τ )α|a|,l (τ ) =

|a|−1

ξi ξj αi+j,l (τ )

for l |a| − 1.

i,j =0

Let u be a new indeterminate and R(t, tr , u) be the resultant of the system of polynomials |a|−1 2 bl (t) + b|a| (t)α|a|,l (t) x|a| − xi xj αi+j,l (t)

(0 l < |a|),

i,j =0

(7)

|a|−1

xi tri − x|a| u

i=0

with respect to the variables x0 , . . . , x|a| . We shall prove that R(t, tr , u) = 0. |a| By a known property of resultants (see [4], p. 11) the coefficient of u2 in R is the resultant R0 of the system obtained from (7) by substitution x|a| = 0. If R0 were 0, the system of homogeneous equations |a|−1

(8)

ξi∗ ξj∗ αi+j,l (t) = 0

i,j =0

would have nontrivial solutions ξi∗ in the algebraic closure of Q(t). However, it then follows from (4), (5), (6), and (8) that (9)

0≡

|a|−1 i=0

ξi∗ tri

2

mod a(t, tr ).

58


Since a(t, tr ) is square free, (9) implies |a|−1

ξi∗ tri ≡ 0 mod a(t, tr );

i=0

which is impossible since |a(t, tr )| = |a|. Therefore R0 = 0 and moreover R0 ∈ Q(t). Let m be chosen so that G(t, tr , u) = a0 (t)m R(t, tr , u) ∈ Z[t, tr , u]. Then a0 (t)m R0 (t) is the leading coefficient of G with respect to u. Let G(t, tr , u) = g0 (t)

q

G (t, tr , u)

=1

where g0 ∈ Z[t], G ∈ Z[t, tr , u] and G are irreducible over Q of positive degree and with leading coefficient g (t) with respect to u. We can order G so that G is of degree 1 in u for p and of degree at least 2 for > p. If for all p we have H (t, tr ) = G (t, tr , 0)2 − b(t, tr )g (t)2 ≡ 0 mod a(t, tr ) then let the leading coefficient of the remainder from division of H by a(t, tr ) in the ring Q(t)[tr ] be f (t)a0 (t)−m , where f ∈ Z[t]. By Hilbert’s irreducibility theorem there 0 such that the polynomials G (τ 0 , t , u) are irreducible and exist integers τ10 , . . . , τr−1 r a0 (τ 0 )c0 (τ 0 )D(τ 0 )

p =1

f (τ 0 )

q

g (τ 0 ) = 0.

=0

Clearly τ 0 is in T . It follows from (5) and (6) that for t = τ 0 , u = βτ 0 (tr ) the system of polynomials (7) has a common zero ξ0 , . . . , ξ|a|−1 , 1 . Since this zero is non-trivial we get successively R τ 0 , tr , βτ 0 (tr ) = 0, G τ 0 , tr , βτ 0 (tr ) = 0 and G τ 0 , tr , βτ 0 (tr ) = 0 for a certain q. Since G τ 0 , tr , u is irreducible of degree at least 2 in u for > p we get p g (τ 0 )β0 (tr ) + G (τ 0 , tr , 0) = 0. Hence by (4) g (τ 0 )2 b(τ 0 , tr ) − G (τ 0 , tr , 0)2 ≡ 0 mod a(τ 0 , tr ) and f (τ 0 ) contrary to the choice of τ 0 . The obtained contradiction shows that for a certain p g (t)2 b(t, tr ) − G (t, tr , 0)2 ≡ 0 mod a(t, tr ). Reducing G (t, tr , 0)g (t)−1 modulo a(t, tr ) in the ring Q(t)[tr ] we find a


59

β(t, tr ) ∈ Q(t)[tr ] such that b(t, tr ) ≡ β(t, tr )2 mod a(t, tr )

(10) and

|β| < |a| or

(11)

β = 0.

We write β 2 (t, tr ) − b(t, tr ) = h−2 (t)a(t, tr )A(t, tr ) where h(t) ∈ Z[t] and A ∈ Z[t, tr ]. In particular h(t)β(t, tr ) ∈ Z[t, tr ]. If A(t, tr ) ≡ 0 identically, we can satisfy (2) by taking x(t, tr ) = 0,

y(t, tr ) = h(t),

z(t, tr ) = h(t)β(t, tr ).

If A(t, tr ) is not identically 0, we have by (11) that |A| < |a|. We now prove the hypotheses of the lemma are satisfied for the polynomials A(t, tr ),

b(t, tr ),

C(t, tr ) = a(t, tr )h(t)c(t, tr )A(t, tr ).

We know that for all integers τ1 , . . . , τr such that C(τ , τr ) = 0, the equation (1) has a solution in integers x, y, z, not all 0. Taking X = a(τ , τr )x, Y = h(τ ) z − yβ(τ , τr ) , Z = h(τ ) b(τ , τr ) y − β(τ , τr )z we obtain

A(τ , τr )X 2 + b(τ , τr )Y 2 − Z 2 = h(τ )2 β(τ , τr )2 − b(τ , τr ) (ax 2 + by 2 − z2 ) = 0.

Also X, Y, Z are integers not all 0, since a(τ , τr )h(τ )A(τ , τr ) = 0. The inductive hypothesis applies to the polynomials A(t, tr ), b(t, tr ), C(t, tr )

since

|A| + |b| < |a| + |b| = n.

Hence there exist polynomials X(t, tr ), Y (t, tr ), Z(t, tr ) with integral coefficients and not all identically zero, such that A(t, tr )X(t, tr )2 + b(t, tr )Y (t, tr )2 ≡ Z(t, tr )2 identically in t, tr . Putting x(t, tr ) = A(t, tr )X(t, tr ), y(t, tr ) = h(t) β(t, tr )Y (t, tr ) + Z(t, tr ) , z(t, tr ) = h(t) b(t, tr )Y (t, tr ) + β(t, tr )Z(t, tr ) we obtain (2). Further x(t, tr ), y(t, tr ), z(t, tr ) do not all vanish identically since neither A(t, tr ) nor b(t, tr ) − β 2 (t, tr ) vanish identically. Remark. The argument following formula (11) is implicit in Skolem’s paper [8].

3. Theorem 1. Let a(t1 , . . . , tr ), b(t1 , . . . , tr ) be polynomials with integral coefficients. Suppose that for all r-tuples of arithmetical progressions P1 , . . . , Pr there exist integers τi ∈ Pi such that the equation (1) has a solution in integers x, y, z not all 0. Then

60


there exist polynomials x(t1 , . . . , tr ), y(t1 , . . . , tr ), z(t1 , . . . , tr ) with integral coefficients, not all identically 0, such that (2) holds identically in t1 , . . . , tr . Proof. It is enough to show that the assumption of the theorem implies the assumption of the lemma. Now take any r-tuple of integers τ1 , . . . , τr , an arbitrary prime p and a positive integer m. By the assumption of the theorem the arithmetical progressions pm t + τ1 , . . . , pm t + τr contain integers τ10 , . . . , τr0 respectively such that the equation a(τ10 , . . . , τr0 )x 2 + b(τ10 , . . . , τr0 )y 2 = z2 has a solution in integers not all 0. Hence it has a solution x0 , y0 , z0 with (x0 , y0 , z0 ) = 1 and we get a(τ1 , . . . , τr )x02 + b(τ1 , . . . , τr )y02 ≡ z02 mod p m . By Theorem 2 of §5 of [1] it follows that (1) is soluble nontrivially in the field of p-adic numbers. By Lemma 2 in §7 ibidem it follows that (1) is soluble nontrivially also in real numbers, hence by Theorem 1 of §7 ibidem it is soluble nontrivially in integers. Added in proof. Slightly different proof of Theorem 1 valid for arbitrary number fields will appear in a forthcoming book [7] of the second author. Theorem 2. Let F (x, y, t1 , . . . , tr ) be any polynomial with integral coefficients which is of degree at most 2 in x and y. Suppose that for all r-tuples of arithmetical progressions P1 , . . . , Pr there exist integers τi ∈ Pi such that the equation F (x, y, τ1 , . . . , τr ) = 0 is soluble in rationals x, y. Then there exist rational functions x(t1 , . . . , tr ), y(t1 , . . . , tr ) with rational coefficients such that F x(t1 , . . . , tr ), y(t1 , . . . , tr ), t1 , . . . , tr ≡ 0 identically in t1 , . . . , tr . Proof. Theorem 2 follows from Theorem 1 for r > 1 in exactly the same way as Theorem B was derived from Theorem A (see [3]). In the argument (page 357) where the Corollary to Theorem 1 of [2] is used, one has instead to apply Theorem 2 of [6]. M. Fried has observed that Theorem B implies an analogous result for curves of genus 0 defined over Q(t). The remark applies, mutatis mutandis, to Theorem 2. One can moreover extend it to equations that define a finite union of curves of genus 0 over the algebraic closure of Q(t). As to the curves of genus 1 it follows from the so-called Selmer’s conjecture in the theory of rational points on such curves that for every integer t there is a rational solution of the equation (12)

x 4 − (8t 2 + 5)2 = y 2

(see [9]). On the other hand, suppose that rational functions x(t), y(t) in Q(t) satisfy (12). There exist infinitely many integer pairs u, v such that 5u2 + 8v 2 is a prime p. Take u, v such that for τ = 5u/8v, x(τ ), y(τ ) are defined. The equation (12) gives 2 2 4vx(τ ) − 100p 2 = 16v 2 y(τ ) .


61

But, by a theorem of Nagell [5] the Diophantine equation X 4 − 100p 2 = Y 2

(p prime ≡ 1 mod 4)

has no rational solutions.

References [1] Z. I. Borevich, I. R. Shafarevich, Number Theory. Academic Press, New York 1966. [2] H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107–116; this collection: A6, 27–35. [3] −−, −−, −−, Quadratic Diophantine equations with a parameter. Acta Arith. 11 (1966), 353–358. [4] F. S. Macaulay, The Algebraic Theory of Modular Systems. Reprint, New York and London 1964. [5] T. Nagell, Zahlentheoretische Notizen I–IV. Vid. Skrifter, I Math. Naturv. Kl. 1923, No. 13, Kristiania 1924. [6] A. Schinzel, On Hilbert’s Irreducibility Theorem. Ann. Polon. Math. 16 (1965), 333–340; this collection: F1, 839–845. [7] −−, Selected Topics on Polynomials. The University of Michigan Press, Ann Arbor 1982. [8] Th. Skolem, Über die Lösung der unbestimmten Gleichung ax 2 + by 2 + cz2 = 0 in einigen einfachen Rationalitätsbereichen. Norsk. Mat. Tidsskr. 10 (1928), 50–62. [9] N. M. Stephens, Congruence properties of congruent numbers. Bull. London Math. Soc. 7 (1975), 182–184.

Originally published in Bulletin of the London Mathematical Society 14 (1982), 345–348


Selmer’s conjecture and families of elliptic curves with J. W. S. Cassels (Cambridge)

1. In this note we consider the elliptic curve (1.1)

A : y 2 = x{x 2 − (1 + t 4 )2 }

over the field C(t) of rational functions with √ complex coefficients. It turns out that all C(t)-rational points on A satisfy x, y ∈ Q( 2, i, t). As a consequence, the curve (1.2)

B : y 2 = x{x 2 − (7 + 7t 4 )2 }

has only points of order 2 over Q(t). On the other hand, the famous conjecture of Selmer [7] is shown to imply that for every rational r the curve (1.3)

Br : y 2 = x{x 2 − (7 + 7r 4 )2 }

has infinitely many rational points. To place this phenomenon in a more general setting, let R = Q[t] and let F ∈ R[x, y] be irreducible over the algebraic closure of Q(t). If x, y are regarded as coordinates of affine space, the equation F = 0 defines an affine curve over Q(t) and also a family of affine curves F (x, y, r) = 0 over Q, as r runs over the elements of Q such that F (x, y, r) is absolutely irreducible (the remaining values of r form a finite set E). Let C be the projective curve defined over Q(t) obtained from normalization of the completion of the curve F = 0 and let Cr be the normalization of the completion of the affine curve F (x, y, r) = 0 for r ∈ Q \ E. If C is a curve of genus 1, then for all but finitely many values r ∈ Q \ E the curve Cr is also of genus 1. Without further comment we ignore the exceptional values of r. Let C Q(t) denote the set of Q(t)-rational points on C , and Cr (Q) the set of Q-rational points on Cr . If C Q(t) and Cr (Q) are nonempty, they have natural structure as groups when one point is selected as origin. They are finitely generated by virtue, respectively, of the theorems of Néron ([6], 3) and of Mordell. Let g(C ) (respectively g(Cr )) be the Théorème reduced rank of C Q(t) (respectively Cr (Q)). Néron has shown (ibid., Théorème 6) that there are infinitely many r ∈ Q such that g(Cr ) g(C ). Our example shows that there can be strict inequality for every rational r.

A11. Selmer’s conjecture and families of elliptic curves

63

An analogous result where r is restricted to values in Z has been obtained by D. J. Lewis and A. Schinzel [4]. The curve C given by y 2 = x 4 −(8t 2 +5)2 has g(C ) = 0, but Selmer’s Conjecture implies that g(Cr ) 1 for every integer r. In this example, however, g(Cr ) = 0 for infinitely many rational r. The following fundamental problem of M. Fried remains open. If, in the language above, C Q(t) is empty, does there exist r ∈ Q for which Cr (Q) is empty? The original version of this note was written by the second author (A. Schinzel). It contained proofs of the statements about the behaviour of B over Q(t) and of Br over Q but not the more general result about A . During the preparation of that version he was visiting the University of California, Irvine. He wishes to thank the University for its hospitality and Dr. L. J. Colliot-Thélène and Professor M. Fried for their valuable suggestions, which have been partially incorporated in the present version.

2. In this section we prove Theorem 1. If x, y ∈ C(t) satisfy (2.1) then x, y ∈ Q(i,

√

y 2 = x{x 2 − (1 + t 4 )2 }, 2, t). All solutions may be described explicitly.

Proof. We extend the base field to C(s, t), where s 2 = 1 + t 4 . Then η2 = ξ(ξ 2 − 1), where η = y/s 3 ,

ξ = x/s 2 = x/(1 + t 4 ).

On putting (2.2)

u = t 2 − s,

v = 21/2 tu,

we have s = − 21 (u + u−1 ),

t = v/21/2 u,

and v 2 = u(u2 − 1). We therefore work on the elliptic curve (2.3)

Y 2 = X(X2 − 1),

whose ring of complex multiplications is Z[i]. By hypothesis, ξ, η ∈ C(s, t) = C(u, v).

64


Hence in terms of group addition on (2.3) we have (2.4)

(ξ, η) = α(u, v) + (u0 , v0 ),

where α ∈ Z[i] and (u0 , v0 ) with u0 , v0 ∈ C is a constant point. Here we use the fact that any rational map of an elliptic curve into itself is a complex multiplication followed by a translation. (For the corresponding result for abelian varieties of any dimension cf. Mumford ([5], p. 43, Corollary 1), Swinnerton-Dyer ([9], Theorem 32) or Lang ([3], Chapter II, Theorem 4).) Let ( ) denote the automorphism of C(s, t)/C(t). Then u = −1/u,

v = −v/u2 ,

and so (2.5)

(u, v) = −(u, v) + (0, 0).

Here (0, 0) is of order 1 + i on (2.3). We require that x, y ∈ C(t), and so ξ = ξ,

η = −η;

that is (2.6)

(ξ, η) = (ξ, −η) = −(ξ, η).

On comparing (2.4), (2.5), (2.6) we see that a necessary and sufficient condition that x, y ∈ C(t) is (2.7)

α(0, 0) = 2(u0 , v0 ).

Hence for given α there are precisely 4 values of (u0 , v0 ) ∈ C2 which will do. More precisely, either (i) α(0, 0) is the point 0 at infinity, in which case (u0 , v0 ) is 0 or a point of order 2 or (ii) α(0, 0) = (0, 0), in which case (u0 , v0 ) is one of the points i, ±(1 − i) , −i, ±(1 + i) . Hence there are 4 solutions of (2.1) defined over C(t) for every α ∈ Z[i]. Since the complex √ multiplication is defined over Q(i) and the only other irrationality we have used is the 2 in (2.2), the truth of the theorem follows. The proof of the statement in §1 about (1.2) follows almost immediately. More generally, we have the Corollary. Let d be a positive integer and suppose that μ2 = λ λ2 − d 2 (1 + t 4 )2 has a solution λ, μ ∈ Q(t) with μ = 0. Then d is a square or twice a square. Proof. The point x√= d −1 λ, y = d −3/2 μ is a solution of (2.1) defined over C(t). It is defined over Q(i, 2, t) only if μ = 0 or if d is of the shape specified.

A11. Selmer’s conjecture and families of elliptic curves

65

3. In this section we prove the assertions in §1 about (1.3): Theorem 2. Selmer’s Conjecture [7] implies that for any r ∈ Q the Mordell-Weil rank of (3.1) y 2 = x x 2 − (7 + 7r 4 )2 is odd, and so there are infinitely many rational points on (3.1). Proof. Let r = l/m with l, m ∈ Z, (l, m) = 1. Then X = m4 x, Y = m6 y satisfy (3.2)

Y 2 = X(X2 − n2 ),

where n = 7l 4 + 7m4 ≡ 6, 7 (mod 8). We have now only to invoke the result of Stephens [8] that, subject to the Selmer conjecture, the rank of (3.2) is odd whenever n ≡ 5, 6, 7 (mod 8) is a positive integer.

4. We conclude with three remarks. (i) We have been unable to find an equianharmonic curve with similar properties to (1.2). However C : x 3 + y 3 = 27(2t + 1)2 + 1 has only finitely many points over Q(t), whereas on Selmer’s Conjecture the specializations Cr for r ∈ Z all have odd rank. (ii) The ideas of §2 rapidly show that the only solutions x, y ∈ C(t) of (4.1) y 2 = x x 2 − (1 + t 2 )2 have y = 0. One extends the ground field to C(s, t), where (4.2)

s2 = 1 + t 2.

Here (4.2) has genus 0 and Lüroth’s Theorem applies. (iii) Presumably the general techniques of Christie [2] (cf. also [1]) for elliptic curves over C(t) could be used to prove Theorem 1, but not so simply.

References [1] J. W. S. Cassels, A Diophantine equation over a function field. J. Austral. Math. Soc. Ser. A 25 (1978), 489–496. [2] M. R. Christie, Positive definite functions of two variables which are not sums of three squares. J. Number Theory 8 (1976), 224–232.

66


[3] S. Lang, Abelian Varieties. Interscience Publ., New York and London 1959. [4] D. J. Lewis, A. Schinzel, Quadratic Diophantine equations with parameters. Acta Arith. 37 (1980), 133–141; this collection: A10, 54–61. [5] D. Mumford, Abelian Varieties. Oxford Univ. Press, London 1970. [6] A. Néron, Problèmes arithmétiques et géométriques rattachés à la notion de rang d’une courbe algébrique dans un corps. Bull. Soc. Math. France 80 (1952), 101–166. [7] E. Selmer, A conjecture concerning rational points on cubic curves. Math. Scand. 2 (1954), 49–54. [8] N. M. Stephens, Congruence properties of congruent numbers. Bull. London Math. Soc. 7 (1975), 182–184. [9] H. P. F. Swinnerton-Dyer, Analytic Theory of Abelian Varieties. Cambridge Univ. Press, London– New York 1974.


Originally published in Acta Arithmetica XL (1982), 399–420

Families of curves having each an integer point

H. Davenport, D. J. Lewis and the writer [3] proved that if an equation with integral coefficients F (x, y, t) = 0 quadratic in x and y is solvable in rational x, y for at least one integer t from every arithmetic progression, then the equation is solvable in rational functions x(t), y(t) ∈ Q(t). The question has been raised whether the solvability of F (x, y, t) = 0 in integers x, y for all integers t implies the solvability of the equation in polynomials. It is the aim of the present paper to study this question in a more general context. We shall prove Theorem 1. If L ∈ Z[x, t] is of degree at most four in x, M ∈ Z[t] and every arithmetic progression contains an integer t ∗ such that L(x, t ∗ ) = M(t ∗ )y is solvable in integers x, y then there exist polynomials X, Y ∈ Q[t] such that L X(t), t = M(t)Y (t). The theorem is no longer true in general if the degree of L is greater than four. Also for L of degree non-exceeding four the conclusion cannot in general be strenghtened to assert the existence of integer valued polynomials X, Y . The relevant examples will be given after the proof of Theorem 1. Theorem 1 easily implies Theorem 2. If F ∈ Z[x, y, t], the highest homogeneous part F0 of F with respect to x, y is quadratic and singular and every arithmetic progression contains an integer t ∗ such ∗ that F (x, in integers x, y, then there exist polynomials X, Y ∈ Q[t] such y, t ) is solvable that F X(t), Y (t), t = 0. It seems likely that if we assume the solvability of F (x, y, t) = 0 in integers x, y for all t ∗ ∈ Z, the conclusion remains true provided F0 is reducible over Q(t). However, in general the conclusion fails as it is shown by the following Theorem 3. The equation x 2 −(4t 2 +1)3 y 2 = −1 is solvable in integers x, y for all t ∗ ∈ Z, but there exist no polynomials X, Y ∈ Q[t] such that X(t)2 − (4t 2 + 1)3 Y (t)2 = −1. Prompted by a question from Professor J. Leicht I have studied the possibility of modifying the assumptions of Theorem 1 so that they would imply the existence of integer valued polynomials X, Y . The result is the following Theorem 4. Let n be a positive integer ≡ 0 (mod 8), A, B, M ∈ Z[t]. If every arithmetic progression contains an integer t ∗ such that A(t ∗ )x n + B(t ∗ ) = M(t ∗ )y is solvable in integers x, y, then there exist integer valued polynomials X, Y such that A(t)X(t)n + B(t) = M(t)Y (t).

68


The condition n ≡ 0 (mod 8) cannot be relaxed, as I shall show by an example. For n = 1 Theorem 4 is contained in a more general result of Skolem [8] concerning polynomials in many variables. According to Skolem the polynomials A, B, M may have any number of variables provided A and M have no common zero. I shall show by an example that already for n = 2 the corresponding statement is false. The possibility of extending Theorems 1, 2, and 4 to polynomials in many variables will be studied in a subsequent paper. For the proof of the above theorems we need several lemmata. Lemma 1. Let D be a Dedekind domain, f, g, h ∈ D[x], p be a prime ideal of D, f (x) ≡ g(x)h(x) (mod p). If g, h are relatively prime mod p and the leading coefficient of g is 1, then for every integer n 0 there exist polynomials gn , hn ∈ D[x] such that (1)

f (x) ≡ gn (x)hn (x) (mod pn+1 ),

(2)

the degree of gn equals the degree of g, the leading coefficient of gn is 1, gn (x) ≡ g(x),

(3)

hn (x) ≡ h(x) (mod p).

Proof. This lemma is closely related to Hensel’s lemma and can be derived by following the proof of Hensel’s lemma given by Hasse [6] up to the point where the solvability of the congruence gn−1 zn + hn−1 yn ≡ fn (mod p),

fn ∈ D[x]

in polynomials yn , zn ∈ D[x] is needed. Then since gn−1 , hn−1 are relatively prime mod p we use the fact that D/p is a field. Lemma 2. Let D be a principal ideal domain, a, b, c ∈ D, (p) be a prime ideal of D. If p /| 2a and d = b2 − 4ac the congruence ax 2 + bx + c ≡ 0 (mod p ν )

(4)

is solvable in x ∈ D if and only if either ordp d ν or ordp d = δ ≡ 0 (mod 2) and the congruence z2 ≡ dp−δ (mod p) is solvable in D. Proof. The congruence (4) is equivalent to (2ax + b)2 ≡ d (mod 4apν ) and since p /| 2a, it is solvable if and only if y 2 ≡ d (mod pν ) is. If δ = ordp d ν it is enough to take y = 0. If δ < ν, the congruence implies δ = 2 ordp y ≡ 0 (mod 2) and y = p δ/2 z, z2 ≡ dp −δ (mod p ν−δ ). Thus the necessity of the condition given in the lemma follows. On the other hand, if the condition is satisfied 2 −δ (mod p), z ∈ D, we can apply Lemma 1 with c and z ≡ dp 0 0 f (x) = x 2 − dp −δ ,

g(x) = x − z0 ,

h(x) = x + z0 ,

The congruence c

x 2 − dp −δ ≡ gn (x)hn (x) (mod p n+1 ),

n = ν − δ − 1.

A12. Families of curves having each an integer point

69

where deg gn = deg g = 1 and the leading coefficient of gn is 1, implies that x 2 ≡ dp−δ δ δ c (mod p ) has solutions ±gn (0). Multiplying by p , we get δ/2 2 p gn (0) ≡ d (mod pν ). Remark. The lemma can easily be modified so that it would apply to all Dedekind domains. It is also possible, although not so easy, to prove analogous statements about congruences of degree three and four. For instance, if D is a principal ideal domain, a, b ∈ D, (p) is a prime ideal of D, p /| 3, then the congruence x 3 + ax + b ≡ 0 (mod p ν ) is solvable in x ∈ D if and only if either ordp b ν or 2 ordp b > 3 ordp a or 3 | β = ordp b < ν, 2β 3 ordp a and the congruence z3 + ap −2β/3 + bp −β ≡ 0 (mod p) is solvable. c

Lemma 3. If A, B ∈ Z[t], (A, B) = 1, then for sufficiently large primes p the divisibility p | A(t ∗ ), t ∗ ∈ Z, implies p /| B(t ∗ ). Proof. Let R be the resultant of A and B. Since (A, B) = 1, we have R = 0 and there exist polynomials U, V ∈ Z[t] such that AU + BV = R. Now, if p /| R we have either p /| A(t ∗ ) or p /| B(t ∗ ).

Lemma 4. Let K be an algebraic number field, F ∈ K[x] be of degree at most four. If the congruence F (x) ≡ 0 (mod p) is solvable for almost all prime ideals of degree 1 in K c then the equation F (x) = 0 is solvable in K. Proof. If F (x) is irreducible in K then the lemma follows from the more general result of Hasse [5]. If F (x) is reducible in K but has no zero there then its degree must be four. If now the congruence F (x) ≡ 0 (mod p) is solvable for almost all ideals p of K rather than for almost all prime ideals p of degree 1 in K then the assertion holds in virtue of Proposition 2 in Fujiwara [4]. However, in the proof of this proposition only prime ideals of degree 1 are needed. Lemma 5. Let Ai , Bi , Ci ∈ Z[t] (i = 1, 2), let P ∈ Z[t] be a primitive irreducible polynomial, A1 A2 ≡ 0 (mod P ) and the polynomials Ai (t)x 2 + Bi (t)x + Ci (t) (i = 1, 2) be prime mod P (t). If for all sufficiently large primes p and all integers t ∗ such that p P (t ∗ ) the congruence (5)

2

Ai (t ∗ )x 2 + Bi (t ∗ )x + Ci (t ∗ ) ≡ 0 (mod pμ )

i=1

is solvable in x ∈ Z then the congruence (6)

2

Ai (t)x 2 + Bi (t)x + Ci (t) ≡ 0 mod P (t)μ

i=1

is solvable in Q[t].

70


Proof. Let Di (t) = Bi (t)2 − 4Ai (t)Ci (t) = P (t)δi Ei (t), where Pi /| Ei (i = 1, 2). (If D1 = 0 or D2 = 0 (6) is clearly solvable.) If for an i 2, δi μ the congruence Ai (t)x 2 + Bi (t)x + Ci (t) ≡ 0 mod P (t)μ is solvable in virtue of Lemma 2 applied with D = Q[t], hence (6) is solvable also. Let P (ϑ) = 0, K = Q(ϑ), p be a prime ideal of degree 1 in K with norm p assumed sufficiently large. Choose t ∗ ≡ ϑ (mod p). Then P (t ∗ ) ≡ 0 (mod p), P (t ∗ + p) ≡ 0 (mod p), P (t ∗ + p) − P (t ∗ ) ≡ pP (t ∗ ) (mod p 2 ). Since (P , P ) = 1, we have by Lemma 3 P (t ∗ ) ≡ 0 (mod p), thus P (t ∗ ) ≡ 0 (mod p2 ) or P (t ∗ + p) ≡ 0 (mod p2 ). Replacing t ∗ by t ∗ + p if necessary we may assume that P (t ∗ ) ≡ 0 (mod p2 ), and that (5) holds for a suitable x = x ∗ ∈ Z. 2 Let R(t) be the resultant of Ai (t)x + Bi (t)x + Ci (t) (i = 1,∗2) with respect to x. By the assumption we have P (t), R(t) = 1 and by Lemma 3 R(t ) ≡ 0 (mod p). On the other hand, if we had Ai (t ∗ )x ∗ 2 + Bi (t ∗ )x ∗ + Ci (t ∗ ) ≡ 0 (mod p) it would follow that

R(t ∗ )

(i = 1, 2)

≡ 0 (mod p). Thus there exists an i 2 such that

Ai (t ∗ )x ∗ 2 + Bi (t ∗ )x ∗ + Ci (t ∗ ) ≡ 0 (mod pμ ). Since (P , Ei ) = 1, we have by Lemma 3 p /| Ei (t ∗ ). Thus ordp Di (t ∗ ) = δi and by D (t ∗ )p −δi i Lemma 2 applied with D = Z we have δi ≡ 0 (mod 2) and = 1, whence p E (t ∗ ) i = 1. p E (ϑ) i Now Ei (t ∗ ) ≡ Ei (ϑ) (mod p) and we get = 1. p Take in Lemma 4 2 1 + (−1)δi 2 F (x) = x − Ei (ϑ) . 2 i=1

We infer that for almost all prime ideals p of degree 1 in K the congruence F (x) ≡ 0 (mod p) is solvable in K. Hence by Lemma 4 F (x) has a zero in K and since E1 (ϑ)E2 (ϑ) = 0, it follows that for an i 2 we have δi ≡ 0 (mod 2) and Ei (ϑ) = G(ϑ)2 where G ∈ Q[t]. Hence Ei (t) ≡ G(t)2 mod P (t) and by Lemma 2 the congruence Ai (t)x 2 + Bi (t)x + Ci (t) ≡ 0 mod P (t)μ is solvable in Q[t]. Lemma 6. Let L ∈ Z[x, t] be of degree at most 4 in x, let P ∈ Z[t] be irreducible and ∗ primitive. If for all sufficiently large primes p and all integers t ∗ such that p P (t μ) the ∗ μ congruence L(x, t ) ≡ 0 (mod p ) is solvable in Z then L(x, t) ≡ 0 mod P (t) is solvable in Q[t].


71

Proof (by induction on μ). We set K = Q(ϑ), where P (ϑ) = 0. μ = 1. Let p be a prime ideal of degree 1 in K with norm p assumed sufficiently large, t ∗ ≡ ϑ (mod p). The argument used in the proof of Lemma 5 shows that without loss of generality we may assume p P (t ∗ ). Hence L(x ∗ , t ∗ ) ≡ 0 (mod p) for an x ∗ in Z, L(x ∗ , ϑ) ≡ 0 (mod p) and by Lemma 4 L(x, ϑ) has a zero in K. Denoting this zero by X(ϑ), X ∈ Q[t], we infer from L X(ϑ), ϑ = 0 that L X(t), t ≡ 0 mod P (t) . The inductive step. Suppose that the lemma is true for exponents less than μ 2 and all polynomials L satisfying the assumptions. Let the congruence L(x, t ∗ ) ≡ 0 (mod pμ ) be solvable in Z for all sufficiently large primes p and all integers t ∗ such that p P (t ∗ ). By the case μ = 1, L(x, ϑ) has a zero in K. If L(x, ϑ) = 0 identically then L(x, t) = P (t)L1 (x, t), L1 ∈ Z[x, t]. For all sufficiently large primes p and all integers t ∗ such that p P (t ∗ ) the congruence L(x, t ∗ ) ≡ 0 (mod pμ−1 by the ) is solvable. Henceμ−1 inductive assumption there exists an X ∈ Q[t] such that L X(t), t ≡ 0 mod p (t) and then L X(t), t ≡ 0 mod P μ (t) . If L(x, ϑ) has a simple zero we have L(x, ϑ) = G(x, ϑ)H (x, ϑ) c

where G, H ∈ Q[x, t], both the degree and the leading coefficient of G with respect to x are 1 and G(x, ϑ), H (x, ϑ) = 1. Hence L(x, t) ≡ G(x, t)H (x, t) mod P (t) , G, H relatively prime mod P and by Lemma 1 applied with D = Q[t], p = P (t) we infer that L(x, t) ≡ Gμ−1 (x, t)Hμ−1 (x, t) mod P μ (t) , where Gμ−1 (x, t) is of degree 1 in x with the leading coefficient 1. Therefore, L −Gμ−1 (0, t), t ≡ 0 mod P μ (t) . If L(x, ϑ) is a product of two coprime quadratic factors we have L(x, t) ≡ G(x, t)H (x, t) mod P (t) , G, H ∈ Q[x, t], where G, H are quadratic in x, relatively prime mod P (t) and we may assume without loss of generality that theleading coefficient of G with respect to x is 1. By Lemma 1 applied with D = Q[t], p = P (t) we have (7) L(x, t) ≡ Gμ−1 (x, t)Hμ−1 (x, t) mod P μ (t) , where polynomials Gμ−1 , Hμ−1 ∈ Q[t] are quadratic with respect to x and relatively prime mod P (t), moreover their leading coefficients are not divisible by P (t). For a suitable integer d = 0 we have dGμ−1 (x, t), dHμ−1 (x, t) ∈ Z[x, t]

c

and

d 2 P −μ (t) L(x, t) − Gμ−1 (x, t)Hμ−1 (x, t) ∈ Z[x, t].

Hence the solvability of the congruence L(x, t ∗ ) ≡ 0 (mod pμ ), for p P (t ∗ ) implies the

72


solvability of the congruence dGμ−1 (x, t ∗ )dHμ−1 (x, t ∗ ) ≡ 0 (mod pμ ). In virtue of Lemma 5 there exists an X ∈ Q[t] such that dGμ−1 X(t), t dHμ−1 X(t), t ≡ 0 mod P μ (t) and then by (7) L X(t), t ≡ mod P (t)μ . There remains only the case where L(x, ϑ) = c(x − a)r , a, c ∈ K, c = 0, r 2. Let c = C(ϑ), a = A(ϑ), where A, C ∈ Q[t]. We have r L(x, t) ≡ C(t) x − A(t) mod P (t) , (P , C) = 1 and the congruence L(x ∗ , t ∗ ) ≡ 0 (mod p) for p P (t ∗ ) implies x ∗ ≡ A(t ∗ ) (mod p). (Note that C(t ∗ ) ≡ 0 (mod p) by Lemma 3.) Hence x ∗ ≡ A(t ∗ ) + P (t ∗ )y ∗ (mod p μ ), y ∗ ∈ Z and we have (8) L A(t ∗ ) + P (t ∗ )y ∗ , t ∗ ≡ 0 (mod pμ ). Let L1 (y, t) = L A(t) + P (t)y, t /P (t). We have for a suitable integer l = 0 lL1 (y, t) ∈ Z[y, t]. The congruence (8) together with p P (t ∗ ) implies that lL1 (y ∗ , t ∗ ) ≡ 0 (mod pμ−1 ).

By the inductive assumption there exists a polynomial Y ∈ Q[t] such that lL1 Y (t), t ≡ 0 mod P (t)μ−1 and then L A(t) + P (t)Y (t), t ≡ 0 mod P μ (t) . Proof of Theorem 1. If M(t) = 0 the theorem follows from Theorem 1 of [2]. If M(t) = 0 let M(t) = m

k

Pi (t)μi

i=1

be the canonical factorization of M into polynomials irreducible and primitive. Take an index i k, a prime p and integer t1∗ such that p Pi (t ∗ ). The arithmetic progression p μi u + t ∗ contains an integer t1∗ such that for suitable x ∗ , y ∗ ∈ Z we have L(x ∗ , t1∗ ) = M(t ∗ )y ∗ . Clearly L(x ∗ , t ∗ ) ≡ L(x ∗ , t1∗ ) ≡ 0 (mod pμ ). Hence by Lemma 6 there exists a polynomial Xi ∈ Q[t] such that μ L Xi (t), t ≡ 0 mod Pi i (t) . By the Chinese Remainder Theorem there exists a polynomial X ∈ Q[t] satisfying k μ μ X ≡ Xi (t) mod Pi i (t) (1 i k). We get L X(t), t ≡ 0 mod Pi i (t) , hence L X(t), t = M(t)Y (t),

i=1

Y (t) ∈ Q[t].

Here is an example showing that Theorem 1 fails for polynomials L of degree 5 in x.

73


Example 1. Let L(x, t) = (x 2 + 3)(x 3 + 3), M(t) = 3t + 1. For every integer t ∗ we k l β have M(t ∗ ) = piαi qj j , where pi are primes ≡ 1 (mod 3), qj are primes ≡ 2 i=1

j =1

β

(mod 3). The congruences x 2 + 3 ≡ 0 (mod piai ) and x 3 + 3 ≡ 0 (mod qj j ) are solvable for all i k, j l. Denoting their solutions by xi and xj , respectively, we can satisfy the β

equation L(x, t ∗ ) = M(t ∗ )y by taking x ≡ xi (mod piαi ), x ≡ xj (mod qj j ) (1 i k, 1 j l). On the other hand, the equation L X(t), t = M(t)Y (t) where X, Y ∈ Q[t] 3 2 √ √ would imply X − 13 + 3 = 0 or X − 13 + 3 = 0 hence −3 ∈ Q or 3 −3 ∈ Q, which is impossible. (The idea comes from van der Waerden [12].) The next example shows that the conclusion of Theorem 1 cannot be sharpened to assert the existence of integer valued polynomials X(t), Y (t) satisfying L X(t), t = M(t)Y (t). Example 2. Let L(x, t) = (2x + 1)(3x + 1), M(t) = 5t + 1. For every integer t ∗ we have M(t ∗ ) = 2α N , N odd. The congruences 2x + 1 ≡ 0 (mod N ), 3x + 1 ≡ 0 (mod 2α ) are both solvable. Denoting their solutions by x1 and x2 respectively we can satisfy the α equation L(x, t ∗ ) = M(t ∗ )y by taking x ≡ x1 (mod N ), x ≡ x2 (mod 2 ). Suppose now that X(t), Y (t) are integer valued polynomials satisfying L X(t), t = M(t)Y (t). Then clearly either

(9)

2X(t) + 1 = (5t + 1)Y1 (t)

or

3X(t) + 1 = (5t + 1)Y2 (t), Y1 , Y2 ∈ Q[t].

Let m be a positive integer such that mi Yi ∈ Z[t] (i = 1 or 2) and let 2δ1 m1 , 3δ2 m2 . Solving the congruence 5t + 1 ≡ 0 (mod 2δ1 +1 ) if i = 1 or 5t + 1 ≡ 0 (mod 3δ2 +1 ) if i = 2, we get from (9) a contradiction (the idea comes from Skolem [10]). Proof of Theorem 2. By the assumption we have 2 F (x, y, t) = C(t) A(t)x + B(t)y + D(t)x + E(t)y + F (0, 0, t) where A, B, C, D, E ∈ Z[t] and we can assume without loss of generality that (A, B) = 1. Let R be the resultant of A and B and let G, H ∈ Z[t] be such that A(t)G(t) + B(t)H (t) = R. We set A(t)x + B(t)y = u,

c

−H (t)x + G(t)y = v

and obtain c

RF (x, y, t) = RC(t)u2 + D(t)G(t) + E(t)H (t) u + A(t)E(t) − B(t)D(t) v + RF (0, 0, t) = 0.

Moreover, if x, y ∈ Z we have u, v ∈ Z. The assumptions of Theorem 1 are satisfied with L(u, t) = RC(t)u2 + D(t)G(t) + E(t)H (t) u + RF (0, 0, t), M(t) = A(t)E(t) − B(t)D(t).

74


By the said theorem there exist polynomials U, V ∈ Q[t] such that identically L U (t), t = M(t)V (t). Setting 1 [G(t)U (t) − B(t)V (t)], R 1 Y (t) = [H (t)U (t) − A(t)V (t)], R we get X, Y ∈ Q[t] and F X(t), Y (t), t = 0. X(t) =

Proof of Theorem 3. Setting ∗ 2 4t ∗2 +1 2t + 4t ∗ + 1 = x + y(4t ∗ 2 + 1) 4t ∗ 2 + 1, we get for every integer t ∗ integers x, y satisfying x 2 − (4t ∗ 2 + 1)3 y 2 = −1. On the other hand, it has been proved already by Abel [1] that all solutions of an equation 2 U 2 (t) − F (t)V by convergents of the continued fraction √ (t) = const = 0 are given expansion of F (t). Since for F (t) = 4t 2 + 1 F (t) = 2t +

1 1 + + ..., 4t 4t

we infer from the equation X(t)2 − (4t 2 + 1)3 X(t)2 = −1 that

n X(t) + (4t 2 + 1) 4t 2 + 1 Y (t) = c 2t ± 4t 2 + 1 ,

n 0.

Hence n(2t)n−1 ≡ 0 (mod 4t 2 + 1), n = 0, X(t) = c, Y (t) = 0 and c2 = −1 contradicting X ∈ Q[t].

For the proof of Theorem 4 we need four lemmata. Lemma 7. Let M(t) = m c c

k i=1

μ

Pi i (t) where polynomials Pi (t) are coprime, irreducible

and primitive, μi > 0. Under the assumptions of the theorem and the conditions BM = 0, (A, M) = 1 there exist polynomials X0 , Y0 ∈ Q[t] such that A(t)X0 (t)n + B(t) = M(t)Y0 (t), k X0 (t) ≡ 0 mod Pi (t)−[−βi /n] , i=1

where Pi (t)βi B(t).


75

β

c

Proof. By the assumption, Pi /| A (1 i k). Set B = Pi i Bi , where Pi /| B. By the Chinese Remainder Theorem for the ring Q[t] it is sufficient to show the solvability in this ring of each congruence μ (10) A(t)X(t)n + B(t) ≡ 0 mod Pi i (t) (1 i k). Let Pi (ϑ) = 0, K = Q(ϑ) and let p be a prime ideal of degree 1 in K with the norm p sufficiently large. We have ϑ ≡ t0 (mod p) for a suitable t0 ∈ Z and Pi (t0 ) ≡ 0 (mod p). Choosing t1 = t0 or t0 + p we can achieve that every t ∗ ≡ t1 (mod p 2 ) satisfies p Pi (t ∗ ). Moreover, since p is sufficiently large we have by Lemma 3 p /| ABi (t ∗ ), whence pβi B(t ∗ ). If βi μi the congruence (10) has the solution X = 0. If βi < μi the equality A(t ∗ )x n + B(t ∗ ) = M(t ∗ )y implies that βi ≡ 0 (mod n), x ≡ 0 (mod pβi /n ) and n xPi (t ∗ )−βi /n ≡ −Bi (t ∗ )/A(t ∗ ) (mod p). However, Bi (t ∗ ) Bi (ϑ) ≡ (mod p) ∗ A(t ) A(ϑ) and thus −Bi (ϑ)/A(ϑ) is an nth power residue for almost all prime ideals of degree 1 in K. In virtue of Flanders’ theorem [3a] −Bi (ϑ)/A(ϑ) = C(ϑ)n , where C ∈ Q[t] and thus A(t)C(t)n + Bi (t) ≡ 0 mod Pi (t) . Hence

A(t)x n + Bi (t) ≡ x − C(t) H (x, t) mod Pi (t) . Clearly, H C(t), t ≡ 0 mod Pi (t) ; thus by Lemma 1 applied with D = Q[t] there exists a Cμi −1 ∈ Q[t] such that μ A(t)Cμi −1 (t)n + Bi (t) ≡ 0 mod Pi i (t) . Now we can satisfy (10) by taking βi /n

X(t) = Cμi −1 (t)Pi

(t).

Lemma 8. Let under the assumptions of Lemma 7 Π (t) =

k

Pi (t)max{−[−μi /n],μi +(n−1)[−βi /n]} ,

i=1

A(t)X(t, v)n + B(t) . M(t) If d, e ∈ Z, dX0 ∈ Z[t], eY0 ∈ Z[t] then dX(t, v) ∈ Z[t, v], [d n m, e]Y (t, v) ∈ Z[t, v]. X(t, v) = X0 (t) + vΠ (t),

Y (t, v) =

Proof. The statement concerning X(t, v) is obvious and that concerning Y (t, v) follows

76


from the identity Y (t, v) = Y0 (t) +

n n ν=1

ν

A(t)X0 (t)n−ν Π (t)ν M(t)−1 v ν .

Indeed, ordPi X0n−1 Π −(n − 1)[−βi /n] + max{−[−μi /n], μi + (n − 1)[−βi /n]} μi = ordPi M, ordPi Π n n max{−[−μi /n], μi + (n − 1)[−βi /n]} μi = ordPi M; hence for each ν = 1, 2, . . . , n X0 (t)n−ν Π (t)ν Pi (t)−μi ∈ Q[t]. Since dX0 ∈ Z[t] and Pi is primitive we have d n−ν X0 (t)n−ν Π (t)ν Pi (t)−μi ∈ Z[t], (11)

md n−ν X0 (t)n−ν Π (t)ν M(t)−1 ∈ Z[t].

Lemma 9. Let P ∈ Z[t] be a primitive polynomial with discriminant D = 0, t ∗ ∈ Z, p a prime. If ordp D = d, ∞ > ordp P (t ∗ ) = e > 2d + 1 then there exists a t0 ≡ t ∗ (mod p e−d−1 ) such that ordp P (t0 ) = ordp P (t ∗ ) − 1. Proof. If P is of degree 1 then P (t ∗ ) ≡ 0 (mod p) implies P (t ∗ ) ≡ 0 (mod p), P being primitive. Therefore, it is enough to take t0 = t ∗ + p e−1 . If P is of degree > 1 then we have for suitable polynomials U , V ∈ Z[t] that P U +P V = D (see Rédei [7], Satz 275). Hence e > 2d +1 implies δ = ordp P (t ∗ ) d. Take t0 = t ∗ + p e−δ−1 . From the Taylor formula we get P (t0 ) ≡ P (t ∗ ) + P (t ∗ )p e−δ−1 (mod p 2(e−δ−1) ). By the assumption 2(e − δ − 1) 2(e − d − 1) > e − 1 = ordp P (t ∗ )p e−δ−1 . Hence ordp P (t0 ) = e − 1.

Remark. It may be that the lemma holds for e > d + 1, but the writer could not prove it.

c

Lemma 10. Under the assumptions of Lemmata 7 and 8 for every prime p there exist an integer c and an integer valued function w(τ ) defined on the set {0, 1, . . . , p2c − 1} such that if t ∗ ∈ Z, v ∗ ∈ Q, t ∗ ≡ τ (mod p 2c ), pc v ∗ ≡ w(τ ) (mod p2c ) then X(t ∗ , v ∗ ) and Y (t ∗ , v ∗ ) are p-adic integers.

Proof. Let nonnegative integers ξ, η be chosen so that pξ X0 , pη Y0 have integral p-adic coefficients. Let Rij for i, j = 1, . . . , k be the resultant of Pi and Pj if i = j and the discriminant of Pi if i = j . Moreover, let Ri0 be the resultant of Pi and A, Ri,k+1 the −βi (1 i k). c resultant of Pi and Bi = BPi (t)


77

Put ij = ordp Rij . Clearly, ij = j i min ordp Pi (t ∗ ), ordp Pj (t ∗ ) for every t ∗ ∈ Z and i = j (1 i, j k). Put further ci = i0 +

k

μj ij + 2ii + 2i,k+1 + nξ + 2η + 2 ordp n + ordp m,

j =1

c

c=

k

ci μi + ξ + ordp m.

i=1

For every nonnegative integer τ < p 2c the arithmetic progression τ + p c−ξ +1 u contains an integer tτ such that for suitable integers xτ , yτ we have A(tτ )xτn + B(tτ ) = M(tτ )yτ ,

(12)

M(tτ ) = 0

(integers tτ , xτ , yτ are not determined uniquely, but any choice will do). If for all i k we have ordp Pi (τ ) ci then o = ordp Π (τ )

k

μi ordp Pi (τ ) c − ξ.

i=1

We define w(τ ) = pc−o−ξ w, where w is a root of the congruence w

Π (τ ) + p ξ X0 (τ ) ≡ xτ p ξ (mod p c ). po

If t ∗ ≡ τ (mod p2c ) and p c v ∗ ≡ w(τ ) (mod p2c ) then we have ordp M(τ ) = ordp m + c

k

μi ordp Pi (τ ) c − ξ,

i=1

hence ordp M(t ∗ ) = ordp M(τ ) = ordp M(tτ ), w(τ ) X(t ∗ , v ∗ ) ≡ X0 (τ ) + c Π (τ ) (mod p c ). p By the definition of w(τ ) w Π (τ ) ≡ xτ (mod p c−ξ ), p o+ξ A(t ∗ )X(t ∗ , v ∗ )n + B(t ∗ ) ≡ A(tτ )xτn + B(tτ ) ≡ M(tτ )yτ (mod p c−ξ ), X(t ∗ , v ∗ ) ≡ X0 (τ ) +

c

hence

ordp A(t ∗ )X(t ∗ , v ∗ )n + B(t ∗ ) min c − ξ, ordp M(tτ ) = ordp M(t ∗ ).

This shows that X(t ∗ , v ∗ ) and Y (t ∗ , v ∗ ) are both p-adic integers. c If, for a certain i k, ordp Pi (τ ) > ci then, since ci ij , we have for all j = i c (1 j k) ord p Pj (τ ) ij cj thus i is uniquely determined. We have the following

78


possibilities: βi μi ≡ 0 (mod n), βi μi ≡ 0 (mod n) and βi < μi which we consider successively. 1. βi μi ≡ 0 (mod n). Here we set ζi = max{ξ, i0 }, −μi /n

Πi (t) = Π (t)Pi

(t),

−μi /n

X0i (t) = X0 (t)Pi

Mi (t) = M(t)Pi (t)

−μi

(t),

.

We have oi = ordp Πi (τ )

μj ordp Pj (τ )

j =1 j =i

c

c

k

k

μj ij ci − ζi c − ζi .

j =1 j =i

Moreover from (12) and βi μi we infer that μ

Pi i (tτ ) | A(tτ )xτn and since

ordp Pi (tτ ) min ordp Pi (τ ), c − ξ + 1 > ci i0 ,

we get ordp A(tτ ) i0 nζi , n(ordp xτ + ζi ) μi ordp Pi (tτ ). We define w(τ ) = pc−oi −ζi w, where w is a root of the congruence xτ p ζi Πi (τ ) w + p ζi X0i (τ ) ≡ μ /n (mod p c ). o i p i Pi (tτ ) If t ∗ ≡ τ (mod p2c ), p c v ∗ ≡ w(τ ) (mod p2c ) we have ordp Mi (τ ) = ordp m +

k

μj ordp Pj (τ ) ci − ζi μi (ci − ζi ) c − μi ζi − ξ,

j =1 j =i

hence ordp Mi (t ∗ ) = ordp Mi (τ ) = ordp Mi (tτ ). On the other hand, taking Xi (t ∗ , v ∗ ) = X0i (t ∗ ) + v ∗ Πi (t ∗ ), we get Xi (t ∗ , v ∗ ) ≡ X0i (τ ) +

w(t) Πi (τ ) (mod p c ) pc

and by the definition of w(τ ) p ζi Xi (t ∗ , v ∗ ) ≡ pζi X0i (τ ) +

w xτ p ζi Π (τ ) ≡ (mod pc ). i μi /n p oi Pi (tτ )


Since μi /n

ordp Pi

79

(t ∗ ) min ordp Pi (τ ), 2c > ci ζi

we infer that X(t ∗ , v ∗ ) = Xi (t ∗ , v ∗ )Pi i (t ∗ ) is a p-adic integer. Moreover p nζi A(t ∗ )Xi (t ∗ , v ∗ )n + Pi (t ∗ )βi −μi Bi (t ∗ ) μ /n

≡ c

A(tτ )p nζi xτn + pnζi Pi (t ∗ )βi −μi Bi (tτ ) (mod p c−ξ ). μ Pi i (tτ )

By (12) the right hand side equals p nζi Mi (tτ )yτ , hence ordp A(t ∗ )Xi (t ∗ , v ∗ )n + Pi (t ∗ )βi −μi Bi (t ∗ ) min c − ξ − nζi , ordp Mi (tτ ) = ordp Mi (t ∗ ) and

ordp A(t ∗ )X(t ∗ , v ∗ )n + B(t ∗ ) μi ordp Pi (t ∗ ) + ordp Mi (t ∗ ) = ordp M(t ∗ ).

Thus Y (t ∗ , v ∗ ) is a p-adic integer. 2. βi μi ≡ 0 (mod n). Here we set w(τ ) = 0. If βi > μi we have X0 (t) ≡ 0 mod Pi (t)−[−βi /n] , A(t)X0 (t)n + B(t) ≡ 0 mod Pi (t)μi +1 , hence Y0 (t) ≡ 0 mod Pi (t) . Moreover, since Pi (t) is primitive, we have (13)

p ξ X0 (t)Pi (t)−1 ∈ Zp [t],

pη Y0 (t)Pi (t)−1 ∈ Zp [t],

where Zp is the ring of p-adic integers. If t ∗ ≡ τ (mod p2c ), p c v ∗ ≡ w(τ ) (mod p2c ), we have v ∗ ≡ 0 (mod pc ), X(t ∗ , v ∗ ) ≡ X0 (t ∗ ) (mod p c ), n n A(t ∗ )X0 (t ∗ )n−ν Π (t ∗ )ν ∗ ν ∗ ∗ ∗ Y (t , v ) = Y0 (t ) + v . ν M(t ∗ ) ν=1

c

Since ordp Pi

(t ∗ )

> ci max{ξ, η} we infer from (13) that X0 (t ∗ ), Y0 (t ∗ ) ∈ Zp .

c

On the other hand, by (11) X0 (t)n−ν Π (t)ν ∈ Zp [t] (ν = 1, 2, . . . , n). M(t) Since ordp v ν c ξ(n − 1) + ordp m we conclude that X(t ∗ , v ∗ ), Y (t ∗ , v ∗ ) are p-adic integers. If βi = μi we have as before X0 (t) ≡ 0 mod Pi (t) . If t ∗ ≡ τ (mod p2c ), p c v ∗ ≡ w(τ ) ≡ 0 (mod p2c ), then X(t ∗ , v ∗ ) ≡ X0 (t ∗ ) (mod p c ) is again a p-adic integer and Y (t ∗ , v ∗ ) ∈ Zp if and only if Y0 (t ∗ ) ∈ Zp . The latter condition is satisfied for all t ∗ ≡ τ (mod p2c ) if it is satisfied for one such t ∗ . Since we cannot have Pi (t ∗ ) = 0 for all t ∗ ≡ τ (mod p2c ) we may assume that Pi (t ∗ ) = 0. If η = 0, Y0 (t ∗ ) ∈ Zp . If η > 0 we have ordp Pi (t ∗ ) > ci 2ii + 1, hence by Lemma 9 mp ξ(n−ν)

80 c


there exists a t0 ≡ t ∗ (mod p ci −ii ) such that ordp Pi (t0 ) = ordp Pi (t ∗ ) − 1. On the other hand ordp A(t ∗ ) i0 < ci − ii , ordp Bi (t ∗ ) i,k+1 < ci − ii ,

c

ordp Mi (t ∗ ) ordp m +

k

μj ij < ci − ii − nξ,

j =1 j =i

hence ordp A(t0 ) = ordp A(t ∗ ) < ∞,

ordp Bi (t0 ) = ordp Bi (t ∗ ) < ∞,

ordp Mi (t0 ) = ordp Mi (t ∗ ) < ∞. Since μi ≡ 0 (mod n) we cannot have simultaneously μi ordp Pi (t ∗ ) + ordp Bi (t ∗ ) ≡ ordp A(t ∗ ) (mod n) and μi ordp Pi (t0 ) + ordp Bi (t0 ) ≡ ordp A(t0 ) (mod n) thus taking t1 = t ∗ or t0 we can achieve that ordp B(t1 ) = μi ordp Pi (t1 ) + ordp Bi (t1 ) ≡ ordp A(t1 ) (mod n),

∞ > ordp Pi (t1 ) ci − ii ,

a = max ordp A(t1 ), ordp B(t1 ), ordp M(t1 ) < ∞. The arithmetic progression t1 + pa+1 u contains an integer t2 such that for suitable integers x2 , y2 we have A(t2 )x2n + B(t2 ) = M(t2 )y2 . Since ordp A(t2 ) = ordp A(t1 ), ordp B(t2 ) = ordp B(t1 ),

ordp Pi (t2 ) = ordp Pi (t1 ), ordp M(t2 ) = ordp M(t1 ),

we have ordp A(t2 )x2n ≡ ordp A(t2 ) ≡ ordp B(t2 ) (mod n). It follows that ordp B(t2 ) ordp M(t2 )y2 ordp M(t2 )

and

ordp B(t1 ) ordp M(t1 ).


81

On the other hand we have ordp Pi (t1 ) ci − ii > ij for all j k + 1, hence ordp Pj (t1 ) ij (j = i),

ordp Mi (t1 ) ordp m +

k

ij μj ci − nξ − ii ;

j =1 j =i

ordp A(t1 )X0 (t1 )n −n[−βi /n] ordp Pi (t1 ) − nξ (μi + 1) ordp Pi (t1 ) − nξ μi ordp Pi (t1 ) + ci − nξ − ii μi ordp Pi (t1 ) + ordp Mi (t1 ) = ordp M(t1 ). Since A(t1 )X0 (t1 )n + B(t1 ) = M(t1 )Y0 (t1 ) we get ordp Y0 (t1 ) 0. However pη Y0 (t1 ) ≡ pη Y0 (t ∗ ) (mod p ci −ii ). Since ci − ii η, Y0 (t ∗ ) is a p-adic integer and so is Y (t ∗ , v ∗ ). 3. βi < μi . Here we have βi ≡ 0 (mod n), Pi (t)βi /n X0 (t),

Pi (t)μi −

n−1 n βi

Π (t).

Let X0i (t) = X0 (t)Pi (t)−βi /n ,

Πi (t) = Π (t)Pi (t)

n−1 n βi −μi ,

Mi (t) = M(t)Pi (t)−μi . We have μi −βi

A(t)X0i (t)n + Bi (t) = Pi

(t)Mi (t)Y0 (t)

and ordp Pi (τ ) > ci i,k+1 + η ordp Bi (τ ) + η,

c

hence μi −βi

ordp Pi

(τ )Mi (τ )Y0 (τ ) > ordp Bi (τ )

and ordp A(τ )X0i (τ )n = ordp Bi (τ ). We get ordp X0i (τ )

c

1 1 ordp Bi (τ ) i,k+1 n n

and c

(14) oi = ordp

nX0i (τ )n−1 Πi (τ ) ordp n + i,k+1 − ordp m ci − η c − η. Mi (τ )

We define w(τ ) = pc−oi −η w, where w is a root of the congruence nX0i (τ )n−1 Πi (τ ) w + p η Y0 (τ ) ≡ 0 (mod pη ). Mi (τ )p oi

82


If t ∗ ≡ τ (mod p2c ) and p c v ∗ ≡ w(τ ) (mod p2c ) we have ∗

ordp Pi (t ) > ci

k

βj ij + i,k+1 + η.

j =1

If Pi (t ∗ ) = 0 then (15)

ordp B(t ∗ ) βi ordp Pi (t ∗ ) + i,k+1 < μi ordp Pi (t ∗ ) − η

c

ordp M(t ∗ ) − η ordp M(t ∗ )Y0 (t ∗ ).

Therefore, (16)

ordp B(t ∗ ) = ordp A(t ∗ )X0 (t ∗ )n .

Moreover, A(t ∗ )B(t ∗ )M(t ∗ ) = 0. Let b = max ordp A(t ∗ ), ordp B(t ∗ ), ordp M(t ∗ ) and let t0 be an integer in the arithmetic progression t ∗ + p b+1 u such that for suitable x0 , y0 ∈ Z A(t0 )x0n + B(t0 ) = M(t0 )y0 . c

We have ordp A(t0 ) = ordp A(t ∗ ), ordp B(t0 ) = ordp B(t ∗ ), ordp M(t0 ) = ordp M(t ∗ ) and by (15) ordp B(t0 ) < ordp M(t0 ) ordp M(t0 )y0 . Hence ordp B(t0 ) = ordp A(t0 )x0n and by (16) (17)

ordp X0 (t ∗ ) = ordp x0 0.

If Pi (t ∗ ) = 0 there exists a t ≡ t ∗ (mod p 2c ) such that Pi (t ) = 0. Since X0 (t ∗ ) ≡ X0 (t ) (mod p 2c−ξ ) we have (17) in every case. On the other hand w(τ ) ordp v ∗ Π (t ∗ ) min c, ordp Π (τ ) pc min c, ordp Π (τ ) − oi − η min{c, ci − oi − η} 0, hence ordp X(t ∗ , v ∗ ) 0. It remains to prove that Y (t ∗ , v ∗ ) is a p-adic integer. We have Y (t ∗ , v ∗ ) = Y0 (t ∗ ) +

nX0i (t ∗ )n−1 Πi (t ∗ ) ∗ v Mi (t ∗ ) n n X0i (t ∗ )n−ν Πi (t ∗ )ν + Pi (t ∗ )(ν−1)(μi −βi ) v ∗ ν . ν Mi (t ∗ ) ν=2


83

Now pη Y0 (t ∗ ) ≡ pη Y0 (τ ) (mod p 2c ), p(n−1)ξ X0i (t ∗ )n−1 Πi (t ∗ ) ≡ p(n−1)ξ X0i (τ )n−1 Πi (τ ) (mod p 2c ), Mi (t ∗ ) ≡ Mi (τ ) (mod p 2c ),

(18) ordp Mi (τ )

k

μj ij + ordp m < ci − nξ + ordp m < 2c.

j =1 j =i

c

Hence pη

nX0i (t ∗ )n−1 Πi (t ∗ ) ∗ v Mi (t ∗ ) ≡ pη

nX0i (τ )n−1 Πi (τ ) w(τ ) mod p c−(n−1)ξ +η−ordp Mi (τ ) c Mi (τ ) p

and since c ci + ordp m we have by the definition of w(τ ) c

p η Y0 (t ∗ ) + p η

nX0i (t ∗ )n−1 Πi (t ∗ ) ∗ v Mi (t ∗ ) ≡ pη Y0 (τ ) +

nX0i (τ )n−1 Πi (τ ) ≡ 0 (mod pη ). Mi (τ )p oi

Thus Y0 (t ∗ ) +

nX0i (t ∗ )n−1 Πi (t ∗ ) ∗ v Mi (t ∗ )

is a p-adic number. Now take ν 2 and consider the term n X0i (t ∗ )n−ν Πi (t ∗ )ν Pi (t ∗ )(ν−1)(μi −βi ) v ∗ ν . Eν (t ∗ , v ∗ ) = ν Mi (t ∗ ) We have by (18) X0i (t ∗ )n−ν Πi (t ∗ )ν μj ij − ordp m, −(n − ν)ξ − ∗ Mi (t ) k

ordp

j =1 j =i

c

while by the congruence p c v ∗ ≡ w(τ ) (mod p2c ), by the definition of w(τ ) and by (14) ordp v ∗ −oi − η − ordp n − i,k+1 + ordp m − η,

c

finally ordp Pi (t ∗ )μi −βi > ci c

k j =1 j =i

μj ij + 2i,k+1 + nξ + 2η + 2 ordp n.

84


Hence ∗

∗

ordp Eν (t , v ) > −(n − ν)ξ −

k

μj ij − ordp m − ν ordp n

j =1 j =i

− νi,k+1 + ν ordp m − νη + (ν − 1)

k

μj ij

j =1 j =i

+ 2(ν − 1)i,k+1 + (ν − 1)nξ + 2(ν − 1)η + 2(ν − 1) ordp n = n(ν − 2)ξ + νξ + (ν − 2)

k

μj ij + (ν − 1) ordp m

j =1 j =i

+ (ν − 2) ordp n + (ν − 2)i,k+1 + (ν − 2)η + (ν − 2) ordp n 0.

c

Thus Eν

(t ∗ , v ∗ )

is a p-adic integer and so is Y (t ∗ , v ∗ ).

Proof of Theorem 4. Suppose first that BM = 0, (A, M) = 1. Let X0 , Y0 have the meaning of Lemma 7 and let d be chosen so that dX0 ∈ Z[t], dY0 ∈ Z[t]. Let further X(t, v), Y (t, v) have the meaning of Lemma 8. In virtue of Lemma 10 for every prime p | dm there exists an integer cp and an integer valued function wp (τ ) defined on the set {0, 1, . . . , pcp −1} such that for all t ∗ ∈ Z, v ∗ ∈ Q the congruences t ∗ ≡ τ (mod p2cp ), pc v ∗ ≡ wp (τ ) (mod p 2cp ) imply that X(t ∗ , v ∗ ) and Y (t ∗ , v ∗ ) are p-adic integers. By a result of Skolem [8] there exists an integer valued polynomial Wp (t) such that t ∗ ≡ τ (mod p 2cp ) implies Wp (t ∗ ) ≡ wp (τ ) (mod p 2cp ). Now take Wp (t) 2cp V (t) = q ϕ(p )cq c p p q |dm q=p

p|dm

where p, q run over primes and set X(t) = X t, V (t) , c

Y (t) = Y t, V (t) .

We assert that X(t), Y (t) belong to the set I of integer valued polynomials. Indeed, by Lemma 8, dX(t, v) ∈ Z[t, v], Moreover, since V (t)

d n mY (t, v) ∈ Z[t, v].

p cp ∈ I

p|dm

c

and X(t, v), Y (t, v) are in v of degrees 1 and n respectively, we have p cp ∈ I , Y (t)d n m p ncp ∈ I. X(t)d c

p|dm

p|dm

85


Thus for t ∗ ∈ Z the values X(t ∗ ) and Y (t ∗ ) are p-adic integers for each p /| dm. On the other hand if p | dm and t ∗ ≡ τ (mod p2cp ), 0 τ < p2cp we have Wq (t ∗ ) ϕ(q 2cq1 )cq 2cp 1 2 q2 1 + Wp (t ∗ ) q ϕ(p )cq pcp V (t ∗ ) ≡ pcp cq1 q1 |dm q1 q2 |dm q |dm q1 =p

q2 =q1

∗

≡ Wp (t ) ≡ Wp (τ ) (mod p

q=p

2cp

),

hence by the property of the polynomials X(t, v), Y (t, v) stated above X(t ∗ ) = X t ∗ , V (t ∗ ) , Y (t ∗ ) = Y t ∗ , V (t ∗ ) are p-adic integers. Suppose now that BM = 0 and (A, M) = 1. Then there exists a primitive polynomial D ∈ Z[t] such that A = DA1 , M = DM1 , A1 , M1 ∈ Z[t] and (A1 , M1 ) = 1. Every arithmetic progression contains an integer t ∗ such that A(t ∗ )x n + B(t ∗ ) = M(t ∗ )y is solvable in integers x, y hence D(t ∗ ) | B(t ∗ ). It follows that D | B and since D is primitive B = DB1 , where B1 ∈ Z[t]. Every arithmetic progression P contains a progression P1 such that D(t ∗ ) = 0 for t ∈ P1 . Therefore, for t ∗ ∈ P1 A(t ∗ )x n + B(t ∗ ) = M(t ∗ )y

implies

A1 (t ∗ )x n + B1 (t ∗ ) = M1 (t ∗ )y

and from the already proved case of the theorem we infer the existence of integer valued polynomials X, Y such that A1 (t)X(t)n + B1 (t) = M1 (t)Y (t). Clearly, A(t)X(t)n + B(t) = M(t)Y (t). It remains to consider the case BM = 0. If B = 0 we can take X = Y = 0. If B = 0 and M = 0 Theorem 1 of [2] implies the existence of a polynomial X ∈ Q[t] such that A(t)X(t)n +B(t) = 0. By the assumption every arithmetic progression contains an integer t ∗ such that either B(t ∗ ) = 0 or X(t ∗ )n is an integer. Since B = 0, the former term of the alternative can be omitted. Let a positive integer d be chosen so that dX ∈ Z[t] and let τ be an arbitrary integer. The arithmetic progression τ + du contains an integer t ∗ such that X(t ∗ ) is an integer. We have dX(τ ) ≡ dX(t ∗ ) (mod d), hence d | dX(τ ) and X(τ ) is an integer. Thus X is an integer valued polynomial and the proof is complete. Now we shall show by an example that the condition n ≡ 0 (mod 8) cannot be omitted from the assumptions of Theorem 4. Example 3. Take n = 8, A(t) = 1, B(t) = −16, M(t) = 2t + 1. For every integer t ∗ we k have M(t ∗ ) = ± piαi , where pi are odd primes. For every i k the congruence i=1

x 8 ≡ 16 (mod piαi ) is solvable (cf. Trost [11]). Denoting a solution of this congruence by xi and using the Chinese Remainder Theorem we find x ≡ xi (mod piαi ) (1 i k), which satisfies x 8 − 16 ≡ 0 (mod 2t ∗ + 1). On the other hand, the existence of polynomials X, Y ∈ Q[t] 8 2 satisfying X(t)8 − 16 = (2t + 1)Y (t) would imply X − 21 = 16, X − 21 = 2, a contradiction.

86


The next example shows that in Theorem 4 polynomials in one variable cannot be replaced by polynomials in two variables even if A = 1 and M is irreducible. 2 + (4t 2 + 1)2 . For every Example 4. Take n = 2, A(t, u) = B(t, u) = 1, M(t, u) = u ∗ ∗ 2 ∗ pair of integers t , u thecongruence x + 1 ≡ 0 mod M(t , u∗ ) is solvable. Indeed, ∗ ∗ α piαi , where α = 0 or 1 and pi ≡ 1 (mod 4). On the other hand c we have M(t , u ) = 2 suppose that polynomials X, Y ∈ Q[t, u] satisfy

X(t, u)2 + 1 = M(t, u)Y (t, u). We get u2 X(t, u)2 ≡ (4t 2 + 1)2 mod M(t, u) and since M is irreducible uX(t, u) ≡ ±(4t 2 + 1) mod M(t, u) . The substitution u = 0 gives

4t 2 + 1 ≡ 0 mod (4t 2 + 1)2 ,

a contradiction.

References √ [1] N. H. Abel, Über die Integration der Differential Formel dx/ R wenn R und ganze Functionen sind. J. Reine Angew. Math. 1 (1826), 185–221; French transl. in Oeuvres choisies, Christiania 1881, T. I, 104–144. [2] H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107–116; this collection: A6, 27–35. [3] −−, −−, −−, Quadratic Diophantine equations with a parameter. Acta Arith. 11 (1966), 353–358. [3a] H. Flanders, Generalization of a theorem of Ankeny and Rogers. Ann. of Math. (2) 57 (1953), 392–400. [4] M. Fujiwara, Hasse principle in algebraic equations. Acta Arith. 22 (1973), 267–276. [5] H. Hasse, Zwei Bemerkungen zu der Arbeit „Zur Arithmetik der Polynome” von U. Wegner in den Math. Ann. 105, S. 628–631. Math. Ann. 106 (1932), 455–456. [6] −−, Zahlentheorie. Akademie-Verlag, Berlin 1969. [7] L. Rédei, Algebra I. Akademische Verlagsgesellschaft, Geest & Portig, Leipzig 1959. [8] T. Skolem, Über die Lösbarkeit gewisser linearer Gleichungen in Bereiche der ganzwertigen Polynome. Kong. Norske Vid. Selsk. Forh. 9 (1937), no. 34. [9] −−, Einige Sätze über Polynome. Avh. Norske Vid. Akad. Oslo I 1940, no. 4. [10] −−, Unlösbarkeit von Gleichungen, deren entsprechende Kongruenz für jeden Modul lösbar ist. Ibid. 1942, no. 4. [11] E. Trost, Zur Theorie von Potenzresten. Nieuw Arch. Wisk. 18 (1934), 58–61. [12] B. L. van der Waerden, Noch eine Bemerkung zu der Arbeit „Zur Arithmetik der Polynome” von U. Wegner in Math. Ann. 105, S. 628–631. Math. Ann. 109 (1934), 679–680.


Originally published in Studia Mathematica LXXVII (1983), 103–109

Hasse’s principle for systems of ternary quadratic forms and for one biquadratic form To Professor Jan Mikusiński on the occasion of the 70th birthday Abstract. Let K be an algebraic number field and f1 , . . . , fk ternary quadratic forms over K. If f1 , . . . , fk have a common non-trivial zero in every completion of K except at most one then— it is proved here—they have a common non-trivial zero in K. Besides an example is given of an absolutely irreducible n-ary biquadratic form (n 3) that represents 0 in every completion of Q but not in Q.

Let K be an algebraic number field and f1 , . . . , fk ∈ K[x1 , . . . , xn ] quadratic forms. Hasse’s principle asserts that if the forms f1 , . . . , fk have a common non-trivial zero in every completion of K they have a common non-trivial zero in K. The principle holds for k = 1, it trivially holds for n = 1, 2, and it fails for K = Q, k = 2, n 4 (see [2]). Thus it remains to consider the case n = 3. We shall prove Theorem 1. If quadratic forms f1 , . . . , fk ∈ K[x, y, z] have a common non-trivial zero in every completion of K except at most one then they have a common non-trivial zero in K. As to biquadratic forms over Q it is easy to give an example of a reducible ternary form for which Hasse’s principle fails (see [1], p. 72). An example of an irreducible ternary biquadratic form with the same property can be constructed by using results of Hilbert [4], namely √ √ norm(x + y 5 + z −31). c

This form, however, is reducible over the complex field. Mordell [6] has left open the question whether there exists an absolutely irreducible ternary biquadratic form not fulfilling Hasse’s principle. The question is answered by Theorem 2. The absolutely irreducible biquadratic form x 4 − 2y 4 − 16y 2 z2 − 49z4 represents 0 in every completion of Q but not in Q; for all n 4 the absolutely irreducible biquadratic form x14 − 17x24 − 2(x32 + . . . + xn2 )2 represents 0 in every completion of Q but not in Q.

88


Lemma 1. If a binary form over K of degree not exceeding 4 represents 0 in all but finitely many completions of K it represents 0 in K.

Proof. See Fujiwara [3].

Lemma 2. Let R(x, y; u1 , . . . , uk , v1 , . . . , vk ) be the resultant of

k i=1

ui f i ,

k

vi fi with

i=1

respect to z (ui , vi are indeterminates). If (1)

R(a, b; u1 , . . . , uk , v1 , . . . , vk ) = 0,

a, b ∈ K, a, b = 0, 0

then either fi have a common non-trivial zero in K or (bx − ay)2 | R(x, y; u1 , . . . , uk , v1 , . . . , vk ) and the forms fi (at, bt, z) differ from their highest common divisor by a constant factor. Proof. If fi are all of degree less than 2 with respect to z then they have a common nontrivial zero, namely If at least one of the forms fi is of degree 2 with respect k 0, 0, 1. to z then both i=1 ui fi and ki=1 vi fi are of degree 2 with respect to z with the leading coefficients independent of x, y. Therefore (1) implies that k i=1

ui fi (a, b, z)

and

k

vi fi (a, b, z)

i=1

have a common factor over the field K(u1 , . . . , uk , v1 , . . . , vk ), hence also over the ring K[u1 , . . . , uk , v1 , . . . , vk ]. The factor must be independent of u1 , . . . , uk , v1 , . . . , vk . If it is of degree 1 in z it has a zero c ∈ K and we have fi (a, b, c) = 0 (1 i k). If it is of degree 2 in z we consider the Sylvester matrix S(x, y; u1 , . . . , uk , v1 , . . . , vk ) of the polynomials i=1 ui fi , ki=1 vi fi . In virtue of a well-known theorem (see [7], Satz 114) the rank of the matrix S(a, b; u1 , . . . , uk , v1 , . . . , vk ) must be 2. Hence all the minors of degree 3 of this matrix vanish and all the minors of degree 3 of the matrix S(x, y; u1 , . . . , uk , v1 , . . . , vk ) are divisible by bx − ay. On the other hand, there are minors of degree 2 of the latter matrix not divisible by bx − ay, in fact independent of x, y. Hence by a very special case of theorem of Rédei [8] k

R(x, y; u1 , . . . , uk , v1 , . . . , vk ) = det S(x, y; u1 , . . . , uk , v1 , . . . , vk ) is divisible by (bx − ay)2 . The last assertion of the lemma follows from the remark that if polynomials fi (a, b, z) (1 i k) have a common factor of degree 2 they differ from this common factor by a constant factor.

A13. Hasse’s principle

89

Proof of Theorem 1. Let us consider the resultant R(x, y; u1 , . . . , uk , v1 , . . . , vk ) of vi fi with respect to z. Viewed as a polynomial in x, y it is either 0 or ui fi and a quartic form. In the first case fi (1 i k) have a common factor, say d. If d is of degree 2 then for each i k we have fi = ci d, ci ∈ K. The solvability of fi (x, y, z) = 0 (i k) in a completion Kv of K implies the solvability of d(x, y, z) = 0 in Kv , and if it holds for all but one completion then by the product formula and Hasse’s principle for one quadratic form we get solvability in K of d(x, y, z) = 0 and hence of fi (x, y, z) = 0 (1 i k). If d is of degree 1 then it has again a non-trivial zero in K and the same conclusion holds. If R(x, y; u1 , . . . , uk , v1 , . . . , vk ) is not identically 0, let r(x, y) be the highest common v1 , . . . , vk . If divisor of its coefficients when viewed as a form in u1 , . . . , uk , fi (x, y, z) (1 i k) have a common non-trivial zero av , bv , cv in Kv , ui fi and vi fi have it also, hence R(av , bv ; u1 , . . . , uk , v1 , . . . , vk ) = 0, which implies (2)

r(av , bv ) = 0.

(Here we use the fact that the coefficients of R are forms in x, y). If av = bv = 0 we have cv = 0; hence the coefficient of z2 in fi is 0 for each i k and the forms fi (1 i k) have in K a common non-trivial zero 0, 0, 1. If av , bv = 0, 0 for each valuation v of K except at most one then by Lemma 1 r has in K a zero, say a, b = 0, 0. Thus bx − ay | r(x, y), bx − ay | R(x, y; u1 , . . . , uk , v1 , . . . , vk ) and by Lemma 2 either fi have a common non-trivial zero in K or (3)

(bx − ay)2 | R(x, y; u1 , . . . , uk , v1 , . . . , vk )

and the forms fi (at, bt, z) (1 i k) differ from their highest common divisor by a constant factor. In the latter case, by (3) (bx − ay)2 | r(x, y). Let (4)

r(x, y) = (bx − ay)α s(x, y),

where α 2, s(a, b) = 0, deg s = deg r − α 2. For every valuation v of K except at most one we have by (2) and (4) bav − abv = 0

or

s(av , bv ) = 0.

The first equation implies av = at, bv = bt for a t ∈ Kv∗ ; thus F (t, u) = s(t, u) h.c.d. fi (at, bt, u) 1ik

has a non-trivial zero in Kv . Since by (4) deg F = deg s + 2 = deg r + 2 − α 4, we infer from Lemma 1 that F has in K a zero, say c, d = 0, 0. If this is a zero of the

90


h.c.d.1ik fi (at, bt, u), then fi (ac, bc, d) = 0 (1 i k),

ac, bc, d = 0, 0, 0.

If, on the other hand, s(c, d) = 0 then by (4) r(c, d) = 0; thus R(c, d; u1 , . . . , uk , v1 , . . . , vk ) = 0 and by Lemma 2 either fi have a common non-trivial zero in K or (dx − cy)2 | R(x, y; u1 , . . . , uk , v1 , . . . , vk ) and fi (ct, dt, z) (1 i k) differ by a constant factor from their highest common divisor. In the latter case (dx − cy)2 | r(x, y) and by (4) r(x, y) = e(bx − ay)2 (dx − cy)2 . For every valuation v of K except at most one we have by (2) bav − abv = 0

c

thus for a suitable t ∈ quartic form

Kv∗

or

dav − cbv = 0;

either av = at, bv = bt or av = ct, bv = dt. It follows that the

G(t, u) = h.c.d. fi (at, bt, u) · h.c.d. fi (ct, dt, u) 1ik

1ik

has a non-trivial zero in Kv . By Lemma 1 G(t, u) has in K a zero, say t0 , u0 = 0, 0. If t0 , u0 is a zero of the h.c.d.1ik fi (at, bt, u) then fi (at0 , bt0 , u0 ) = 0 (1 i k),

at0 , bt0 , u0 = 0, 0, 0;

if t0 , u0 is a zero of the h.c.d.1ik fi (ct, dt, u) then c

fi (ct0 , dt0 , u0 ) = 0 (1 i k),

ct0 , dt0 , u0 = 0, 0, 0.

The proof is complete. For the proof of Theorem 2 we need three lemmata. Lemma 3. The equation u4 − 17v 4 = 2w2 has no solutions in Q except 0, 0, 0. Proof. See Lind [5] or Reichardt [10].

Lemma 4. Let F (x1 , . . . , xn ) be a polynomial with integer p-adic coefficients and γ1 , . . . , γn p-adic integers. If for an i n we have F (γ1 , . . . , γn ) ≡ 0 (mod p2δ+1 ), ∂F (γ1 , . . . , γn ) ≡ 0 (mod pδ ), ∂xi ∂F (γ1 , . . . , γn ) ≡ 0 (mod pδ+1 ) ∂xi

A13. Hasse’s principle

91

(δ a nonnegative integer) then there exist p-adic integers θ1 , . . . , θn such that F (θ1 , . . . , θn ) = 0 and θ1 ≡ γ1 (mod p δ+1 ), . . . , θn ≡ γn (mod p δ+1 ).

Proof. See [1], p. 42. c

Lemma 5. f (x, y, z) = x 4 − 2y 4 − 16y 2 z2 − 49z4 is irreducible over every field of characteristic different from 2 and 17. Proof. Let k be a field of this kind. It is enough to show that f (x, y, z) is irreducible as a polynomial in x over k(y, z). If it were not, then by Capelli’s theorem (see [9], Satz 428) ±(2y 4 + 16y 2 z2 + 49z4 ) would have to be a square in k(y, z). This condition implies that 162 − 4 · 2 · 49 = −8 · 17 = 0, which is possible only if char k = 2 or char k = 17.

Proof of Theorem 2. f (x, y, z) = x 4 − 17z4 − 2(y 2 + 4z2 )2 ; hence by Lemma 3 if f (x, y, z) = 0 and x, y, z ∈ Q we have x = y 2 + 4z2 = 0 and thus x = y = z = 0. Also x14 − 17x24 − 2(x32 + . . . + xn2 )2 = 0 implies x1 = x2 = . . . = xn = 0 for xi ∈ Q. It remains to show that f (x, y, z) represents 0 in every field Qp including Q∞ = R. We verify this first using Lemma √ 3 for p = ∞, 2, 5, 7, 13 and 17. For p = ∞ we take x = 4 2, y = 1, z = 0. c For p = 2 we use Lemma 4 with γ1 = 3, γ2 = 2, γ3 = 1, δ = 2, i = 1. For p = 5 we use Lemma 4 with γ1 = 0, γ2 = 2, γ3 = 1, δ = 0, i = 2. For p = 7 we use Lemma 4 with γ1 = 2, γ2 = 1, γ3 = 0, δ = 0, i = 1. For p = 13 we use Lemma 4 with γ1 = 1, γ2 = 2, γ3 = 3, δ = 0, i = 1. For p = 17 we use Lemma 4 with γ1 = 0, γ2 = 1, γ3 = 2, δ = 0, i = 2. For p = 2, 5, 7, 13, 17 we have either p 37 or for a suitable sign(±7 | p) = 1. In the latter case the congruence f (x, 0, z) = (x 2 − 7z2 )(x 2 + 7z2 ) ≡ 0 (mod p) is solvable non-trivially, and denoting its solution by γ1 , γ3 we use Lemma 4 with γ2 = 0, δ = 0, i = 1. It remains to consider primes p 37. For such primes f is by Lemma 5 absolutely irreducible over Fp . Moreover, it has no singular zeros. Indeed, the equations 4x 3 = 0, c

−8y 3 − 32yz2 = 0,

−32y 2 z − 196z3 = 0

imply x = 0 and either y = 0, 196z3 = 0 or y 2 + 4z2 = 0, 68z3 = 0; thus in any case x = y = z = 0. By the Riemann-Hurwitz formula the curve f (x, y, z) = 0 is over Fp of genus 3. Therefore by Weil’s theorem the number of points on this curve with coordinates in Fp √ is greater than p + 1 − 6 p, i.e., at least one. Since all points are non-singular, Lemma 4 applies with δ = 0 and a suitable i.

92


Note added in proof. I have learned that already in 1981 A. Bremner, D. J. Lewis and P. Morton found the example 3x 4 + 4y 4 − 19z4 of a ternary biquadratic form for which Hasse’s principle fails, but they did not publish it.

References [1] Z. I. Borevich, I. R. Shafarevich, Number Theory. Academic Press, New York 1966. [2] J.-L. Colliot-Thélène, D. Coray, J.-J. Sansuc, Descente et principe de Hasse pour certaines variétés rationnelles. J. Reine Angew. Math. 320 (1980), 150–191. [3] M. Fujiwara, Hasse principle in algebraic equations. Acta Arith. 22 (1973), 267–276. [4] D. Hilbert, Über Diophantische Gleichungen. Nachr. Königl. Gesell. Wiss. Göttingen 1897, 48–54. [5] C. E. Lind, Untersuchungen über die rationale Punkte der ebenen kubischen Kurven vom Geschlecht Eins. Doctoral dissertation, Uppsala 1940. [6] L. J. Mordell, The Diophantine equation Ax 4 + By 4 + Cz4 = 0. Proc. Cambridge Philos. Soc. 68 (1970), 125–128. [7] O. Perron, Algebra I. Walter de Gruyter, Berlin 1951. [8] L. Rédei, Über die Determinantenteiler. Acta Math. Acad. Sci. Hungar. 3 (1952), 143–150. [9] −−, Algebra I. Akademische Verlagsgesellschaft, Geest & Portig, Leipzig 1959. [10] H. Reichardt, Einige im Kleinen überall lösbare, im Grossen unlösbare diophantische Gleichungen. J. Reine Angew. Math. 184 (1942), 12–18.


Originally published in Sets, Graphs and Numbers, Budapest (Hungary), 1991 Colloquia Mathematica Societatis János Bolyai 60 North-Holland, Amsterdam 1992, 329–356

On Runge’s theorem about Diophantine equations with A. Grytczuk (Zielona Góra)

In 1887, C. Runge [14] proved the following theorem. Let f ∈ Z[X, Y ] be an irreducible polynomial. If the equation f (x, y) = 0 has infinitely many integer solutions then the following conditions hold(∗ ): (C1 ) the highest powers of X and Y in f occur as isolated terms aX m and bY n ; (C2 ) for every term cX Y σ of f we have n + mσ mn; (C3 ) the sum of all terms of f for which n + mσ = mn is up to a constant factor a power of an irreducible polynomial (the last condition is stated by Runge in a little weaker form, but his proof gives what is asserted here). In the course of the proof Runge established also under the same assumption the following condition which together with (C1 ) is stronger than (C2 )–(C3 ): (C4 ) there is only one system of conjugate Puiseux expansions at x = ∞ for the algebraic function y = y(x) defined by f (x, y) = 0. An essential feature of Runge’s theorem is that if his conditions are not fulfilled, all integral solutions of the equation f (x, y) = 0 can be found effectively. The special case, where all real roots of the leading form of f (X, Y ) are simple and rational = 0 was rediscovered by E. Maillet [11] and an algorithm to find bounds for the size of solutions has been given by him in [12] under the additional assumption that all roots of the leading form of f are distinct. In the general case, bounds have been given by D. L. Hilliker and E. G. Straus [4]. If d0 = min{degX f, degY f } = 1 and d = max{degX f, degY f } > 2 their bound given in Theorem 3.3 is false (see below). But for d = 1 or d0 > 1 they have proved that if Runge’s conditions (C1 )–(C3 ), which they formulate differently, are violated then all integral solutions of the equation f (x, y) = 0 satisfy 4(f + 1)2 if d = 1, max{|x|, |y|} < 2d 3 if d0 > 1, (8df )d (∗ )

Capital letters X, Y, . . . , Ξ, H, . . . denote indeterminates, small letters x, y, . . . , ξ, η, . . . denote elements of the relevant fields.

94


where f is the height of f (see [4], Theorem 3.2 and Theorem 4.9). Although they do not say so the same inequality follows from their argument if (C4 ) is violated (then in their notation degy Q∗ se < degy f , hence (Q∗ , f ) = 1). Many special equations to which Runge’s theorem applies have been considered. Thus for f of degree 2 the second author gave the essentially best possible bound max |x|, |y| f 4 (see [16]) and for equations y 2 = f (x), f — a monic quartic polynomial, D. W. Masser [13] gave the essentially best possible bound |x| f 3 (essentially means here up to a multiplicative constant). D. W. Masser has also called our attention to the fact that the bound of Hilliker and Straus can be improved by means of a recent result of W. Schmidt [18]. Indeed, when one combines the argument of Hilliker and Straus with Theorem 3 of [18] one obtains the following assertion. Let f ∈ Z[X, Y ] be irreducible of height f . Let degX f = m, degY f = n, d0 = min{m, n}, d = max{m, n} and let integers x, y satisfy f (x, y) = 0. If (C1 ) is not satisfied by the variable X or (C4 ) does not hold, then 16mn6 (m+2)(n+2) |x| 217 m3 n6 f 16m2 n5 (m+2)(n+2) |y| 217 m3 n6 f . If (C1 ) holds, but (C2 ) or (C3 ) does not, then 16d 5 dn(m+2)(n+2) 0 |x| 217 d03 d 6 f 16d 5 dm(m+2)(n+2) 0 |y| 217 d03 d 6 f . Hilliker and Straus have used the original Runge’s approach based on the Puiseux expansions of the algebraic function defined by f (x, y) = 0. In 1922 Skolem [20] gave another proof of Runge’s theorem based on elimination theory. In the present paper we use Skolem’s approach to prove a bound for max |x|, |y| which is better for d > 1 than the bound of Hilliker and Straus, and often better than the assertion above, but which does not apply if only (C4 ) is violated. In the course of the proof we fill in a gap that occurs in Skolem’s paper. We prove the following Theorem. Let f ∈ Z[X, Y ] be irreducible of height f , m = degX f , n = degY f , d0 = min{m, n}, d = max{m, n} and let integers x, y satisfy f (x, y) = 0. (i)

If (C1 ) is not satisfied by the variable X, then 2n(mn+1)3 , |x| (m + 1)(n + 1)(mn + 1)2/n f 2(mn+1)3 . |y| (m + 1)(n + 1)(mn + 1)2/n f

A14. Runge’s theorem about Diophantine equations

(ii)

95

If (C1 ) holds, but (C2 ) does not, then 96m3 n4 (m,n)−4 d 4 +m−1 d0 −1 0 , |x| < (4mnd0 )8mn(m,n) f 96m4 n3 (m,n)−4 d 4 +n−1 d0 −1 0 . |y| < (4mnd0 )8mn(m,n) f

(iii) If (C1 ) and (C2 ) hold, but (C3 ) does not, then 5 m3 n4 (m,n)4 +m−1 (m,n)2 −1 128 , |x| < (mn)3mn(m,n) f 5 m4 n3 (m,n)4 +n−1 (m,n)2 −1 128 . |y| < (mn)3mn(m,n) f No attention should be attached to the coefficients of f , which in the cases (ii)–(iii) probably can be much improved. On the other hand, an effort has been made to obtain the exponents as small as the method allows. Corollary. In the notation of the theorem either (C1 )–(C3 ) hold or (45f )250 if d = 2, max |x|, |y| < 96d 11 2 3 8d if d > 2. (4d ) f The example of the equation xy − ty − tx d = 0 with a solution x = t d+1 + t, y = (t d + 1)d shows that the exponents 250 and 96d 11 in the corollary cannot be lowered below 4 or d 2 respectively. The same example shows that the bound d(f + 1)2d given by Hilliker and Straus in their Theorem 3.3 for the case d0 = 1 is false for d > 2. On the 11 can be replaced by 16d 7 (d + 2)2 . c other hand, the assertion on p. 93 implies that 96d The proof of the theorem falls into three main steps. (A) (Lemmas 1 to 4). We consider the special case, where the leading form of f is a monomial divisible by XY . For equations of this special type if f (x, y) = 0 and, say, |x| max{x0 , |y|} we have such good control of y as a function of x that we can construct a polynomial F ∈ Z[X, Y ] prime to f , which vanishes for all pairs (x, y) in question; this bounds the solution. (B) (Lemmas 5 to 9). Under the general assumptions of the theorem we find polynomials G, H ∈ Z[X, Y ] such that the minimal equation connecting G(x, y), H (x, y) subject to f (x, y) = 0 is of the special type considered in (A). (C) (Lemmas 10, 11 and the proof of the theorem in the strict sense). We deduce the estimates in the general case from those obtained in (A) by deducing bounds for x, y from those for G(x, y), H (x, y). It should be mentioned that the condition (C3 ) has been sharpened by the second author [15] and the condition (C4 ) by M. Ayad [1]. However their proofs do not lead to bounds for the size of possible solutions of the equation f (x, y) = 0, since they use an ineffective theorem of Siegel [19]. N and N0 denote the set of positive integers and of nonnegative integers, respectively. For a real number x, x is the least integer greater than or equal to x. For a polynomial F with coefficients ai ∈ C we shall put 1/ν |ai |ν (ν = 1, 2). F = max |ai |, F ν = i

i

96


We conclude the introduction by expressing our thanks to Professor J. W. S. Cassels for the remarks incorporated in the present version of the paper and to Mr. K. Stefański for his remarks on an early draft. Lemma 1. Let P , Q ∈ Z[X, y], (P , Q) = 1, degX P = p1 , degY P = p2 , and similarly for Q. If P (x, y) = Q(x, y) = 0

(1) then

q2 p2 |x| P (p1 + 1) p2 + 1 Q(q1 + 1) q2 + 1 .

Proof. Since (P , Q) = 1 the resultant R(X) of P and Q with respect to Y is not zero and clearly (1) implies that R(x) = 0. Since the leading coefficient of R is in absolute value 1 we have by the inequality of Landau [6] |x| R2 .

(2) Now

R2 max |R(ξ )|

(3)

|ξ |=1

and using the Hadamard inequality for the determinant of the Sylvester matrix we obtain ! p2 ! q2 2 2 " P (p1 + 1) (p2 + 1) Q2 (q1 + 1)2 (q2 + 1) . R2 j =1

j =1

The inequalities (2) and (3) imply the lemma.

Lemma 2. For all integers n > ν > 0, r 1, t 0, μ 0, λ 1, k = 0 for every polynomial f ∈ Z[X, Y ] with the leading form equal to kX ν Y n−ν there exist polynomials Br,t,μ , Cr,t,λ ∈ Z[X] with the following properties deg Br,t,μ μ,

(4)

Br,t,μ n2(r−1) f r ,

(5) (6)

deg Cr,t,λ νr + λ + t − 1,

(7)

Cr,t,λ 2n2(r−1) f r ;

if f (x, y) = 0, x = 0 then y

n−ν+t

=

r(ν−1)+t μ=0

Br,t,μ (x) n−1+(r−1)(ν−1)+t−μ Cr,t,λ (x) n−ν−λ y + y . (kx ν )r (kx ν )r n−ν

λ=1


97

Proof. We define the polynomials Br,t,μ , Cr,t,λ by induction on r as follows t+ν−1

B1,t,μ (X)Y n−1+t−μ +

μ=0

n−ν

C1,t,λ (X)Y n−ν−λ

λ=1 n−1 = −Y t f (X, Y ) − kX ν Y n−ν := − Aj (X)Y n−1−j +t j =0

and

Br+1,t,μ = −

(8)

h+j =μ r(ν−1)+th0 n>j 0

(9)

Cr+1,t,λ = kXν Cr,t,λ −

(μ (r + 1)(ν − 1) + t),

Br,t,h Aj

(1 λ n − ν).

Br,t,h Aj

h+j =(r+1)(ν−1)+t+λ r(ν−1)+th0 n>j 0

All assertions of the lemma follow by induction on r. Indeed, we have −Aμ if 0 μ min{n − 1, t + ν − 1}, B1,t,μ = 0 otherwise, hence deg B1,t,μ deg Aμ μ, B1,t,μ Aμ f and by (8) deg Br+1,t,μ max {deg Br,t,h + deg Aj } max {h + j } = μ, h+j =μ

h+j =μ

Br+1,t,μ n max Br,t,h Aj n f max Br,t,h 2

h+j =μ

n ·n 2

Similarly

C1,t,λ =

2(r−1)

hμ

f

r+1

−Aν+λ+t−1 0

= n f 2r

r+1

.

if 1 λ n − ν − t, otherwise,

hence (6) and (7) follow from (9) by induction on r. Indeed, these formulae are true for r = 1 and assuming that (6), (7) hold for a fixed r we find from (4), (5) and (9) deg Cr+1,t,λ max ν + deg Cr,t,λ , (r + 1)(ν − 1) + t + λ max ν(r + 1) + λ + t − 1, (r + 1)(ν − 1) + t + λ = ν(r + 1) + λ + t − 1,

98


Cr+1,t,λ kCr,t,λ + n k

c

max Bh,r,t Aj h+j =(r+1)(ν−1)+t+λ · 2n2(r−1) f r + n2 f max Bh,r,t h(r+1)(ν−1)+t+λ 2(r−1) r+1 2 2(r−1) r

2n f + n f n 2r n + 2n2(r−1) f r+1

f

2n2r f r+1 . The remaining assertion of the lemma is true by definition of B1,t,μ , C1,t,λ for r = 1 and assuming its truth for a fixed r we infer by (8) and (9) from f (x, y) = 0 y n−ν+t =

r(ν−1)+t

Br,t,μ (x) n−1+(r−1)(ν−1)+t−μ y (kx ν )r μ=0 r(ν−1)+t n−ν Br,t,μ (x) Cr,t,λ (x) n−ν−λ + y = y (ν−1)r+t−μ ν r (kx ) (kx ν )r μ=0 λ=1 n−1 n−ν Aj (x) n−1−j Cr,t,λ (x) n−ν−λ · − y + y ν kx (kx ν )r

c

=−

j =0 (r+1)(ν−1)+t μ=0

c

− c

=

λ=1

y n−1+r(ν−1)+t−μ (kx ν )r+1

n−ν

Br,t,h (x)Aj (x)

h+j =μ 0j ν > 0, k = 0 and every polynomial f ∈ Z[X, Y ] with the leading form equal to kX ν Y n−ν there exists a polynomial F ∈ Z[X, Y ] with the following properties (10) c (11) (12)

the leading form of F is a monomial in Y, degX F < degY F < n2 , 2n2 F < n2 f .

(13)

If x, y ∈ Z, |x| |y| and

(131 )

f (x, y) = 0

then either F (x, y) = 0 or 2n3 |x| < n2 f .


99

Proof. By Lemma 2 with r = n − ν + t (131 ) implies for x = 0 y

n−ν+t

=

(n−ν+t)(ν−1)+t μ=0

(14) + c

n−ν λ=1

Bn−ν+t,t,μ (x) n−1+(n−ν+t−1)(ν−1)+t−μ y (kx ν )n−ν+t

Cn−ν+t,t,λ (x) n−ν−λ y . (kx ν )n−ν+t

Now let us write (15)

Cn−ν+t,t,λ = Dt,λ X ν(n−ν+t) + Et,λ X ν(n−ν+t)−n+ν+λ+1 + Ft,λ ,

where (16)

Dt,λ , Et,λ , Ft,λ ∈ Z[X], deg Et,λ < n − ν − λ − 1 if λ < n − ν − 1, Et,n−ν−1 = Et,n−ν = 0, deg Ft,λ < ν(n − ν + t) − n + min{λ + ν + 1, n}.

It follows by (6) that deg Dt,λ λ + t − 1.

(17) We have from (14) (ky)n−ν+t =

(n−ν+t)(ν−1)+t μ=0

c

(18)

+ +

n−ν λ=1 n−ν λ=1

Bn−ν+t,t,μ (x) ν(n−ν+t)−μ y x ν(n−ν+t)

Dt,λ (x)y n−ν−λ +

n−ν Et,λ (x) n−ν−λ y x n−ν−λ−1 λ=1

Ft,λ (x) n−ν−λ y . x ν(n−ν+t)

t For n − ν 3 let us denote all the quotients y s /x n−ν−1 , where 0 < t < s n − ν − 1 in and let whatever order by Θ1 , . . . , ΘN , where N = 2

(19)

n−ν N Et,λ (x) n−ν−λ y = at,j Θj . x n−ν−λ−1 j =1

λ=1

c

By (16) we have at,j ∈ Z, by (15) and Lemma 2 (20)

|at,j | < 2n2(n−ν+t−1) f n−ν+t .

By virtue of the Bombieri-Vaaler theorem [2] there exists an integer solution [C0 , C1 , . . . , C2N−1 ] of the system of equations (21)

2N−1 t=0

Ct at,j = 0

(1 j N )

100


satisfying the condition 0 < max |Ct |

√

2N

0t 2, then |(K2 OF )2 | 8 unless d = p or 2p, p ≡ ±3 mod 8 a prime, in which case (K2 OF )2 = Z/2Z ⊕ Z/2Z. Proof. If d ≡ ±1 mod 8 we have K2 OF /K2 F 2 = Z/2Z ⊕ Z/2Z by the result of [1], on the other hand r2 (K2 F ) 1 unless d = p or 2p, p ≡ ±3 mod 8 a prime.

C8. Sylow 2-subgroups of K2 OF for quadratic number fields

259

Corollary 2. If d ≡ ±1 mod 8, then |(K2 OF )2 | 16 unless d = pq, p ≡ q ≡ 3 mod 8 primes or d = p = u2 − 2w 2 , u > 0, u ≡ 3 mod 4, w ≡ 0 mod 4 in which case (K2 OF )2 = Z/2Z ⊕ Z/2Z ⊕ Z/2Z. Proof. If d ≡ ±1 mod 8 we have |K2 OF /K2 F | 8 by the result of [1]. On the other hand r2 (K2 F ) 1 unless d = pq, p ≡ q ≡ 3 mod 8 primes or d = p = u2 − 2w 2 , u > 0, u ≡ 3 mod 4, w ≡ 0 mod 4, and then we apply [1] again. / N F , −2 ∈ / NF , Corollary 3. e4 1 provided either −1 ∈ or −1 ∈ NF , −2 ∈ / N F , d ≡ 1 mod 8, or −1 ∈ N F , −2 ∈ N F , d = u2 − 2w 2 , u > 0, u ≡ 1 mod 4, w ≡ 0 mod 4. Proof. In these cases r2 (K2 OF ) < r2 (K2 OF /K2 F ) + r2 (K2 F ).

Corollary 4. e8 1 provided d = pq, p ≡ −q ≡ 3 mod 8 primes or p ≡ −1 mod 8, a prime. Proof. In both cases r2 (K2 OF ) = 2, (K2 OF )2 16 by Corollary 2, and e4 1.

Corollary 5. Sylow 2-subgroup of K2 OF is generated by symbols if and only if d = 2, p or 2p, p ≡ −3 mod 8 a prime. Proof. The only symbols in K2 OF are 1, {−1, −1}, {−1, ε}, {−1, −ε}, where ε is the fundamental unit in F . Thus from Corollaries 1 and 2 it follows that if K2 OF is generated by symbols, then d = 2, p or 2p, p ≡ −3 mod 8 a prime. Evidently {−1, −1} = 1 and in the case p ≡ −3 mod 8 we have ±2 ∈ / N F . Consequently 2ε ∈ / F ∗2, ε ∈ / Δ, {−1, ε} = 1. If d = p or 2p, p ≡ 3 mod 8, then −2 ∈ N F . Since 2 ramifies in F , we have ε ∈ 2F ∗ 2 , i.e. {−1, ε} = 1 and there are only two symbols in K2 OF . √ Theorem 3. If the fundamental unit ε = α + β d of F satisfies N ε = 1, α, β ∈ Z, α ± 1 is not the square of an integer, and (4)

α = 2k·2

r−1 −1

· c2

r

for some positive integers k, r, c,

then in K2 OF there is an element of order 2r+1 . Proof. Suppose that 2ε ∈ F ∗ 2 . Then 2 ramifies in F and we have d ≡ 2 or 3 mod 4. √ 2 √ Hence 2α + 2β d = γ + δ d for some γ , δ ∈ Z. Consequently 2α = γ 2 + δ 2 d, 4 = (γ 2 − δ 2 d)2 , ±2 = γ 2 − δ 2 d, α ± 1 = γ 2 contrary to the assumption. It follows that 2ε ∈ / F√∗ 2 , and hence ε ∈ / Δ, i.e. {−1, ε} = 1. Evidently ε = α + β d satisfies ε2 − 2αε + 1 = 0. Hence 1 = −ε2 + 2αε. From the properties of symbols {x, y} it follows that {−1, 2} = 1, {ε2 , ε} = {ε, ε}2 = {ε, −1}2 = 1. Consequently 1 = {−ε2 , 2αε} = {−1, ε} {−1, α} {ε2 , 2α}

260

C. Algebraic number theory

and hence {−1, ε} = {−1, α} {ε2 , 2α}. Now from (4) it follows that {−1, α} = {−1, 2}s {−1, c}2r = 1 where s = k · 2r−1 − 1, and {ε 2 , 2α} = {ε2 , 2k·2

r−1

r

r

r

· c2 } = {ε, 2}k·2 {−ε 2 , c}2 ,

r

because {−1, c2 } = 1. Moreover {ε, 2} {−ε2 , c} ∈ Ker τ , because −ε2 = 1 − 2αε ≡ 1 mod α and hence 2 −ε ≡ 1 mod c. It follows that every tame symbol (−ε 2 , c)v is trivial. Therefore 2r (5) {−1, ε} = {ε, 2k } {−ε2 , c} . Since the element {−1, ε} has order 2, from (5) it follows that the element {ε, 2k } {−ε2 , c} ∈ K2 OF

has order 2r+1 .

The next Lemma enables us to give examples of fields satisfying assumptions of Theorem 3. √ Lemma. A unit of the form 2a h + 22a h2 − 1, where a 1, h 1, is fundamental in its field provided h < 22a+2 − 3. Proof. Suppose that 2a h +

√ n 22a h2 − 1 = u + v d ,

where n > 1, u, v ∈ Z, u, v > 0. Clearly n is odd and u2 − dv 2 = 1. It follows that u | 2a h, v 2 d | 22a h2 − 1. Hence vd ≡ 1 mod 2, u ≡ 0 mod 2. If u = 2s m, m odd, then √ n

√ n 1 ≡ 2s mod 2s+1 , 2a h = u+v d + u−v d 2 thus s a, and u 2s 2a . We have √ n 1 √ 3

√ n √ 3 1 2a h = u+v d + u−v d u+v d + u−v d 2 2 = u3 + 3uv 2 d = 4u3 − 3u 23a+2 − 3 · 22 . Hence h 22a+2 − 3, a contradiction.

Corollary 6. If a1 , h1 = a2 , h2 , h1 , h2 odd, hi < 22ai +2 − 3, then the fields ' Q 22ai · h2i − 1 are different. √ 4a − 1 , a = 2r−1 − 1, r 4, then in K2 OF there is an element

Corollary 7. If F = Q of order 2r+1 .


261

√ Proof. Taking in Lemma h = 1 we infer that the unit 2a + 4a − 1 is fundamental in F . It satisfies the assumptions of Theorem 3 provided 2a ± 1 is not the square of an integer. The latter condition is equivalent in virtue of an old theorem of Gerono to a = 1, 3, which holds for r 4.

4. Imaginary quadratic fields √ d , d < −2 square-free have t odd prime factors. Then t if 2 ∈ N F, e2 = t − 1 if 2 ∈ / N F, if d ≡ 1 mod 8, e2 e4 r2 (K2 F ) = e2 − 1 if d ≡ 1 mod 8.

Theorem 4. Let F = Q (6) (7)

Proof. If r is the 2-rank of Cl(F )/Cl2 (F ), then 2r is the number of classes C ∈ Cl(F ) satisfying the condition C 2 ∈ Cl2 (F ) and distinct mod Cl2 (F ). If N1 is the number of solutions of C 2 ∈ Cl2 (F ), then 2r · |Cl2 (F )| = N1 , on the other hand

Hence

N1 = Cl2 (F ) ∩ Cl(F )2 · 2r2 (Cl(F )) . 2r2 (Cl(F ))−r = Cl2 (F ) : Cl2 (F ) ∩ Cl(F )2 .

Now e2 = g(2) + r − 1,

t +1 r2 Cl(F ) + g(2) = t

Thus (6) is equivalent to the formula (8)

q = Cl2 (F ) : Cl2 (F ) ∩ Cl(F )

2

=

2 1

if d ≡ 5 mod 8, if d ≡ 5 mod 8.

if d ≡ 5 mod 8, 2 ∈ / N F, otherwise.

In order to prove this formula observe that by Gauss’s theorem the class of an ideal a belongs to Cl(F )2 if and only if N a ∈ N F . Therefore Cl2 (F ) ⊂ Cl(F )2 if and only if either d ≡ 5 mod 8 (then Cl2 (F ) is trivial) or 2 ∈ N F . This confirms (8). In order to prove (7) we proceed to write out all elements {−1, a} of order 2 in K2 OF . √ If 2 ∈ NF√, by the reciprocity law for the norm residue symbols d ∈ N Q( 2) √ we have and since Z[ 2] is the unique factorization domain d ∈ N Z[ 2], d = u2 − 2w 2 .

262


Consider now all elements {−1, γ δ}, where ⎧ ⎪ if 2 ∈ / N F, ⎨1 √ γ = 1 or u + d if d = u2 − 2w 2 , u > 0, d ≡ 2 or 3 mod 4, √ ⎪ ⎩ 1 1 or 2 (u + d) if d = u2 − 2w 2 , u > 0, d ≡ 1 mod 4, and δ is an odd divisor of d positive or negative, but not divisible by a fixed prime factor of d. The number of such elements is 2t+1 if 2 ∈ N F , 2t if 2 ∈ / N F , that is 2e2 +1 . They 2 belong to Ker τ since (γ δ) = aq , where a | 2. Moreover, different elements γ δ are distinct mod F ∗ 2 ∪ 2F ∗ 2 . Since F ∗ 2 ∪ 2F ∗ 2 is a subgroup of index 2 in Δ, the set of numbers γ δ can contain at most two elements from each class of equivalence mod Δ. But the number of equivalence classes is 2e2 , hence among numbers γ δ each equivalence class mod Δ is represented exactly twice. Therefore 2r2 (K2 F )+1 is the number of elements {−1, γ δ} in question that belong to K2 F , i.e. satisfy [−1, γ δ]v = 1 for every non-complex valuation v of F. Since F is imaginary there are no real valuations. If v is a discrete valuation then we infer in the same way as for real quadratic fields in the proof of Theorem 2 that [−1, γ δ]v = 1 4 if either d ≡ 1 mod 8 or d ≡ 1 mod 8, γ δ ≡ 1 mod . (2, γ )2 4 If d ≡ 1 mod 8 and γ δ ≡ 1 mod then there exists a discrete valuation v and a (2, γ )2 quadratic Hilbert symbol such that (9)

[−1, γ δ]v = 1.

Hence if d ≡ 1 mod 8 we have r2 (K2 F ) = e2 . If d ≡ 1 mod 8 the number of δ’s (positive or negative) in each residue class ±1 mod 4 is 2t−1 . Hence 2r2 (K2 F )+1 = 2e2 ,

r2 (K2 F ) = e2 − 1.

Moreover since the symbol [−1, γ δ]v that satisfies (9) is quadratic, we have e4 r2 (K2 F ). The proof is complete. Corollary 8. (K2 OF )2 2 unless d = −1, −2, −p or −2p, p ≡ ±3 mod 8 a prime, in which case |K2 OF | is odd. Proof. |K2 OF | odd is equivalent in virtue of Theorem 4 to t = 1, 2 ∈ / NF .

Corollary 9. If d ≡ 1 mod 8, then (K2 OF )2 4 unless d = −p (p ≡ −1 mod 8 a prime) or d = −pq (p ≡ −q ≡ 3 mod 8 primes), in which case (K2 OF )2 = Z/2Z. Proof. (K2 OF )2 < 4 implies e2 = 0 or 1. The first case is excluded by Corollary 8, the second case gives t = 1, 2 ∈ N F or t = 2, 2 ∈ / N F . Together with d ≡ 1 mod 8 this condition is equivalent to d = −p, p ≡ −1 mod 8 a prime or d = −pq, p, q primes satisfying p ≡ −q ≡ 3 mod 8. However by (7) e4 = 0, hence in these cases (K2 OF )2 = Z/2Z.


263

Added in proof. In view of recent results of B. Mazur and A. Wiles, and J. Hurrelbrink it can be deduced from Corollary 5 that K2 OF is generated by symbols iff d = 2, 5 or 13.

References [1] J. Browkin, The functor K2 for the ring of integers of a number field. In: Universal Algebra and Applications (Warsaw, 1978), Banach Center Publ. 9, PWN, Warsaw 1982, 187–195. [2] H. Garland, A finiteness theorem for K2 of a number field. Ann. of Math. (2) 94 (1971), 534–548. [3] H.-M. Gebhardt, Zur Berechnung des Funktors K2 von einigen euklidischen Ringen. Schr. Math. Inst. Univ. Münster (2) Heft 13 (1977). [4] H. Hasse, Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper, Teil II: Reziprozitätsgesetz. Physica-Verlag, Würzburg–Wien 1965. [5] D. Quillen, Higher algebraic K-theory I. In: Algebraic K-theory I. Higher K-theories (Seattle, 1972), Lecture Notes in Math. 341, Springer, Berlin 1973, 85–147. [6] J. Tate, Relations between K2 and Galois cohomology. Invent. Math. 36 (1976), 257–274. [7] −−, Appendix to: H. Bass, J. Tate, The Milnor ring of a global field. In: Algebraic K-theory II. “Classical” Algebraic K-theory, and Connections with Arithmetic (Seattle, 1972), Lecture Notes in Math. 342, Springer, Berlin 1973, 429–446.

Originally published in Tatra Mountains Mathematical Publications 11 (1997), 35–42


A class of algebraic numbers

To Władysław Narkiewicz on his sixtieth birthday Abstract. It is shown that a nonnegative real number is a K-number in the sense of Korec [2] if and only if it is an algebraic integer not less in absolute value than any of its conjugates.

I. Korec [2] has introduced a class of K-numbers defined as follows. A K-system is a vector α1 , . . . , αm ∈ Rm + such that for all i m and j m we have αi αj =

(1)

m

cij k αk ,

where cij k ∈ N0 ,

k=1

(R+ is the set of positive real numbers, N0 the set of nonnegative integers); α is a K-number if there exists a K-system α1 , . . . , αm such that α=

m

a k αk ,

ak ∈ N0 .

k=1

It follows that α ∈ R+ ∪ {0}. Korec proved that all √ K-numbers are algebraic integers and asked several questions about them, e.g., if 3 − 2 is a K-number. The following characterization of K-numbers allows one to decide whether a given algebraic integer is a K-number. Theorem. A nonnegative real number is a K-number if and only if it is an algebraic integer and it is not less in absolute value than any of its conjugates. The proof presented here depends only on Dirichlet’s approximation theorem. Lemma 1. If (2)

α=

m

ak αk , where α1 , . . . , αm is a K-system,

k=1

then α is an algebraic integer and for each l ∈ N we have (3)

αl =

m k=1

akl αk ,

akl ∈ N0 .

ak ∈ N0

C9. A class of algebraic numbers

265

Proof. The formula (1) shows that αi is a characteristic root of the matrix (cij k )j k . Thus αi is a zero of a polynomial over Z with the leading coefficient ±1, hence αi is an algebraic integer for all i m and so is α by (2). To prove (3) we proceed by induction on l. For l = 1, (3) follows from (2) with akl = ak . Assume that (3) holds for an exponent l. Then by (2) and (1) α l+1 = =

m i=1 m

ai α i

m

ai aj l

i,j =1

m a j l αj = (ai aj l )αi αj

j =1 m

i,j =1

cij k αk

k=1

and (3) holds for the exponent l + 1 with ak,l+1 =

m

ai aj l cij k .

i,j =1

Lemma 2. Let k1 , k2 ∈ N0 ,

for j k1 , for k1 < j k1 + k2 ,

x + aj x 2 + aj x + b j

fj (x) =

where aj ∈ R+ , bj ∈ R+ and let k1 +k2

fj (x) =

j =1

l

Aλ x l−λ .

λ=0

Then A0 = 1 and for λ < l we have (4)

0
|αie | or α e = αie . Proof. Put β=

n

αi2 .

i=1 |αi |=α

We have β ∈ R+ , because in the above product every non-real αi appears together with its complex conjugate. Hence we have β=

n

|αi |2 = α e ,

e = 2#{1 i n : |αi | = α}.

i=1 |αi |=α

On the other hand, if σ is any isomorphism of Q(α1 , . . . , αn ) we have α e |α σ |e and α e = |α σ |e implies n β = α e = (αiσ )2 = |β σ |. i=1 |αi |=α

Since |αiσ | α, it follows that |αiσ | = α whenever |αi | = α and thus {αi : |αi | = α} = {αiσ : |αi | = α},

β = βσ .

Remark. This lemma follows easily from a theorem of Boyd [1], however, the above proof is shorter than Boyd’s proof. Lemma 5. Let α, α1 , . . . , αn have the meaning of Lemma 4. If α > 1, α |αi | (1 i n), then there exist positive integers l and L such that for each i (5)

either α l − L > |αil − L| and #(αil − L) < 0 or αil = α l .

Proof. Using Lemma 4 and replacing, if necessary, α by α e , and n by the number of conjugates of α e , we may assume that n > 1, α > |αi | (2 i n). Since also α > 1, there is a positive integer l such that l α > 4 and α l 8. max |αi | 2in

268


Then 1 l αl α − max |αi |l − max |αi |l > 1 2in 2in 2 8 and there exists a positive integer L such that 1 l α − max |αi |l > L > max |αi |l . 2in 2in 2 The numbers l and L satisfy the conditions of the lemma, because for each i > 1 α l − L > L + max |αi |l |αil − L| 2in

and #(αil − L) = #αil − L max |αi |l − L < 0. 2in

Proof of the Theorem. We begin by proving the necessity of the condition. Assume that α is a K-number and (2) holds. By Lemma 1, α is an algebraic integer. Suppose that α is less in absolute value than α σ , where σ is an embedding of Q(α1 , . . . , αm ) into C. Clearly α = 0. Choose l so large that σ l |αkσ | |α | > max . 1km αk α Then from (3) we get a contradiction. Indeed, (α σ )l =

m

akl αkσ

k=1

and since α = 0, at least one akl is positive, hence it follows that σ l σ l m m |α | |α | |α σ |l akl |αkσ | < akl αk = αl = |α σ |l . α α k=1

k=1

The obtained contradiction shows that α |α σ | for all embeddings σ of Q(α) into C. We proceed to prove the sufficiency of the condition. Let α1 = α, α2 , . . . , αn be all the conjugates of α. Assume that α is an algebraic integer and α |αi | (1 i n). If n = 1, α clearly is a K-number. If n > 1, we have α = 1 and since α < 1 would imply n 0 < αi < 1, i=1

which contradicts c

n i=1

αi = NQ(α)/Q α ∈ Z, we have α > 1. By Lemma 5 there exist

positive integers l and L such that for each i n, (5) holds. Let β = α l − L, β2 , . . . , βk be all conjugates of β different from it. By (5) we have β > |βj | and

#βj < 0

(2 j k).

269


We may assume without loss of generality that βj is real for j r, and complex for r < j r + 2s = k, βr+s+j = β r+j . By Lemma 3 there exists a ν ∈ N such that #βjν 21 |βj |ν−1 #βj < 0

(6)

(2 j k)

and, moreover, (7)

β max |βj |

ν

r+s−1 j =1

2j k

$ # |βj +1 | . max 2, −#βj +1

Now we apply Lemma 2 with k1 = r − 1, k2 = s fj (x) = x − βjν+1 fj (x) = =

(1 j r − 1),

(x − βjν+1 )(x − βjν+s+1 ) 2 x − 2#βjν+1 x + |βj +1 |2ν

(r j r + s − 1).

By (6), the assumptions of Lemma 2 are satisfied and we infer that r+s−1

fj (x) =

j =1

k−1

Aλ x k−1−λ ,

λ=0

where A0 = 1 and for λ < k − 1 # $ r+s−1 |βj +1 |2ν Aλ+1 ν max −2#βj +1 , . 0< Aλ −2#βjν+1

(8)

j =1

However, by (6) $ $ # # |βj +1 |2ν |βj +1 | ν ν |βj +1 | max 2, , max −2#βj +1 , −2#βjν+1 −#βj +1 hence by (7) and (8) 0
klν − l,

hence by induction on i + j : α i β j is a combination of γp with coefficients in N0 for all i, j ∈ N0 , and thus γi γj is such a combination for all i, j klν. It follows that α is a K-number. Corollary 1. If α = 0 is a K-number, it is an element of a K-system consisting of elements of Z[α]. Proof. This follows from the construction of a K-system in the second part of the proof of Theorem. Corollary 2. If α, β are K-numbers then Q(α + β) = Q(α, β). Proof. Let α1 = α, α2 , . . . , αn be all the conjugates of α, β1 = β, β2 , . . . , βp be all the conjugates of β. By the Theorem, α |αi |, β |βj |, hence if α + β = αi + βj , we have |αi | = α, |βj | = β. However, since α, β are real, we have also #αi = α, #βj = β, thus αi = α, βj = β. This shows that [Q(α + β) : Q] = [Q(α, β) : Q], and since α + β ∈ Q(α, β), it follows that Q(α + β) = Q(α, β). Corollary 3. If α, β are K-numbers and αβ ∈ N, then there exists an e ∈ N such that α e ∈ N, β e ∈ N. Proof. Since αβ ∈ N, we have for every embedding σ of Q(α, β) in C: α σ β σ = αβ, and because by the Theorem α |α σ |, β |β σ |, it follows that α = |α σ |, β = |β σ |. By e = |α σ |e implies α e = (α e )σ , and c Lemma 4 there exists an exponent e ∈ N such that α


271

β e = |β σ |e implies β e = (β e )σ . Hence α e ∈ Q, β e ∈ Q and since α, β are positive algebraic integers, Q can be replaced by N. Remark. Corollaries 2 and 3 establish the conjectures made by Korec (cf. [3]).

References [1] D. W. Boyd, Irreducible polynomials with many roots of maximal modulus.ActaArith. 68 (1994), 85–88. [2] I. Korec, Irrational speeds of configuration growth in generalized Pascal triangles. Theoret. Comput. Sci. 112 (1993), 399–412. [3] −−, Open problems more or less related to generalized Pascal triangles. Preprint, June 1995.

Originally published in Acta Arithmetica 113 (2004), 401–408


On values of the Mahler measure in a quadratic field (solution of a problem of Dixon and Dubickas) To Robert Tijdeman at the occasion of his 60th birthday

For an algebraic number α, let M(α) be the Mahler measure of α and let M = {M(α) | α ∈ Q}. No method is known to decide whether a given algebraic integer β is in M. Partial results have been obtained by Adler and Marcus [1], Boyd [2]–[4], Dubickas [6]–[8] and Dixon and Dubickas [5], but the problem has not been solved even for β of degree two. The following theorem, similar to, but not identical with Theorem 9 of [5], is an easy consequence of [7]. Theorem 1. A primitive real quadratic integer β is in M if and only if there exists a rational integer a such that β > a > |β | and a | ββ , where β is the conjugate of β. If the condition is satisfied, then β = M(β/a) and a = N (a, β), where N denotes the absolute norm. There remain to be considered quadratic integers that are not primitive. The following theorem deals with the simplest class of such numbers. Theorem 2. Let K be a quadratic field with discriminant Δ > 0, β, β be conjugate primitive integers of K and p a prime. If (1)

pβ ∈ M,

then either there exists an integer r such that (2) pβ > r > p β and r | ββ , p /| r or (3)

β ∈ M and p splits in K.

Conversely (2) implies (1), while (3) implies (1) provided either # 1 + √Δ 2 $ (4) β > max −4β , 4 or √ (5) p > Δ.

C10. Mahler measure in a quadratic field

273

Remark 1. (2) implies β > pβ|β |/r p. Theorem 2 answers two questions raised in [5]. √ 5

Corollary 1. For all primes p we have p 3+2 or p ≡ ±1 (mod 5).

∈ M if and only if either p = 2, or p = 5,

Corollary 2. For every real quadratic field K there is an irreducible polynomial f ∈ Z[x], basal in the sense of [5], such that M(f ) ∈ K, but the zeros of f do not lie in K. Corollary 3. In every real quadratic field K there are only finitely many integers pβ, where p is prime, while β is primitive and totally positive, for which the condition pβ ∈ M is not equivalent to the alternative of (2) and (3). Proof of Theorem 1. Necessity. Let β = M(α), let f be the minimal polynomial of α over Z, a > 0 its leading coefficient, D its degree, and α1 , . . . , αD all its zeros. By Lemma 2 of [7] applied with d = 2, (6)

ββ = a 2

D

αi = (−1)D af (0).

i=1

Moreover, by formula (3) of [7], D = 2s, where s is the number of i D with |αi | > 1. Without loss of generality we may assume that |αi | > 1 precisely for i s. For some η ∈ {1, −1} we have (7)

s

αi = ηβ/a,

i=1

hence, by (6), (8)

D

αi = ηβ /a,

i=s+1

which gives (9)

β > a > β .

Also, by (6), (10)

a | ββ .

Sufficiency. Assume the existence of an integer a satisfying (9) and (10) and consider the polynomial g(x) = ax 2 − (β + β )x + ββ /a. If g is not primitive, there exists a prime p such that p | a, p | β +β and p | ββ /a. However, ββ then p2 | ββ and β/p is a zero of the polynomial x 2 − β+β p x + p2 belonging to Z[x],

274


contrary to the assumption that β is primitive. Therefore, g is the minimal polynomial of β/a over Z and β = M(β/a). Also, (a) | (a 2 , aβ, aβ , ββ ) | (a 2 , a(β + β ), ββ ) = (a), hence (a) = (a 2 , aβ, aβ , ββ ) = (a, β)(a, β ).

The proof of Theorem 2 is based on three lemmas. Lemma 1. If an integer β of K is the Mahler measure of an algebraic number whose minimal polynomial over Z has leading coefficient a, then a is the norm of an ideal of K. Proof. In the notation of the proof of Theorem 1 (necessity part) we have (7) and (8). Since ηβ /a is the only conjugate of ηβ/a, every automorphism of the splitting field of f that sends an αi (i s) to an αj (j > s) sends the set {α1 , . . . , αs } onto {αs+1 , . . . , αD } (compare the proof of Lemma 2 in [7]). Hence {α1 , . . . , αs } and {αs+1 , . . . , αD } are blocks of imprimitivity of the Galois group of f and the coefficients of the polynomials P (x) =

s

(x − αi ),

D

P (x) =

i=1

(x − αi )

i=s+1

belong to a quadratic field, which clearly is K. Let the contents of P and P be a−1 and a −1 , where a and a are conjugate ideals of K. Since f is primitive, we have (1) = cont f = cont(aP P ) = (a)/aa and, since a > 0, a = N a.

Lemma 2. If the dash denotes conjugation in K, δ, ε are elements of K such that δ > 1 > δ > −1/2, (1, δ) | ε, ε = ε , √ |ε − ε | + 1 < 4 δ,

(11) (12) (13)

while p is an ideal of K, then there exists γ ∈ K such that (14) (15)

(1, δ) (1, γ , δ) = , p √ |γ | < 2 δ, γ < 1 + δ .

Proof. Take an integer α of K divisible by p(1, δ)−1 . Applying Theorem 74 of [9] with a=

(α)(1, δ) , p

b=

p (1, δ)

we find an integer ω of K, such that (α, ω) = a, hence ω (1, δ) (16) 1, = . α p

275


Taking b=

0 its leading coefficient, and D its degree. By (6) and (7) with β replaced by pβ, we have (20) (21)

p2 ββ = (−1)D af (0), pβ > max a, |f (0)| min a, |f (0)| > p|β |.

Let p μ a, p ν ββ . If μ = 0 or μ = ν + 2, then (2) follows with r = a or r = |f (0)|, respectively. Therefore, assume (22)

1 μ ν + 1.

276


Let a = pμ b. By (20) and (22), pμ−1 b | ββ , while by (21), β > p μ−1 b > |β |. By Theorem 1 we have β ∈ M. If ν > 0, then p | ββ and since β is primitive, p splits in K. If ν = 0 we have, by (22), μ = 1 and since, by Lemma 1, a is the norm of an ideal of K, p splits in K. This proves (3). In the opposite direction, (2) implies pβ = M(pβ/r) ∈ M. Indeed, the minimal polynomial of pβ/r is rx 2 − p(β + β )x + ββ /r, where (r, β + β , ββ /r) = 1, since β is primitive (see the proof of Theorem 1). Assume now that (3) holds. By Theorem 1 we have β = M(β/b), where b ∈ N,

(23)

β > b > |β |,

b = N (b, β).

Replacing b by β|β |/b, if necessary, we may assume (24) b β|β |. First, assume (4). Since β is primitive all prime ideal factors of (b, β) are of degree one and no two of them are conjugate. Hence there exists c ∈ Z such that √ Δ+ Δ (25) ω := ≡ −c (mod (b, β)). 2 We put δ = β/b, ε = (c + ω)/b. In order to apply Lemma 2 we have to check the assumptions. Now, (11) follows from (23), (24) and β > −4β , (12) follows from (25), and (13) is equivalent to the inequality √ √ √ Δ/ b + b < 4 β. The left hand side considered as a function of b on takes its √ √ β] √ √ the interval [1, √ maximum √ at an end of the√ interval. We have Δ + 1 < 4 β by (4) and Δ/ β + β < 4 β since β (1 + Δ)/2. The assumptions of Lemma 2 being satisfied there exists γ ∈ K such that (26)

(1, γ , δ) =

(b, β) 1 = , (b)p (b, β )p

√ |γ | < 2 δ,

|γ | < 1 + δ .

Let us consider the polynomial P (x) = x 2 + γ x + δ. The discriminant of P , γ 2 − 4δ,√is negative, hence P is irreducible over the real field K, moreover its zeros are equal to δ > 1 in absolute value. On the other hand, the zeros of the polynomial P (x) = x 2 + γ x + δ


277

are less than 1 in absolute value. This is clear if γ 2 − 4δ < 0, since |δ | < 1 and if γ 2 − 4δ 0 the inequality ' |γ | + γ 2 − 4δ √ ' β 2 δ(1 + δ ) 4 =2 (b + β ) 2 β 4 β|β | + sgn β |β |3 /β . R := N (1, δ) b

If β > 0 we clearly have R > 1, if β < 0 we have, by (26), R = 2 4 β|β | β − |β | 4 4 β|β |/ β + |β | . √ √ If |β | 21 β, it follows that R 4 β|β | β > 1, √ √ while if |β | > 21 β, it follows that √ √ β 4 R > √ √ = 2 > 1; 2 2 β thus (27) implies

√ min N (1, δ), 2 δ(1 + δ ) = N (1, δ)

and (19) follows from (5). Consider now the case β=

1+

√

4e + 1 . 2

By (23), b2 +b > e > b2 −b, b | e, which implies e = b2 . On the other hand, 4e+1 = f 2 Δ

278


for some f ∈ N. The inequality √ p> Δ=

√ 4b2 + 1 f

implies by a tedious computation p

√ √ 2b + 1 Δ N (1, δ) Δ > ' = √ , f min N (1, δ), 2 δ(1 + δ ) 2 βb (b + β )

hence (19) holds. The assumptions of Lemma 3 being satisfied there exists γ ∈ K satisfying (26) and arguing as before we obtain pβ = M(α), where α is a zero of

x2

+ γ x + δ.

√ Proof of Corollary 1. For β = (3 + 5)/2 the condition (4) is satisfied. Now, (2) is fulfilled by p = 2 only, and (3) is fulfilled by p = 5 and p ≡ ±1 (mod 5) only. Proof of Corollary 2. Take a totally positive unit ε > 1 of K and a prime p > ε that splits in K. Then by Theorem 2, pε ∈ M. Assume that the basal irreducible polynomial f of pε has all its zeros in K. Hence pε pε f (x) = a x ± x± , pε > a > pε , a ∈ N a a and the condition p2 /a ∈ Z together with p > ε imply a = p. However, for a = p, f is not primitive. √ Example 1. For K = Q( 2) we can take √ pε = 21 + 14 2 = M(7x 4 + 2x 3 + 41x 2 + 22x + 7). Proof of Corollary 3. There are only finitely many totally positive integers β of K, which are Perron numbers, but do not satisfy (4). Remark 2. By a more complicated argument one can show that for β totally positive, (3) implies (1) unless √ √ Δ Δ 1 4 4 4 β and p < 1 + √ . Nβ + √ Nβ + √ 4 4 2 β Nβ Nβ Example 2. Theorem 2 does not allow us to decide whether 1 + is open, as is a more general question, whether (3) implies (1).

√ 17 ∈ M. This question


279

References [1] R. L. Adler, B. Marcus, Topological entropy and equivalence of dynamical systems. Mem. Amer. Math. Soc. 20 (1979), no. 219. [2] D. W. Boyd, Inverse problems for Mahler’s measure. In: Diophantine Analysis (ed. J. H. Loxton and A. J. van der Poorten), London Math. Soc. Lecture Note Ser. 109, Cambridge Univ. Press, Cambridge 1986, 147–158. [3] −−, Perron units which are not Mahler measures. Ergodic Theory Dynam. Systems 6 (1986), 485–488. [4] −−, Reciprocal algebraic integers whose Mahler measures are non-reciprocal. Canad. Math. Bull. 30 (1987), 3–8. [5] J. D. Dixon, A. Dubickas, The values of Mahler measures. Mathematika 51 (2004), 131–148. [6] A. Dubickas, Mahler measures close to an integer. Canad. Math. Bull. 45 (2002), 196–203. [7] −−, On numbers which are Mahler measures. Monatsh. Math. 141 (2004), 119–126. [8] −−, Mahler measures generate the largest possible groups. Math. Res. Lett. 11 (2004), 279–283. [9] E. Hecke, Lectures on the Theory of Algebraic Numbers. Springer, New York–Berlin 1981.

Part D Polynomials in one variable


Commentary on D: Polynomials in one variable by Michael Filaseta

√ D1. If q0 = (1 + 5)/2, then 1, q02 and q03 form a three term arithmetic progression. K. Zarankiewicz posed the problem of determining whether an irrational number q exists such that the infinite sequence 1, q, q 2 , q 3 , . . . contains four numbers that are in arithmetic progression. In this paper, Schinzel resolves the problem by showing that for any complex number q, if four numbers from the sequence 1, q, q 2 , q 3 , . . . are in arithmetic progression, then the four numbers are equal and are either all 0 or a root of unity. The problem of Zarankiewicz can be found, for example, in [43]. For integers a, b and c satisfying 0 a < b < c and q = 0, observe that q a , q b and q c are in arithmetic progression precisely when q is a root of x c−a − 2x b−a + 1. Schinzel consequently looks at the factorization of f (x) = x n − 2x m + 1. It is not difficult to see that both α and 1/α are roots of f (x) if and only if α is a root of x d − 1 where d = gcd(n, m). A method of W. Ljunggren in [23] then applies to determine the factorization of f (x)/(x d − 1). Schinzel thus obtains the factorization of f (x)/(x d − 1) given in Theorem 1. The factorization of f (x)/(x d −1), as described above, is of significance to the problem of Zarankiewicz as whenever q a , q b , q c and q d (with 0 a < b < c < d integers) form a four term arithmetic progression and q = 0, the number q must simultaneously be a root of x c−a −2x b−a +1 and x d−b −2x c−b +1. Knowing the factorization of f (x)/(x d −1) allows c−a Schinzel to determine the precise value of gcd x − 2x b−a + 1, x d−b − 2x c−b + 1 . In particular, if q is a root of unity, then q c−a = q b−a = q d−b = q c−b = 1 and, consequently q a = q b = q c = q d . If, on the other hand, q is not a root of unity, then one is led to a contradiction as a, b, c and d will not themselves form a four term arithmetic progression and Theorem 2 implies the only common roots of x c−a − 2x b−a + 1 and x d−b − 2x c−b + 1 are gcd(c − a, b − a, d − b, c − b) roots of unity. D2. This paper sets a foundation for future investigations done by Schinzel, in particular for his work on trinomials and his sequence of papers on the reducibility of lacunary polynomials. The paper is an initial attempt at describing effectively the canonical factorization of polynomials with fixed coefficients and variable exponents by associating each factorization with a corresponding factorization of a polynomial from a finite list of polynomials in several variables. The canonical factorization is not of the polynomials

284

D. Polynomials in one variable

themselves but rather come in two forms. Underlying the literature on this subject and the forms of the canonical factorizations obtained is some important notation. These are explained thoroughly in Schinzel’s own work. In particular, the reader can consult D4 or Schinzel’s book [37]. A brief, but not completely adequate, explanation is that KF denotes the polynomial F with every (multivariate) cyclotomic factor removed, and LF denotes the polynomial F with every (multivariate) self-inverse factor removed. The polynomials KF and LF should also be normalized to have the same leading coefficient as F . This paper concerns the canonical factorization of KF which in its most general form has proved to be more difficult to handle than the canonical factorization of LF . The latter is given a satisfactory answer by Schinzel in D4; the former is not given a satisfactory answer until 30 years later, with additional assumptions, in D12. The paper contains a conjecture concerning the factorization of KF which motivates much of Schinzel’s later work on the reducibility of lacunary polynomials. The present paper gives a partial answer to the conjecture in the case that the number of variables involved is 2 and the polynomials considered are over the field of rational numbers. This leads to an application to the factorization of K(ax n +bx m +c), that is a determination of how ax n +bx m +c factors after its cyclotomic factors are removed. In addition, Schinzel describes the cyclotomic factors of the trinomials ax n +bx m +c. As indicated, D2 considers the case where the coefficients of the polynomials are fixed and the exponents are variable. The reverse situation, with fixed exponents and variable coefficients, is considered in later papers; in particular, the factorization of trinomials in this manner is done rather thoroughly in his works D10, D13, D14. Theorem 5 is a consequence of the main result in the paper combined with Capelli’s theorem and gives information about the factorization of polynomials of the form x n +f (x), which later played a crucial role in his finding a connection between the factorization of x n + f (x) and covering systems of congruences in D3. An alternative approach for establishing Theorem 5 has been found by M. Filaseta, K. Ford and S. Konyagin in [14]. D3. Using the notation of D0 (f ) and e0 (later below) as given in Lemma 4, one can extend Theorem 1 to include the additional equivalent statement: A . For every polynomial f (x) with integer coefficients such that f (0) = 0, f (1) = −1 and f (x) = 1, there exists an integer n > max{D0 (f ), deg f } such that x n + f (x) is irreducible over the rationals. In other words, each of A, A and B are equivalent. To see this, note that A clearly implies A so that, by Theorem 1, one needs only show that A implies A. Suppose then that A holds. Fix f (x) and n as in A . The proof that B implies A begins by showing that the conditions in A must hold if f (x) is of the form 4h(x)4 . Therefore, we consider only the case that f (x) is not of this form. Furthermore, we define the set M as in the sentence containing the display (31). As noted M is finite. We take M to be the product of all μ where (μ, α) ∈ M. In particular, for all positive integers m and m , if ζm is a root of x m + f (x), then m divides M . We consider two cases depending on whether (n, e0 ) = 1 or (n, e0 ) = 1. If (n, e0 ) = 1, we appeal to Lemma 4. As both x n + f (x) and K(x n + f (x)) are irreducible, we deduce that x n + f (x) = K(x n + f (x)). In particular, x n + f (x) cannot

285

Polynomials in one variable

have a cyclotomic factor. For any positive integer t, we deduce that x n+tM e0 + f (x) does not have a cyclotomic factor. Hence, Lemma 4 implies that x n+tM e0 + f (x) is irreducible for every positive integer t, and A follows. In the case that (n, e0 ) = 1, we obtain a contradiction by showing that in fact x n +f (x) is reducible. As e0 = 1, there is a prime p dividing e0 and we can write f (x) = −w0 (x)p for some polynomial w0 (x) ∈ Z[x]. Further, n = pv for some positive integer v. Thus, x n + f (x) can be written as a product of x v − w0 (x) and w1 (x) = x (p−1)v + x (p−2)v w0 (x) + . . . + x v w0 (x)p−2 + w0 (x)p−1 . As pv = n > deg f = p · deg w0 , we see that deg(x v − w0 (x)) > 0. On the other hand, if deg w1 = 0, then there is a constant c ∈ Z such that x pv − w0 (x)p = c x v − w0 (x) , which implies deg w0 = v and w0 (x) is monic. Setting d = deg(x v − w0 (x)), we see that the left side above has degree (p − 1)v + d and the right side has degree d. This leads to v = 0 and w0 (x) = 1, contradicting that f (1) = −1. This completes the argument that A can be added as an additional equivalence in Theorem 1. The odd covering problem, that is the problem of finding a finite covering system of the integers consisting of distinct odd moduli > 1, goes back over 40 years and has an interesting history. P. Erd˝os and J. Selfridge after discussing the problem had different opinions about whether such a covering should exist. Erd˝os felt that such a covering probably does exist and offered $25 to anyone who could prove him wrong, that is for a proof that no such covering exists. Selfridge thought (at the time) that no such covering exists and offered $300 for an explicit example of such an odd covering (an example that would demonstrate that he was wrong). No monetary award was offered for a nonconstructive proof that there is an odd covering. Selfridge over the years raised the amount he was willing to give for an explicit example of an odd covering. In an email message dated November 16, 1998, to M. Filaseta, Selfridge raised the offer to $2000. The problem remains open. Turán’s conjecture, as mentioned in the introduction, is to show that for every polynomial f (x) ∈ Z[x], there is a polynomial g(x) ∈ Z[x] which is irreducible over the rationals, satisfies deg g deg f , and is close to f (x) in the sense that the sum of the absolute values of the coefficients of f (x) − g(x) is bounded by an absolute constant, say C. This form of the conjecture remains open. However, in D5, Schinzel shows the existence of such a g(x) with C = 3 provided one removes the requirement that deg g deg f . A natural first attempt at constructing such a g(x) is to consider the polynomials of the form x n + f (x) or of the form x n + f (x) ± 1 with the hope of possibly showing that one of these must always satisfy the requirements imposed on g(x). The present paper shows that if one can get away with taking g(x) in one of these forms, even without the requirement that deg g deg f , then condition B in Theorem 1 holds and that this in turn implies that there is an odd covering of the integers. Given the long history with the odd covering problem, this paper speaks to the difficulty of one likely being able to determine if g(x) can be taken to be of one of the forms x n + f (x) and x n + f (x) ± 1.

286


With the requirement that deg g deg f , Turán’s conjecture has been verified with C = 4 for all polynomials f (x) ∈ Z[x] of degree 24 by A. Bérczes and L. Hajdu in the two papers [2] and [3]. The example at the beginning of the paper of an f0 (x) ∈ Q[x] for which x n + f0 (x) has a cyclotomic factor for every positive integer n can be demonstrated by considering when each of the cyclotomic polynomials Φm (x) divides f0 (x) where m runs through the divisors > 1 of 12. This example was already noted in D2. Here, 12f0 (x) ∈ Z[x]. A similar example, say f1 (x), with different cyclotomic factors, is shown to exist with 4f1 (x) ∈ Z[x] by M. Filaseta in [13]. There one can also find a different exposition of the connection between the irreducibility of x n + f (x) and the odd covering problem. Schinzel’s problem suggested by D in this paper, to determine whether in every finite covering system there must exist two distinct moduli m and n with m dividing n, has become a popular open problem associated with finite covering systems; see, for example, Richard Guy’s book [19], Section F13 (this is an expanded third edition, but the problem appears in the same section in the second edition). The paper ends with a comment that Selfridge had an argument for C implying D. We note that this is fairly simple. Suppose aj (mod mj ), for j = 1, 2, . . . , r, is a finite covering system that does not satisfy the conditions in D. Write mj = 2ej m j with m j odd. Then the m j are distinct (otherwise, some modulus would divide another). It follows that either aj (mod m j ), for j = 1, 2, . . . , r, is a finite odd covering of the integers with moduli > 1, and C does not hold, or for some j , we have m j = 1 and ej 1. In the latter case, we may suppose j = 1. Then we claim aj (mod m j ), for j = 2, . . . , r, is a finite odd covering of the integers with moduli > 1. Indeed, if x ≡ a1 (mod m1 ), then x + m 2 · · · m r ≡ a1 (mod m1 ) which implies there must be some j ∈ {2, . . . , r} such that x ≡ x + m 2 · · · m r ≡ aj (mod m j ). D4, D6. To put this work in perspective, consider that there was at the time a bit of literature concerning classes of lacunary polynomials having a few prescribed coefficients for which the factorization was rather well understood. More than a decade earlier, E. S. Selmer in [42] established that x n − x − 1 is irreducible for all integers n > 1 and that x n + x + 1 is irreducible for n 1 unless n > 2 and n ≡ 2 (mod 3) (when it is divisible by x 2 + x + 1). Shortly afterwards, H. Tverberg in [46] generalized the approach of Selmer to handle the factorization of polynomials of the more general form x n ± x m ± 1; and at the same time, W. Ljunggren in [23] worked out an elegant approach to determine the factorization of polynomials of the form x n ± x m ± 1 and x n ± x m ± x k ± 1. Later, in [24], Ljunggren also determined with the same method the factorization of x n ± x m ± x k ± p, where p is an arbitrary prime. In particular, in each case, the polynomial is irreducible unless it has a cyclotomic factor. Schinzel D1 himself had already taken advantage of Ljunggren’s approach to obtain information about the factorization of x n − 2x m + 1, and A. T. Jonassen [22] had applied Ljunggren’s approach to the trinomials x n ±x m ±4. We note that some time later, W. H. Mills in [26] observed that although Ljunggren’s approach was correct, the paper that developed the approach had an error. Ljunggren incorrectly claimed that a quadrinomial x n ± x m ± x k ± 1 removed of its cyclotomic factors is either 1 or irreducible. Mills himself used Ljunggren’s approach to obtain a correct result, classifying


287

the cases where the quadrinomials removed of their cyclotomic factors are reducible. Also, later, M. Filaseta and I. Solan in [17] pointed out Ljunggren’s approach easily applies to the quintics f (x) = x n + x m + x k + x $ + 1 allowing one to show that Lf , in the notation of D4, is always 1 or irreducible in this case. The ability to determine when particular classes of polynomials, containing fixed coefficients and variable exponents, are reducible was underway at the time of Schinzel’s work in D4, but only with regard to specific cases where the coefficients were few and small in absolute value. This remarkable paper of Schinzel’s went far beyond what was done in the above papers and got to the core of these investigations. By taking, for example, F (x1 , x2 , . . . , xk ) = a0 + a1 x1 + . . . + ak xk in Theorem 2 in D4, where a0 , a1 , . . . , ak are arbitrary fixed nonzero integers, one can effectively determine a precise classification of the positive integers n1 < . . . < nk for which Lf is reducible and, in these cases, the factorization of Lf into k a product of irreducibles, where f (x) = aj x nj . Thus, in theory, any result similar to j =0

the above stated theorems can be obtained, at least when dealing with the factorization of Lf . Some of the results above address instead the issue of factoring Kf , but a look at these results shows that in those instances it is not difficult to analyze the contribution cyclotomic factors make to the factorization of f . Indeed, in those instances, one can deduce the cyclotomic factors through consideration of f (x) − x nk f (1/x). In general, it is more difficult to obtain information about the factorization of Kf . For a different approach to obtaining the above consequence of Theorem 2 see [12]. Note that the interesting corollary to Theorem 2 in D4 is improved in a later paper by Schinzel on the subject, specifically in the corollary to Lemma 1 in D7. Further investigations into this subject can be found in the commentary for D7. D5, D7, D8, D12. A discussion of Turán’s problem and the implication of D5 already appears in the commentary for D3. The result achieved by Schinzel in D5 remains the strongest published result toward the validity of Turán’s conjecture for arbitrary f ∈ Z[x]. In D7, Schinzel improves on work in D2 concerning the factorization of KF . In parn ticular, D7 describes the canonical factorization for KF (x1n1 , . . . , xk k ) (see Theorem 2). This result is extended to the field of coefficients being any finitely generated field in Theorem 43 of Schinzel’s book [37]. In D7, Schinzel also obtains information about the number of irreducible factors, counted with multiplicity, of KF based on the now classical estimates for roots off the unit circle in the complex plane by P. E. Blanksby and H. L. Montgomery [4], by C. J. Smyth [44] and (in an added note at the end of the paper) by E. Dobrowolski [9]. This subject has been explored further, specifically by Schinzel in [33] and in [35], as well as by C. Pinner and J. Vaaler in [27], [28] and [29]. Also, note that D7 obtains information about the factorization of non-reciprocal quadrinomials. The factorization of non-reciprocal quadrinomials is handled in detail in [36] (see also Theorem 78 of Schinzel’s book [37]). In D8, Schinzel’s main theorem bounds the number of reciprocal non-cyclotomic factors of an arbitrary polynomial f ∈ Z[x] under conditions which hold, for example, provided the coefficients are not too large and f has one term that has degree sufficiently large compared to the degrees of its other terms. In particular, this result implies that the

288 polynomial

D. Polynomials in one variable n j =0

j

x 2 has no reciprocal non-cyclotomic factors. He uses this to resolve a

problem posed by K. Mahler in [25]. In D12, Schinzel shows how work of E. Bombieri and U. Zannier (see Schinzel’s book [37] and, in particular, the appendix written by Zannier there) implies improvements on various of the prior results in his serious of papers on the reducibility of lacunary polynomials. In particular, the canonical factorization of KF (x n1 , . . . , x nk ) is dealt with in a rather general form for the first time and in an elegant way here. One nice example of an improvement in prior work is Corollary 2 which (even in its previous form) implies that one can typically expect that a lacunary (or sparse) reducible polynomial removed of its cyclotomic factors is irreducible. To put the result in a more precise perspective, fix integers A 2 and r 3 and consider all polynomials f in Z[x] with coefficients in absolute value bound by A, with f (0) = 0, with f consisting of r terms and with deg f N. Then as N tends to infinity, the density of such f having a cyclotomic factor (and even simply divisible by x − 1) is positive whereas Corollary 2 is asserting in a strong sense that the density of such f for which Kf is reducible is zero. It follows then that a typical lacunary reducible polynomial not divisible by x is divisible by a cyclotomic polynomial. D9. This problem concerns a problem of Rényi dating back to around 1949. The problem itself has an illusion of seeming like it should have a simple solution, but the problem is apparently quite difficult and is resolved for the first time, after approximately 40 years, in this paper. In its simplest form, the problem asks whether Qk tends to infinity where Qk is defined as the minimum number of terms (by definition, non-zero) that can appear in the square of a polynomial with exactly k terms. Here, one can take the polynomials to have real coefficients, though other variations of the problem are possible. Schinzel’s argument that log log k Qk > , log 2 gives not only the strongest lower bound on Qk but also still provides the only published argument for a lower bound that tends to infinity with k. That Qk can be < k is not even completely obvious. The simple example w(x) = 1 + 2x − 2x 2 + 4x 3 + 4x 4 shows that w(x)2 can have the same number of terms as w(x). Observe that if one takes h(x) = w(x)2 = 1 + 4x + 28x 4 + 32x 7 + 16x 8 , then it is clear that w(x) · w(x 7 ) has 5 · 5 = 25 terms whereas h(x) · h(x 7 ) has fewer than 25 terms (since the product of the constant term of either h(x) or h(x 7 ) times the term of degree 7 in the other combine to give one term of degree 7 with coefficient 36 in the product). Thus, w(x) · w(x 7 ) is an example of a polynomial whose square contains fewer terms than itself. One can check that this polynomial has 25 terms and its square has 21 terms. Examples involving smaller values of k exist. D. Coppersmith and J. Davenport

289


in [8] give the example (125x 6 + 50x 5 − 10x 4 + 4x 3 − 2x 2 + 2x + 1)(−110x 6 + 1), which is a polynomial of degree 12 having 13 non-zero terms, and the square of this polynomial has 12 non-zero terms. J. Abbott in [1] has shown using computations with Gröbner bases that every polynomial in Z[x] of smaller degree than 12 cannot have a square with fewer terms than itself. It is still possible that a polynomial in Z[x] with fewer than 13 non-zero terms (apparently of degree at least 12) could have a square with fewer terms than itself, and it would be of interest to determine what is the fewest non-zero terms such a polynomial can have. There is an extensive history concerning Rényi’s problem. A. Rényi’s initial work associated with the problem is [30], though he notes some contributions on the size of Qk given by L. Rédei and L. Kalmár. It is noted that Q29 28. Based on this example and a lemma, Rényi shows that the average of Qk /k, where k ranges from 1 to n tends to 0. This implies that lim inf Qk /k = 0. Rényi conjectures here that lim Qk /k = 0. k→∞

k→∞

P. Erd˝os in [11] resolves this problem of Rényi and indicates that Rényi has also asked for a proof that lim Qk = ∞, the problem resolved by Schinzel in D9. Erd˝os also shows, as k→∞

mentioned in Schinzel’s paper, that Qk < c1 k 1−c2 for some positive constants c1 and c2 . W. Verdenius in [47] showed that one can take c2 = log13 8. Further investigations were made by D. Coppersmith and J. Davenport in the paper quoted above. In particular, they establish an analog to Erd˝os’s result for higher powers of polynomials. These authors, other than Schinzel, restricted attention to the case that the polynomials have real coefficients. R. Freud in [18] extended the results of Erd˝os to polynomials with integer coefficients. The interested reader should consult Schinzel’s book [37]. There rather general theorems over arbitrary fields are discussed. There has been recent very nice progress in this area due to U. Zannier [48]. He has shown that for each positive integer k, there is a computable number B(k) such that if g(x) and h(x) are non-constant polynomials in C[x] with g(h(x)) having k terms, then h(x) has B(k) terms. In particular, as the number of terms of h(x) ∈ C[x] tends to infinity, the number of terms of g(h(x)) must also tend to infinity and uniformly over non-constant g(x) ∈ C[x]. D11. This solution to an approximately 30 year old problem has led to several further investigations. H. P. Schlickewei and C.Viola in [39] showed that the bound given by (1) can be replaced by 244000 (deg p)1000 . Their methods would easily allow for smaller exponents. In particular, they make use of a preprint of H. P. Schlickewei and W. M. Schmidt which later appeared in print as [38]. This paper has some slight improvements over the preprint that was used. H. P. Schlickewei and C. Viola, in [41] obtain a result which replaces the role of Schlickewei and Schmidt’s when conducting analogous investigations for k-nomials which is treated by H. P. Schlickewei and C. Viola in [40]. The problem posed by Gy˝ory and Schinzel at the end of the introduction was resolved negatively in the case that k 6 by L. Hajdu in [20]. Further investigations of this type can be found in L. Hajdu and R. Tijdeman’s paper [21].

290


D10, D13, D14. The motivation of these papers is explained in D10. The factorization of binomials has been well understood; in particular, over any field K, the binomial x n − a ∈ K[x] is reducible if and only if either a is a pth power in K for some prime p dividing n or 4 | n and a = −4b4 for some b ∈ K. For a proof of this result, one can consult Section 13 of Schinzel’s book [34] or Section 2.1 of Schinzel’s more recent book [37]. In D10, Schinzel offers corresponding results for the factorization of trinomials x n + ax m + b. As elaborated on in D13, the condition n 2m is made as the trinomials x n + ax m + b and x n + ab−1 x n−m + b−1 are either both irreducible or both reducible and either n 2m or n 2(n − m) must hold. Consequence 1 of D10 implies that for a field K with characteristic π 0, there is a constant C depending only on K such that if n and m are integers such that π /| nm(n − m) and n/(n, m) > C, then for every non-zero a and b in K, the trinomial x n + ax m + b is either irreducible or it has a non-trivial binomial or trinomial factor. This motivates in part the goals of D13 and D14 where we find Schinzel going beyond examining which trinomials x n + ax m + b are reducible or irreducible and determining instead more detailed information about the factorization of these trinomials. Specifically, D13 provides necessary and sufficient conditions for the trinomial x n + ax m + b divided by a certain binomial factor to be reducible, and D14 provides necessary and sufficient conditions for the trinomial x n + ax m + b divided by a certain proper trinomial factor to be reducible. There is a tremendous amount of literature surrounding the factorization of trinomials which would be impossible to cover in any detail here. Some discussions of these results occur elsewhere in the commentaries (for example, the commentaries on D2, D4, D11, D12, D16). An intriguing problem in A. Schinzel [32] is to determine whether or not there is an absolute constant C such that every trinomial in Q[x] has an irreducible factor with C terms. J. Abbott (private communication) has noted that the trinomial 64x 20 +7x 2 +4 factors as the product of the two irreducible polynomials f (x) and f (−x), where f (x) = 8x 10 + 16x 9 + 16x 8 + 8x 7 + 4x 4 + 8x 3 + 8x 2 + 5x + 2. This shows that if C exists, then C 9, which is the current best known lower bound. For more on this problem, see A. Bremner [6], or A. Choudhry and A. Schinzel [7]. D15, D16. Having dealt with the greatest common divisor of two trinomials of a rather specific form in D1, it is natural that Schinzel would take up the task of resolving a related intriguing problem posed by P. Weinberger in 1976. The problem, as indicated in D15, is to determine, for fixed positive integers r and s, the supremum of the number of terms that can occur in gcd(f, g), where f and g run over all polynomials in K[x], for some given field K, with f consisting of r terms and g consisting of s terms. In D15, the problem is resolved entirely except for the case that K has characteristic 0 and r = s = 3. In D16, results related to the latter case are obtained, but the problem in this situation remains open. It is possible that gcd(f, g) always consists of 6 terms in this case. D17. Beginning with some measure L(f ) of the size of a polynomial f , one can ask to determine the infimum $(P ) on the sizes of the multiples of a given polynomial P or, more precisely, on L(P G) where G ranges over a specific collection of polynomials. In this paper, L(f ) denotes the sum of the absolute values of the coefficients of f , P ∈ R[x]


291

and G(x) is restricted to monic polynomials in R[x]. The condition that G is monic is necessary to avoid the infimum trivially being 0. One focus of the paper is to determine for what P the value of $(P ) is attained, that is to determine for what P ∈ R[x], there exists a monic polynomial G(x) ∈ R[x] that minimizes the value of L(P G). Another focus of the paper is to obtain an effective approach for computing $(P ). Complete answers are given in many instances; as examples, $(P ) is attained and computable in the case that P has no roots on the unit circle and $(P ) is computable for all quadratic P . An intriguing question is raised as to how to establish the value of $(P ) in the specific case that P = 2x 3 + 3x 2 + 4. The problem was investigated previously by A. Dubickas [10]. Similar problems have been considered by others. A result due to G. Szegö [45] in 1914 is that the infimum above is precisely the Mahler measure of P if one uses as a measure of the size of the polynomial the squareroot of the sum of the squares of the absolute values of the coefficients. For Szegö’s result, one can consider P ∈ R[x] and multiples of the form P G where G(x) is monic in R[x] or one can consider the analog with P and G in C[x]. I. Z. Ruzsa [31] has generalized Szegö’s result to polynomials in several variables (see also Theorem 38 on page 227 of [37]). The analogous problem with P ∈ Z[x] and the requirement that G ∈ Z[x], G = 0 with G not necessarily monic, has been investigated by M. Filaseta, M. L. Robinson and F. S. Wheeler in [15] and by M. Filaseta and I. Solan in [16]. The first of these papers handles the case where P is irreducible, and the second handles the case where P is not divisible by a cyclotomic polynomial. In particular, in these cases, the infimum is attained and is computable. The methods apply to rather general notions for the size of a polynomial and, in particular, one can use the two sizes described above; the proofs, however, are restricted to the norm given by the squareroot of the sum of the squares of the coefficients. We note that finding multiples of polynomials with restrictions on the coefficients and, in particular, with small norm has applications to the Prouhet–Tarry–Escott Problem. Let n be a positive integer. Suppose P (x) = (x −1)n and that there is a polynomial G(x) ∈ R[x] such that L(f ) = 2n where f = P G and L(f ) is the sum of the absolute values of the coefficients of f . Suppose further that the coefficients of f are restricted to {0, ±1}. Then f (1) = 0 implies that f has n coefficients that are 1 and n coefficients that are −1. It follows that there are 2n distinct nonnegative integers a1 , . . . , an and b1 , . . . , bn such that f (x) =

n i=1

x − ai

n

x bj .

j =1

As (x − 1)n is a factor of f , we have f (k) (1) = 0 for k ∈ {1, 2, . . . , n − 1}. A simple argument implies that n i=1

aik =

n

bjk

for k ∈ {1, 2, . . . , n − 1}.

j =1

Observe also that if we begin with 2n distinct nonnegative integers aj and bj that satisfy this last string of equalities and define f (x) as the difference of the two sums, as above, with these exponents, then f (x) is a multiple of P (x) = (x − 1)n with L(f ) = 2n. It is

292


unknown whether or not such aj and bj exist for n 13. The example f (x) = 1 + x 11 + x 24 + x 65 + x 90 + x 129 + x 173 + x 212 + x 237 + x 278 + x 291 + x 302 − x 3 − x 5 − x 30 − x 57 − x 104 − x 116 − x 186 − x 198 − x 245 − x 272 − x 297 − x 299 , divisible by (x − 1)12 , is based on a combination of observations by Nuutti Kuosa, Jean-Charles Meyrignac and Chen Shuwen. Whether such aj and bj exist for general n is a form of the Prouhet–Tarry–Escott Problem. Thus, it is of interest to know whether an algorithm exists for finding a multiple f of P as above, but the methods described in the papers here do not resolve this issue. For a somewhat recent paper on this subject, we mention the work of P. Borwein, P. Lisonˇek, and C. Percival [5].

References [1] J. Abbott, Sparse squares of polynomials. Math. Comp. 71 (2002), 407–413. [2] A. Bérczes, L. Hajdu, Computational experiences on the distances of polynomials to irreducible polynomials. Math. Comp. 66 (1997), 391–398. [3] −−, −−, On a problem of P. Turán concerning irreducible polynomials. In: Number Theory (Eger, 1996), de Gruyter, Berlin 1998, 95–100. [4] P. E. Blanksby, H. L. Montgomery, Algebraic integers near the unit circle. Acta Arith. 18 (1971), 355–369. [5] P. Borwein, P. Lisonˇek, C. Percival, Computational investigations of the Prouhet–Tarry–Escott problem. Math. Comp. 72 (2003), 2063–2070. [6] A. Bremner, On reducibility of trinomials. Glasgow Math. J. 22 (1981), 155–156. [7] A. Choudhry, A. Schinzel, On the number of terms in the irreducible factors of a polynomial over Q. Glasgow Math. J. 34 (1992), 11–15. [8] D. Coppersmith, J. Davenport, Polynomials whose powers are sparse. Acta Arith. 58 (1991), 79–87. [9] E. Dobrowolski, On a question of Lehmer and the number of irreducible factors of a polynomial. Acta Arith. 34 (1979), 391–401. [10] A. Dubickas, Arithmetical properties of powers of algebraic numbers. Bull. London Math. Soc. 38 (2006), 70–80. [11] P. Erd˝os, On the number of terms of the square of a polynomial. Nieuw Arch. Wiskunde (2) 23 (1949), 63–65. [12] M. Filaseta, On the factorization of polynomials with small Euclidean norm. In: Number Theory in Progress, Vol. 1 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 143–163. [13] −−, Coverings of the integers associated with an irreducibility theorem of A. Schinzel. In: Number Theory for the Millennium, II (Urbana, 2000), A K Peters, Natick 2002, 1–24. [14] M. Filaseta, K. Ford, S. Konyagin, On an irreducibility theorem of A. Schinzel associated with coverings of the integers. Illinois J. Math. 44 (2000), 633–643.


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[15] M. Filaseta, M. L. Robinson, F. S. Wheeler, The minimal Euclidean norm of an algebraic number is effectively computable. J. Algorithms 16 (1994), 309–333. [16] M. Filaseta, I. Solan, Norms of factors of polynomials. Acta Arith. 82 (1997), 243–255. [17] −−, −−, An extension of a theorem of Ljunggren. Math. Scand. 84 (1999), 5–10. [18] R. Freud, On the minimum number of terms in the square of a polynomial. Mat. Lapok 24 (1973), 95–98 (Hungarian). [19] R. K. Guy, Unsolved Problems in Number Theory, third edition. Problem Books in Math., Springer, New York 2004. [20] L. Hajdu, On a problem of Gy˝ory and Schinzel concerning polynomials. Acta Arith. 78 (1997), 287–295. [21] L. Hajdu, R. Tijdeman, Polynomials dividing infinitely many quadrinomials or quintinomials. Acta Arith. 107 (2003), 381–404. [22] A. T. Jonassen, On the irreducibility of the trinomials x m ± x n ± 4. Math. Scand. 21 (1967), 177–189. [23] W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 (1960), 65–70. [24] −−, On the irreducibility of certain lacunary polynomials. Norske Vid. Selsk. Forh. (Trondheim) 36 (1963), 159–164. [25] K. Mahler, On the zeros of a special sequence of polynomials. Math. Comp. 39 (1982), 207–212. [26] W. H. Mills, The factorization of certain quadrinomials. Math. Scand. 57 (1985), 44–50. [27] C. G. Pinner, J. D. Vaaler, The number of irreducible factors of a polynomial I. Trans. Amer. Math. Soc. 339 (1993), 809–834. [28] −−, −−, The number of irreducible factors of a polynomial II. Acta Arith. 78 (1996), 125–142. [29] −−, −−, The number of irreducible factors of a polynomial III. In: Number Theory in Progress, Vol. 1 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 395–405. [30] A. Rényi, On the minimal number of terms of the square of a polynomial. Hungarica Acta Math. 1 (1947), 30–34. [31] I. Z. Ruzsa, On Mahler’s measure for polynomials in several variables In: Number Theory in Progress, Vol. 1 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 431–444. [32] A. Schinzel, Some unsolved problems on polynomials. In: Neki nerešeni problemi u matematici, Matematiˇcka Biblioteka 25, Beograd 1963, 63–70; this collection: E1, 703–708. [33] −−, On the number of irreducible factors of a polynomial. in: Topics in Number Theory, Colloq. Math. Soc. János Bolyai 13, North-Holland, Amsterdam 1976, 305–314. [34] −−, Selected Topics on Polynomials. University of Michigan Press, Ann Arbor 1982. [35] −−, On the number of irreducible factors of a polynomial, II. Ann. Polon. Math. 42 (1983), 309–320, [36] −−, Reducibility of lacunary polynomials VI. Acta Arith. 47 (1986), 277–293. [37] −−, Polynomials with Special Regard to Reducibility. Encyclopaedia of Mathematics and its Applications 77, Cambridge Univ. Press, Cambridge 2000. [38] H. P. Schlickewei, W. M. Schmidt, The number of solutions of polynomial-exponential equations. Compositio Math. 120 (2000), 193–225.

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[39] H. P. Schlickewei, C. Viola, Polynomials that divide many trinomials. Acta Arith. 78 (1997), 267–273. [40] −−, −−, Polynomials that divide many k-nomials. In: Number Theory in Progress, Vol. 1 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 445–450. [41] −−, −−, Generalized Vandermonde determinants. Acta Arith. 95 (2000), 123–137. [42] E. S. Selmer, On the irreducibility of certain trinomials. Math. Scand. 4 (1956), 287–302. [43] W. Sierpiński, Remarques sur les progressions arithmétiques. Colloquium Math. 3 (1954), 44–49. [44] C. J. Smyth, On the product of the conjugates outside the unit circle of an algebraic integer. Bull. London Math. Soc. 3 (1971), 169–175. [45] G. Szegö, Ein Grenzwertsatz über die Toeplitzschen Determinanten einer reellen positiven Funktion. Math. Ann. 76 (1915), 490–503. [46] H. Tverberg, On the irreducibility of the trinomials x n ± x m ± 1. Math. Scand. 8 (1960), 121–126. [47] W. Verdenius, On the number of terms of the square and the cube of polynomials. Indag. Math. 11 (1949), 546–565. [48] U. Zannier, On composite lacunary polynomials and the proof of a conjecture of Schinzel. Preprint.

Originally published in Colloquium Mathematicum IX (1962), 291–296


Solution d’un problème de K. Zarankiewicz sur les suites de puissances consécutives de nombres irrationnels

Dédié à la mémoire de K. Zarankiewicz

K. Zarankiewicz a posé le problème : existe-il un nombre irrationnel q tel qu’on puisse extraire de la suite q, q 2 , . . . quatre termes formant une progression arithmétique ? (cf. [2], p. 44, P 115). La réponse négative à ce problème (même quand on admet q complexe, les quatre termes correspondants étant distincts) est une conséquence immédiate du théorème 2, qui va suivre. En appliquant la méthode ingénieuse de W. Ljunggren [1], le théorème suivant sera d’abord établi : Théorème 1. Les nombres n et m étant des entiers tels que n > m > 0 et n = 2m, le polynôme g(x) =

x n − 2x m + 1 x (n,m) − 1

est irréductible, à l’exception des cas n = 7k, m = 2k et n = 7k, m = 5k, dans lesquels g(x) est un produit de deux facteurs irréductibles, à savoir (x 3k + x 2k − 1)(x 3k + x k + 1) et (x 3k + x 2k + 1)(x 3k − x k − 1) respectivement. Lemme. Soit (1) c

f (x) = x n − 2x m + 1 = ϕr (x)ψs (x) où r + s = n,

ϕr (x) et ψs (x) étant des polynômes unitaires de degré r et s respectivement, et aux coefficients entiers. Soit en outre 7k, 2k = n, m = 7k, 5k. Alors au moins l’un des deux facteurs de (1) est un polynôme réciproque. Correction: Colloq. Math. XII (1964), 289.

296


Démonstration du lemme. Il suffit de considérer le cas où n > 2m. En posant f1 (x) = x r ϕr (x −1 )ψs (x) =

(2)

n

ci x n−i

et

f2 (x) = x s ψs (x −1 )ϕr (x),

i=0

il vient f2 (x) = x n f1 (x −1 ) =

(3)

n

cn−i x n−i ,

i=0

f1 (x)f2 (x) = (x − 2x + 1)(x n − 2x n−m + 1). n

(4)

m

En comparant les coefficients de x 2n et de x n dans (2) et (3), nous trouvons c0 cn = 1 et c02 + c12 + c22 + . . . + cn2 = 6, d’où c

cn = c0 = ±1

(5)

et

2 c12 + c22 + . . . + cn−1 = 4.

Il résulte de (5) que deux cas sont possibles : Cas I. L’un des nombres ci , où i = 1, 2, . . . , n − 1, soit ck , est égal à ±2 et les autres sont égaux à 0. Cas II. Quatre des nombres ci , soit ck1 , ck2 , ck3 et ck4 , où k1 < k2 < k3 < k4 , sont égaux à ±1 et les autres sont égaux á 0. Considérons ces deux cas successivement. Dans le cas I, on peut admettre sans restreindre la généralité que n 2k. On a d’après (2) et (3) pour le polynôme réciproque f1 (x)f2 (x) (6)

f1 (x)f2 (x) = x 2n + c0 ck x 2n−k + c0 ck x n+k + 6x n + c0 ck x n−k + c0 ck x k + 1

et d’après (4) (7)

f1 (x)f2 (x) = x 2n − 2x 2n−m − 2x n+m + 6x n − 2x n−m − 2x m + 1. En comparant (6) à (7), on trouve k = m et c0 ck = −2, d’où f2 (x) = c0 (x n − 2x m + 1) = c0 f (x)

et

x s ψs (x −1 ) = c0 ψs (x) ;

ainsi ψs (x) est un polynôme réciproque. Dans le cas II, on peut admettre sans restreindre la généralité que n k1 + k4 . Alors le polynôme réciproque f1 (x)f2 (x) se réduit d’après (2) et (3) à la somme f1 (x)f2 (x) = x 2n + c0 ck4 x n+k4 + c0 ck3 x n+k3 + c0 ck2 x n+k2 + c0 ck1 x n+k1 (8)

+ c0 ck1 x 2n−k1 + ck1 ck4 x n+k4 −k1 + ck1 ck3 x n+k3 −k1 + ck1 ck2 x n+k2 −k1 + c0 ck2 x 2n−k2 + ck2 ck4 x n+k4 −k2 + ck2 ck3 x n+k3 −k2 + c0 ck3 x 2n−k3 + ck3 ck4 x n+k4 −k3 + c0 ck4 x 2n−k4 + 6x n + . . . .

c

En comparant (7) à (8), on constate que chaque exposant sauf 2n se présente dans (8) un nombre pair de fois. L’exposant 2n−k1 n’apparaissant qu’une seule fois lorsque 2n−k1 > n + k4 , on a donc 2n − k1 = n + k4 , d’où k4 = n − k1 . On peut admettre sans restreindre la généralité que n k2 + k3 . La condition que l’exposant max(n + k3 , n + k4 − k1 , 2n − k2 )

D1. Puissances consécutives de nombres irrationnels

297

figure dans la somme (8) nécessairement un nombre pair de fois entraîne deux possibilités : 1◦ k3 = n − k2 et k2 < 2k1 , 2◦ k2 = 2k1 et k3 < n − 2k1 . Considérons ces deux possibilités successivement. Dans 1◦ , les relations min(n + k4 , n + k3 , 2n − k1 , 2n − k2 ) > n + k4 − k1 > max(n + k3 − k1 , n + k2 − k1 , n + k4 − k2 , n + k3 − k2 , n + k4 − k3 ), n + k2 = 2n − k3

et

n + k1 = 2n − k4

entraîneraient que l’exposant n + k4 − k1 doit se présenter un nombre impair de fois, ce qui vient d’être constaté comme impossible. Dans 2◦ , les relations min(n + k4 , n + k3 , n + k2 , 2n − k1 , n + k4 − k1 , n + k3 − k1 , 2n − k2 , n + k4 − k2 , 2n − k3 , n + k4 − k3 ) > n + k1 = n + k2 − k1 = 2n − k4 entraînent que n + k1 = n + k3 − k2 , d’où k3 = 3k1 . On a maintenant n + k4 = 2n − k1 , n + k2 = n + k3 − k1 , n + k1 = n + k2 − k1 = n + k3 − k2 = 2n − k4 , n + k4 − k1 = 2n − k2 et n + k4 − k2 = 2n − k3 . Il en résulte que n + k3 = n + k4 − k3 , c’est-à-dire n = k1 + k4 = k1 + 2k3 = 7k1 et (8) se réduit à la forme

(9)

f1 (x)f2 (x) = x 14k1 + c0 ck4 + c0 ck1 x 13k1 + ck1 ck4 + c0 ck2 x 12k1 + c0 ck3 + ck2 ck4 x 11k1 + c0 ck3 + ck3 ck4 x 10k1 + c0 ck2 + ck1 ck3 x 9k1 + c0 ck1 + ck1 ck2 + ck2 ck3 + c0 ck4 x 8k1 + 6x 7k1 + . . . .

Enfin, la comparaison de (7) à (9) montre qu’il y a encore deux éventualités à considérer, à savoir (A) m = k1

et

(B)

m = 3k1 ,

l’égalité m = 2k1 étant exclue par l’hypothèse. Si l’on a (A), il vient en comparant les coefficients dans (7) et (9), c0 ck4 = c0 ck1 = −1, ck1 ck4 = −c0 ck2 , c0 ck3 = −ck2 ck4 , c0 ck1 + ck1 ck2 + ck2 ck3 + c0 ck4 = −2, d’où ck4 = ck3 = ck2 = ck1 = −c0 , ce qui entraîne c0 ck1 + ck1 ck2 + ck2 ck3 + c0 ck4 = 0, donc une contradiction avec la formule précédente.

298


Si l’on a (B), la comparaison de ces coefficients donne c0 ck4 = −c0 ck1 ,

ck1 ck4 = −c0 ck2 ,

c0 ck3 = ck2 ck4 = −1,

c0 ck3 + ck3 ck4 = −2,

c

d’où ck4 = ck3 = −c0 , ce qui entraîne c0 ck3 + ck3 ck4 = 0, donc également une contradiction avec la formule précédente. Ainsi le cas II est démontré impossible et le lemme se trouve établi.

Démonstration du théorème 1. Si n, m = 7k, 2k ou bien 7k, 5k, on a (n, m) = k et on vérifie aisément que g(x) = x 3k + x 2k − 1 x 3k + x k + 1 ou x 3k + x 2k + 1 x 3k + x k − 1 respectivement. Or les polynômes x 3k + x 2k ± 1 et x 3k ± x k ± 1 sont irréductibles en vertu du théorème 3 de Ljunggren (voir [1]). En supposant, par contre, que 7k, 2k = n, m = 7k, 5k, au moins l’un des facteurs du membre droit de (1) aurait en vertu du lemme la propriété suivante : si λ est une racine de ce facteur, il en est de même de λ−1 . La réductibilité du polynôme g(x) entraînerait donc que g(λ) = g(λ−1 ) = 0

(10)

pour un certain λ complexe. Il en résulte que λn − 2λm + 1 = 0 et λn − 2λn−m + 1 = 0, d’où successivement λn−2m = 1, λ2m − 2λm + 1 = 0, λm = 1 et λn = 1, donc λ(n,m) = 1. En même temps, l’hypothèse n = 2m entraîne nλn−1 = 2mλm−1 ; par conséquent λ serait une racine simple de x n − 2x m + 1. Vu que λ(n,m) = 1, on aurait donc g(λ) =

λn − 2λm + 1 = 0, λ(n,m) − 1

contrairement à (10). La démonstration du théorème 1 est ainsi achevée.

Théorème 2. Les nombres n, m, p et q étant des entiers tels que n > m > 0, p > q > 0 et n, m = p, q, on a x (n,m,p,q) − 1 si n, p = 2m, 2q, n m p q 2 (x − 2x + 1, x − 2x + 1) = (m,q) −1 si n, p = 2m, 2q. x Démonstration. On a pour r > s x r − 2x s + 1 = gr,s (x) = x (r,s) − 1

r/(r,s)−1 i=s/(r,s)

x

(r,s)i

−

s/(r,s)−1

x (r,s)i ;

i=0

si r = 2s, les polynômes gr,s (x) et les facteurs irréductibles de g7k,2k (x) et de g7k,5k (x) sont donc deux à deux distinct et différents des facteurs irréductibles de x l − 1, quel que

D1. Puissances consécutives de nombres irrationnels

299

soit l’entier l. Il en résulte que l’on a pour n = 2m et p = 2q (x n − 2x m + 1, x p − 2x q + 1) = x (n,m) − 1, x (p,q) − 1 = x (n,m,p,q) − 1, et on voit aisément que la même formule subsiste lorsqu’on a l’une des égalités n = 2m et p = 2q. Par contre, pour n, p = 2m, 2q, il vient 2 (x n −2x m +1, x p −2x q +1) = (x m −1)2 , (x q −1)2 = (x m −1, x q −1)2 = x (m,q) −1 ,

ce qui achève la démonstration du théorème 2.

Pour en déduire la solution du problème précité de Zarankiewicz, il suffit de remarquer que si les nombres q α , q β , q γ et q δ , où α < β < γ < δ, forment une progression arithmétique, on a q γ −α − 2q β−α + 1 = 0

et

q δ−β − 2q γ −β + 1 = 0 ;

si, au contraire, q (γ −α,β−α,δ−β,γ −β) − 1 = 0, on a qα = qβ = qγ = qδ. Le théorème suivant peut être établi d’une façon tout à fait analogue que le théorème 1 : Théorème 3. Les nombres n et m étant des entiers tels que n > m > 0, le polynôme f (x) = x n + 2ε1 x m + ε2 , où ε1 , ε2 = ±1, est un produit de deux facteurs dont le premier a pour racines précisément celles des racines de f qui sont des racines de l’unité, et le second, soit g(x), satisfait aux conditions : (α) si ε1 = −ε2 , n = 7k et m = 2k, on a g(x) = (x 3k + ε2 x 2k + ε1 )(x 3k + x k + ε2 ) ; c

(β) si ε1 = −1, n = 7k et m = 5k, on a g(x) = (x 3k + ε2 x 2k + ε2 )(x 3k − x k − ε2 ) ; (γ ) si ε1 = 1, n = 7k et m = 3k, on a g(x) = (x 3k − ε2 x 2k + ε2 )(x 4k + ε2 x 3k + x 2k + 1) ; (δ) si ε1 = ε2 , n = 7k et m = 4k, on a g(x) = (x 3k − x k + ε2 )(x 4k + x 2k + ε2 x k + 1) et hors des cas (α)–(δ) le polynôme g(x) est irréductible. Notons enfin que toutes les racines de f (x) qui sont en même temps celles de l’unité (s’il en existe) (a) sont simples à l’exception du cas où n = 2m et ε2 = 1, dans lequel elles sont doubles,

300


(b) satisfont à l’équation x = d

1 si ε1 = −1 et ε2 = 1, −1 si ε1 = (−1)n1 +m1 +1 et ε2 = (−1)n1 ,

dans laquelle d = (n, m), n = dn1 et m = dm1 . Cela élargit partiellement les résultats de Ljunggren (voir [1]), qui a étudié la réductibilité des polynômes f (x) = x n + ε1 x m + ε2 x p + ε3

où

n>m>p>0

et

ε1 , ε2 , ε3 = ±1.

Travaux cités [1] W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 (1960), 65–70. [2] W. Sierpiński, Remarques sur les progressions arithmétiques. Colloquium Math. 3 (1954), 44–49.

Originally published in Acta Arithmetica XI (1965), 1–34


On the reducibility of polynomials and in particular of trinomials

1. In the course of this paper reducibility means reducibility over the rational field Q unless stated to the contrary. Constants are considered neither reducible nor irreducible. A factorization of a polynomial into a product of a constant and of coprime powers of c irreducible polynomials is called its standard form. For a given polynomial f (x) = 0, Kf (x) denotes the factor of f (x) of the greatest possible degree, whose no root is 0 or a root of unity and whose leading coefficient is equal to the leading coefficient of f (x). Clearly Kf (x) =

f (x), x d

f (x)

(x δ − 1)d

,

ϕ(δ)d

where d is the degree of f (x). The paper is emerged from the efforts to solve the following problem formulated in [4]: Do there exist integers a, b = 0 such that for infinitely many rational r one can find integers m, n satisfying (i) m/n = r, (ii) K(x n + ax m + b) is reducible? The negative answer to this problem follows at once from Theorem 3 below; however, more general results are obtained. To state them I use the following notation: k I α If Φ(x1 , . . . , xk ) is a rational function of the form ai xj i,j , where ai = 0, αi,j i=0 j =1 / 0 are integers and the systems αi,1 , . . . , αi,k are all different for i I , then J Φ(x1 , . . . , xk ) = Φ(x1 , . . . , xk )

k

− mini αi,j

xj

.

j =1

Corrected following Errata, Acta Arith. 11 (1965), 491; and Corrigenda, ibid. 16 (1969), 159.

302


It is clear that J Φ(x1 , . . . , xk ) is a polynomial and that the operation J as well as K is distributive with respect to multiplication. I prove Theorem 1. For every irreducible polynomial F (x) not dividing x δ − x (δ > 1) and every positive integer n there exists an integer ν satisfying the following conditions: (i) 0 < ν C(F ); (ii) n = νu, u integer; (iii) if F (x ν ) = F1 (x)F2 (x) · · · Fr (x) is a standard form of F (x ν ), then F (x n ) = F1 (x u )F2 (x u ) · · · Fr (x u ) is a standard form of F (x n ). C(F ) is an effectively computable constant independent of n. Theorem 2. For every irreducible polynomial F (y, z) satisfying J F (y, z) = ±J F (y −1 , z−1 ) and for every pair of positive integers n, m there exists an integral nonsingular matrix * ) ν 1 μ1 ν2 μ2 satisfying the following conditions: (i) 0 νi C1 (F ), 0 μi C1 (F ) (i = 1, 2); (ii) n = ν1 u + ν2 v, m = μ1 u + μ2 v, u, v integers 0; (iii) if c

J F (y ν1 zν2 , y μ1 zμ2 ) = const F1 (y, z)e1 F2 (y, z)e2 · · · Fr (y, z)er is a standard form of J F (y ν1 zν2 , y μ1 zμ2 ), then either KF (x n , x m ) = const KF1 (x u , x v )e1 KF2 (x u , x v )e2 · · · KFr (x u , x v )er is a standard form of KF (x n , x m ) or max{n, m} C0 (F )(n, m). C0 (F ) and C1 (F ) are effectively computable constants independent of n, m. For every polynomial F (x) to which Theorem 1 applies the number of irreducible factors of F (x n ) remains bounded as n tends to infinity. On the other hand, if F (x) is any cyclotomic polynomial Xk (x) and (n, k) = 1, then Xkd (x); F (x n ) = Xk (x n ) = d |n

F (x n )

thus the number of irreducible factors of can be arbitrarily large. Therefore, the condition in Theorem 1 that F (x) does not divide x δ − x is necessary. On the other hand, it seems that the condition in Theorem 2: J F (y, z) = ±J F (y −1 , z−1 ), is too strong and could be replaced by the condition that F (y, z) does not divide yzJ (y δ1 zδ2 − 1) for any integers δ1 , δ2 not both zero. Moreover, the following conjecture seems to me plausible.

D2. On the reducibility of polynomials

303

Conjecture. Let F (y1 , . . . , yk ) be an irreducible polynomial which does not divide y1 · · · yk J (y1δ1 y2δ2 · · · ykδk − 1) for any integers δ1 , . . . , δk not all zero. For every system of k positive integers n1 , . . . , nk there exists an integral non-singular matrix [νi,j ] (1 i k, 1 j k) satisfying the following conditions: 0 νi,j C1 (F ) (1 i k, 1 j k); k (ii) ni = νi,j uj (1 i k), uj integers 0 (1 j k);

(i)

j =1

(iii) if JF

k j =1

ν

yj 1,j ,

k

ν

yj 2,j , . . . ,

j =1

k

ν yj k,j = const F1 (y1 , . . . , yk )e1 · · · Fr (y1 , . . . , yk )er

j =1

is a standard form of J F

k

ν

j =1

yj 1,j ,

k j =1

ν

yj 2,j , . . . ,

k j =1

ν yj k,j , then either

KF (x , . . . , x ) = const KF1 (x , . . . , x ) · · · KFr (x u1 , . . . , x uk )er n1

nk

u1

uk e1

is a standard form of KF (x n1 , . . . , x nk ) or α1 n1 + . . . + αk nk = 0, where αi are integers not all zero and |αi | C0 (F ) (1 i k). C0 (F ) and C1 (F ) are constants independent of n1 , . . . , nk . The method of proof in Theorem 1 permits us to obtain an analogous result for reducibility in an arbitrary algebraic number field. The method of proof in Theorem 2 is c valid only for totally real number fields and their totally complex quadratic extensions (in the latter case the condition J F (y, z) = ±J F (y −1 , z−1 ) should be replaced by −1 , z−1 )). I do not know, however, any algebraic number field c J F (y, z) = const J F (y in which the Conjecture could be disproved. The following theorem can easily be inferred from Theorems 1 and 2. Theorem 3. For any given non-zero integers a, b, c there exist two effectively computable constants A(a, b, c) and B(a, b, c) such that if n > m > 0 and K(ax n + bx m + c) is reducible, then (i) n/(n, m) A(a, b, c), (ii) there exist integers ν and μ such that m/μ = n/ν is integral, 0 < μ < ν B(a, b, c) and if K(ax ν + bx μ + c) = const F1e1 (x) · · · Frer (x) is a standard form of K(ax ν + bx μ + c), then K(ax n + bx m + c) = const F1e1 (x n/ν ) · · · Frer (x n/ν ) is a standard form of K(ax n + bx m + c). In order to complete the investigation of trinomials I also prove

304


Theorem 4. If a, b, c are integers = 0, 0 < m < n, d = (m, n), m = dm1 , n = dn1 , ε, η denote ±1, then ax n + bx m + c K(ax n + bx m + c) ⎧ 2d x + ε m1 ε n1 ηm1 x d + 1, ⎪ ⎪ ⎪ ⎨ x d − (−ε)m1 εη 2 , = ⎪ x d − (−ε)m1 εη, ⎪ ⎪ ⎩ 1,

if c = εa = ηb, n1 + m1 ≡ 0 (mod 3), ε m1 = ηn1 , if c = εa + ηb, (−ε)m1 = (−η)n1 , anε + bmη = 0, if c = εa + ηb, (−ε)m1 = (−η)n1 , anε + bmη = 0, otherwise.

Theorems 3 and 4 generalize the results of papers [1], [3] and [2], in which the case |a| = |c| = 1 has been considered. The results of those papers could be expressed in the present language in the form A(1, ±1, ±1) = 0, A(1, ±2, ±1) = B(1, ±2, ±1) = 7, 2 A(1, ±p, ±1) 4p (p prime > 2), respectively. The ideas of papers [1] and [2] are fundamental for the proof of Theorem 2. The Conjecture formulated above would give a result similar to Theorem 3 but concerning (k + 1)-nomials. As a second application of Theorem 2 I prove Theorem 5. Let f (x) ≡ ±1 be a polynomial such that f (0) = 0. There exist two constants D0 (f ) and D1 (f ) = 0 such that if n > D0 (f ) and n, D1 (f ) = 1, then K x n + f (x) is irreducible. It seems natural to ask whether the irreducibility of K x n + f (x) cannot be replaced in Theorem 5 by the irreducibility of x n + f (x) provided f (1) = −1. The example 1 (3x 9 + 8x 8 + 6x 7 + 9x 6 + 8x 4 + 3x 3 + 6x + 5) 12 shows that it is impossible. In fact, f0 (1) = −1 and x n + f0 (x) has for every n a factor in common with x 12 − 1. I do not know whether a similar phenomenon can occur for polynomials with integral coefficients. f0 (x) =

2. Lemma 1. Let Ω be an algebraic number field, and α an element of Ω which is not 0 or a root of unity. There exist only finitely many integers e such that α = wβ e , where w is a root of unity, β ∈ Ω. The greatest of them, e(α, Ω), satisfies the following relations: (1) e(α, Ω) (exp 2N 2 ) log N H (α) , where N is the degree of Ω and H (α) is the height of the irreducible primitive polynomial of α, (2)

e(α n , Ω) = ne(α, Ω) (n = 1, 2, . . . ).


305

Proof. We note first that if γ ∈ Ω is an algebraic integer and γ (i) (i = 1, . . . , n) are all its conjugates, then N 1 max γ (i) H (γ ) 1 + max γ (i) . (3) 1iN N 1iN This obvious inequality implies the following one: (4) max β (i) > exp exp(−2N 2 ), 1iN

which holds for all integers β ∈ Ω which are not 0 or roots of unity. Indeed, assuming the contrary we would clearly have N > 1 and for all k exp 2N 2 max (β k )(i) exp 1. 1iN

Hence, by (3) applied to γ = β k , H (β k ) (1 + exp 1)N

(1 k exp 2N 2 ).

Now there are no more integers of degree N and height H than N (2H +1)N ; thus there are no more integers of degree N and height (1 + exp 1)N than N 2 N 2(1 + exp 1)N + 1 < 3N (1 + exp 1)N < exp 2N 2 (N > 1). It follows that among the numbers β k (1 k exp 2N 2 ) at least two are equal, whence β is a root of unity. The contradiction obtained proves (4). Now we show that the equality α = wβ e ,

(5)

where w is a root of unity, β ∈ Ω, e 1, implies (6) e (exp 2N 2 ) log N H (α) . This will prove the existence of e(α, Ω) and inequality (1). Let α be a zero of a primitive irreducible polynomial a0 x m + . . . + am−1 x + am , where m | N, ai rational integers, a0 > 0, H (α) = max |ai |. 0im

If a0 = 1, α is an integer, and by (5), β is also an integer which is neither 0 nor a root of unity. It follows from (5) that log max α (i) = e log max β (i) , 1iN

1iN

and hence by (3) applied to γ = α and by (4) log N H (α) e exp(−2N 2 ), which gives (6). If a0 > 1, a0 α is an integer but α is not. Therefore, there exists a prime ideal p such that pλ a0 , pμ a0 α and μ < λ. Let pν a0 β. It follows from (5) that (λ − μ) = (λ − ν)e, and

306


since λ − μ > 0, we get λ − ν > 0 and e λ − μ λ. On the other hand, (norm p)λ | a0N , hence 2λ a0N H (α)N . This gives eλ

N log H (α) < (exp 2N 2 ) log N H (α) , log 2

i.e. again (6). In order to prove (2) we put e(α, Ω) = e, e(α n , Ω) = f , (n, f ) = d and assume (7)

α = w1 β e ,

α n = w2 γ f ,

where w1 , w2 are roots of unity, β, γ ∈ Ω. Clearly α n = w1n β ne , whence f ne. On the other hand, there exist integers p, q such that pn − qf = d and it follows from (7) that d p α(α q γ −p )f/d = α d+qf γ −pf = α pn (w2 α −n )p = w2 ; c

thus α(α q γ −p )f/d is a root of unity, say w3 . We get α = w3 (α −q γ p )f/d and, by the definition of e, e f/d f/n e. This gives f = ne and completes the proof. Lemma 2. Let Ω be an algebraic number field and α an element of Ω which is not 0 or a root of unity. For every positive integer n we put ν = ν(α, Ω, n) = n, 2e(α,Ω)−1 e(α, Ω)! . If g(x) is a monic polynomial irreducible over Ω and g(x) | x n − α, then g(x) = G(x n/ν ), where G(x) is a polynomial over Ω. Proof. We proceed by induction with respect to e(α, Ω). If e(α, Ω) = 1, then neither α = β p , p > 1, nor α = −4β 4 , β ∈ Ω; thus, in view of a theorem of Capelli (for the proof and references see [5], pp. 288–294), x n − α is irreducible in Ω and g(x) = x n − α. The lemma holds with G(x) = x − α. Assume that the lemma is true for all Ω and α with e(α , Ω ) < m (m > 1) and let e(α, Ω) = m, g(x) | x n − α. If x n − α is irreducible, then the lemma is trivially true with G(x) = x ν − α. If it is reducible, then by the theorem of Capelli (A) α = β p , where p | n, p prime > 1, β ∈ Ω, or (B) α = −4β 4 , where 4 | n, β ∈ Ω. We consider these cases successively, using the following notation: ζq is a primitive qth root of unity, Ωq = Ω(ζq ), dq is the degree of Ωq over Ω, NΩq /Ω is the norm of elements of Ωq or polynomials over Ωq relative to Ω.

307


(A) We have here g(x) | x n − β p = (x n/p − β)

(8)

p−1

(x n/p − ζpr β).

r=1

If g(x) | x n/p − β

our inductive assumption applies directly, since by (A) and Lemma 1 m = e(α, Ω) = pe(β, Ω) > e(β, Ω).

(9)

Putting ν0 = ν(β, Ω, n/p) we have ν0 | n/p, 2m−2 (m − 1)! ,

g(x) = G0 (x n/pν0 ),

G0 (x) ∈ Ω[x] and it is sufficient to take G(x) = G0 (x ν/pν0 ). If g(x) /| x n/p − β, let h(x) be a monic factor of g(x) irreducible over Ωp . By (8) p−1 n/p h(x) | g(x) (x − ζpr β); r=1

thus for some positive r < p h(x) | x n/p − ζpr β.

(10)

Let h(1) (x) = h(x), . . . , h(dp ) (x) be all the conjugates of h(x) relative to Ω. It follows from (10) that (i) (j )r h (x), h(j ) (x) β ζp(i)r − ζp (1 i j dp ); thus h(i) (x) (i = 1, 2, . . . , dp ) are relatively prime in pairs. Since h(i) (x) | g(x), it follows that (11) g(x) = NΩp/Ω h(x) . On the other hand, we have e ζpr β, Ωp = e1 < m. Indeed, if ζpr β = wγ e1 , then α = β p = wp γ pe1

w a root of unity, γ ∈ Ωp ,

pe and α dp = NΩp /Ω (w p ) NΩp /Ω (γ ) 1 .

It follows by Lemma 1 that dp m = e(α dp , Ω) > pe1 and, since dp p − 1, e1 < m. Applying the inductive assumption to (10) and putting ν1 = ν ζpr β, Ωp , n/p , we get (12)

ν1 | n/p, 2m−2 (m − 1)! ,

h(x) = H (x n/pν1 ),

H (x) ∈ Ωp [x].

308

c


Since p | m by (9), we have ν1 p | (n, 2m−1 m!) = ν and it is sufficient to put G(x) = NΩp /Ω H (x ν/pν1 ) . Indeed, by (11) and (12)

g(x) = NΩp /Ω H (x n/pν1 ) = G(x n/ν ).

(B) We have here g(x) | x n + 4β 4 =

x n/4 − (ε + ηζ4 )β .

εη=±1

Let h(x) be a monic factor of g(x) irreducible over Ω4 . There is a pair of integers ε, η such that εη = ±1, h(x) | x n/4 − (ε + ηζ4 )β.

(13)

It follows, like (11) from (10), that

g(x) = NΩ4 /Ω h(x) . On the other hand, e (ε + ηζ4 )β, Ω4 = e2 < m. Indeed, if

(14)

(ε + ηζ4 )β = wγ e2 , then α = −4β 4 = w4 γ 4e2

w a root of unity, γ ∈ Ω4 ,

4e and α d4 = NΩ4 /Ω (w 4 ) NΩ4 /Ω (γ ) 2 .

It follows by Lemma 1 that d4 m = e(α d4 , Ω) > 4e2 and since d4 2, e2 < m. Applying the inductive assumption to (13) and putting ν2 = ν(ε + ηζ4 , Ω4 , n/4) we get (15)

ν2 | n/4, 2m−2 (m − 1)! ,

h(x) = H (x n/4ν2 ),

H (x) ∈ Ω4 [x].

By Lemma 1 and (B) m = e(a, Ω) = 2e(2β 2 , Ω). Thus 2 | m and by (15) 4ν2 | (n, 2m−1 m!) = ν. Now put G(x) = NΩ4 /Ω H (x)ν/4ν2 . By (14) and (15)

g(x) = NΩ4 /Ω H (x)n/4ν2 = G(x n/ν ),

which completes the inductive proof.

Proof of Theorem 1. Put C(F ) = exp (2N 2 + log log N H )(exp 2N 2 ) log N H , where N is the degree of F and H its height. Let a0 be the leading coefficient of F ,α any of its zeros and Ω = Q(α). For any given n, we put ν = n, 2e(a,Ω)−1 e(a, Ω)! . Clearly


309

ν e(a, Ω)e(a,Ω) and by Lemma 1, ν C(F ). Besides, u = n/ν is an integer; thus parts (i) and (ii) of Theorem 1 are proved. In order to prove (iii) assume that F (x ν ) = F1 (x) · · · Fr (x)

(16)

is a standard form of F (x ν ) (since F (x) is irreducible, there are no multiple factors). Clearly Fj (x u ) (1 j r) are relatively prime in pairs and it remains to show that they are all irreducible. Let fj (x) be a monic irreducible factor of Fj (x u ). Clearly fj (x) | F (x n ).

(17)

We now use the following Lemma of Capelli (cf. [5], pp. 288–290): if xn − α =

(18)

l

gi (x)

i=1

is a decomposition of x n − α into monic factors irreducible over Q(α) = Ω and NΩ/Q denotes the norm relative to Q, then F (x n ) = a0

(19)

l

NΩ/Q gi (x)

i=1

is the decomposition of into monic factors irreducible over Q. It follows from (17) and (19) that for some i l F (x n )

(20)

fj (x) = NΩ/Q gi (x).

On the other hand, it follows from (18) and Lemma 2 that (21)

gi (x) = Gi (x u ),

where Gi (x) is a polynomial over Ω. By (20), (21) and the choice of fj (x) (22)

fj (x) = NΩ/Q Gi (x u ) | Fj (x u );

thus NΩ/Q Gi (x) | Fj (x). Since Fj (x) is irreducible, Fj (x) = const NΩ/Q Gi (x); thus by (22) Fj (x u ) = const fj (x) and by the choice of fj (x), Fj (x u ) is irreducible. This completes the proof.

310 c


Corollary to Theorem 1. For every polynomial F (x) = 0 and every positive integer n there exists an integer ν satisfying the following conditions: 0 ν C (F );

(23)

n = νu,

(24)

u—an integer;

if KF (x ν ) = const F1 (x)e1 F2 (x)e2 · · · Fr (x)er is a standard form of KF (x ν ), then KF (x n ) = const F1 (x u )e1 F2 (x u )e2 · · · Fr (x u )er is a standard form of KF (x n ). ε

Proof. Let KF (x) = const Φ1ε1 (x)Φ2ε2 (x) · · · Φ (x) be a standard form of KF (x). Since each polynomial Φi (x) (1 i ) satisfies the conditions of Theorem 1, there exists for each i a positive integer νi satisfying the following conditions: 0 < νi C(Φi ); n = νi ui , ui —an integer; if Φi (x νi ) = Φi,1 (x)Φi,2 (x) · · · Φi,ri (x) is a standard form of Φi (x νi ), then Φi (x n ) = Φi,1 (x ui )Φi,2 (x ui ) · · · Φi,ri (x ui ) is a standard form of Φi (x n ). We put ν = [ν1 , . . . , ν ],

C (F ) =

max C(Φi ) !.

1i

Conditions (23) and (24) are clearly satisfied. Since νi | ν we have u | ui and the irreducibility of Φi,j (x ui ) implies the irreducibility of Φi,j (x ui /u ) (1 i , 1 j ri ). Since polynomials Φi,j are relatively prime in pairs, it follows that KF (x ) = const ν

ri

Φi,j (x ui /u )εi

i=1 j =1

is a standard form of

KF (x ν )

and

KF (x n ) = const

ri

Φi,j (x ui )εi

i=1 j =1

is a standard form of

KF (x n ).

This completes the proof.

3. c

Lemma 3. For any two relatively prime polynomials G(y, z) = 0, H (y, z) = 0 there exist two constants B0 (G, H ) = B0 1 and B1 (G, H ) such that if n, m are positive integers, then J G(x n , x m ) J H (x n , x m ) (n,m)B0 (25) − 1)B0 , , x KG(x n , x m ) KH (x n , x m ) KG(x n , x m ), KH (x n , x m ) = 1 (26) unless max{n, m} B1 (G, H )(n, m).


311

Proof. Let R(z) be the resultant of polynomials G(y, z) and H (y, z) with respect to y and S(y) their resultant with respect to z, and D = max{degree R, degree S}. Let e2π ir1 , e2πir2 , …, e2πirk be all the roots of unity which are zeros of R and 1 , 2 , . . . , k their respective multiplicities, and similarly let e2πis1 , e2πis2 , . . . , e2πisl be all the roots of unity which are zeros of S and σ1 , σ2 , . . . , σl their respective multiplicities. Let d be the least common denominator of 1, r1 , . . . , rk , s1 , . . . , sl . We put B0 = B0 (G, H ) = d max 1, max i , max σj , B1 (G, H ) = DC (R)C (S), 1ik

where

C

whence

Since (27)

1j l

is a constant from the Corollary to Theorem 1. Clearly J R(z) B0 J S(y) B0 (z − 1)B0 , (y − 1)B0 , KR(z) KS(y) J R(x m ) mB0 J S(x n ) nB0 B0 − 1) , − 1)B0 ; (x (x KR(x m ) KS(x n ) J R(x m ) J S(x n ) (n,m)B0 , (x − 1)B0 . KR(x m ) KS(x n ) G(x n , x m ), H (x n , x m ) R(x m ), S(x n ) ,

(25) follows. In order to prove (26), assume that f (x) is an irreducible polynomial such that f (x) | KG(x n , x m ), KH (x n , x m ) . By (27) f (x) | KR(x m )

and f (x) | KS(x n ).

Now by the Corollary to Theorem 1 there exist a μ C (R) and a polynomial F1 (x) such that (28)

f (x) = F1 (x m/μ ) and

F1 (x) | KR(x μ ).

Similarly there exist a ν C (S) and a polynomial F2 (x) such that (29)

f (x) = F2 (x n/ν )

and F2 (x) | KS(x ν ).

Let d1 , d2 be the degrees of F1 and F2 respectively. It follows from (28) and (29) that m n d1 = d2 , d1 Dμ, d2 Dν. μ ν Hence max{n, m}/(n, m) Dμν DC (R)C (S) = B1 (G, H ), which completes the proof. Lemma 4. Let ki (0 i l) be an increasing sequence of integers. Let kj1 − ki1 , . . . , kjP − kiP (P 0) be all the numbers besides kl − k0 which appear only once in

312


the double sequence kj − ki (0 i < j l). Suppose that for each pair p, q, where 1p i . a and b are solutions of the system (31) and satisfy the following conditions: (32) aj = 0 = al ,


313

: 9 (33) all ai are in the interval ai , aj , ai = ai for i < i , ai = aj for i i and for i > j ,

(34) bj = 0 = bj = bl ,

9 : (35) all bi are in the interval bi , bj , bi = bi for i < i , bi = bj for i i and for i > j .

0 / 0 / Now,/ (32) 0and (33) imply that i , j is for some p P identical with ip , jp . Indeed, by (32), i , j = 0, l, whence we would have in the opposite case / 0 / 0 aj − ai = ah − ag , where g, h = gi ,j , hi ,j = i , j . It follows from (33) that ah = aj , ag = ai , whence g i , h j . On the other hand, kj − ki = kh − kg and since ki are increasing, / 0 g / = i , 0h = j , which gives a contradiction. Similarly, (34) and (35) imply that i ,/ j is0 iq ,/ jq , where 1 q P. 0 Moreover, by (34) i , j = i , j . Thus p = q and without loss of generality we = c , we get from (30), (32) may assume p < q. Putting for brevity cp,q = c and cp,q and (34)

(36)

cp,q (kl − k0 ) + c (kj − ki ) + c (kj − ki ) = 0,

(37)

c (aj − ai ) + c (aj − ai ) = 0,

(38)

c (bj − bi ) + c (bj − bi ) = 0.

Since kl > k0 , it follows from (36) that c = 0 or c = 0. Now in view of (33) and (37) |c | |c |, and in view of (35) and (38) |c | |c |. Hence c = ±c = 0. If c = −c , (35) and (38) imply that bj = bj , which contradicts (34). If c = c , (33) and (37) imply that ai = aj , i > i . Similarly (35) and (38) imply that bi = bj , i > i . The contradiction obtained proves that system (31) has at most two linearly independent solutions. Therefore, the rank of the matrix M of system (31) is at least l − 1. If this rank is l, solving the system by means of Cramer’s formulae we get xi = xμ Di /D c

(0 i l, μ fixed),

where D and Di are determinants of order l and, as can easily be seen from the form of matrix M, the sum of the absolute values of integers standing in any line of Di does not exceed 5c. Hence |Di | < (5c)l and a fortiori |Di |/(D0 , . . . , Dl ) < (5c)l . Since (kμ − k0 )Di = ki − k0 (0 i l), D (kμ − k0 )(D0 , . . . , Dl )/D is an integer and the lemma holds with s = (kμ − k0 )(D0 , . . . , Dl )/D, κi = Di /(D0 , . . . , Dl ),

λi = 0

t = 0, (0 i l).

314


If the rank of M is l − 1, we get similarly xi = (xμ Di + xν Di )/D c

where D,

Di

(39)

and

Di

(0 i l, μ, ν fixed),

are determinants of order l − 1,

|Di | < (5c)l−1 ,

|Di | < (5c)l−1

(0 i l).

Integral vectors [xμ , xν ] such that all numbers (xμ Di + xν Di )/D (0 i l) are integers form a module, say M. Clearly [0, |D|] ∈ M and [|D|, 0] ∈ M. Let ξ1 be the least positive integer such that for some η1 , [ξ1 , η1 ] ∈ M and let η2 be the least positive integer such that [0, η2 ] ∈ M. Clearly [ξ1 , η1 ] and [0, η2 ] form a basis for M and without loss of generality we may assume 0 η1 < η2 . On the other hand, η2 |D| and ξ1 |D|. Since ki are integers, [kμ − k0 , kν − k0 ] ∈ M; thus there are integers s, t such that kμ − k0 = ξ1 s, kν − k0 = η1 s + η2 t. Putting κi = (ξ1 Di + η1 Di )/D, λi = η2 Di /D, we get for i l ki − k0 = κi s + λi t and by (39) |κi |

c

ξ1 η1 |D | + |D | 2(5c)l−1 < (5c)l , |D| i |D| i η2 |λi | |D | < (5c)l−1 . |D| i

This completes the proof. c c

Remark. For a given finite linear set, denote by the number of distances linearly independent over Q and by ρ0 the number of linearly independent distances which appear only once. It follows from the lemma that if 0 2, then 2. It can easily be found from remark 1 at the end of paper [2] that if 0 = 1 then = 1. The equality = 0 suggests itself, but I am unable to prove it. Definition. For a given integral matrix A, h(A) will denote the maximum of absolute values of the elements of A. Lemma 5. Let Γ be any given integral matrix 2×2. For arbitrary positive integers d, n, m there exists an integral matrix * ) ν μ1 M= 1 ν2 μ2 satisfying the conditions: (40) (41) (42)

2 0 νi (2d 2 )! ,

2 0 μi (2d 2 )! (i = 1, 2), |M| > 0, [n, m] = [u, v]M, u, v integers 0,

and with the following property. If (43)

[n, m]Γ = [s, t]Δ,


315

where s, t are integers, Δ is an integral matrix, |Δ| = 0 and h(Δ) d,

(44) then

MΓ = T Δ and [s, t] = [u, v]T ,

(45)

where T is an integral matrix and 2 h(T ) 4d (2d 2 )! h(Γ ).

(46)

Proof. Let S be the set of all integral matrices Δ satisfying (43) and (44). Integral vectors [x, y] such that for all Δ ∈ S and suitable integers sΔ , tΔ , [x, y]Γ = [sΔ , tΔ ]Δ form a module, say M. By (44) (2d 2 ) |Δ| = 0, whence |Δ| divides (2d 2 )!. It follows that [(2d 2 )!, 0] ∈ M and [0, (2d 2 )!] ∈ M. Let ξ1 be the least positive integer such that, for some η1 , [ξ1 , η1 ] ∈ M and let η2 be the least positive integer such that [0, η2 ] ∈ M. Clearly [ξ1 , η1 ] and [0, η2 ] form a basis for M and we may assume without loss of generality that 0 η1 < η2 . Hence 0 < ξ1 (2d 2 )!,

(47)

0 η1 < η2 (2d 2 )!.

Let 1 1 1 η1 = − − ... − η2 b1 b2 br be the expansion of η1 /η2 into a continued fraction, where bp are integers > 1 (1 p r); if η1 = 0 let r = 0. Put A−1 = −1,

Ap+1

c

c

B−1 = 0; A0 = 0, B0 = 1; = bp+1 Ap − Ap−1 , Bp+1 = bp+1 Bp − Bp−1 (0 p < r).

It follows that the sequences Ap , Bp are increasing and for 0 p r Ap Bp−1 − Bp Ap−1 = 1,

(48) (49) c

(50)

0 Ap η1 ,

0 < Bp η2 ,

Ap /Bp < Ar /Br = η1 /η2

(if p < r).

Since m > 0, n > 0, we have η1 m ξ1 Ar − · < . η2 n η2 Br Let q be the least non-negative integer which can be substituted for r in the last inequality. Assuming A−1 /B−1 = −∞ we have therefore (51)

Aq Aq−1 η1 m ξ1 − · < . Bq−1 η2 n η2 Bq

Let us put * ) *) * ) * ) −Aq Bq η1 − A q η2 Bq ξ1 η1 Bq ξ1 ν μ1 = = . M= 1 ν2 μ2 Bq−1 −Aq−1 0 η2 Bq−1 ξ1 Bq−1 η1 − Aq−1 η2

316


Inequalities (47), (49) and (50) imply (40). By (48) ξ η |M| = 1 1 = ξ1 η2 > 0. 0 η2 Moreover, the vectors [ν1 , μ1 ], [ν2 , μ2 ] form a basis for M. Since [n, m] ∈ M, there are integers u, v satisfying (42). We have * ) 1 B η − Aq−1 η2 −Bq η1 + Aq η2 [n, m] q−1 1 [u, v] = [n, m]M −1 = c −Bq−1 ξ1 Bq ξ1 ξ1 η2 =

1 Bq−1 (nη1 − mξ1 ) − Aq−1 η2 , Aq η2 − Bq (nη1 − mξ1 ) . ξ1 η 2

It follows from (51) that u 0, v 0. In order to prove the last statement of the lemma suppose that for some integral matrix Δ (43) and (44) hold. Thus Δ ∈ S and since [νi , μi ] ∈ M (i = 1, 2) there are integers σi , τi such that [νi , μi ]Γ = [σi , τi ]Δ (i = 1, 2). Putting * ) σ τ (52) T = 1 1 σ2 τ2 we get MΓ = T Δ. On the other hand, (42) and (43) imply [u, v]MΓ = [s, t]Δ.

(53)

Since |Δ| = 0 by (44), we get (45) from (52) and (53). Finally, by (52), (40) and (44) 2 h(T ) = h(MΓ Δ−1 ) 4h(M)h(Γ )h(Δ) 4d (2d 2 )! h(Γ ).


Lemma 6. Let f (x) be an irreducible polynomial not dividing x δ − x (δ > 1), α, β integers, α > 0 or β > 0. For arbitrary positive integers n, m such that αn + βm > 0 there exists an integral matrix ) * ν μ1 M= 1 ν2 μ2 satisfying the conditions (54) (55) (56) (57)

0 νi C(f, α, β),

0 μi C(f, α, β) (i = 1, 2), |M| > 0,

[n, m] = [u, v]M,

u, v integers 0,

ανi + βμi 0 (i = 1, 2),

and having the following property: if f y αν1 +βμ1 zαν2 +βμ2 = f1 (y, z) · · · fr (y, z)


317

is a standard form of f y αν1 +βμ1 zαν2 +βμ2 , then f (x αn+βm ) = f1 (x u , x v ) · · · fr (x u , x v ) is a standard form of f (x αn+βm ). C(f, α, β) is an effectively computable constant, independent of n and m. Proof. By Theorem 1, there exists a positive integer ν C(f ) such that αn + βm = νw, w integer, and having the following property: if (58)

f (x ν ) = f1 (x) · · · fr (x)

is a standard form of f (x ν ), then (59)

f (x αn+βm ) = f1 (x w ) · · · fr (x w )

is a standard form of f (x αn+βm ). Now we distinguish two cases, αβ 0 and αβ < 0. If αβ 0 we put in Lemma 5: ) * α α Γ = , d = C(f ). β β Let

) M=

ν1 μ1 ν2 μ2

*

be an integral matrix whose existence for n, m is asserted in that lemma. It follows from (40) that 2 2 2 2 2C (f ) ! , 0 μ 2C (f ) ! (i = 1, 2); (60) 0 ν i i c 2 thus (54) is satisfied with C(f, α, β) = 2C 2 (f ) ! and, in view of α 0, β 0, (57) holds. Formulae (55) and (56) follow from (41) and (42). We apply the last statement of Lemma 5 with ) * ν 0 [s, t] = [w, w], Δ = . 0 ν In virtue of that statement there exists an integral matrix T such that *) * ) * ) ν 0 ν1 μ1 α α =T , ν2 μ 2 β β 0 ν whence (61)

ν | ανi + βμi

(i = 1, 2).

If αβ < 0, we may assume without loss of generality α > 0, β < 0. We put in Lemma 5: ) * αn + βm ν ν , m = m , d = α, n = Γ = −β −β ν

318 c


(the “prime” is added to avoid a confusion in notation). In virtue of that lemma there exists an integral matrix * ) ν μ 1 M = 1 ν2 μ 2 such that (62)

2 0 νi (2α 2 )! ,

2 0 μ i (2α 2 )!

(63)

|M | > 0,

(64)

[(αn + βm)/ν, m] = [u, v]M ,

We apply the last statement of Lemma 5 with [s, t] = [w, w],

(i = 1, 2),

u, v integers 0. )

* α 0 Δ= . 0 α

In virtue of that statement there exists an integral matrix T such that ) *) * ) * ν1 μ 1 ν ν α 0 =T , 0 α ν2 μ 2 −β −β whence α | ννi − βμ i (i = 1, 2). We put νi = (ννi − βμ i )/α,

μi = μ i

(i = 1, 2).

(62) implies (57) and the inequality 2 2 (i = 1, 2); 0 νi (2α 2 )! C(f ) + |β| , 0 μi (2α 2 )! 2 2 thus (54) is satisfied with C(f, α, β = (2α )! C(f ) + |β| . Formulae (55) and (56) c follow from (63) and (64); besides we have (61). In order to prove the last property of the matrix M postulated in the lemma, we put (65) fj (y, z) = fj y (αν1 +βμ1 )/ν z(αν2 +βμ2 )/ν (1 j r ). By (56) fj (x u , x v ) = fj (x w ), whence by (59), fj (x u , x v ) is irreducible. We show that fj (y, z) is not reducible. Denote by δ the degree of fj (x) and suppose that fj (y, z) = g(y, z)h(y, z),

(66)

where g is of degree γ1 in y, γ2 in z; h is of degree χ1 in y, χ2 in z and γ1 + γ2 > 0, χ1 + χ2 > 0. By (65) we have δ(ανi + βμi )/ν = γi + χi

(i = 1, 2).

On the other hand, fj (x w ) = g(x u , x v )h(x u , x v ). The degree of fj (x w ) equals δ(αn + βm)/ν = δu(αν1 + βμ1 )/ν + δv(αν2 + βμ2 )/ν = u(γ1 + χ1 ) + v(γ2 + χ2 ).


319

The degree of g(x u , x v )h(x u , x v ) can be equal to uγ1 + uχ1 + vγ2 + vχ2 only if the degree of g(x u , x v ) equals uγ1 + vγ2 and the degree of h(x u , x v ) equals uχ1 + vχ2 . Since fj (x w ) is irreducible we get uγ1 + vγ2 = 0 or uχ1 + vχ2 = 0, whence u = 0 and γ2 χ2 = 0 or v = 0 and γ1 χ1 = 0 or u = v = 0. The last case is impossible by (56), and in view of symmetry it is enough to consider u = 0, γ2 = 0. Thus g(y, 0) = g(y, z) is not constant and, since fj (0) = 0, it follows from (65) and (66) that αν2 + βμ2 = 0. This gives αn + βm = u(αν1 + βμ1 ) + v(αν2 + βμ2 ) = 0. The contradiction obtained proves that no fj (y, z) (1 j r ) is reducible. Since fj (y, z) are also not constant, and by (58) f y αν1 +βμ1 zαν2 +βμ2 = f1 (y, z) · · · fr (y, z), it follows that the polynomials fj (y, z) (1 j r ) and fj (y, z) (1 j r), after a suitable permutation, differ only by constant factors. Since the polynomials fj (x u , x v ) = fj (x w ) (1 j r ) are irreducible and coprime, the same applies to fj (x u , x v ) (1 j r), which completes the proof.

4. Proof of Theorem 2. It is clear that if Φ(x) is a polynomial, then KΦ(x) = KJ Φ(x). We take this equality as a definition of KΦ(x), where Φ(x) is a rational function of the form I ai x αi (αi integers). i=0

Now let F (y, z) =

I

ai y αi zβi ,

i=0

where ai are integers = 0 and the pairs αi , βi (0 i I ) are all different (it is clearly sufficient to prove the theorem for polynomials with integral coefficients). Let be the rank of the matrix ) * α1 − α0 α2 − α0 . . . αI − α0 . β1 − β 0 β2 − β 0 . . . β I − β 0 We consider separately two cases, = 1 and = 2. Case = 1. In this case there exist integers α, β and γi (0 i I ) such that α > 0 or β > 0 and αi − α0 = αγi ,

βi − β0 = βγi

(0 i I ).

Put f (x) = J

I

ai x γi .

i=0

Clearly (67)

J F (y, z) = Jf (y α zβ ).

320


Since F (y, z) is irreducible and is different from ay, az, both f (x) and Jf (x −1 ) = I J ai x −γi are irreducible. If we had for some δ > 1, f (x) | x δ − x, this would imply i=0

f (x) = ±Jf (x −1 ), whence J F (y, z) = ±J F (y −1 , z−1 ), against the assumption. Thus −1 both and Jf f (x) (x ) satisfy the conditions of Lemma 6 and constants C(f, α, β), −1 C Jf (x ), α, β are well defined. We put C0 (F ) = max{|α|, |β|}; C1 (F ) = max C(f, α, β), C Jf (x −1 ), α, β . If αn + βm = 0, we have max{n, m} C0 (F )(n, m). If αn + βm < 0, we can replace in (67) f by Jf (x −1 ), α by −α, β by −β, which will not affect the inequality C(f, α, β) C1 (F ).

(68)

We may therefore assume without loss of generality that αn + βm > 0 and (68) holds. Let * ) ν μ1 M= 1 ν2 μ2 be an integral matrix, whose existence for n, m is asserted in Lemma 6. Since by (57) and (67) J F (y ν1 zν2 , y μ1 zμ2 ) = f (y αν1 +βμ1 zαν2 +βμ2 ), KF (x n , x m ) = f (x αn+βm ), Theorem 2 follows in this case ( = 1) from Lemma 6 and (68). Case = 2. We may assume without loss of generality that α1 − α0 α2 − α0 β1 − β0 β2 − β0 = 0. Let ξ be any irrational number. Clearly the numbers (α1 − α0 ) + ξ(β1 − β0 ) and (α2 − α0 ) + ξ(β2 − β0 ) are incommensurable; thus there are incommensurable distances in the set of points αi + ξβi (0 i I ). By remark 1 at the end of paper [2] (cf. also the remark after Lemma 4) there are in this set two incommensurable distances which appear only once in the double sequence αj − αi + ξ(βj − βi ) (0 / i < j I ). This0 , i , j , j such that α − α , β − β means that there exist 4 non-negative integers i j/ i j i 0 / 0 and αj − αi , βj − βi appear only once in the double sequence αj − αi , βj − βi (0 i < j I ) and αj − αi αj − αi βj − βi βj − βi = 0. We put in Lemma 5 (69)

)

* αj − αi αj − αi Γ = , βj − βi βj − βi

where N = max max{αi , βi }, A = 0iI

I i=0

ai2 .

d = 2(10N 2 )A ,


Let

)

ν μ1 M= 1 ν2 μ2

321

*

be an integral non-singular matrix, whose existence for n, m is asserted in that lemma. Thus we have by (40) and (42) 2 2 (70) (i = 1, 2), 0 νi (8 · 102A N 4A )! , 0 νi (8 · 102A N 4A )! [n, m] = [u, v]M,

(71)

u, v integers 0.

We see that assertions (i) and (ii) of Theorem 2 are satisfied with C1 (F ) = (8 · 2A N 4A )! 2 . c 10 Moreover, by (70) and (71) max{n, m} 2C1 (F ) max{u, v},

(72)

(n, m) (u, v).

Let (73)

J F (y ν1 zν2 , y μ1 zμ2 ) = const F1 (y, z)e1 F2 (y, z)e2 · · · Fr (y, z)er

be a standard form of J F (y ν1 zν2 , y μ1 zμ2 ). In order to prove (iii) we have to show that either KF (x n , x m ) = const KF1 (x u , x v )e1 KF2 (x u , x v )e2 · · · KFr (x u , x v )er is a standard form of KF (x n , x m ) or max{n, m} C0 (F )(n, m), where C0 (F ) is a constant independent of n, m. In view of (70) it is sufficient to prove the same with C0 (F ) replaced by C0 (F, M), a constant depending only on F and M. define C0 (F, M) we notice that by the assumption In order to = 1 and by Lemma 3 there exist two constants, J F (y, z), J F (y −1 , z−1 ) B00 = B0 J F (y, z), J F (y −1 , z−1 ) 1 and B01 = B1 J F (y, z), J F (y −1 , z−1 ) , such that J F (x n , x m ) J F (x −n , x −m ) (n,m)B00 − 1)B00 , (x (74) KF (x n , x m ) KF (x −n , x −m ) and (75)

KF (x n , x m ), KF (x −n , x −m ) = 1

unless max{n, m} B01 (n, m). Since for an arbitrary polynomial f (x) (76) we get from (74) (77) c

Jf (x)/Kf (x) = Jf (x −1 )/Kf (x −1 ), J F (x n , x m ) (n,m)B00 − 1)B00 . (x KF (x n , x m )

322


Let Sj (1 j r) be the set of all the polynomials not divisible by Fj (y, z) which l l ci y σi zτi , where ci are integers = 0, ci2 = A and are of the form J i=0

i=0

max{|σi |, |τi |} 16 · 10 N 2A

4A+1

C1 (F )

(0 i l).

Clearly the sets Sj (1 j r) are finite and effectively computable. Moreover, for each H ∈ Sj there exists by Lemma 3 a constant B1 (Fj , H ) such that (78) KFj (x u , x v ), KH (x u , x v ) = 1 unless max{u, v} B1 (Fj , H )(u, v). Finally for each pair i, j , where 1 i < j r, there exists by Lemma 3 a constant B1 (Fi , Fj ) such that (79) KFi (x u , x v ), KFj (x u , x v ) = 1 unless max{u, v} B1 (Fi , Fj )(u, v). We put 2 C0 (F, M) = max 8N C12 (F )B00 , B01 ,

2C1 (F ) max max B1 (Fj , H ), 2C1 (F ) max B1 (Fi , Fj ) . 1j r H ∈Sj

1i<j r

If for any pair i, j , where 1 i < j r, KFi (x u , x v ), KFj (x u , x v ) = 1 we have by (72) and (79)

max{n, m}/(n, m) 2C1 (F )B1 (Fi , Fj ) C0 (F, M). It remains to prove that if any polynomial KFj (x u , x v ) (1 j r) is not irreducible, then max{n, m} C0 (F, M)(n, m). We shall do that in two steps assuming first that KF1 (x u , x v ) is constant and secondly that it is reducible (the treatment of F1 (x u , x v ) instead of Fj (x u , x v ) does not affect generality and simplifies a little the notation). c

1. Assume that KF1 (x u , x v ) is constant. Let (80)

F1 (y, z) =

k

bj y γj zδj

/ 0 (bj = 0, γj , δj all different)

j =0

and let 1 be the rank of the matrix * ) γ 1 − γ0 . . . γk − γ0 . δ1 − δ 0 . . . δ k − δ 0


323

It follows from (70) and (73) that (81)

0 γj N (μ1 + ν1 ) 2N C1 (F ), 0 δj N (μ2 + ν2 ) 2N C1 (F )

(0 j k)

and 1 = 1 or 2. If 1 = 1, we have (82)

F1 (y, z) = Jf (y γ zδ ),

where f is a polynomial in one variable, γ , δ are integers (cf. p. 319) and by (81) (83)

0 < max{|γ |, |δ|} 2N C1 (F ).

KF1 (x u , x v ) = const implies Kf (x γ u+δv ) = const ; thus γ u + δv = 0 or Kf (x) = const . In the first case, by (83) (84)

max{u, v} 2N C1 (F )(u, v);

in the second case by (76) Jf (x) = ±Jf (x −1 ) and by (82) Jf (y γ zδ ) = F1 (y, z) = ±J F1 (y −1 , z−1 ). The last equality implies by (73) Jf (y γ zδ ) J F (y ν1 zν2 , y μ1 zμ2 ), J F (y −ν1 z−ν2 , y −μ1 z−μ2 ) . By a substitution y = ημ2 ζ −ν2 , z = η−μ1 ζ ν1 we get (85) Jf (ηγ μ2 −δμ1 ζ −γ ν2 +δν1 ) J F (η|M| , ζ |M| ), J F (η−|M| , ζ −|M| ) . However (J F (y, z), J F (y −1 , z−1 ) = 1, and thus J F (η|M| , ζ |M| ), J F (η−|M| , ζ −|M| ) = 1 and (85) implies Jf (ηγ μ2 −δμ1 ζ −γ ν2 +δν1 ) = const . Since by (82) Jf (x) = const , we get γ μ2 − δμ1 = 0, −γ ν2 + δν1 = 0. Since

μ2 −μ1 −ν2 ν1 = |M| = 0,

the last system of equations gives γ = δ = 0, against (83). The contradiction obtained proves (84). If 1 = 2 we may assume without loss of generality that γ1 − γ0 γ2 − γ0 δ1 − δ0 δ2 − δ0 = 0.

324


On the other hand, by (73)

J F1 (x u , x v ) J F (x n , x m ) KF1 (x u , x v ) KF (x n , x m )

and since KF1 (x u , x v ) = const , we get from (77) J F1 (x u , x v ) (x (n,m)B00 − 1)B00 . It follows by (80) that either (86)

uγi + vδi = uγj + vδj

for

i = 1 or 2 and some j k

or 2 |u(γi − γ0 ) + v(δi − δ0 )| B00 (n, m)

(87)

(i = 1, 2).

(81) and (86) imply max{u, v} 2N C1 (F )(u, v),

(88) (81) and (87) imply

2 max{u, v} 4N C1 (F )B00 (n, m).

(89)

In view of (72) it follows from (84), (88) and (89) that max{n, m} C0 (F, M)(n, m). 2. Assume that KF1 (x u , x v ) is reducible. Let f (x) be an irreducible primitive factor u v n m c of it. Since by (71) and (73) KF1 (x , x ) | F (x , x ), we have F (x n , x m ) = f (x)g(x),

(90)

where g(x) is a polynomial with integral coefficients. It follows from (75) that KF1 (x u , x v ), KF (x −n , x −m ) = 1 unless max{n, m} B01 (n, m), whence by (90) (91)

f (x) = const KF1 (x u , x v ), K[f (x)g(x −1 )]

or max{n, m} B01 (n, m).

(92)

In order to calculate the right hand side of (91) we put (93)

f (x)g(x −1 ) =

l i=0

ci x ki

(ci integers = 0, k0 < k1 < . . . < kl )


325

and consider two expressions for F (x n , x m )F (x −n , x −m ): F (x n , x m )F (x −n , x −m ) =

I

ai2 +

ai aj x nαj +mβj −(nαi +mβi ) ,

0i,j I i=j

i=0

(94)

l ci2 + ci cj x kj −ki . f (x)g(x −1 ) f (x −1 )g(x) =

0i,j l i=j

i=0

If for any pair i, j : (95)

i = j

and nαj + mβj − (nαi + mβi ) = 0,

we get n(αj − αi ) + m(βj − βi ) = 0, whence max{n, m} N (n, m).

(96) Similarly, if for any pair i, j (97)

0 / i, j = i , j

and

nαj + mβj − (nαi + mβi ) = nαj + mβj − (nαi + mβi )

or (98)

0 / i, j = i , j

and nαj + mβj − (nαi + mβi )

= nαj + mβj − (nαi + mβi ), 0 / 0 we get by the choice of i , j , i , j (p. 320) a linear homogeneous equation on m and n with non-zero coefficients absolutely 2N, whence /

(99)

max{n, m} 2N (n, m).

If no pair i, j satisfies (95), (97) or (98), it follows from (94) that (100)

l i=0

ci2

=

I

ai2 = A,

i=0

(101) the numbers nαj + mβj − (nαi + mβi ) and nαj + mβj − (nαi + mβi ) appear among the differences kj − ki (0 i l, 0 j l), (102) each number kj − ki which appears only once in the double sequence kj − ki (0 i < j l) has a value nγ + mδ, where |γ | N , |δ| N . Let kj1 − ki1 , kj2 − ki2 , . . . , kjP − kiP (P 0) be all the numbers mentioned in (102) besides kl − k0 .

326


If P 2, 1 p < q P , it follows from kl − k0 = γ0 n + δ0 m, kjp − kip = γp n + δp m, kjq − kiq = γq n + δq m, ⎤ ⎡ γ0 δ0 p,q = rank of ⎣γp δp ⎦ , γq δq that cp,q (kl − k0 ) + cp,q (kjp − kip ) + cp,q (kjq − kiq ) = 0,

where

@ ⎧? γ δ γ δ δ γ ⎪ p p q q 0 0 ⎪ ⎪ , , ⎨ γq δq γ0 δ0 γp δp [cp,q , cp,q , cp,q ]= ⎪ [γp , −γ0 , 0] ⎪ ⎪ ⎩ [δp , −δ0 , 0]

if p,q = 2, if p,q = 1 and γ0 = 0, if p,q = 1 and δ0 = 0.

Clearly 0 < max{|cp,q |, |cp,q |, |cp,q |} 2N 2

(1 p < q P ).

Therefore, the assumptions of Lemma 4 are satisfied with c 2N 2 and we get from that lemma (103)

ki − k0 = sκi + tλi

(0 i l),

where s, t, κi , λi (0 i l) are integers, |κi | (10N 2 )l , |λi | (10N 2 )l . Since by (93) and (100) l < A, we have (104)

|κi | < (10N 2 )A ,

|λi | < (10N 2 )A

(0 i l).

Now by (101), (103) and (104) (105)

nαj + mβj − (nαi + mβi ) = κ s + λ t, nαj + mβj − (nαi + mβi ) = κ s + λ t,

where κ , λ , κ , λ are integers and (106) 0 < max |κ |, |λ | < 2(10N 2 )A ,

0 < max |κ |, |λ | < 2(10N 2 )A .

(κ = λ = 0 or κ = λ = 0 would imply (96).) We put ) * κ κ Δ = . λ λ It follows from (69), (105) and (106) that (107)

[n, m]Γ = [s, t]Δ and h(Δ) < 2(10N 2 )A = d.

327


We distinguish two cases, |Δ| = 0 and |Δ| = 0. If |Δ| = 0, let * ) ξ1 η1 −1 = |Γ |ΔΓ ξ2 η2 (by the choice of Γ , Γ −1 exists). It follows from (106) that (108)

0 < max{|ξ1 |, |η1 |} 2h(Δ)h(Γ ) < 4 · 10A N 2A+1 .

On the other hand, since ξ1 η2 − η1 ξ2 = 0, we get from (107) ξ1 m − η1 n = 0, and thus by (108) (109)

max{n, m}/(n, m) < 4 · 10A N 2A+1 < C1 (F ).

If |Δ| = 0, we can apply to (107) the last statement of Lemma 5. In virtue of that statement there exists an integral matrix T such that [s, t] = [u, v]T

(110) and (111)

2 h(T ) 8 · 10A N 2A+1 (8 · 102A N 4A )! = 8 · 10A N 2A+1 C1 (F ).

Put (112)

T

) * ) * σ κi = i λi τi H (y, z) = J

(113)

(0 i l), l

ci y σi zτi .

i=0

We have by (103) and (110) ki − k0 = uσi + vτi

(0 i l);

thus by (93), (113) and (91) l ci x ki −k0 = KH (x u , x v ), K f (x)g(x −1 ) = K i=0

(114)

f (x) = const KF1 (x u , x v ), KH (x u , x v ) .

If we had F1 | H , it would imply KF1 (x u , x v ) | KH (x u , x v ), whence KF1 (x u , x v ) | f (x), against the choice of f (x). Thus (F1 , H ) = 1. On the other hand, by (104), (111) and (112) max{|σi |, |τi |} 16 · 102A N 4A+1 C1 (F )

(0 i l);

thus by (100), (113) and the choice of S1 (p. 322), H ∈ S1 . It follows by (78) that KF1 (x u , x v ), KH (x u , x v ) = 1

328


or max{u, v}/(u, v) B1 (F1 , H ) max B1 (F1 , H ).

(115)

H ∈S1

A comparison with (114) shows that (115) holds. In view of (72) it follows from (92), (96), (99), (109) and (115) that max{n, m} C0 (F, M)(n, m).

The proof is complete.

5. Proof of Theorem 3. Put in Theorem 2 F (y, z) = ay + bz + c. Since c = 0 we have J F (y, z) = ±J F (y −1 , z−1 ); thus the assumptions of the theorem are satisfied and the constant C0 (ay + bz + c) exists. Put A(a, b, c) = C0 (ay + bz + c), (116)

B(a, b, c) = A(a, b, c)

max

(α,β)∈S,α>β

C (ax α + bx β + c),

where C is a constant from the Corollary to Theorem 1 and S consists of all pairs of relatively prime positive integers A(a, b, c). Now, assume that n, m are positive integers, n > m and K(ax n +bx m +c) is reducible. Let * ) ν1 μ1 M= ν2 μ2 be an integral non-singular matrix whose existence for F, n, m is asserted in Theorem 2. Without loss of generality we may assume that |M| > 0. It follows from part (iii) of Theorem 2 that if K(ax n + bx m + c) is reducible then either J (ay ν1 zν2 + by μ1 zμ2 + c) is reducible or n/(n, m) C0 (F ) = A(a, b, c).

(117)

We prove that the former eventuality is impossible. Suppose that J (ay ν1 zν2 + by μ1 zμ2 + c) = G1 (y, z)G2 (y, z),

(118)

where G1 , G2 are polynomials. By a substitution y = ημ2 ζ −ν2 , z = η−μ1 ζ ν1 we get J (aη|M| + bζ |M| + c) = J G1 (ημ2 ζ −ν2 , η−μ1 ζ ν1 )J G2 (ημ2 ζ −ν2 , η−μ1 ζ ν1 ). However, J (aη|M| + bζ |M| + c) = aη|M| + bζ |M| + c = aη|M| + D(ζ ) is irreducible by the theorem of Capelli applied to the binomial a|η|M +D(ζ ) in the function field Q(ζ ); in fact, ±D(ζ ) is not a power of any element of Q(ζ ) with exponent > 1. It


329

follows that J Gi (ημ2 ζ −ν2 , η−μ1 ζ ν1 ) = const , whence by a substitution η =

Y ν1 Z ν2 ,

ζ =

i = 1 or 2,

Y μ1 Z μ2

J Gi (Y |M| , Z |M| ) = const ,

i = 1 or 2.

Since |M| = 0 and by (118) Gi (0, 0) = 0, we get Gi (y, z) = const (i = 1 or 2). This shows that J (ay ν1 zν2 + by μ1 zμ2 + c) is irreducible and consequently (117) holds. Part (i) of Theorem 3 is thus proved. In order to prove part (ii) we notice that by the Corollary to Theorem 1 there exists an integer δ satisfying the following conditions: (119) 0 < δ C ax n/(n,m) + bx m/(n,m) + c ; (n, m) = δu, u integer; if

K ax nδ/(n,m) + bx mδ/(n,m) + c = const F1 (x)e1 F2 (x)e2 · · · Fr (x)er is a standard form of K ax nδ/(n,m) + bx mδ/(n,m) + c , then K(ax n + bx m + c) = const F1 (x u )e1 F2 (x u )e2 · · · Fr (x u )er is a standard form of K(ax n + bx m + c). We put ν = nδ/(n, m), μ = mδ/(n, m). Clearly n/ν = m/μ is integral. Further by (117) and the definition of S: n/(n, m), m/(n, m) ∈ S, and thus by (116) and (119) ν A(a, b, c)δ B(a, b, c).


6. Proof of Theorem 4. Put (ax n + bx m + c)/K(ax n + bx m + c) = g(x). Clearly g(x) | ax n + bx m + c and g(x) | cx n + bx n−m + a, whence g(x) | (cx n + a)(ax n + bx m + c) − bx m (cx n + bx n−m + a) a 2 + c2 − b2 n = ac x 2n + x +1 , ac g(x) | (cx m + b)(ax n + bx m + c) − ax m (cx n + bx n−m + a) b2 + c2 − a 2 m = bc x 2m + x +1 . bc If g(x) = 1, it follows that a 2 + c2 − b2 a 2 + c2 − b2 x2 + x + 1 = K x 2 + x+1 , ac ac 2 + c2 − a 2 2 + c2 − a 2 b b x2 + x + 1 = K x 2 + x+1 . bc bc

(120)

330


On the other hand, the only monic reciprocal quadratic polynomials which have roots of unity as zeros are x 2 + 1, x 2 ± x + 1, x 2 ± 2x + 1. It follows that a 2 + c2 − b2 = εrac, b2 + c2 − a 2 = ηsbc, where |ε| = |η| = 1; r = 0, 1 or 2; s = 0, 1 or 2. Hence 2c2 = εrac + ηsbc, 2c = εra + ηsb, and

2a 2 − 2b2 = εrac − ηsbc; 4a 2 − 4b2 = (εra)2 − (ηsb)2 ,

a 2 4 − (εr)2 = b2 4 − (ηs)2 .

The last inequality implies (εr)2 = (ηs)2 , and thus r = s. Since c = 0, it is impossible that s = r = 0; thus two cases remain: (121)

r = s = 1,

a 2 = b2 ,

r = s = 2,

(122)

c = εa = ηb;

c = εa + ηb.

In the first case, by (120) g(x) | x 2n + εx n + 1, and since all zeros of x 2n + εx n + 1 are roots of unity g(x) = (ax n + bx m + c, x 2n + εx n + 1). On the other hand, by (121) c(x 2n + εx n + 1)x m = η(ax n + bx m + c) + c(x m+n − εη)(x n + ε) and since (x 2n + εx n + 1, x n + ε) = 1, it follows that g(x) = (ax n + bx m + c, x 2n + εx n + 1) = (x 2n + εx n + 1, x m+n − εη). In the second case, by (120) g(x) | x 2n + 2εx n + 1 = (x n + ε)2 , and since all zeros of x n + ε are roots of unity, g(x) = ax n + bx m + c, (x n + ε)2 . On the other hand, by (122) ax n + bx m + c = a(x n + ε) + b(x m + η), and every multiple factor of ax n + bx m + c divides nax n + bmx m = na(x n + ε) + mb(x m + η) − (anε + mbη). It follows that in the second case (x n + ε, x m + η)2 g(x) = (x n + ε, x m + η)

if anε + bmη = 0, if anε + bmη = 0.

In order to complete the proof it remains to calculate (x 2n + εx n + 1, x m+n − εη)

and

(x n + ε, x m + η).


c

331

This is easily done by factorization in cyclotomic fields (cf. [1], p. 69, where in Theorem 3 εm ε n should be replaced by ε m1 ε n1 ).

7. Proof of Theorem 5. Put in Theorem 2 F (y, z) = y + f (z). Since f (z) ≡ ±1 and f (0) = 0, J F (y, z) = ±J F (y −1 , z−1), and thus the assumptions of the theorem are satisfied and the constants C0 y + f (z) , C1 y + f (z) exist. We put D0 (f ) = max C0 y + f (z) , C1 y + f (z) , D1 (f ) = the greatest common divisor of multiplicities of all the zeros of f (z). By Theorem 2 for every n there exists an integral matrix * ) ν1 μ1 ν2 μ2 satisfying the following conditions (123) 0 νi C1 y + f (z) , (124) (125)

0 μi C1 y + f (z) (i = 1, 2),

n = ν1 u + ν2 v, 1 = μ1 u + μ2 v, u, v integers 0; if K x n + f (x) is reducible, then either J y ν1 zν2 + f (y μ1 zμ2 ) is reducible or n = max{n, 1}/(n, 1) C0 y + f (z) D0 (f ).

It follows from (123) and (124) that μ1 u = 0, μ2 v = 1 or μ1 u = 1, μ2 v = 0. In view of symmetry it is enough to consider the first possibility. We then have μ2 = v = 1 and u = 0 or μ1 = 0. If u = 0, then (126) n = ν2 C1 y + f (z) D0 (f ). If μ1 = 0, then

J y ν1 zν2 + f (y μ1 zμ2 ) = y ν1 zν2 + f (z).

By the theorem of Capelli applied to the binomial y ν1 + z−ν2 f (z) in the function field Q(z), y ν1 zν2 + f (z) is reducible only if −z−ν2 f (z) = g(z)p and p | ν1 or z−ν2 f (z) = 4g(z)4 and 4 | ν1 , where g(z) is a rational function and p is a prime. Since f (0) = 0, it follows that for some prime p p | D1 (f ), By (124), this implies (127)

p | ν1

and

n, D1 (f ) = 1.

p | ν2 .

Therefore, if K x n + f (x) is reducible, at least one of the inequalities (125), (126) and (127) is satisfied. This completes the proof.

332


Note added in proof. 1. H. Zassenhaus and the writer have proved (cf. [4a]) the following improvement of inequality (4): max |β (i) | > 1 + 2−N−4 . Hence inequality (1) can be improved as 1iN

follows: e(a, Ω) (2N+4 + 1) log(NH (α)). 2. E. G. Straus has proved the equality = 0 conjectured on p. 314. His general proof (to appear in [4b]) specialized to the case 0 = 2 would lead also to a simpler proof of Lemma 4 than that given in the present paper.

References [1] W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 (1960), 65–70. [2] J. Mikusiński, A. Schinzel, Sur la réductibilité de certains trinômes. Acta Arith. 9 (1964), 91–95. [3] A. Schinzel, Solution d’un problème de K. Zarankiewicz sur les suites de puissances consécutives de nombres irrationnels. Colloq. Math. 9 (1962), 291–296; Correction, ibid. 12 (1964), 289; this collection: D1, 295–300. [4] −−, Some unsolved problems on polynomials. In: Neki nerešeni problemi u matematici, Matematiˇcka Biblioteka 25, Beograd 1963, 63–70; this collection: E1, 703–708. [4a] A. Schinzel, H. Zassenhaus, A refinement of two theorems of Kronecker. Michigan Math. J. 12 (1965), 81–85; this collection: C1, 175–178. [4b] E. G. Straus, Rational dependence in finite sets of numbers. Acta Arith. 11 (1965), 203–204. [5] N. Tschebotaröw, Grundzüge der Galoisschen Theorie. Übersetzt und bearbeitet von H. Schwerdtfeger, Noordhoff, Groningen–Djakarta 1950.

Originally published in Acta Arithmetica XIII (1967), 91–101


Reducibility of polynomials and covering systems of congruences

The following problem has been proposed by Professor P. Turán in an oral communication: n ai x n−i Does there exist a constant C such that for every polynomial f (x) = (ai integers, a0 = 0), there is a polynomial g(x) = over the rationals and satisfying

n

n

i=0

bi

x n−i

(bi integers) irreducible

i=0

|bi − ai | C ?

i=0

This problem, apparently very difficult, becomes simpler if one removes the condition that the degree of g should not exceed the degree of f . Then it seems plausible that for polynomials f (x) with f (0) = 0 the value of C can be taken 1, i.e. for a suitable n and a suitable sign the polynomial ±x n + f (x) is irreducible. I have treated the irreducibility of x n + f (x) in [3] and I have proved (Theorem 5) that for every polynomial f (x) with rational coefficients such that f (0) = 0, f (1) = −1 and f (x) ≡ 1, there exist infinitely many n’s for which x n + f (x) has exactly one irreducible factor that is not a cyclotomic polynomial (the precise formulation of the theorem says a little more). The example f0 (x) =

1 (3x 9 + 8x 8 + 6x 7 + 9x 6 + 8x 4 + 3x 3 + 6x + 5) 12

shows that x n + f (x) may have cyclotomic factors for any n. In this example, however, the coefficients of f (x) are not integers. The aim of the present paper is to investigate the irreducibility of x n + f (x), where f (x) has integer coefficients and to show its connection with the so called covering systems of congruences. A system of congruences ai mod mi is called covering if every integer satisfies one of the congruences (cf. [1] and the papers quoted there). The precise formulation of the results is given below, but their most striking consequence is that if there are no covering systems with distinct odd moduli > 1 (the conjecture of Selfridge), then for every polynomial f (x) with integer coefficients such that f (0) = 0, f (1) = −1, x n + f (x) is irreducible for infinitely many n.

334


Theorem 1. The following two propositions are equivalent. A.

For every polynomial f (x) with integer coefficients such that f (0) = 0, f (1) = −1 and f (x) ≡ 1, there exists an arithmetical progression N such that if ν ∈ N then x ν + f (x) is irreducible over the rationals.

B.

In every finite covering system of congruences ai mod mi (mi > 1) at least one of the quotients mj /mi equals q α (q prime, α 0), aj ≡ ai mod mi and either q > 2 or mi ≡ 1 mod 2 or aj ≡ ai mod (mi /2).

Theorem 2. There is an implication C → B → D, where C and D are the following propositions. C.

In every finite covering system of congruences ai mod mi (mi > 1) either there are two equal moduli or there is a modulus even.

D.

In every finite covering system of congruences ai mod mi (mi > 1) at least one modulus divides another one.

Notation. Z is the ring of integers, Q the field of rationals, a monic polynomial means a polynomial with the highest coefficient ±1. Xn (x) is the nth cyclotomic polynomial, ζn is a primitive nth root of unity. For any polynomial f (x), Kf (x) is the factor of f (x) of the greatest possible degree whose no root is 0 or a root of unity and whose leading coefficient is equal to that of f (x). Lemma 1. Let Fi (x), ai (x) ∈ Z[x] (i = 1, 2, . . . , r). If the polynomials Fi (x) (i = 1, 2, . . . , r) are monic and relatively prime in pairs modulo every prime, then there exists a polynomial f (x) ∈ Z[x] such that f (x) ≡ ai (x) mod Fi (x),

(1)

degree f (x) < degree

(2)

r

Fi (x).

i=1

Proof. For each i r consider the polynomials Fi (x) and Gi (x) = Fi (x)

−1

r

Fi (x).

i=1

Since they are monic and relatively prime mod 2 they are relatively prime over Q and there exist polynomials Ui (x), Vi (x) ∈ Z[x] such that degree Ui < degree Gi ,

degree Vi < degree Fi

and Fi (x)Ui (x) + Gi (x)Vi (x) = Ri = 0. ui Ui∗ (x),

Let Ui (x) = Vi (x) = vi Vi∗ (x), where ui , vi are integers and Ui∗ (x), Vi∗ (x) are primitive polynomials.

D3. Reducibility of polynomials and covering systems of congruences

335

If Ri /(ui , vi ) has any prime factor p, we have vi ui or p /| . either p /| (ui , vi ) (ui , vi ) Without loss of generality we may assume the former. Since Fi (x) and Gi (x) are relatively prime mod p, it follows from (3)

Fi (x)

ui vi Ri U ∗ (x) + Gi (x) V ∗ (x) = (ui , vi ) i (ui , vi ) i (ui , vi )

that

ui Ui∗ (x) ≡ 0 mod p, Gi (x) . (ui , vi )

Since ui = 0 and Gi (x) is monic, the degree of Ui∗ (x) is less than the degree of Gi (x) also mod p, thus we get a contradiction. Therefore, Ri /(ui , vi ) has no prime factors; equals εi = ±1 and it follows from (3) that vi 1 mod Fi (x), ∗ Gi (x)Vi (x) ≡ (4) εi (ui , vi ) 0 mod Gi (x). Now, put r

(5)

i=1

vi ai (x)Gi (x)Vi∗ (x) = q(x) Fi (x) + f (x), (ui , vi ) r

εi

i=1

where q(x) ∈ Q[x] and degree f (x) < degree

r

Fi (x).

i=1

Since

r

Fi (x) is monic, f (x) ∈ Z[x]. By (4) and (5) (1) holds.

i=1

Remark. Without the condition (2) the lemma is true also if polynomials Fi (x) are not monic, but the proof is much more complicated. Lemma 2. If q is a prime, then Xm (x), Xn (x) (m n) are relatively prime mod q except if n/m = q α (α 0), in which case Xn (x) ≡ Xm (x)ϕ(n)/ϕ(m) mod q.

(6)

Proof. Let m = q μ m1 , n = q ν n1 , where q /| m1 n1 . We have by the properties of cyclotomic polynomials μ

(7)

Xm (x) =

Xm1 (x q ) Xm1

μ−1 (x q )

≡ Xm1 (x)q

μ −q μ−1

mod q

(μ 1)

336


and similarly ν

(8)

Xn (x) =

Xn1 (x q ) Xn1

ν−1 (x q )

≡ Xn1 (x)q

ν −q ν−1

(ν 1).

mod q

If n1 = m1 , the polynomials Xm1 (x), Xn1 (x) are relatively prime over Q, and both divide x n1 m1 − 1. Thus their resultant R divides the discriminant of x n1 m1 − 1 and since n1 m1 ≡ 0 mod q we get R ≡ 0 mod q. Hence Xm1 (x) and Xn1 (x) are relatively prime mod q and by (7) and (8) the same is true about Xm (x) and Xn (x). If n1 = m1 , (6) follows from (7) and (8) after taking into account the case μ = 0. Lemma 3. For every odd c 1 and integer α 1 the polynomial α−1 (9) D2α c (x) = 21 X2α c (x) − Xc (x 2 ) belongs to Z[x], is monic and relatively prime mod every prime to X2β c (x), where β < α. Proof. We have X2α c (x) = Xc (−x 2

(10)

α−1

),

thus D2α c (x) ∈ Z[x]. If c = 1, D2α c (x) = 1, thus the lemma is true. If c > 1 and c∗ is the product of all distinct prime factors of c, we have ∗

∗

Xc (x) = Xc∗ (x c/c ) = x ϕ(c) − μ(c∗ )x ϕ(c)−c/c + . . . , whence D2α c (x) = μ(c∗ )x 2

α−1 (ϕ(c)−c/c∗ )

+ ...

and D2α c (x) is monic. Since Xc (x 2

α−1

)=

α−1

X2β c (x),

β=0

it follows from Lemma 2 that D2α c (x) and X2β c (x) (β < α) are relatively prime modulo every odd prime. In order to prove that they are relatively prime mod 2 consider their resultant R. We have R= D2α c (ζ ), where ζ runs through all primitive roots of unity of degree 2β c. By (9) and (10) β α−1 R = 2−ϕ(2 c) Xc (−ζ 2 ). When ζ runs through all primitive roots of unity of degree 2β c, ζ 2 through all primitive roots of unity of degree c. Therefore, β β γ ϕ(2 ) Xc (−ζc ) . R = 2−ϕ(2 c) (γ ,c)=1

α−1

runs ϕ(2β ) times


Since

Xc (x) =

337

(x − ζcδ ),

(δ,c)=1

we have

γ

Xc (−ζc ) =

(γ ,c)=1

Thus R=

γ

(ζc + ζcδ ) = 2ϕ(c)

(γ δ,c)=1

γ

(ζc + ζcδ )ϕ(2

β)

(γ δ,c)=1

γ

(ζc + ζcδ ).

(γ δ,c)=1 γ =δ

≡

γ

(ζc − ζcδ )ϕ(2

β)

β

≡ d ϕ(2 ) mod 2,

(γ δ,c)=1 γ =δ

where d is the discriminant of Xc (x). Since d is odd, R is also odd and the proof is complete. c

c

Lemma 4. Let f (x) be a polynomial satisfying the assumptions of Proposition A. Let e0 be the greatest integer e such that −f (x) = g(x)e , g(x) ∈ Z[x]. There exists a constant D0 (f ) such that if n > D0 (f ), (ν, e0 ) = 1 and ν ≡ 0 mod 4 in the case f (x) = 4h(x)4 , h(x) ∈ Z[x], then K x ν + f (x) is irreducible over Q. Proof. Put in Theorem 2 of [3]: F (y, z) = y + f (z), n = ν, m = 1. By that theorem there * ) ν1 μ1 with the following properties: exists an integral matrix M = ν2 μ2 (11) (12)

0 νi C1 (F ),

0 μi C1 (F )

[ν, 1] = [u, v]M,

(i = 1, 2),

(u, v integers 0),

(13) if y ν1 zν2 +f (y μ1 zμ2 ) = const F1 (y, z)e1 F2 (y, z)e2 · · · Fr (y, z)er is a decomposition of y ν1 zν2 + f (y μ1 zμ2 ) into factors irreducible over Q, then either K x ν + f (x)) = const KF1 (x u , x v )e1 KF2 (x u , x v )e2 · · · KFr (x u , x v )er is a decomposition of K x ν + f (x) into factors irreducible over Q or ν C0 (F ). C0 (F ) and C1 (F ) are constants independent of ν. We take D0 (f ) = max{C0 (F ), C1 (F )} and assume ν > D0 (f ), (ν, e0 ) = 1 and ν ≡ 0 mod 4 if f (x) = 4h(x)4 , h(x) ∈ Z[x]. It follows from (12) that (14)

ν = ν1 u + ν2 v,

1 = μ1 u + μ2 v,

thus by (11) μ1 u = 1, μ2 v = 0

or

μ1 u = 0, μ2 v = 1.

338 c


In view of the symmetry we may assume the latter. Thus μ2 = v = 1 and either u = 0 of μ1 = 0. If u = 0, then ν = ν2 D0 (f ) against the assumption. If μ1 = 0, y ν1 zν2 + f (y μ1 zμ2 ) = y ν1 zν2 + f (z). By the theorem of Capelli (cf. [3], p. 6(1 )) the last polynomial can be reducible over Q only if (15)

−f (z)z−ν2 = k(z)p ,

where

p | ν1 and k(z) ∈ Q(z)

f (z)z−ν2 = 4k(z)4 ,

where

4 | ν1 and k(z) ∈ Q(z).

or (16)

Since f (0) = 0, (15) implies that p | e0 and p | ν2 , thus by (14) (ν, e0 ) = 1 against the assumption. Similarly (16) implies that f (z) = 4h(z)4 , h(z) ∈ Z[z] and ν ≡ 0 mod 4, ν1 ν2 μ1 μ2 again contrary ν to the assumption. Thus y z + f (y z ) is irreducible over Q and by (13) K x + f (x) is also irreducible. Proof of Theorem 1. Implication A → B. Assume B is false, thus there exists a finite set S of integral pairs (m, a) with m > 1 and with the following properties. (17) For every integer ν there exists a pair (m, a) ∈ S such that ν ≡ a mod m (the system a mod m, (m, a) ∈ S is covering). (18) If (m, a) ∈ S, (n, b) ∈ S and n/m = q α (q prime, α 0), then either b ≡ a mod m m or q = 2, m ≡ 0 mod 2 and b ≡ a mod . 2 Let S0 be a subset of S irreducible with respect to property (17). If (m, a) ∈ S0 , (n, b) ∈ S0 , (m, a) = (n, b) and m | n, then b ≡ a mod m; otherwise, S0 \ {(n, b)} would also have property (17). Property (18) is hereditary, but in view of the last remark it takes for S0 the following simpler form. If (m, a) ∈ S0 , (n, b) ∈ S0 , (m, a) = (n, b) and n/m = q α (q prime, α 0), then m q = 2, α > 0, m ≡ 0 mod 2, b ≡ a mod and b ≡ a mod m. 2 Divide the set S0 into classes assigning two pairs (m, a) and (n, b) to the same class if n/m = 2α (a 0 or < 0). We obtain the decomposition of S0 S0 =

(19)

r .

Ci ,

i=1

and the pairs in any class Ci can be represented in the form (2αij ci , aij ) (j = 1, 2, . . . , ki ), α c where ci is odd and either ki = 1, 2 i1 ci > 1 or (20)

(1 )

0 < αi1 < αi2 < . . . < αiki = αi , aij ≡ aiki mod 2αij −1 ci , This collection, page 306.

aij ≡ aiki mod 2αij ci

(1 j < ki ).


339

For each i such that ki > 1 consider the system of congruences (21)

g(x) ≡ 0 mod

k i −1 j =1

X2αij ci (x),

g(x) ≡ −x aiki mod D2αi ci (x).

By Lemma 3 for each j < ki , X2αij ci (x) is relatively prime to D2αi ci (x) mod every k i −1 prime, thus the same is true about X2αij ci (x). By Lemma 1 for each i such that ki > 1, j =1

there exists a polynomial gi (x) ∈ Z[x] satisfying the system (21). Now, put for each i r: ⎧ aik ⎨ gi (x) + x i X αi (x) − x aiki , if k > 1, i 2 ci D2αi c (x) (22) fi (x) = ⎩ ai1 i if ki = 1, −x , and consider the system of congruences f (x) ≡ fi (x) mod

(23)

ki j =1

X2αij ci (x)

(i = 1, 2, . . . , r).

By Lemma 2 the moduli are relatively prime in pairs mod every prime, thus by Lemma 1 there exists a polynomial fr+1 (x) ∈ Z[x] satisfying (23) and such that (24)

degree fr+1 (x) < degree

ki r i=1 j =1

X2αij ci (x).

We claim that (25)

fr+1 (x) ≡ −x aij mod X2αij ci (x)

(1 i r, 1 j ki ).

This is clear by (22), if ki = 1. On the other hand, if ki > 1, Xci (x 2

αi −1

)=

α i −1

X2β ci (x),

β=0

thus 2D

2αi ci

(x) ≡ X

2αi c

i

(x) mod

k i −1 j =1

X2αij ci (x)

and it follows from (21) (with g replaced by gi ), (22) and (23) (with f replaced by fr+1 ) that

(26)

fr+1 (x) ≡ x aiki mod

k i −1 j =1

X2αij ci (x),

fr+1 (x) ≡ −x aiki mod X2αi ci (x).

340


By (20) aiki ≡ aij mod 2αij −1 ci , but aiki ≡ aij mod 2αij ci , hence x 2aiki ≡ x 2aij mod X2αij ci (x), x aiki ≡ x aij mod X2αij ci (x), thus x aiki ≡ −x aij mod X2αij ci (x)

(27)

(1 j < ki ).

Now (25) follows from (26) and (27). Put (28) t = max 1, 2 − fr+1 (0), −fr+1 (1) and consider the polynomial f0 (x) = fr+1 (x) + t

(29)

ki r i=1 j =1

X2αij ci (x).

By (20) 2αij ci > 1, thus we have f0 (0) = fr+1 (0) + t 2,

f0 (1) fr+1 (1) + t 0

and the polynomial f0 (x) satisfies the assumptions of Proposition A. On the other hand, by the choice of S0 and (19) for every integer ν 0 there exist i r and j ki such that ν ≡ aij mod 2αij ci . Hence x ν ≡ x aij mod X2αij ci (x) and by (25) and (29) (30)

x ν + f0 (x) ≡ 0 mod X2αij ci (x).

ki r However, by (24) and (28) f0 (x) has the degree equal to that of X2αij ci (x) and the i=1 j =1 r leading coefficient positive. Since ki > 1, the degree of x ν + f0 (x) is greater than i=1

that of X2αij ci (x) and it follows from (30) that x ν + f0 (x) is reducible. Thus we have proved more than was necessary, namely the existence of a polynomial f (x) satisfying the assumptions of Proposition A and such that x ν + f (x) is reducible for all ν 0. Implication B → A. Let f (x) be a polynomial satisfying the assumptions of A and let e0 be the greatest integer e such that −f (x) = g(x)e ,

g(x) ∈ Z[x].

Consider first the case, where f (x) = 4h(x)4 , h(x) ∈ Z[x]. Then let r0 be the least c number r such that (r, 2e0 ) = 1 and r > D0 (f ). The arithmetical progression N : 2e0 t +r0 (t = 0, 1, . . . ) has the property asserted inA. Indeed, if ν ∈ N then ν > D0 (f ), (ν, e0 ) = 1 and ν ≡ 0 mod 4, thus, by Lemma 4, K x ν + f (x) is irreducible. But no root of unity, ζm say, can be a zero of x ν + f (x), since it would follow that ζmν + 4h(ζm )4 = 0, which is impossible.

ζmν ≡ 0 mod 4,


341

Assume now that f (x) = 4h(x)4 , h(x) ∈ Z[x]. Let P be the set of all pairs (p, 0), where p is a prime and p | e0 . Let M be the set of all pairs (μ, α), where 0 α < μ and ζμα + f (ζμ ) = 0.

(31)

M is finite. Indeed, it follows from (31) that f (ζμ )f (ζμ−1 ) = 1, thus ζμ is a root of the equation x d f (x)f (x −1 ) − x d = 0, where d is the degree of f (x), and we get ϕ(μ) 2d. Since f (1) = −1 we have μ > 1 for all (μ, α) ∈ M. We claim that the system of congruences a mod m, where (m, a) ∈ P ∪ M, does not satisfy the condition for covering system asserted in B. Indeed, suppose that (m, a) ∈ P ∪ M, (n, b) ∈ P ∪ M, n (32) = q α (q prime, α 0), b ≡ a mod m, m m q > 2 or m ≡ 1 mod 2 or b ≡ a mod . (33) 2 (m, a) ∈ P , (n, b) ∈ P impossible in view of (32). Consider first the case (m, a) ∈ P , (n, b) ∈ M. By the definition of P , m is a prime and −f (x) = k(x)m ,

k(x) ∈ Z[x].

On the other hand, by the definition of M ζnb + f (ζn ) ≡ 0. Thus, we get k(ζn )m = ζnb and (34)

β , k(ζn ) = ζmn

where

(β, n) = (b, n).

k(ζn ) is a primitive root of unity of degree mn/(β, mn), but k(ζn ) ∈ Q(ζn ), thus by a known theorem (cf. e.g. [2], p. 536) 2(β, mn) mn 2n ; m (2, n) (β, mn) (2, n) and (35)

2β m . (2, n)

Since by the first part of (32) m | n, it follows from (34) and (35) that b ≡ 0 mod m, which contradicts the second part of (32). Consider next the case (m, a) ∈ M, (n, b) ∈ P . Then since n is a prime and m > 1, it follows from (32) that n = m, thus we can interchange the roles of m and n and apply the preceding case. Consider finally the case (m, a) ∈ M, (n, b) ∈ M. We have (36)

x a + f (x) ≡ 0 mod Xm (x), x b + f (x) ≡ 0 mod Xn (x).

342


Since by Lemma 2, Xn (x) ≡ Xm (x)ϕ(n)/ϕ(m) mod q, we get from (36) x b − x a ≡ 0 mod q, Xm (x) . Since x and Xm (x) are relatively prime mod q, it follows that (37) x |b−a| − 1 ≡ 0 mod q, Xm (x) . Put Δ = |b − a| = q δ Δ1 , where q /| Δ1 . We have q δ δ Xd (x)q mod q. (38) x Δ − 1 ≡ x Δ1 − 1 ≡ d |Δ1

It follows from (37), (38) and Lemma 2 that for some d1 | Δ1 and some β 0, m = q β , Xm (x) ≡ Xd1 (x)ϕ(m)/ϕ(d1 ) mod q, d1 and

δ Xd1 (x)q ≡ 0 mod q, Xd1 (x)ϕ(m)/ϕ(d1 ) .

The last congruence implies qδ

(39)

ϕ(m) = ϕ(q β ). ϕ(d1 )

If β = 0 or q > 2, it follows from (39) that δ β, thus Δ ≡ 0 mod m and b ≡ a mod m m contrary to (32). If β > 0 and q = 2 we get from (39) that δ β − 1, thus Δ ≡ 0 mod 2 m and b ≡ a mod contrary to (33). 2 By the proposition B, the system a mod m, where (m, a) ∈ P ∪ M, is not covering, thus there exist numbers D1 and r1 such that if ν ≡ r1 mod D1 , then ν ≡ a mod m for any (m, a) ∈ P ∪ M. c Let r2 be the least integer r such that r ≡ r1 mod D1 and r > D0 (f ). The arithmetical progression N : D1 t + r2 (t = 0, 1, . . . ) has the property asserted if ν ∈ N ν in A. Indeed, c then ν > D0 (f ) and (ν, e0 ) = 1, hence by Lemma 4 K x + f (x) is irreducible. ν On the other νhand,

no root of unity can be a zero of x + f (x), since this would imply m, ν − m ∈ M for a suitable m. m Proof of Theorem 2. The implication B → D being obvious, it is enough to prove C → B. Assume B is false, thus (compare the proof of Theorem 1, implication A → B) there exists a covering system aij mod 2αij ci (1 i r, 1 j ki ), where ci are odd and distinct and for ki > 1 (20) holds. Consider the system of congruences (40)

aiki mod ci

(41)

αi0 j

ai0 j mod 2

(1 i r, ci > 1),

(1 j ki0 ),

where ci0 = 1.

If C is true, system (40) is not covering, thus there exists an integer ν1 such that ν1 ≡ aiki mod ci ,

for any i r with ci > 1.


343

On the other hand, system (41) is not covering, since by (20) ai0 j (1 j ki0 ) are 1 distinct and αi j < 1. Thus there exists an integer ν2 such that j 2 0 ν2 ≡ ai0 j mod 2αi0 j

for any

j ki0 .

By the Chinese Remainder Theorem there exists ⎧ r ⎪ ⎨ν mod ci , 1 ν0 ≡ i=1 ⎪ ⎩ ν2 mod 2αi0 . By the choice of ν1 and ν2 , ν0 does not satisfy any of the congruences (40) and (41), thus system (40)–(41) and a fortiori the system aij mod 2αij ci is not covering and we get a contradiction. Remark. The implication C → D was first proved in a similar way by J. L. Selfridge.

References [1] P. Erd˝os, Some recent advances and current problems in number theory. In: Lectures on Modern Mathematics, vol. 3, Wiley, New York 1965, 196–244. [2] H. Hasse, Zahlentheorie. Akademie-Verlag, Berlin 1963. [3] A. Schinzel, On the reducibility of polynomials and in particular of trinomials. Acta Arith. 11 (1965), 1–34; Errata, ibid., 491; this collection: D2, 301–332.

Originally published in Acta Arithmetica XVI (1969), 123–159


Reducibility of lacunary polynomials I

To the memory of my teachers Wacław Sierpiński and Harold Davenport

1. The present paper is in close connection with [9], the notation of that paper is used and extended (for a result which requires little notation see Corollary to Theorem 2). Reducibility means reducibility over the rational field Q. Constants are considered neither reducible nor irreducible. If f (x1 , . . . , xk ) = 0 is a polynomial, then can

f (x1 , . . . , xk ) = const

s

fσ (x1 , . . . , xk )eσ

σ =1

means that polynomials fσ are irreducible and relatively prime in pairs. k xiαi where f is a polynomial, αi are integers and If Φ(x1 , . . . , xk )=f (x1 , . . . , xk ) i=1 f (x1 , . . . , xk ), x1 · · · xk = 1 then J Φ(x1 , . . . , xk ) = f (x1 , . . . , xk ) (this definition is equivalent to one given in [9]). Let can

J Φ(x1 , . . . , xk ) = const

s

fσ (x1 , . . . , xk )eσ .

σ =1

We set KΦ(x1 , . . . , xk ) = const LΦ(x1 , . . . , xk ) = const

1

2

fσ (x1 , . . . , xk )eσ , fσ (x1 , . . . , xk )eσ ,

where 1 is extended over these fσ which do not divide J (x1δ1 · · · xkδk − 1) for any [δ1 , . . . , δk ] = 0, 2 is extended over all fσ such that (∗)

Jfσ (x1−1 , . . . , xk−1 ) = ±fσ (x1 , . . . , xk ). Corrigenda: Acta Arithmetica XIX (1971), 201; ibid. XXXIV (1978), 265.

345

D4. Reducibility of lacunary polynomials I

The leading coefficients of KΦ and LΦ are assumed equal to that of J Φ. In particular for k = 1, KΦ(x) equals J Φ(x) deprived of all its cyclotomic factors and LΦ(x) equals J Φ(x) deprived of all its monic irreducible reciprocal factors (a polynomial f (x) −1 ) = ±f (x)). J 0 = K0 = L0 = 0. Note that (∗) implies c is reciprocal if Jf (x −1 −1 Jfσ (x1 , . . . , xk ) = const fσ (x1 , . . . , xk ). The operations J, K, L are distributive with respect to multiplication, besides for k = 1, J and K are commutative with the substitution x → x n (n 0), L does not share this property and is always performed after the substitution. We have KJ = J K = K, LJ = J L = L, LK = KL = L; the first two formulae follow directly from the definitions, the last one requires a proof (see Lemma 11). The paper has emerged from unsuccessful efforts to prove the conjecture formulated in [9] concerning the factorization of KF (x n1 , . . . , x nk ) for given F . The operation L has turned out more treatable and the analogue of the conjecture for LF (x n1 , . . . , x nk ) appears below as Lemma 12. For a polynomial F (x1 , . . . , xk ) F is the sum of squares of the absolute values of the coefficients of F ; if F = 0, |F | is the maximum of the degrees of F with respect to xi (1 i k), |F |∗ = max{|F |2 , 2} + 2, exp1 x = exp x, expj x = exp(expj −1 x). From this point onwards all the polynomials considered have integral coefficients unless stated to the contrary. The highest common factor of two polynomials is defined only up to a constant; the formulae involving it should be suitably interpreted; we set (0, 0) = 0. Theorem 1. For any polynomial F = 0 and any integer n = 0 there exist integers ν and u such that 2 (i) 0 ν exp 10|F | log |F |∗ log F , n = uν,

(ii) (iii)

can

KF (x ν ) = const

s

can

Fσ (x)eσ implies KF (x n ) = const

σ =1

c

s

J Fσ (x u )eσ .

σ =1

This is a quantitative formulation of Corollary to Theorem 1 [9] and a generalization of that theorem. Theorem 2. For any polynomial F (x1 , . .. , xk ) and any integral vector n = [n1 , . . . , nk ] = 0 such that F x n1 , . . . , x nk = 0 there exist an integral matrix N = [νij ] ir of rank r and an integral vector v = [v1 , . . . , vr ] such that j k

(i)

max |νij | cr (F ),

(ii)

n = vN ,

346 (iii) LF


r

yiνi1 , . . . ,

i=1

r

yiνik

can

= const

s

Fσ (y1 , . . . , yr )eσ

implies

σ =1

i=1

s can e LFσ x v1 , . . . , x vr σ . LF x n1 , . . . , x nk = const σ =1

Moreover

c

⎧ ⎪ if r = k, exp 9k · 2F −5 ⎪ ⎨ 2 cr (F ) = exp 5 · 2F −4 2F log |F |∗ if r + k = 3, ⎪ ⎪ ⎩ F −1 ∗ exp(k−r)(k+r−3) 10k|F | log F otherwise.

Corollary. For any polynomial f (x) = 0 the number of its irreducible non-reciprocal factors except x counted with their multiplicities does not exceed expf 2 −5f +7 f + 2 (a bound independent of |f |). Theorem 2 is the main result of the paper. An essential role in the proof is played by a result of Straus [11]. It is an open question equivalent to the conjecture from [9] whether a similar theorem, possibly with greater constants cr (F ), holds for the operation K instead of L. The case k = 1 is settled by Theorem 1, for k = 2 a partial result is given by Theorem 3. For any polynomial F (x1 , x2 ) such that KF (x1 , x2 ) = LF (x1 , x2 ) and any integral vector n = [n1 , n2 ] = 0 such that F x n1 , x n2 = 0 there exist an integral matrix N = [νij ] ir of rank r and an integral vector v = [v1 , vr ] such that j 2

exp 9 · 2F −4 if r = 2, max |νij | 2F +1 i,j if r = 1, exp 500F 2 2|F |∗

(i) (ii) (iii) KF

r i=1

yiνi1 ,

r i=1

yiνi2

n = vN , s can Fσ (y1 , yr )eσ = const

implies

σ =1 s can e KFσ x v1 , x vr σ . KF x n1 , x n2 = const σ =1

This theorem is closely related to Theorem 2 of [9] but is both quantitative and more general, since it does not assume the irreducibility of F . Theorem 4. If k 2, a0 = 0, aj = 0 and nj (1 j k) are integers then either k nj aj x L a0 + j =1


c

347

is irreducible or there is an integral vector [γ1 , . . . , γk ] such that ⎧ 2 2 ⎪ ⎪ ⎪24 j =0 aj2 +5 log aj2 if k = 2, ⎪ ⎪ ⎨ j =0 0 < max |γj | k ⎪ j k ⎪ aj2 +3 2 ⎪ j =0 ⎪exp2k−4 k2 if k > 2 log aj ⎪ ⎩ j =0

and k

γj nj = 0.

j =1

Theorem 5. If a, b, c, n, m are integers, n > m > 0, abc = 0 then either K(ax n +bx m +c) is irreducible or n/(n, m) 24(a

2 +b2 +c2 )+5

log(a 2 + b2 + c2 )

and there exist integers ν and μ such that m/μ = n/ν is integral, 0 < μ < ν exp(a 2 + b2 + c2 )2 24(a

2 +b2 +c2 )+11

and can

K(ax ν + bx μ + c) = const

s

Fσ (x)eσ

σ =1

implies can

K(ax n + bx m + c) = const

s

Fσ (x n/ν )eσ .

σ =1

This is a quantitative formulation of Theorem 3 of [9]. The proofs of Theorems 1, 2, 3, 4, 5 are given in §§ 2, 3, 4, 5, 5 respectively. Some of the proofs could be simplified at the cost of increasing the order of cr (F ) and of other similar constants. Since however simplifications would not be great and the constants already are, I did as much as I could not to increase their order. On the other hand I have refrained from making generalizations to algebraic number fields. The method of proof of Theorem 1 works in any algebraic number field, while the method of proof of Theorems 2 and 3 works only in totally real fields and their totally complex quadratic extensions. The fields of these two types share the property that the trace of a square of the absolute value of any non-zero element is positive. In the case of totally complex fields, the definition of LΦ(x1 , . . . , xk ) must be modified, namely condition (∗) is to be replaced by Jfσ (x1−1 , . . . , xk−1 ) = const fσ (x1 , . . . , xk ). c

A generalization to function fields over totally real fields is also possible. The following notation is used through the paper in addition to that introduced already. 1.

|Ω| is the degree of a field Ω.

348 c


2. 3.

ζq is a primitive root of unity of order q. If Ω is a field and α ∈ Ω, α = 0, then 0 if α = ζq for some q, e(α, Ω) = maximal e such that α = ζq β e with some q and β ∈ Ω, otherwise.

4.

h(M) is the maximum of the absolute values of the elements of a matrix M (the height of M).

M T and M A are matrices transposed and adjoint to M, respectively. The same notation applies to vectors treated as matrices with one row. The elements of a vector denoted by a bold face letter are designated by the same ordinary letter with indices. Bold face capital letters represent matrices except Ω that is a field.

2. Lemma 1. Let Ω be an algebraic number field and α = 0 an element of Ω satisfying an equation f (α) = 0, where f is a polynomial. Then ⎧ 2 ∗ ⎪ ⎨20|Ω| log |Ω| log f always, (1) e(α, Ω) 25 |Ω| log f if α is not conjugate to α −1 , ⎪ ⎩ −1 (2 log 2) |Ω| log f if α is not an integer. Besides, for any algebraic number field Ω 1 ⊃ Ω (2)

e(α, Ω 1 )

|Ω 1 | e(α, Ω). |Ω|

Proof. If α is a root of unity, the lemma follows from the definition of e(α, Ω). Assume that α is not a root of unity and let (3)

α = ζq β e ,

β ∈ Ω,

e = e(α, Ω).

If α is an integer, β is also. It follows that (4)

log α = e log β ,

where α is the maximal absolute value of the conjugates of α. Now by a recent result of Blanksby and Montgomery [1] and by a slight refinement of a theorem of Cassels [3] (see p. 379 of the present paper) −1 always, 40|Ω|2 log |Ω|∗ − 1 β 1+ −1 if α is not conjugate to α −1 . 5|Ω| − 1 Hence (5)

1 40|Ω|2 log |Ω|∗ log β 5|Ω|

always, if α is not conjugate to α −1 .


c

349

[The second inequality for β and hence the second inequality for 1/ log β are not justified, the inequality (1) is nevertheless true, see the paper D6.] On the other hand α does not exceed the maximal absolute value of the zeros of f and by the inequality of Carmichael– Masson (see [5], p. 125) α f 1/2 , hence log α

(6)

1 2

log f .

The first part of the lemma follows now from (4), (5) and (6). Assume that α is not an integer and let a0 be the leading coefficient of f . Since f (α) = 0, a0 α is an integer. Therefore there exists a prime ideal p of Ω such that − ordp a0 ordp α < 0. It follows from (3) that ordp α = e ordp β and e − ordp α ordp a0 . On the other hand, taking norms N from Ω to Q we get |Ω| N (p)ordp a0 a0 , whence e ordp a0 |Ω|

log f 5 log |a0 | |Ω| < |Ω| log f , log 2 2 log 2 2

which proves (1). In order to prove (2), assume that α = ζr β1e1 ,

β1 ∈ Ω 1 ,

e1 = e(α, Ω 1 )

and take norms N1 from Ω 1 to Ω. We get a d = N1 (ζr )N1 (β1 )e1 ;

e1 e(α d , Ω),

where d = |Ω 1 |/|Ω|. Since by Lemma 1 of [9] e(α d , Ω) = de(α, Ω)

(2) follows.

Lemma 2. If Φ(x) is any irreducible polynomial not dividing x δ − x (δ = 1), α is any of its zeros, Ω = Q(α), n is an integer = 0, ν = n, 2e(α,Ω)−1 e(α, Ω)! , then can

Φ(x ν ) = Φ1 (x) · · · Φr (x)

350


implies can

J Φ(x n ) = J Φ1 (x n/ν ) · · · J Φr (x n/ν ).

c

Proof. The proof for n > 0 does not differ from the proof of Theorem 1 of [9]. The case n < 0 can be reduced to the former in view of the identity J Φ(x n ) = Ψ (x −n ), where Ψ (x) = J Φ(x −1 ). Proof of Theorem 1. Let

can

KF (x) = const

Φi (x)εi .

i=1

For each Φi we denote by αi , Ω i , νi the relevant parameters from Lemma 2 and set ν = n, max 2e(αi ,Ω i )−1 e(αi , Ω i )! , u = nν −1 . 1i

We may assume that either F 5 or |F | 3, F 3 because otherwise s = 0. Since 2m−1 m! mm and |Ω i | |F | (i = 1, . . . , ) we get by Lemma 1 ν exp 20|F |2 log |F |∗ log F (log 20|F |2 + log2 |F |∗ + log2 F ) 2 exp 10|F | log |F |∗ log F , which proves (i). (ii) is clear. In order to prove (iii) we notice that 2m1 −1 m1 ! | 2m2 −1 m2 ! for m1 m2 , thus νi | ν for i . By Lemma 2 can

Φi (x ) = νi

ri

Φij (x)

j =1

implies can

ri

can

j =1 ri

Φi (x ν ) = J Φi (x n ) =

Φij (x ν/νi ), J Φij (x n/νi ),

j =1

whence can

ri

can

i=1 j =1 ri

KF (x ) = ν

KF (x n ) =

Φij (x ν/νi )εi , J Φij (x n/νi )εi .

i=1 j =1


351

Denoting the polynomials Φij (x ν/νi ) (1 i , 1 j ri ) by F1 , . . . , Fs we obtain (iii).

3. Lemma 3. Let P (x1 , . . . , xk+1 ) = 0, Q(x1 , . . . , xk+1 ) = 0 be polynomials with complex coefficients, (P , Q) = G and P = GT , Q = GU . The resultant of T , U with respect to xi divides a certain nonvanishing minor of Sylvester’s matrix R of P , Q formed with respect to xi (|R| being the resultant of P , Q). Proof. Consider polynomials A(x), B(x), C(x) of degrees |A| > 0, |B| > 0, |C| with indeterminate coefficients a0 , . . . , b0 , . . . , c0 , . . . , the resultant D of A, B and any minor S c of order |A| + |B| + |C| of Sylvester’s matrix R of AC, BC. Since D is absolutely irreducible and prime to a0 b0 (see [6], Satz 120), we have either S = DV , where V is a polynomial in the coefficients of A, B, C, or there exist complex values of the coefficients such that D = 0 and a0 b0 c0 S = 0 (cf. [6], Satz 136). A(x) and B(x) with these coefficients have a common factor of degree > |C| and by a well known theorem ([6], Satz 114) the rank of R is less than |A| + |B| + |C|. The contradiction obtained with S = 0 proves that (7)

S = DV

for any minor S of order |A| + |B| + |C| of R. Now, if neither T nor U is constant with respect to xi we set A(xi ) = T (x1 , . . . , xk+1 ), B(xi ) = U (x1 , . . . , xk+1 ), C(xi ) = G(x1 , . . . , xk+1 ). Since (AC, BC) = C, it follows from the quoted theorem that at least one of the c minors of order |A| + |B| + |C| of R does not vanish. By (7) this minor has the property asserted in the lemma. If T , say, is constant with respect to xi and the relevant degree of U is u, the principal c minor S of order u has the said property (if u = 0 we take S = 1). c

Lemma 4. Let T (x1 , x2 ), U (x1 , x2 ) be polynomials with complex coefficients, (T , U ) = 1. The number of pairs η, ϑ such that T (η, ϑ) = U (η, ϑ) = 0 does not exceed the degree of the resultant of T , U with respect to xi (i = 1, 2). Remark. The lemma must be notorious but it is not readily found in the literature. Proof. It suffices to consider i = 2. Let t, u be the degrees of T , U with respect to x2 and for a given η let tη , uη be the degrees of T (η, x2 ), U (η, x2 ). Let R(x1 ) be Sylvester’s matrix of T , U formed with respect to x2 , R(x1 ) its determinant and R η Sylvester’s matrix of T (η, x2 ), U (η, x2 ). If tη = t, uη = u then R η = R(η), otherwise R η can be obtained from R(η) by crossing out step by step row i, column i (1 i u − uη ), row u + i, column i (u − uη < i (u − uη ) + (t − tη )). At each step all non-zero elements crossed out are in a row, thus the rank diminishes by at most one. We get rank of R η rank of R(η) − (t − tη ) − (u − uη ).

352


Now if there are kη different ϑ such that T (η, ϑ) = U (η, ϑ) = 0, T (η, x2 ), U (η, x2 ) have a common factor of degree at least kη , thus ([6], Satz 114) rank of R η tη + uη − kη . It follows that the rank of R(η) does not exceed t + u − kη , whence by differentiation (x1 − η)kη | R(x1 ). Giving η all the possible values, we obtain kη |R|.

Lemma 5. Let P (x1 , . . . , xk+1 ) = 0, Q(x1 , . . . , xk+1 ) = 0 be polynomials and S = 0 a minor of their Sylvester’s matrix formed with respect to xi (1 i k + 1). The following inequalities hold |S| 2|P | |Q|, S P 2|Q| Q2|P | . Proof. We assume without loss of generality i = k + 1 and set P =

m

m−i Pi (x1 , . . . , xk )xk+1 ,

Q=

n

n−j

Qj (x1 , . . . , xk )xk+1 .

j =0

i=0

Since m |P |, n |Q| and Sylvester’s matrix of P , Q is ⎤ ⎫ ⎡ P0 P1 . . . Pm ⎬ ⎢ . . . . . . . . . . . . . . . . . . . . . ⎥ n times ⎥ ⎭ ⎢ ⎢ P0 P1 . . . Pm ⎥ ⎥⎫ ⎢ ⎥⎬ ⎢Q0 Q1 . . . Qn ⎥ ⎢ ⎣ . . . . . . . . . . . . . . . . . . . . . ⎦ m times ⎭ Q0 Q1 . . . Qn it follows that |S| n max |Pi | + m max |Qj | 2|P | |Q|. In order to estimate S we note that + 2π + ··· S = (2π)−k 0

S(eiϕ1 , . . . , eiϕk )2 dϕ1 dϕ2 . . . dϕk

2π

0

(cf. [2], Lemma 6 of Chapter VIII), hence (8)

2 S max S(eiϕ1 , . . . , eiϕk ) . 0ϕ2π

On the other hand, for any polynomial R with integral coefficients 2 (9) max R(eiϕ1 , . . . , eiϕk ) R2 . 0ϕ2π

353


Using (8), Hadamard’s inequality and (9) we obtain n m n m iϕ1 iϕk 2 iϕ1 iϕk 2 S max Pj (e , . . . , e ) Qj (e , . . . , e ) 0ϕ2π

m j =0 m

j =0

j =0

2 max Pj (eiϕ1 , . . . , eiϕk )

0ϕ2π

Pj 2

n n

j =0

P

n n j =0

m Qj 2

j =0

2|Q|

Q

2|P |

m

2 max Qj (eiϕ1 , . . . , eiϕk )

m

0ϕ2π

2n n

Pj

j =0

2m Qj

j =0

.

Lemma 6. If an m-dimensional sublattice of the n-dimensional integral lattice contains m linearly independent vectors v 1 , . . . , v m then it has a basis of the form m

c1j v j , . . . ,

j =1

m

cmj v j ,

j =1

where 0 cij < cjj 1 (i = j ),

cij = 0 (i < j ).

Proof. The proof is obtained by a standard method (see [2], Appendix A). For a more precise result see [7]. Lemma 7. Let ki (0 i l) be an increasing sequence of integers. Let kjp − kip (1 p p0 ) be all the numbers which appear only once in the double sequence kj − ki (0 i j l). Suppose that for each p kjp − kip =

k

cpq nq ,

q=1 c

where cpq are integers, |cpq | c, c positive integer. Then either there exist integral matrices K = [κqi ]qk and Λ = [λqt ]qk il

tk

and an integral vector u such that (10)

(11)

[k1 − k0 , . . . , kl − k0 ] = uK, n = [n1 , . . . , nk ] = uΛ, l/2 h(K) k max{c2 , 2} + 2 , 0 λqt < λtt 2l−1 (q = t),

λqt = 0 (q < t)

or there exists an integral vector γ such that c

(l+1)(k−1)/2 . γ n = 0 and 0 < h(γ ) k k−1 max{kc2 , 2} + 2k

354


0 / Proof. By the assumption for each / pair i, 0 j where 0 i j l and i, j = ip , jp (1 p p0 ) there exists a pair gij , hij = i, j such that kj − ki = khij − kgij . Let us consider the system of linear homogeneous equations x0 = 0,

/ 0 i, j = i1 , j1 , . . . , ip0 , jp0 ,

xj − xi − xhij + xgij = 0, (12) xjp − xip −

k

cpq yq = 0

(1 p p0 )

q=1

satisfied by xi = ki − k0 (0 i l), yq = nq (1 q k). Let A be the matrix of the system obtained from (12) by cancelling the first equation and substituting x0 = 0 in the others, B be the matrix of the coefficients of the x’s, −Γ the matrix of the coefficients of the y’s so that A = B | − Γ in the sense of juxtaposition (the vertical line is added in order to avoid a confusion with the subtraction). We assert that (12) has at most k linearly independent solutions. Indeed, if we had k + 1 such solutions a 1 , . . . , a k+1 then taking as ξ1 , . . . , ξk+1 real numbers rationally k+1 independent we should find a set of reals ami ξm (0 i l), where all the differences m=1

would span over the rationals a space of dimension k + 1, while the differences occurring only once k+1 m=1

(amjp − amip )ξm =

k+1 m=1

ξm

k q=1

cpq am,l+q =

k q=1

cpq

k+1

am,l+q ξm

m=1

would span a space of dimension at most k contrary to the theorem of Straus [11]. It follows that the rank of A is l + , where 0 < k. If the rank of B is l then since one row of B (corresponding to i, j = 0, l) is [0, . . . , 0, 1] there exists a nonsingular c submatrix Δ of B of order l containing this row. Solving the system by means of Cramer formulae we find a system of k linearly independent integral solutions which can be written (horizontally) in the form K | Λ , where elements of K are determinants obtained from Δ by replacing one column by a column of Γ and Λ = DI k , D = |Δ|, I k is the identity c matrix of order k. By Hadamard’s inequality and an inequality for determinants with real entries (see this c collection, paper M4) l/2 l/2 |D| 2l−1 , h(K ) min{ max{c2 , 4} + 2 , (c + 1)l } max{c2 , 2} + 2 . c From K | Λ we obtain by Lemma 6 a fundamental system of integral solutions K | Λ satisfying (11). Since the system is fundamental there exists an integral vector u satisfying (10). If the rank of B is less than l, we find a system of k − linearly independent integral c solutions in the form K | Λ , where elements of Λ are up to a sign minors of A of order

355


l + . The rank of Λ is less than k, otherwise the equality BK T = Γ Λ would imply T T T T Γ = BK (Λ )−1 , A = B | − Γ = B I l | − K (Λ )−1 T

and the rank of A would be less than l, which is impossible. By Hadamard’s inequality and the inequality for determinants with real entries quoted above (l+)/2 (l+)/2 c h(Λ ) min{ max{kc2 , 4} + 2 , (kc + 1)l+ } max{kc2 , 2} + 2k . c

By a well known lemma ([2], Lemma 3 of Chapter VI) there exists an integral vector γ = 0 such that Λ γ T = 0 and (k−max{,1})/ max{,1} (l+1)(k−1)/2 c k k−1 max{kc2 , 2} + 2k . h(γ ) h(Λ )k Since n = u Λ (u not necessarily integral) we get γ n = nγ T = u Λ γ T = 0.

Remark. The proof of Straus can be transformed into a proof that (12) has at most k linearly independent solutions, which does not use any irrationalities and is in this respect nearer to the proof of Lemma 4 in [9]. Suppose that a 1 , . . . , a k+1 are solutions, a m = [0, am1 , . . . , aml , am,l+1 , . . . , am,l+k ]. There exist integers b1 , . . . , bk+1 not all zero such that k+1

bm am,l+q = 0

(1 q k).

m=1

Consider a vector a =

k+1

bm a m = [0, a1 , . . . , al , 0, . . . , 0]. It is also a solution of (12).

m=1

Set i = the least i such that ai = min aj or max aj , j

0j l

0j l

= the greatest i such that ai = min aj + max aj − ai . 0j l

0j l

/ The 0 equality aj − ai = ah − ag implies ai = ag/, a j = 0 ah , i g, j h and/ either0 i , j = g, h or kj − ki > kh − kg . It follows that i , j is identical with some ip , jp (1 p p0 ) and we get

aj − ai =

k

cpq al+q = 0.

q=1

Hence ai = 0 (0 i l + k) and k+1 m=1

bm a m = 0.

356


Lemma 8 (L8k ). Let P (x1 , . . . , xk ) = 0, Q(x1 , . . . , xk ) = 0 be polynomials and (P , Q) = G. For any integral vector n = [n1 , . . . , nk ] we have either LP (x n1 , . . . , x nk ), LQ(x n1 , . . . , x nk ) = LG(x n1 , . . . , x nk ) or |P | |Q| > 0 and there exists an integral vector β such that (13) (14)

βn = 0,

0 < h(β)
2. exp2k−5 2P

Lemma 9 (L9k ). For any polynomial F (x1 , . . . , xk ) = 0, any integral vector n = [n1 , . . . , nk ] and any irreducible factor f (x) of LF (x n1 , . . . , x nk ) either there exist an integral matrix Λ = [λqt ] of degree k, an integral vector u = [u1 , . . . , uk ] and a polynomial T (z1 , . . . , zk ) such that (15) (16)

0 λqt < λtt 2F −2 (q = t),

λqt = 0 (q < t),

n = uΛ, k k λqk λq1 T (z1 , . . . , zk ) | F zq , . . . , zq , q=1

q=1 uk

f (x) = const LT (x , . . . , x ) u1

or F 3 and there exists an integral vector γ such that (17)

γ n = 0, 2F −1 log F if k = 2, 120 2|F |∗ 0 < h(γ ) < F −1 ∗ log F if k > 2. exp2k−4 7k|F |

We prove these lemmata by induction showing first L82 and then the implications L8k → L9k (k 1), L9k → L8k+1 (k > 1). Since L81 is obvious this argumentation is sufficient. Proof of L82 . If P = GT , Q = GU and LP (x n1 , x n2 ), LQ(x n1 , x n2 ) = LG(x n1 , x n2 ) then for some ξ not conjugate to ξ −1 : T (ξ n1 , ξ n2 ) = 0 = U (ξ n1 , ξ n2 ). Let Ri be the resultant of T (x1 , x2 ), U (x1 , x2 ) with respect to xi and Si a nonvanishing minor of Sylvester’s matrix of P , Q, divisible by Ri , whose existence is asserted in Lemma 3. Set (18)

αi = ξ ni ,

Ω = Q(α1 , α2 ).

|Ω| does not exceed the number of distinct pairs η, ϑ satisfying T (η, ϑ) = U (η, ϑ) = 0 thus by Lemma 4 |Ω| |Ri | |Si |

(i = 1, 2).

357


Since ξ (n1 ,n2 ) ∈ Ω, it follows |Q(ξ )| (n1 , n2 )|Ω|. Moreover R3−i (αi ) = 0, S3−i (αi ) = 0 and if αi is not an integer or ni = 0 we get from (18) and Lemma 1 (19) |ni | e αi , Q(ξ ) (2 log 2)−1 |Q(ξ )| log S3−i (2 log 2)−1 (n1 , n2 )|Si | log S3−i . If αi is an integer and ni = 0, ξ sgn ni is also an integer. It is not conjugate to ξ − sgn ni , thus by the already quoted refinement of Theorem 1 of [3] ξ sgn ni > 1 +

1 ; 5|Q(ξ )| − 1

1 < 5|Q(ξ )|. log ξ sgn ni

On the other hand, by the inequality of Carmichael–Masson αi S3−i 1/2 ;

log αi

1 2

log S3−i .

It follows from (18) that |ni | =

log αi 5 5 < |Q(ξ )| log S3−i (n1 , n2 )|Si | log S3−i . sgn n i 2 2 log ξ

In of Lemma 5 this inequality together with (19) implies L82 on taking β = view n2 −n1 , . (n1 , n2 ) (n1 , n2 ) Proof of the implication L8k → L9k . Let F (x1 , . . . , xk ) =

I

ai x1αi1 · · · xkαik

i=0

where ai are integers = 0 and the vectors α i are all different. Let further J F (x n1 , . . . , x nk ) = f (x)g(x)

c

where f and g have integral coefficients (if necessary we may change f (x) by a constant factor without impairing the assertion of the lemma). We set f (x −1 )g(x) = c

l

ci x ki

(ci integers = 0, k0 < k1 < . . . < kl )

i=0

and consider two expressions for F (x n1 , . . . , x nk )F (x −n1 , . . . , x −nk ): n1

nk

F (x , . . . , x )F (x

−n1

,...,x

−nk

)=

I i=0

ai2 +

ai aj x nα j −nα i ,

0i,j I i=j

l ci2 + ci cj x kj −ki . f (x −1 )g(x) f (x)g(x −1 ) = i=0

0i,j l i=j

358


If for any pair i, j and nα j − nα i = 0

i = j

(20)

we have (17) with h(γ ) |F |. If no pair i, j satisfies (20), it follows that F (x n1 , . . . , x nk ) = 0 l

(21)

ci2 =

i=0

I

ai2 = F ,

l F − 1,

i=0

each number kj −ki which appears only once in the double sequence kj −ki (0 i j l) k nq dq with |dq | |F |. has a value q=1

Applying Lemma 7 with c = |F | we find either integral matrices K = [κqt ], Λ = [λqt ] and an integral vector u satisfying (15), (16) and ki − k0 =

k

h(K) < k|F |∗ F −1

κqi uq ,

q=1

or an integral vector satisfying (17) with 2F −1 log F if k = 2, 120 2|F |∗ F (k−1)/2 2 k−1 ∗ k|F | < h(γ ) < k F −1 ∗ exp2k−4 7k|F | log F if k > 2. c We notice that F 3 since otherwise LF (x n1 , . . . , x nk ) = const. Set P (z1 , . . . , zk ) =

I

ai

i=0

Q(z1 , . . . , zk ) = J

l i=0

k

k

zq

t=1 λqt αit

,

q=1

ci

k

κ

zqqi .

q=1

Clearly |P | k|F |2F −2 ,

|Q| 2k|F |∗ F −1 ,

whence (22)

|P | + |Q| 3k|F |∗ F −1 ,

|P | |Q| k 2 2F −1 |F |∗ F .

The vectors [κ1i , . . . , κki ] (0 i l) are all different since such are the numbers n n ki − k0 . Similarly, by (16) the vectors λ1t αit , . . . , λkt αit (0 i l) are all different since such are the numbers

k

t=1

αit nt . Therefore, by (21)

t=1

(23)

t=1

P = Q = F .


359

We get from L8k that either LP (x u1 , . . . , x uk ), LQ(x u1 , . . . , x uk ) = LG(x u1 , . . . , x uk ) or βu = 0 with β satisfying (14). In the former case Lg(x) = const LF (x n1 , . . . , x nk ), Lf (x −1 )g(x) = const LP (x u1 , . . . , x uk ), LQ(x u1 , . . . , x uk ) = const LG(x u1 , . . . , x uk ), LP (x u1 , . . . , x uk ) LF (x n1 , . . . , x nk ) = = const LT (x u1 , . . . , x uk ), f (x) = Lg(x) const LG(x u1 , . . . , x uk ) where T = P G−1 . In the latter case we have k 2, γn = 0

with

γ = βΛA ,

h(γ ) kh(β)h(ΛA ) k(k − 1)(k−1)/2 h(Λ)k−1 h(β) and we estimate h(γ ) separately for k = 2 and for k > 2, using (14), (15), (22), (23) and |F |∗ 2, F 3. For k = 2 we obtain h(γ ) 2h(Λ) · 5|P | |Q| log P 2|Q| Q2|P | 5 · 2F −1 · 2F +1 |F |∗ F · 12|F |∗ F −1 log F 2F −1 log F . 120 2|F |∗ For k > 2 we use the inequality

k(k − 1)(k−1)/2 h(Λ)k−1 < k k−1 2(k−1)(F −2) < exp2k−4 6k|F |∗ F −1 log F

and obtain h(γ ) k(k − 1)(k−1)/2 h(Λ)k−1 × exp2k−4 6k|F |∗ F −1 log F + log log 5k 2 2F −1 |F |∗ F + log 3 exp22k−4 6k|F |∗ F −1 log F + log 25 k 2 + F log 2|F |∗ + log 3 − 1 < exp2k−4 7k|F |∗ F −1 log F .

Proof of the implication L9k → L8k+1 (k > 1). Let P = GT , Q = GU , let Rj be the resultant of T , U with respect to xj and let Sj be a nonvanishing minor of Sylvester’s matrix of P , Q divisible by Rj , whose existence is asserted in Lemma 3. If LP (x n1 , . . . , x nk+1 ), LQ(x n1 , . . . , x nk+1 ) = LG(x n1 , . . . , x nk+1 ) then |P | |Q| > 0 and there exists an irreducible polynomial f (x) such that f (x) | LT (x n1 , . . . , x nk+1 ), LU (x n1 , . . . , x nk+1 ) .

360


Clearly for each j k + 1 f (x) | Rj (x n1 , . . . , x nk+1 ) | Sj (x n1 , . . . , x nk+1 ), where x nj does not occur among the arguments of Rj and Sj . By L9k either there exist an integral nonsingular triangular matrix Λj with nonnegative entries, an integral vector uj and a polynomial Tj such that h(Λj ) 2Sj −2 ,

(24) (25) (26)

Tj | Sj

k

[n1 , . . . , nj −1 , nj +1 , . . . , nk+1 ] = Λj uj , k λqk λq1 zq , . . . , zq , f (x) = const Tj (x uj 1 , . . . , x uj k )

q=1

q=1

or γ j [n1 , . . . , nj −1 , nj +1 , . . . , nk+1 ] = 0 with

2Sj −1 log Sj if k = 2, 120 2|Sj |∗ 0 < h(γ j ) < Sj −1 ∗ exp2k−4 7k|Sj | log Sj if k > 2.

In the latter case we have βn = 0, where 0 < h(β)

max

1j k+1

h(γ j ).

If k = 2 we obtain from Lemma 5 2Sj −1 log Sj h(β) 120 2|Sj |∗ < exp log(120 log Sj ) + (Sj − 21 ) log(16|P |2 |Q|2 + 8) < exp log log P 2|Q| Q2|P | + P 2|Q| Q2|P | log 16|P |2 |Q|2 + 8 + log 5 < exp 2P 2|Q| Q2|P | log 5|P | |Q| + log 21 . If k > 2 we have similarly h(β) exp2k−4 7k|Sj |∗ Sj −1 log Sj < exp2k−3 21 Sj log(4|P |2 |Q|2 + 2) + log log Sj + log 7k < exp2k−3 P 2|Q| Q2|P | log 5|P | |Q| + log 7k . In the former case we set uk+1 = v = [v1 , . . . , vk ], find f (x) = const LTk+1 (x v1 , . . . , x vk ), Jf (x −1 ) = const LTk+1 (x −v1 , . . . , x −vk ) and (27)

Jf (x −1 ) LTk+1 (x −v1 , . . . , x −vk ) J Tk+1 (x −v1 , . . . , x −vk ) = = . v v f (x) LTk+1 (x 1 , . . . , x k ) J Tk+1 (x v1 , . . . , x vk )


361

Let Tk+1 (z1 , . . . , zk ) =

I

ai z1αi1 z2αi2 · · · zkαik ,

i=0

where ai = 0 (0 i I ) and the vectors α i are all different. Since Sk+1 = 0, |Λk+1 | = 0 we get by (26) h(α i ) k|Sk+1 |h(Λk+1 )

(28) c

(0 i I ).

Let α i v take its minimum for i = m, maximum for i = M. We have J Tk+1 (x v1 , . . . , x vk ) = x −α m v

c

I

ai x α i v ,

i=0

(29)

J Tk+1 (x −v1 , . . . , x −vk ) = x α M v

I

ai x −α i v .

i=0

Since

Jf (x −1 )

= const f (x) we get from (27)

d(x) = am J Tk+1 (x −v1 , . . . , x −vk ) − aM J Tk+1 (x v1 , . . . , x vk ) = 0. By (29) the lowest term in d(x) is of the form ax γ v , where γ = α i − α m or α M − α i so that a = 0;

(30)

γv > 0

and by (28) h(γ ) k|Sk+1 |h(Λk+1 ).

(31) It follows that (32) c

Jf (x −1 ) J Tk+1 (x −v1 , . . . , x −vk ) aM a = ≡ + 2 x γ v mod x γ v+1 . v v 1 k f (x) J Tk+1 (x , . . . , x ) am am

By (25) |Λk+1 |γ v = (γ ΛA k+1 )[n1 , . . . , nk ] and since (33)

γ = γ ΛA k+1 = 0

we have for some j k, γj = 0. Applying (25) and (26) we find as above (34) c

bN b Jf (x −1 ) ≡ + 2 x δv j mod x δvj +1 f (x) bn bn

with (35) (36)

b = 0,

δv j > 0,

h(δ) k|Sj +1 |h(Λj +1 ).

It follows from (30), (32), (34) and (35) that γ v = δv j ,

362


which gives |Λj |γ [n1 , . . . , nk ] = |Λk+1 |δ [n1 , . . . , nj −1 , nj +1 , . . . , nk+1 ] with δ = δΛA j.

(37) Hence j −1

|Λj |γi − |Λk+1 |δi ni + |Λj |γj nj

i=1

+

k

|Λj |γi − |Λk+1 |δi−1 ni + |Λk+1 |γk nk+1 = 0,

i=j +1

which is the desired equality (13) with 0 < h(β) |Λj |h(γ ) + |Λk+1 |h(δ ). It follows from (24), (31), (33), (36), (37) and Lemma 5 that h(β) h(Λj )k k(k − 1)(k−1)/2 h(Λk+1 )k−1 h(γ ) + h(Λk+1 )k k(k − 1)(k−1)/2 h(Λj )k−1 h(δ) k 2 (k − 1)(k−1)/2 h(Λj )k h(Λk+1 )k |Sj | + |Sk+1 | k+3 < exp log k + k Sj + Sk+1 || log 2 + log |Sj | + |Sk+1 | 2 k+3 < exp log k + 2kP 2|Q| Q2|P | log 2 + log 4|P | |Q| . 2 For k = 2 we get

h(β) < exp 2P 2|Q| Q2|P | log 5|P | |Q| + log 21 ,

for k > 2 we use the inequality kx < exp2k−4 x

(x 0)

and obtain

h(β) exp 2kP 2|Q| Q2|P | + k log 4|P | |Q|k < exp2k−3 2P 2|Q| Q2|P | log 5|P | |Q| + log 7k .

Lemma 10. If Q = 0 is a polynomial, J Q(y1−1 , . . . , yk−1 ) = ±J Q(y1 , . . . , yk ) and LQ(x v1 , . . . , x vk ) = const, then (38)

βv = 0 with h(β) 2|Q|.


363

Proof. Let the degree of J Q with respect to yj be qj and aα y1α1 · · · ykαk , J Q(y1 , . . . , yk ) = where the summation is taken over all integral vectors α satisfying 0 αj qj . Clearly aq−α y1α1 · · · ykαk J Q(y1−1 , . . . , yk−1 ) = and there exist integral vectors α j and α −j (1 j k) such that αjj = qj , aα j = 0, α−jj = 0, aα −j = 0.

In view of the condition J Q(y1−1 , . . . , yk−1 ) = ±J Q(y1 , . . . , yk ) we have for some α l , α −l aα l = aq−α l ,

(39)

aα −l = −aq−α −l .

Let the product αv taken over all α for which aα = 0, attains its minimum for α = α m , maximum for α = α n . We have aα x αv , J Q(x v1 , . . . , x vk ) = x −α m v aα x −αv . J Q(x −v1 , . . . , x −vk ) = x α n v All the exponents αv are different unless (38) holds (even with h(β) |Q|). In particular, Q(x v1 , . . . , x vk ) = 0. The equality LQ(x v1 , . . . , x vk ) = const implies J Q(x v1 , . . . , x vk ) = const J Q(x −v1 , . . . , x −vk ) and by the comparison of constant terms aα n J Q(x v1 , . . . , x vk ) = aα m J Q(x −v1 , . . . , x −vk ). Comparing the leading coefficients on both sides we get aα2 n = aα2 m , i.e. aα n = ±aα m , aα x −αv . aα x αv = ±x (α m +α n )v

(40)

In particular, we have for each j k and a suitable β j aα j x α j v = ±aβ j x (α m +α n −β j )v . If α + β j − α m − α n = 0 we get again (38), otherwise αmj + αnj = αjj + βjj αjj = qj .

(41)

Similarly we have for each j k and a suitable β −j aα −j x α −j v = ±aβ −j x (α m +α n −β −j )v ; thus either (38) holds or αmj + αnj = α−jj + β−jj = β−jj qj .

364


The last inequality together with (41) implies αm + αn = q and x (α m +α n )v

aα x −αv =

aq−α x αv .

It follows now from (39) and (40) that with a suitable sign and a suitable integral α α ±l v = αv,

α = a ±l

which gives (38) again. Lemma 11. For any polynomial F (x1 , . . . , xk ) = 0 LKF (x1 , . . . , xk ) = KLF (x1 , . . . , xk ) = LF (x1 , . . . , xk ).

Proof. In view of the definition of the operations K and L it is enough to prove that for any integral vector [δ1 , . . . , δk ] = 0 and any factor Q(y1 , . . . , yk ) of J (y1δ1 · · · ykδk − 1) J Q(y1−1 , . . . , yk−1 ) = ±J Q(y1 , . . . , yk ). Supposing the contrary we apply Lemma 10 with i (1 i k). vi = 4h(δ) + 1 Since the conditions βv = 0, h(β) 2|Q| 2h(δ) imply β = 0, it follows from that lemma LQ(x v1 , . . . , x vk ) = const. On the other hand LQ(x v1 , . . . , x vk ) | L(x vδ − 1) and since all factors of x |vδ| − 1 are reciprocal we get a contradiction.

c

Lemma 12. For any polynomial F (x1 , . . . , xk ) and any integral vector n = [n1 , . . . , nk ] such that F (x n1 , . . . , x nk ) = 0 there exist an integral matrix M = [μij ] of order k and an integral vector v = [v1 , . . . , vk ] such that 0 μij < μjj exp 9k · 2F −5 (i = j ), n = vM,

(42) (43)

μij = 0 (i < j );

and either (44) c

LF

k i=1

μ yi i1 ,

k i=1

μ yi i2 , . . . ,

k

μ yi ik

can

= const

s

Fσ (y1 , . . . , yk )eσ

σ =1

i=1

implies (45)

can

LF (x n1 , . . . , x nk ) = const

s

LFσ (x v1 , . . . , x vk )eσ

σ =1

or F 3 and there exists an integral vector γ such that (46)

γ n = 0,


where

365

max 120(2|F |∗ )2F −1 log F , 8|F | exp 9 · 2F −3 if k = 2, (47) 0 < h(γ ) < exp2k−4 9k|F |∗ F −1 log F if k > 2. c If k = 2 and some LFσ (x v1 , x v2 ) in (45) are allowed to be constants then (47) can be replaced by 0 < h(γ ) < 120(2|F |∗ )2F −1 log F . Proof. If F 2 then by Lemma 11 s = 0, LF (x1n1 , . . . , xknk ) = const and it suffices to take M = I k (the identity matrix). Therefore we assume F 3. c Let S be the set of all integral matrices Λ = [λqt ] of order k satisfying (48)

0 λqt < λtt 2F −2 (q = t), n = uΛ

(49)

λqt = 0 (q < t),

with integral u.

Integral vectors m such that for all Λ ∈ S and a suitable integral vector v Λ m = vΛΛ form a module M, say. By (48) for any Λ ∈ S, |Λ| divides exp kψ(2F −2 ) = μ, c

where ψ is Chebyshev’s function. Clearly vectors [μ, 0, . . . , 0], [0, μ, . . . , 0], . . . , [0, . . . , 0, μ] belong to M. It follows from Lemma 6 that M has a basis μ1 , . . . , μk such that 0 μij < μjj μ (i = j ),

c

μij = 0 (i < j ).

Since by Theorem 12 of [8], ψ(x) < 1.04x < 98 x for all x, the matrix M satisfies (42), since n ∈ M it satisfies also (43). In order to prove the alternative (45) or (46) and (47) we set k k μ μ yi i1 , . . . , yi ik P (y1 , . . . , yk ) = F i=1

(50) can

= const

k

yiαi

i=1

Hi (x1 , . . . , xk ) =

i=1 s 1

Fσ (y1 , . . . , yk )eσ ,

σ =1 k j =1

μij xj

∂F ∂xj

(note that P = 0 since F (x n1 , . . . , x nk ) = 0). It follows s1 k k ∂P μi1 μik −1 ∂Fσ = yi yi = Hi yi , . . . , yi eσ F σ + αi P (51) ∂yi ∂yi c i=1

i=1

σ =1

366


and by (43) (52) x vi

(53)

P (x v1 , . . . , x vk ) = F (x n1 , . . . , x nk ), ∂P v1 (x , . . . , x vk ) = Hi (x n1 , . . . , x nk ). ∂yi

(44) implies J Fσ (y1−1 , . . . , yk−1 ) = ±Fσ (y1 , . . . , yk )

(54)

(σ > s).

Assume now that for some distinct , τ s1 (55) D(x) = LF (x v1 , . . . , x vk ), LFτ (x v1 , . . . , x vk ) = 1. We consider two cases: ∂F ∂Fτ 1. for some j : = 0 and = 0, ∂yj ∂yj ∂F ∂Fτ · = 0. 2. for each i: ∂yi ∂yi 1. Here Hj = 0 and we set G = (F, Hj ). It follows from (50) and (51) that s1 k k ∂P μ μ yi i1 , . . . , yi ik = const P , yj = const P Fσ−1 (y1 , . . . , yk ), G ∂y j c i=1

σ =1

i=1

where the product is taken over all σ satisfying (1 i k) we obtain from (50), (51)

∂Fσ = 0. On substituting yi = x vi ∂yj

k k μi1 vi μik vi v1 vk vj ∂P v1 vk D(x)LG x ,..., x , . . . , x ), Lx (x , . . . , x ) , LP (x ∂yj i=1

i=1

which in view of (43), (52) and (53) gives D(x)LG(x n1 , . . . , x nk ) | LF (x n1 , . . . , x nk ), LHj (x n1 , . . . , x nk ) . By (55) and Lemma 8 we have (46) with 5|F | |Hj | log F 2|Hj | Hj 2|F | 0 < h(γ ) < exp2k−5 2F 2|Hj | Hj 2|F | log 5|F | |Hj | + log 7k 2. Here we have for some h, j ∂F = 0, ∂yh

∂Fτ = 0; ∂yh

∂F = 0, ∂yj

∂Fτ = 0, ∂yj

thus Hh = 0, Hj = 0. We set G = (Hh , Hj ). It follows from (50) and (51) that

(56)

∂P e −1 yh = F Fτeτ U, ∂yh ∂P e yj = F Fτeτ −1 V , ∂yj

U ≡ 0 mod F , V ≡ 0 mod Fτ ,

if k = 2, if k > 2.

367


hence G

k

μ

yi i1 , . . . ,

i=1

On substituting yi =

k

μik

yi

e −1

= F

Fτeτ −1 (U, V )(y1 , . . . , yk ).

i=1

x vi

we obtain from (56)

k k x μi1 vi , . . . , x μik vi D(x)LG i=1

i=1

Lx vh ∂P (x v1 , . . . , x vk ), Lx vj ∂P (x v1 , . . . , x vk ) , ∂yh ∂yj

which in view of (43) and (53) gives D(x)LG(x n1 , . . . , x nk ) | LHh (x n1 , . . . , x nk ), LHj (x n1 , . . . , x nk ) . By (55) and Lemma 8 we have (46) with 5|Hh | |Hj | log Hh 2|Hj | Hj 2|Hh | 0 < h(γ ) < exp2k−5 2Hh 2|Hj | Hj 2|Hh | log 5|Hh | |Hj | + log 7k

if k = 2, if k > 2.

Since for all i: |Hi | |F |, k ∂F 2 2 2 Hi k μij xj ∂x k h(M) |F | F , j j =1

it follows in both cases that if k = 2 0 < h(γ ) < 20|F |3 log 4h(M)2 |F |2 F

< 20|F |3 log 4|F |2 F + 20|F |3 · 9 · 2F −3 < 120 2|F |∗ )2F −1 log F ,

if k > 2

0 < h(γ ) < exp2k−4 4|F | log k 2 h(M)2 |F |2 F + log log 5|F |2 + log 3 < exp2k−4 5|F | log k 2 |F |2 F + |F | · 9k · 2F −2 < exp2k−4 9k|F |∗ F −1 log F .

c

Assume, therefore, that for all distinct , τ s1 (57) LF (x v1 , . . . , x vk ), LFτ (x v1 , . . . , x vk ) = 1 and let f (x) be any irreducible factor of LF (x n1 , . . . , x nk ). By Lemma 9 either (46)–(47) = [λqt ] of order k, an integral vector c hold or there exist an integral matrix Λ u = [u1 , . . . , uk ] satisfying (48)–(49) and a polynomial T such that k k λqk λq1 (58) zq , . . . , zq , T (z1 , . . . , zk ) | F q=1

(59)

q=1 uk

f (x) = const LT (x , . . . , x ). u1

368


Since Λ ∈ S and by the choice of M: μ1 , . . . , μn ∈ M we have for some integral vectors θ 1 , . . . , θ n : μi = θ i Λ, thus ⎡ ⎤ θ1 M = ΘΛ (60) ⎢ ⎥ , Θ = ⎣ ... ⎦ . (61) u = vΘ θn Set W (y1 , . . . , yk ) = J T

k

yiϑi1 , . . . ,

i=1

k

yiϑik

.

i=1

We have by (58) and (60) W (y1 , . . . , yk ) | F

k

μ yi i1 , . . . ,

i=1

k

μ yi ik

,

i=1

by (59) and (61) f (x) = const LW (x v1 , . . . , x vk ). Since f (x) is irreducible, the last two formulae imply in view of (50) f (x) = const LF (x v1 , . . . , x vk )

(62) and since

Jf (x −1 )

for some s1

= ±Jf (x) we have by (54) s. By (57) s1 LFσ (x v1 , . . . , x vk )eσ = 1 f (x), σ =s+1

and because of the arbitrariness of f (x) s1 LFσ (x v1 , . . . , x vk )eσ = 1. LF (x n1 , . . . , x nk ), σ =s+1

Since by (50) and (52) LF (x n1 , . . . , x nk ) = const

s1

LFσ (x v1 , . . . , x vk )eσ ,

σ =1

it follows that LF (x , . . . , x ) = const n1

nk

s

LFσ (x v1 , . . . , x vk )eσ .

σ =1

Moreover, none of the LFσ (x v1 , . . . , x vk ) (σ s) is reducible since taking as f (x) any of its reducible factors we would obtain from (62) a contradiction with (57). It remains to prove that none of LFσ (x v1 , . . . , x vk ) (σ s) is constant unless (46) holds with if k = 2, 8|F | exp 9 · 2F −3 0 < h(γ ) < F −1 ∗ log F if k > 2. exp2k−4 9k|F | c


369

This follows from Lemma 10 on taking Q = Fσ , since (38) implies (46) with γ = βM A and 0 < h(γ ) kh(M A )h(β) k(k − 1)(k−1)/2 h(M)k−1 2|P | 2k 2 (k − 1)(k−1)/2 h(M)k |F | 2k 2 (k − 1)(k−1)/2 |F | exp 9k 2 2F −5 .

Remark. A comparison of Lemma 12 with the conjecture from [9] shows besides the replacement of K by L the two differences: it is not assumed that F is irreducible, it is not assumed that n1 > 0, …, nk > 0 and it is not asserted that v1 0, . . . , vk 0 (instead it is asserted that M is triangular). As to the first difference one may note the fact overlooked in [9] that if F is irreducible all the exponents eσ in (44) are 1. Indeed, in the notation of the preceding proof eσ > 1 implies ∂P ∂P ,..., Fσ (y1 , . . . , yk ) | P (y1 , . . . , yk ), ∂y1 ∂yk hence

J F (x1 , . . . , xk ), H1 (x1 , . . . , xk ), . . . , Hk (x1 , . . . , xk ) = 1.

Since |M| = 0 it follows by the definition of Hi that ∂F ∂F = 1, , . . . , xk J F (x1 , . . . , xk ), x1 ∂x1 ∂xk which for an irreducible F is impossible. As to the second difference it may be noted that the formulation with the assumption n1 0, . . . , nk 0 and the assertion v1 0, . . . , vk 0 (but M not necessarily triangular and h(M) possibly greater) is also true its proof however involves the following theorem of Schmidt [10]. c If M is a full sublattice of the integral k-dimensional lattice and M+ consists of all vectors of M with nonnegative coordinates then there exists a finite subset M0 of M+ such that every vector of M+ is a linear combination of k vectors of M0 with nonnegative integral coefficients. In the proof of Lemma 5 of [9] the truth of this theorem for k = 2 was established together with a bound for the height of the vectors of M0 in terms of M. Such a bound in the general case has been found recently by R. Lee. Proof of Theorem 2. The theorem is true for k = 1 by Lemma 12. Assume that it is true for polynomials in k − 1 variables and consider F (x1 , . . . , xk ). By Lemma 12 either there exist a matrix M and a vector v with the properties (42), (43), (45) or we have F 3 and there exists a vector γ satisfying (46), (47). In the former case the theorem holds with r = k, in the latter case n belongs to the module N of integral vectors perpendicular to γ . If γ = [0, . . . , 0, γν , . . . , γk ] with γν = 0, N contains k − 1 linearly independent vectors [1, 0, . . . , 0], . . . , [0, . . . , 1, 0, . . . , 0], [0, . . . , γν+1 , −γν , 0, . . . , 0], . . . , [0, . . . , γk , 0, . . . , −γν ] and by Lemma 6 it has a basis which written in the form of

370


a matrix Δ = [δtj ] t 1. Here we use the inequality (k − 1)2 exp 9(k − 1)2F

2 −5

c

< exp 11(k − 1)2F −5 < exp2 9k2F −1 < exp2 9k|F |∗ F −1 log F 2

and obtain

c

2 h(N) (k − 1)2 exp2k−4 9k|F |∗ F −1 log F · exp 9(k − 1)2F −5 exp22k−4 9k|F |∗ F −1 log F < exp2k−4 10k|F |∗ F −1 log F . 4. k − 1 > max(r, 2). Here we use the inequality F 2 20k log F 2k 2 |F |∗ exp2k−4 (9k|F |∗ F −1 log F ) 2F 2 < exp2k−4 (9k|F |∗ F −1 log F ) = exp2 exp2k−6 (9k|F |∗ F −1 log F ) + log 2F 2

c

< exp2k−4 (9k|F |∗ F −1 log F + 1)

372


and obtain h(N ) (k − 1)2 exp2k−4 (9k|F |∗ F −1 log F ) × exp(k−r−1)(k+r−4) (10(k − 1)|F |∗

F −1

log F )

< exp2k−4 (10k|F |∗ F −1 log F ) F 2 × exp(k−r−1)(k+r−4) 20k log F 2k 2 |F |∗ exp2k−4 (9k|F |∗ F −1 log F ) < exp2k−3 (9k|F |∗ F −1 log F + 1) × exp(k−r−1)(k+r−4)+2k−4 (9k|F |∗ F −1 log F + 1) < exp2(k−r)(k+r−3) (9k|F |∗ F −1 log F + 1) c

< exp(k−r)(k+r−3) (10k|F |∗ F −1 log F ). Proof of Corollary. Let J F (x) = a0 +

k j =1

aj x nj , where aj = 0, nj distinct > 0. Set in

Theorem 2 F (x1 , . . . , xk ) = a0 +

k

aj xj .

j =1

We have (74)

k F − 1 = f − 1,

|F |∗ = 2.

By Theorem 2, the number l of irreducible factors of Lf (x) equals the number of irreducible factors of r r r LF yiνi1 , yiνi2 , . . . , yiνik i=1

i=1

i=1

(in the notation of the theorem), hence l = 0 if f 2 and l 2rh(N ) otherwise. Thus if k = 2 we get from (i) and (74) l max 2k exp 9k · 2F −5 , max 2r exp(k−r)(k+r−3) (10k|F |∗ F −1 log F ) r 0, (f, g) = (f, h) = (g, h) = 1.

374


We have Q2 =

c

c

U2 = α

∂f/∂x ∂P ∂g/∂x1 ∂h/∂x1 1 = α +β + P = 0, ∂x1 f g h ∂h P f gh G2 = ,h , T2 = , ∂x1 f gh (∂h/∂x1 , h)

∂g h ∂f h h + βf + fg . g · ∂x1 (∂h/∂x1 , h) ∂x1 (∂h/∂x1 , h) (∂h/∂x1 , h)

Since f (ξ n1 , ξ n2 ) = g(ξ n1 , ξ n2 ) it follows T2 (ξ n1 , ξ n2 ) = U2 (ξ n1 , ξ n2 ) = 0. In any case c

(76)

Ti (ξ n1 , ξ n2 ) = Ui (ξ n1 , ξ n2 ) = 0

with suitable i.

Let Rij be the resultant of Ti , Ui with respect to xj and Sij a nonvanishing minor of Sylvester’s matrix of P , Qi divisible by Rij . Since |P | = |F |,

|Qi | |F |,

P = F ,

Qi |F |2 F

we get from Lemma 5 |Sij | 2|F |2 ,

Sij (|F | F )4|F |

(1 i, j 2).

Set Ω = Q(ξ n1 , ξ n2 ). By (76) |Ω| does not exceed the number of distinct pairs η, ϑ satisfying Ti (η, ϑ) = Ui (ϑ, η) = 0 and by Lemma 4 |Ω| |Rij | |Sij |.

c

Since ξ (n1 ,n2 ) ∈ Ω, it follows |Q(ξ )| (n1 , n2 )|Ω|. c

Moreover Ri,3−j (ξ nj ) = 0, Si,3−j (ξ nj ) = 0 and we get by Lemma 1 with Ω 1 = Q(ξ ) |nj |e ξ, Q(ξ ) e ξ nj , Q(ξ ) (n1 , n2 )e(ξ nj , Ω) (n1 , n2 )20|Ω|2 log |Ω|∗ log Si,3−j (n1 , n2 )20|Sij |2 log |Sij |∗ · 4|F | log(|F | F ) (n1 , n2 )120(2|F |∗ )2F −1 log F ,

c

which completes the proof.

Proof of Theorem 3. If F 2 then s = 0, KF (x n1 , x n2 ) = const and it suffices to take N = I 2 . Suppose therefore F 3 and assume first max{|n1 |, |n2 |} > 120(2|F |∗ )2F −1 log F . (n1 , n2 ) We apply Lemmata 12 and 13 to polynomial F and vector [n1 , n2 ]. If M = [μij ] is the


375

matrix of Lemma 12 then [n1 , n2 ] = [v1 , v2 ]M. Moreover μ

μ

μ

μ

can

KF (y1 11 y2 21 , y1 12 y2 22 ) = const

(77)

s

Fσ (y1 , y2 )eσ

σ =1

implies by Lemma 11 μ

μ

μ

μ

can

LF (y1 11 y2 21 , y1 12 y2 22 ) = const

s0

Fσ (y1 , y2 )eσ

σ =1

where

J Fσ (y1−1 , y2−1 )

= ±Fσ (y1 , y2 ) for σ s0 exclusively, and by Lemma 12

LF (x n1 , x n2 ) = const

(78)

s0

LFσ (x v1 , x v2 )eσ ,

σ =1

(x v1 , x v2 )

the polynomials LFσ are relatively prime in pairs and either irreducible or constant. By Lemma 13, KF (x n1 , x n2 ) = LF (x n1 , x n2 ), thus KFσ (x v1 , x v2 ) = LFs (x v1 , x v2 )

(σ s0 )

and we get KF (x , x ) = const n1

n2

s0

KFσ (x v1 , x v2 )eσ .

σ =1

If none of LFσ (x v1 , x v2 ) (σ s0 ) is constant we set N = M. By (42) and (43), (i) and (ii) hold. As to (iii) it remains to prove s0 = s. Supposing contrariwise that Fs (y1 , y2 ) = ±J Fs (y1−1 , y2−1 ) we obtain −μ22 μ21 μ12 −μ11 z2 , z1 z2 ). μ11 μ21 μ12 μ22 On the other hand, by (77), Fs (y1 , y2 ) divides f (y1 y2 , y1 y2 ) where f (x1 , x2 ) is a certain irreducible factor of KF (x1 , x2 ). By the assumption KF (x1 , x2 ) = LF (x1 , x2 ) μ

−μ21

D(z1 , z2 ) = J Fs (z1 22 z2

we have f (x1 , x2 ), Jf (x1−1 , x2−1 ) = 1 μ

−μ12 μ11 z2 )

, z1

and

−μ

= ±J Fs (z1

|M| |M| −|M| −|M| J F (z1 , z2 ), J F (z1 , z2 ) = 1. −μ

μ

On substituting y1 = z1 22 z2 21 , y2 = z1 12 z2 11 we infer that D(z1 , z2 ) divides |M| |M| −|M| −|M| J F (z1 , z2 ) and J F (z1 , z2 ), thus D(z1 , z2 ) = const and since the substitution is invertible (|M| = 0), Fs (y1 , y2 ) = const, a contradiction. If some LF (x v1 , x v2 ) is constant then we have by Lemma 10 (79)

max{|v1 |, |v2 |} 2|Fσ | 4|F | h(M). (v1 , v2 )

In this case we set r = 1,

) N=

n2 n1 , (v1 , v2 ) (v1 , v2 )

*

376


so that (ii) is clearly satisfied. By (42), (43) and (79) h(N ) 8|F | h(M)2 8|F | exp(9 · 2F −3 ), thus (i) holds. Finally by (78) KF (x n1 /(v1 ,v2 ) , x n2 /(v1 ,v2 ) ) = const

s0

KFσ (x v1 /(v1 ,v2 ) , x v2 /(v1 ,v2 ) )eσ ,

σ =1

(x v1 /(v1 ,v2 ) , x v2 /(v1 ,v2 ) )

where the polynomials KFσ are relatively prime in pairs and irreducible or constant simultaneously with KFσ (x v1 , x v2 ). This proves (iii). Assume now that max{|n1 |, |n2 |} (80) 120(2|F |∗ )2F −1 log F = m (n1 , n2 ) and set F (x) = J F (x n1 /(v1 ,v2 ) , x n2 /(v1 ,v2 ) ).

(81) Clearly

|F | 2|F | m and by (8) and (9) F max |F (eiϕ )|2 max |F (eiϑ1 , eiϑ2 )|2 F 2 . 0ϕ2π

0ϑ2π

Let ξ be a zero of F (x). If ξ −1 is not conjugate to ξ , then by Lemma 1 e ξ, Q(ξ ) 25 |F | log F 10|F | m log F . If ξ −1 is conjugate to ξ , then ξ is a zero of KF (x n1 /(n1 ,n2 ) , x n2 /(n1 ,n2 ) ) LF (x n1 /(n1 ,n2 ) , x n2 /(n1 ,n2 ) ) and by Lemma 13

In both cases (82) (83) Put (84)

e ξ, Q(ξ ) m. e ξ, Q(ξ ) 600(2|F |∗ )2F log2 F , log e ξ, Q(ξ ) 3F |F |∗ .

ν = n1 , n2 , max 2e(ξ,Q(ξ ))−1 e ξ, Q(ξ ) ! ,

where the maximum is taken over all zeros ξ of F (x). It follows like in the proof of Theorem 1 that KF (x ν ) = const can

s σ =1

Fσ (x)eσ

(n1 , n2 ) = νv,


377

implies KF (x (n1 ,n2 ) ) = const can

(85)

s

Fσ (x v )eσ

σ =1

(since v > 0, KFσ

(x v )

= J Fσ

= Fσ Set ) * n1 n2 N= , ν. (n1 , n2 ) (n1 , n2 ) (x v )

(x v )).

We get from (80), (82), (83) and (84) e(ξ,Q(ξ )) h(N) m max e ξ, Q(ξ ) exp 3F |F |∗ + 900(2|F |∗ )2F +1 F log2 F exp 500(2|F |∗ )2F +1 F 2 , thus (i) holds. (ii) is clear from (84). Finally by (81) KF (x ν11 , x ν12 ) = KF (x ν ),

KF (x n1 , x n2 ) = KF (x (n1 ,n2 ) )

and (iii) follows from (85).

5. Lemma 14. If k 2, aj = 0 (0 j k) are complex numbers and M = [μij ] is an integral nonsingular matrix of degree k then k k μij J a0 + aj zi j =1

i=1

is absolutely irreducible. Proof. We may assume without loss of generality that |M| > 0. Suppose that there is a factorization k k μ aj zi ij = T (z1 , . . . , zk )U (z1 , . . . , zk ), J a0 + j =1

i=1

where T = const, U = const. Setting zi =

k

μ

yh hi ,

where [μ hi ] = |M| · M −1

|M|

= T (y1 , . . . , yk )U (y1 , . . . , yk ),

h=1

we obtain (86)

a0 +

k j =1

aj yj

378


where

T = JT

k

μ yh h1 , . . . ,

h=1

U = JU

k

k

μ yh hk

h=1 μ yh h1 , . . . ,

h=1

k

μ yh hk

= const, = const.

h=1

However (86) is impossible since as follows from Capelli’s theorem already |M|

a0 + a1 y1

|M|

+ a2 y2

is absolutely irreducible (cf. [14]).

Remark. The following generalization of the lemma seems plausible. If aj = 0 (0 j k) are complex numbers and the rank of an integral matrix [μij ] il

j k

exceeds (k + 1)/2, then l k μ aj zi ij J a0 + j =1

c

i=1

is absolutely irreducible. Proof of Theorem 4. Set in Lemma 12: F (x1 , . . . , xk ) = a0 +

k

aj xj

j =1

and let M be the matrix of that lemma. Since by Lemma 14 k k μ μ yi i1 , . . . , yi ik JF c

i=1

i=1

LF (x n1 , . . . , x nk )

is irreducible, we conclude that either is irreducible or constant or γ n = 0 with if k = 2, 120(2|F |∗ )2F −1 log F 0 < h(γ ) < ∗ F −1 log F ) if k > 2. exp (9k|F | c 2k−4 If however LF (x n1 , . . . , x nk ) is constant we obtain the relation γ n = 0 from Lemma 10. k Taking into account that |F |∗ = 2, F = aj2 , we get the theorem. j =0

Proof of Theorem 5. It follows from Theorem 4 that L(ax n + bx m + c) is irreducible unless max{n, m} 2 2 2 24(a +b +c )+5 log(a 2 + b2 + c2 ). (n, m)


379

On the other hand, by Lemma 13 (with F (x1 , x2 ) = ax1 + bx2 + c) K(ax n + bx m + c) = L(ax n + bx m + c) unless max{n, m} 2 2 2 120 · 42(a +b +c )−1 log(a 2 + b2 + c2 ) (n, m) 24(a

2 +b2 +c2 )+5

log(a 2 + b2 + c2 ).

This proves the first part of the theorem. To obtain the second part we apply Theorem 3 with F (x1 , x2 ) = ax1 +bx2 +c. In view of Lemma 14 and the reducibility of K(ax n +bx m +c), the matrix N is of rank 1 and we have 2 2 2 h(N ) exp 500(2|F |∗ )2F +1 F 2 exp 24(a +b +c )+11 (a 2 + b2 + c2 )2 . Note added in proof. The original result of [1] concerning an algebraic integer α of degree n is α > 1 + (40n2 log n)−1

(n > 1).

This implies the inequality α > 1 + (40n2 log n∗ − 1)−1 used in the proof of Lemma 1 since 40n2 log(n∗ /n) > 1 for n > 1. For completeness we list below the modifications needed in [3] in order to obtain the inequality α > 1 + (5n − 1)−1 used in the same proof. Inequality (2.4) should be replaced by 1 0). Inequality (4.4) and the preceding formula should be replaced by δ = 1+

1 2 10n − 1 −1= , 5n − 1 (5n − 1)2

Π1 (δe1/e )n .

The two inequalities following (4.5) should be replaced by 2n(n−1) 1 +

1 2n(n−1) e2n/5 , 5n − 1

Π1 Π2 < (nδe1/e+2/5 )n < 1 For n = 2 the lemma is true because then α

√

2.

(n > 2).

380


References [1] P. E. Blanksby, H. L. Montgomery, Algebraic integers near the unit circle. Acta Arith. 18 (1971), 355–369. [2] J. W. S. Cassels, An Introduction to Diophantine Approximation. Cambridge Univ. Press, Cambridge 1957. [3] −−, On a problem of Schinzel and Zassenhaus. J. Math. Sci. 1 (1966), 1–8. [4] W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials. Math. Scand. 8 (1960), 65–70. [5] M. Marden, Geometry of Polynomials. Amer. Math. Soc., Providence 1966. [6] O. Perron, Algebra I. Walter de Gruyter, Berlin 1951. [7] R. Remak, Elementare Abschätzungen von Fundamentaleinheiten und des Regulators eines algebraischen Zahlkörpers. J. Reine Angew. Math. 165 (1931), 159–179. [8] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers. Illinois J. Math. 6 (1962), 64–94. [9] A. Schinzel, On the reducibility of polynomials and in particular of trinomials. Acta Arith. 11 (1965), 1–34; Errata, ibid., 491; this collection: D2, 301–332. [10] W. M. Schmidt, A problem of Schinzel on lattice points. Acta Arith. 15 (1968), 199–203. [11] E. G. Straus, Rational dependence in finite sets of numbers. Acta Arith. 11 (1965), 203–204. [12] N. Tschebotaröw, Grundzüge der Galoisschen Theorie. Übersetzt und bearbeitet von H. Schwerdtfeger, Noordhoff, Groningen–Djakarta 1950. [13] H. Tverberg, On the irreducibility of the trinomials x n ± x m ± 1. Math. Scand. 8 (1960), 121–126. [14] −−, A remark on Ehrenfeucht’s criterion for irreducibility of polynomials. Prace Mat. 8 (1964), 117–118.

Originally published in Acta Arithmetica XVI (1970), 371–392


Reducibility of lacunary polynomials II

To the memory of my teachers Wacław Sierpiński and Harold Davenport

This paper is based on the results of [6] and the notation of that paper is retained. In particular |f | is the degree of a polynomial f (x) and f is the sum of squares of the coefficients of f , supposed rational. The aim of the paper is to prove the following theorem. Theorem. For any non-zero integers A, B, and any polynomial f (x) with integral coefficients, such that f (0) = 0 and f (1) = −A − B, there exist infinitely many irreducible polynomials Ax m + Bx n + f (x) with m > n > |f |. One of them satisfies m < exp (5|f | + 2 log |AB| + 7)(f + A2 + B 2 ) . Corollary. For any polynomial f (x) with integral coefficients there exist infinitely many irreducible polynomials g(x) with integral coefficients such that 2 if f (0) = 0, f − g 3 always. One of them, g0 , satisfies |g0 | < exp (5|f | + 7)(f + 3) . The example A = 12, B = 0, f (x) = 3x 9 + 8x 8 + 6x 7 + 9x 6 + 8x 4 + 3x 3 + 6x + 5 taken from [4], p. 4 (1 ), shows that in the theorem above it would not be enough to assume A2 + B 2 > 0. On the other hand, in the first assertion of Corollary the constant 2 can probably be replaced by 1, but this was deduced in [5] from a hypothetical property of covering systems of congruences. Corollary gives a partial answer to a problem of Turán (see [5]). The complete answer would require |g0 | max{|f |, 1}. Lemma 1. If

k ν=1

aν ζlαν = 0, where aν , αν are integers, then either the sum

divided into two vanishing summands or for all μ ν k l | (αμ − αν ) exp ϑ(k). (1 )

Page 304 in this collection.

can be

382


Proof. Thisis the result of Mann [2] stated in a form more convenient for our applications. If cannot be divided into two vanishing summands, the relation = 0 is in Mann’s terminology irreducible. Then according to his Theorem 1 there are distinct primes p1 , p2 , . . . , ps where p1 < p2 < . . . < ps k and p1 p2 · · · ps th roots of unity ην such that ζlαν = ην ζ,

1, . . . , k.

Hence we get l | (αμ − αν )p1 p2 · · · ps

(1 μ ν k)

and since p1 p2 · · · ps | exp ϑ(k) the lemma follows.

Lemma 2. Let A, B, f satisfy the assumptions of the theorem and besides |f | > 0, f (x) = εAx q + ηBx r (ε = ±1, η = ±1). Then there exists an integer d such that d < exp 25 |f |

(1) and

Aζlm + Bζln + f (ζl ) = 0

(2) implies l | d. Proof. Set

d = exp ψ(|f |) exp ϑ(|f | + 3). By the inequality ϑ(x) ψ(x) < 1.04x (see [3], Theorem 12) it follows that d exp 25 |f | for |f | > 7 and for |f | 7 the same can be verified directly. Assume now (2). Setting |f | ai x i we get f (x) = i=0

S = Aζlm + Bζln +

|f |

ai ζli = 0.

i=0

The sum S can be divided into a certain number 1 of vanishing summands for which further such division is impossible. If at least one summand with k terms, say, contains at q least two terms from f (ζl ), aq ζl and ar ζlr (q = r), say, then by Lemma 1 l | (q−r) exp ϑ(k) and since q − r | exp ψ(|f |), k |f | + 3, we get l | d. If each summand contains at most one term from f (ζl ), then since each term is contained in a certain summand the number of terms in f (ζl ) is at most two. Since |f | > 0, f (0) = 0 the number of terms is exactly two, f (x) = aq x q + ar x r ,

q

and Aζlm + aq ζl = Bζln + ar ζlr = 0

(q = r).

It follows hence aq = εA, ar = ηB, ε = ±1, η = ±1; f (x) = εAx q + ηBx r , contrary to the assumption.

383

D5. Reducibility of lacunary polynomials II

Lemma 3. If A, B are integers, 0 < |A| |B|, ε = ±1, η = ±1 and q

Aζlm + Bζln + εAζl + ηBζlr = 0,

(3) then either

q

ζlm + εζl = ζln + ηζlr = 0

(4) or

B = 2θ A (θ = ±1), (5)

ζlm =

q εζl ,

=

ηζlr ,

n−q

{εθ ζl

r−q

, εηθζl

} = {ζ3 , ζ32 }

or B = θ A (θ = ±1), ζln

(6)

q

{ζlm , εζl } = {−θζln , −ηθζlr }.

Proof. Set A = (A, B)A1 , B = (A, B)B1 . By (3) q

A1 (ζlm + εζl ) = −B1 (ζln + ηζlr )

c

q

ϕ(l)

and it follows on taking norms that B1 divides the norm of ζlm + εζl . The latter can be divisible by ϕ(l)th power of a prime only when it is 0 or 2ϕ(l) . Hence we get either (4) or q B1 = ±1 or B1 = ±2, ζlm = εζl . Since |A1 | |B1 | and (A1 , B1 ) = 1 we get besides (4) the two possibilities B = 2θ A

(θ = ±1),

q

ζlm = εζl ,

q

εζl + θζln + θηζlr = 0

or B = θA

(θ = ±1),

q

ζlm + θ ζln + εζl + θηζlr = 0,

q

ζlm + εζl = 0.

Taking the complex conjugates we get in the former case −q

εζl

+ θ ζl−n + θηζl−r = 0,

in the latter case −q

ζl−m + θ ζl−n + εζl

+ θηζl−r = 0.

It follows that the elements of both sets occurring in (5) or (6) have the same non-zero sum and the same sum of reciprocals, hence the sets coincide. Lemma 4. Let A, B, f satisfy the assumptions of the theorem and besides |A| |B|; |f | = 0 or f (x) = εAx q + ηBx r , ε = ±1, η = ±1. Then there exist integers a, b, d such that (7)

d 3|f | + 3

and m > 0, n > 0, m ≡ a, n ≡ b mod d implies (8) K Ax m + Bx n + f (x) = Ax m + Bx n + f (x).

384


Proof. Assume first that f (x) = εAx q + ηBx r , where qr = 0. Since f (1) = −A − B it follows ε=1

(9)

or

η = 1.

(8) holds unless for some l we have (3). Consider separately four cases B = ±A, ±2A,

(10)

B = 2θ A

(11)

(θ = ±1),

(12)

B = −A,

(13)

B = A.

In case (10) by Lemma 3, (3) implies (4) and by (9) l ≡ 0 mod 2. We set d = 2, 1−η a = q + 1, b = r + . If m ≡ a mod d we infer from (4) ε = 1, l ≡ 2 mod 4, 2 l 1+η 1+η n≡r+ · mod l, n ≡ r + mod 2, which contradicts n ≡ b mod d. 2 2 2 1−η . In case (11) by Lemma 3, (3) implies (4) or (5). We set d = 6, a = q +1, b = r + c 2 By the argument given above, (4) is impossible. (5) is impossible also since it implies l ≡ 0, m ≡ q mod 3. If q = r = 0 it is enough to take d = 2, thus (7) holds. In case (12) by Lemma 3, (3) implies (4) or (6). Since f (1) = −A − B = 0 we have ε = −η. In view of symmetry between q and r we assume r = 0 and set 1−ε , 2 3−ε d = 4, a = q , 2 (4) implies l ≡ 0 mod 2 and d = 2,

a=q+

m≡q+ c

1+ε 2 3+ε b=q 2 b=q+

1+ε l · , 2 2

n≡

if

q ≡ 0 mod 2,

if

q ≡ 1 mod 2.

1−ε l · mod l, 2 2

hence if m ≡ a mod 2, then ε = 1, l ≡ 0 mod 4, and n ≡ 0 mod 4, contrary to n ≡ b mod d. (6) implies l ≡ 0 mod 2 and either m ≡ n mod 2 or 1−ε l 1+ε l · , n≡q+ · mod l, 2 2 2 2 hence if n ≡ b mod 2, then either m ≡ b mod 2 or ε = −1, l ≡ 0 mod 4, and m ≡ 0 mod 4, contrary to m ≡ a mod d. In case (13) by Lemma 3, (3) implies (4) or (6). In view of symmetry between q and r we assume r = 0 and set m≡

d = 2, d = 2q, d = 2q,

a = 0, b = q + 1 a=b=1 a =b =q +1

if ε = η = 1, if ε = 1, η = −1, if ε = −1, η = 1

(note that if ε = −η we have q > 0 since f (0) = 0). If ε = η = 1, (4) or (6) implies l ≡ 0 mod 2, m + n ≡ q mod 2 which is incompatible with m ≡ 0, n ≡ q + 1 mod 2.

385


If ε = 1, η = −1, (4) implies l ≡ 0 mod 2, n ≡ 0 mod 2 contrary to n ≡ b mod 2; (6) implies l ≡ 0 mod 2, m ≡ 0 mod 2 or m − n ≡ 2l mod l, q ≡ 0 mod l contrary to m ≡ a mod 2, m − n ≡ 0 mod q (q even). If ε = −1, η = 1, (4) implies l ≡ 0 mod 2, m ≡ q mod l contrary to m ≡ a mod 2; l c (6) implies l ≡ 0 mod 2, n ≡ q mod 2 or m − n ≡ 2 mod l, q ≡ 0 mod l contrary to n ≡ b mod 2, m − n ≡ 0 mod q (q even). Assume now that |f | = 0, f (x) = εA + ηB. Then by Theorem 4 of [4], (8) holds unless c

m1 + n1 ≡ 0 mod 3,

f (x) = εA = ηB,

εn1 = ηm1 ,

where m1 = m/(m, n), n1 = n/(m, n). We set d = 3, a = b = 1. If m ≡ a, n ≡ b mod d we have m + n ≡ 0 mod 3 and m1 + n1 ≡ 0 mod 3. Lemma 5. Let D = {m, n : 0 m < d, 0 n < d} and let l1 , . . . , lk be divisors of d relatively prime in pairs. Set Dlj = {m, n : 0 m < lj , 0 n < lj } (1 j k) and let S(lj ) be a subset of D such that / 0 (14) m, n ∈ S(lj ), m , n ∈ D and

/ 0 m, n ≡ m , n mod lj imply

Then

/

0 m , n ∈ S(lj ).

d −2 |S(lj )| = lj−2 S(lj ) ∩ Dlj , d

k k D S(lj ) = d −2 |S(lj )|,

−2

j =1

j =1

where |S| is the cardinality of S. Proof. Set L = l1 l2 · · · lk ,

D0 = m, n : 0 m < dL−1 , 0 n < dL−1 .

Choose integers aj such that aj ≡ 1 mod lj ,

aj ≡ 0 mod Llj−1

(1 j k).

The formula m, n ≡ m0 , n0 L +

k /

0 mj , nj aj mod d,

/

0 mj , nj ∈ Dlj

j =1

c

settles one-to-one correspondence between D and the Cartesian product D0 × Dl1 × . . . × Dlk in such a way that / 0 m, n ≡ mj , nj mod lj .

386


If χj is the characteristic function of S(lj ) then by (14) χj (m, n) = χj (mj , nj ). Hence d −2 |S(lj )| = d −2

m,n∈D

where

i

χj (m, n) = d −2

1

m0 ,n0 ∈D0

···

k

χj (mj , nj ),

is taken over all mi , ni ∈ Dli and

d −2 |S(lj )| = d −2 |D0 |

k

|Dli |

j

i=1,i=j

c

χj (mj , nj ) = lj−2 S(lj ) ∩ Dlj .

It follows further d

k D S(lj )

−2

j =1

= d −2

k

χj (m, n) = d −2

m,n∈D j =1

= d −2 |D0 |

m0 ,n0 ∈D0

k j =1

j

χj (mj , nj ) = L−2

1

···

k k

χj (mj , nj )

j =1

k k S(lj ) ∩ Dl = d −2 |S(lj )|. j j =1

j =1

Lemma 6. The following inequalities hold (15)

1+

p=3

c

(16)

(17)

∞ p p < 1.376, > 0.3435, 3 2 3 2 p − p − 2p + 1 p − p − 2p + 1 p=3

∞ p=3

c

c

∞

∞

∞ p p 1+ 3 < 1.459, > 0.4165, p − p 2 − 3p + 1 p 3 − p 2 − 3p + 1

1−

p=3

p=3

∞ 2(p2 − 1) p2 − 1 > 0.3683, > 0.3804, p(p 3 − p 2 − 3p + 1) p(p 3 − p 2 − 3p + 1) p=3

where p runs over primes. Proof. We have for p 11 and c = 2 or 3 1 1 1 c+1 p 1 c+2 < 2+ 3+ + 3+ < 3 , p2 p p4 p − p 2 − cp + 1 p p p4

387


hence ∞

p −2 +

p=11

0.3106, − 2p + 1

p > 0.3836, p 3 − p 2 − 3p + 1

hence ∞ c

p=3

log 1 +

p < 0.3190, p 3 − p 2 − 2p + 1

∞ p=3

p > 0.3435, p 3 − p 2 − 2p + 1

388


∞ c

log 1 +

p=3

p < 0.3774, p 3 − p 2 − 3p + 1

∞ p=3

p > 0.4165, p 3 − p 2 − 3p + 1

which implies (15) and (16). In order to prove (17) we notice that for p 11 2(p2 − 1) 2(p 2 − 1) 2 2 13 log 1 − > − >− 2 − 3 − 4, 3 2 3 2 p(p − p − 3p + 1) p(p − p − 3p + 1) p p p p2 − 1 1 1 3 > 2+ 3+ 4, p(p 3 − p 2 − 3p + 1) p p p hence ∞ p=11

log 1 −

∞ ∞ ∞ 2(p 2 − 1) −2 −3 p − 2 p − 13 p −4 > −2 p(p 3 − p 2 − 3p + 1) p=11

p=11

p=11

= −0.066779 − 2ε2 − 2ε3 − 13ε4 > −0.0668,

c

∞ p=11

∞ ∞ ∞ p2 − 1 −2 −3 p + p + 3 p −4 > p(p 3 − p 2 − 3p + 1) p=11

p=11

p=11

= 0.032931 + ε2 + ε3 + 3ε4 > 0.0329.

c

On the other hand, 7

log 1 −

p=3 7 p=3

c

p(p 3

2(p 2 − 1) > −0.9319, 2 − p − 3p + 1)

p2 − 1 > 0.3475, p(p 3 − p 2 − 3p + 1)

whence ∞

log 1 −

p=3

c

∞ p=3

2(p 2 − 1) > −0.9987, p(p 3 − p 2 − 3p + 1)

p2 − 1 > 0.3804, p(p 3 − p 2 − 3p + 1)


Lemma 7. Let A, B, f satisfy the assumptions of the theorem. Then there exist integers a, b, d such that (18)

d 3 exp 25 |f |

and m > 0, n > 0, m ≡ a, n ≡ b mod d implies (19) K Ax m + Bx n + f (x) = Ax m + Bx n + f (x).

389


Proof. In view of symmetry we can assume 0 < |A| |B|. In virtue of Lemma 4 we can suppose that A, B, f satisfy the assumptions of Lemma 2; set d = 2d0 , where d0 is an integer from that lemma. (18) follows from (1) and (19) holds unless we have (2) for some l | d0 . Put D = m, n : 0 m < d, 0 n < d , Dl = m, n : 0 m < l, 0 n < l , El = m, n ∈ D : Aζlm + Bζln + f (ζl ) = 0 . E El then m > 0, n > 0, m ≡ a, n ≡ b mod d implies (19). Since f (1) = If a, b ∈ l |d E −A − B we have E1 = D. We show that E = ∅ separately in each of the cases (10), l |d

(11), (12), (13). In the first two cases we use the inequality D |D \ El |, El |D| − l |d

1 2 − l −2 = 2 − > 0. d −2 El 1 − 6 l |d

1 0. l =2− 36 36 · 6 l=1

In case (12) let β be the least exponent such that f (ζ2β ) = 0 if such equality is possible, otherwise β = ∞, 2−β = 0. In the former case 2β | d, since A(ζ20β − ζ20β ) + f (ζ2β ) = 0. Set El = m, n ∈ D : m ≡ n mod l and

El

=

El \ El El ∪ El

if l = 2β or l is an odd prime, otherwise.

If l has an odd prime factor p then El ∩ Ep \ El ⊂ El ∩ Ep ⊂ El \ Ep = ∅. If l = 2α , where α < β, then by the choice of β El \ El ⊂ El \ El = ∅.

c

If l = 2α , where α β, then El ∩ E2 β \ El ⊂ El ∩ E2 β ⊂ El \ E2 β = ∅. E E E Hence El ⊂ El and it remains to estimate El . With this end we note that l |d

(21)

l |d

l |d

⎧ ⎪0 if l = 2β , ⎨ (D \ El \ E ) ∩ Dl 1 if l = 2, l ⎪ ⎩ (2, l) otherwise.

Indeed, if m, n ∈ D \ El \ El , q, r ∈ D \ El \ El we have (22)

q

A(ζlm − ζln ) + f (ζl ) = A(ζl − ζlr ) + f (ζl ) = 0;

m ≡ n, q ≡ r mod l,

thus (20) holds with B = −A, ζlm − ζln = 0. Hence in virtue of Lemma 3 q

ζlm = ζl , ζln = ζlr

q

or ζlm = −ζlr , ζln = −ζl

and (23)

m, n ≡ q, r mod l or l ≡ 0 mod 2, m, n ≡ r + l/2, q + l/2 mod l.

This gives (21) for l = 2β , 2. If l = 2β then (22) is impossible, thus D \ El \ El = ∅. Finally, if l = 2 (22) implies q ≡ r + 1 mod 2, thus (23) is satisfied by only one residue class m, n mod 2.

391


We have further

E ∩ Dl = l.

(∗) c

l

In virtue of Lemma 5 it follows from (21), (∗) and the definition of El that ⎧ −1 l + l −2 if l is an odd prime, ⎪ ⎪ ⎪ ⎨l −1 if l = 2β , d −2 |D \ El | −1 ⎪ 4 if l = 2 = 2β , ⎪ ⎪ ⎩ otherwise. (2, l)l −2 Set ordp d = op . We get ⎧ ∞ ⎪ −1 ⎪ ⎪ 2 + 21−2α = 23 ⎪ ⎪ o 2 ⎨ α=2 ∞ |D \ E2 α | < d −2 ⎪ −1 −β ⎪ + 2 + 21−2α = 4 α=1 ⎪ ⎪ ⎪ ⎩ α=2

if β = 1, 5 12

+ 2−β − 21−2β
1;

α=β

c

o2 o2 D e2 = d −2 E2 α 1 − d −2 |D \ E2 α | > α=1

1 3

= c2 ,

α=1

op op ∞ D ep = d −2 Ep α 1 − d 2 |D \ Ep α | > 1 − p −1 − p −2 − p −2α c

α=1

α=1

α=2

= On the other hand, D

El

=

l |d

D

Ep α

\

.

pα |d

1

p3

(D

− p2

− 2p + 1 = cp − 1)

p(p 2

\ El ) ∩

D

Ep α

(p > 2).

,

pα |d p/| l

D D D Ep α − Ep α , (D \ El ) ∩ El l |d

where

-

pα |d p/| l

are taken over all divisors l or d except the prime powers. op E The families of sets S(p op ) p|d ∪ S(l) and {S(p op ) p|d , where S(pop ) = Ep α , 1

and

1

pα |d

1

p/| l

α=1

S(l) = D \ El , satisfy the assumptions of Lemma 5, hence D d −2 El ep − d −2 |D \ El | ep = ep 1 − (l, 2)l −2 ep−1 l |d

> c

1

p|d

p|d

ep 1 −

∞ l=2

p|d p/| l

(l, 2)l −2

p|l

cp−1 +

p|d

pα >1

1

(p α , 2)p−2α cp−1 .

p|l

392


The function (l, 2)l −2

p|l

∞

(l, 2)l −2

l=2

cp−1 is multiplicative. Therefore

cp−1 =

p|l ∞

=3

∞

∞

p=2

α=1

1+

(p α , 2)p−2α cp−1 − 1

∞ 1 + cp−1 (p 2 − 1)−1 − 1 = 3 1+

p=3

(p

α

p=3

, 2)p−2α cp−1

pα >1

=

∞

21−2α c2−1 +

α=1

∞ ∞

p −2α cp−1

p − 1, p 3 − p 2 − 2p + 1

= 2+

p=3 α=1

∞ p=3

p . p 3 − p 2 − 2p + 1

In virtue of Lemma 6 we have ∞ ∞ p p 1+ 3 + > 0.2, 4−3 2 3 2 p − p − 2p + 1 p − p − 2p + 1 p=3

p=3

hence

D ep > 0 d −2 El > 0.2d 2 l |d

p|d

and the proof in case (12) is complete. In case (13) let β be the least positive exponent such that f (ζ2β ) = 0. c Since for β 2 β−2 A ζ22β−1 + ζ20β−1 + f (ζ2β−1 ) = 0, c we have 2β−1 | d0 , hence by the choice of d, 2β | d. We set c El = m, n ∈ D : m ≡ n mod l , El = m, n ∈ D : m ≡ n + ⎧ if l = 2α , α < β, El ⎪ ⎪ ⎪ ⎨E if l = 2β , l El = ⎪ E l \ El if l is an odd prime, ⎪ ⎪ ⎩ El ∪ El otherwise. c If l has an odd prime factor p then c

c

El ∩ Ep \ El ⊂ El ∩ Ep ⊂ El \ Ep = ∅. If l = 2α , 0 < α β, m, n ∈ El , then by the choice of β m 2Aζlm = 0 if α < β, A ζl + ζln + f (ζl ) = f (ζl ) = 0 if α = β, thus El \ El = ∅. If l = 2α , α > β, then

c

El ∩ E2 β \ El ⊂ El ∩ E2 β ⊂ E2 β ∩ E2 β = ∅.

l 2

mod l ,

393


Hence

D

and it remains to estimate |

E l |d

El ⊂

l |d

D

El

l |d

El |. With this end we note that (D \ El \ E ) ∩ Dl 2. l

c

(24)

c

Indeed, if m, n ∈ D \ El \ El , q, r ∈ D \ El \ El we have q

c

A(ζlm + ζln ) + f (ζl ) = A(ζl + ζlr ) + f (ζl ) = 0;

m ≡ n + 2l , q ≡ r +

l 2

mod l,

thus (20) holds with A = B, ζlm + ζln = 0. Hence in virtue of Lemma 3 q

ζlm = ζl , ζln = ζlr

q

ζlm = ζlr , ζln = ζl

or

and (25)

m, n ≡ q, r

or

r, q mod l.

|El ∩ Dl | l,

|E2 β ∩ D2β | = 2β .

We have further c

(26)

In virtue of Lemma 5 it follows from (24), (26) and the definition of El that if l composite = 2α (α β), 2l −2 −2 (27) d |D \ El | −1 l + 2l −2 if l prime > 2. On the other hand, since E2 β ⊂ E2 α (α < β) β D E2 α = d −2 |E2 β | = 2−β . d −2 α=1

Set ordp d = op . We get o2 o2 D E2 α d −2 |E2 β | − d −2 |D \ E2 α | 2−β = d −2 |E2 β | e2 = d −2 α=1

c

> 2−β −

∞

α=β+1

21−2α = 2−β −

1 3

· 21−2β = c2 ,

α=β+1 c

and for prime p > 2 op op ∞ D ep = d −2 Ep α 1 − d −2 |D \ Ep α | > 1 − p −1 − 2 p −2α

c

α=1

α=1

α=1

=

p3 − p 2 − 3p + 1 = cp . p(p 2 − 1)

394


If l = 2α l1 , α > 0, l1 odd > 1 then (28) d −2 (D \ El ) ∩ E2 β 21−max(β−α,0) l −2 . c

For α β the inequality follows at once from (27). In order to show it for α < β suppose that q, r ∈ D \ El and set El,l = m, n ∈ D : m, n ≡ q, r , r, q mod l1 , 1 α . El,2 α = m, n ∈ D : m, n ≡ q, r , r, q mod 2 Since m, n ∈ D \ El implies (25) we have D \ El ⊂ (D \ El,l ) ∩ (D \ El,2 α ). 1

The sets S(l1 ) = D \ El,l , 1

S(2β ) = (D \ El,2 α) ∩ E β 2

satisfy the assumptions of Lemma 5, hence d −2 |S(l1 )| = l1−2 |S(l1 ) ∩ Dl1 | 2l1−2 , 0 if q ≡ r mod 2α , −2 β −2β β d |S(2 )| = 2 |S(2 ) ∩ D2β | = −α−β 2 if q ≡ r mod 2α , d −2 (D \ El ) ∩ E2 β d −2 S(l1 ) ∩ S(2β ) = d −2 |S(l1 )| d −2 |S(2β )|, c

which implies (28). Now we have . D D D D (D \ El ) ∩ El = Epα ∩ E2p \ Epα l |d

pα |d

2p|d

1

pα |d p/| l

\

(29)

. 2

(D \ El ) ∩ E2 β ∩

D

Ep α ,

pα |d p/| 2l

D D D D Ep α ∩ E2p d −2 (D \ El ) ∩ Ep α d −2 El d −2 − l |d

pα |d

1

2p|d

−

2

pα |d p/| l

D d −2 (D \ El ) ∩ E2 β ∩ Ep α , pα |d p/| 2l

= 1, pα , 2 and 2 are where 1 and 1 are taken over all l | d such that l ≡ 1 mod 2, l taken over all l | d such that l ≡ 0 mod 2, l = 2p (p is a prime). 1 and 2 are estimated easily. Indeed, the family of sets {S(l)} ∪ {S(p op )}p|d, p/| l ,

where S(l) = Dl \ El , S(p op ) =

op D α=1

Ep α ,

395


satisfies for each l the assumptions of Lemma 5. Hence by (27) Σ1 =

1

d −2 |D \ El |

ep

p|d p/| l

ep

1

p|d

2

2

21−max(β−α,0)−2α l1−2

p|l1

p|l1

cp−1

−

∞ p=3

2−β p −2 cp−1

.

396


Now ∞

2

=

1−max(β−α,0)−2α

α=1

β

∞

21−β−α +

α=1

21−2α

α=β+1

= 21−β − 21−2β +

1 3

· 21−2β = 2c2 2e2 .

On the other hand, by (30) and Lemma 6 c

2l1−2

l1 =3 l1 odd

cp−1 = 2

∞

1+

p=3

p|l1 ∞

p −2 cp−1 =

p=3

p − 2 < 2 · 1.459 − 2 = 0.918, p 3 − p 2 − 3p + 1

∞ p=3

p(p 3

p2 − 1 > 0.3804. − p 2 − 3p + 1)

Hence (32)

Σ2
1. If β = 1 we put

pα |d

2p|d

E21 = m, n ∈ D : m, n ≡ 0, 1 mod 2 , E22 = m, n ∈ D : m, n ≡ 1, 0 mod 2 ,

so that E21 ∪ E22 = E2 ,

(33)

E21 ∩ E22 = ∅.

= ∅ we put further E 1 2 If E2 \ E2p 2p,p = E2p,p = D (p prime 3). If E2 \ E2p = ∅ let q, r ∈ E2 \ E2p . Then also r, p ∈ E2 \ E2p and in view of symmetry we may assume q, r ∈ E21 . We set 1 E2p,p = m, n ∈ D : m, n ≡ q, r mod p , 2 E2p,p = m, n ∈ D : m, n ≡ r, q mod p . implies (25) with l = 2p, we have Since m, n ∈ D \ E2p 2 1 2 = E21 ∩ E2p,p ∪ E2 ∩ E2p,p , E2 ∩ E2p D D D D (34) Ep α ∩ E2p = S1 (p op ) ∪ S2 (p op ), pα |d

2p|d

p|d

p|d

where Si (2 ) = o2

c

E2i

∩

o2 D α=1

E2 α ,

Si (p ) = op

op D α=1

i Ep α ∩ E2p,p

(p 3).

397


The family of sets {Si (p op )}p|d satisfies for i = 1, 2 the assumptions of Lemma 5, and by (33) the two summands in (34) are disjoint, hence D D d −2 Ep α ∩ E2p d −2 |S1 (p op )| + d −2 |S2 (p op )|. = pα |d

2p|d

p|d

p|d

However, by (33) o2 D E2α = d 2 e2 , |S1 (2 )| + |S2 (2 )| = |S1 (2 ) ∪ S2 (2 )| = E2 ∩ o2

o2

o2

o2

α=1 c

and for prime p > 2 op D i d −2 |Si (p op )| d −2 Ep α − d −2 |D \ E2p,p | α=1 −2

= ep − p

i ) ∩ Dp ep − p −2 . (D \ E2p,p

Hence

D D d −2 Ep α ∩ E2p (ep − p −2 ) d −2 |S1 (2o2 )| + |S2 (2o2 )| pα |d

2p|d

p|d p>2

c

=

p|d

c

>

ep ·

(1 − p −2 ep−1 ) >

p|d p>2

p|d

ep ·

∞

(1 − p −2 cp−1 )

p=3

∞ ∞ ep 1 − p −2 cp−1 + 3−2 · 5−2 c3−1 c5−1 > ep 1.014 − p −2 cp−1 . p=3

p|d

p|d

p=3

It follows from (29), (31) and (32) that ∞ ∞ D ep 0.011 − p −2 cp−1 + 2−β e2−1 p −2 cp−1 > 0.011 ep > 0. d −2 El c

l |d

p|d

If β > 1, we put

p=3

p=3

p|d

E21 = m, n ∈ D : m, n ≡ 0, 0 mod 2 , E22 = m, n ∈ D : m, n ≡ 1, 1 mod 2

so that again (33) holds. = D and q, r ∈ D \ E we set If p > 2 is a prime, E2p 2p = m, n ∈ D : m, n ≡ q, r , r, q mod p E2p,p and we assign p into class P0 , P1 or P2 according to whether q, r ∈ / E2 , q, r ∈ E21 or 2 q, r ∈ E2 , respectively. implies (25) with l = 2p, the residue classes of q, r, r, q Since m, n ∈ D \ E2p , P1 , P2 are well mod 2p are determined uniquely up to a permutation and sets E2p,p

398


defined. We have E2 ∩ E2p

Hence E2 ∩

D

⎧ 1 2 ⎪ ⎨E2 ∪ E2 ∩ E2p,p if p ∈ P1 , = E21 ∪ E22 ∩ E2p,p if p ∈ P2 , ⎪ ⎩ otherwise. E2

D D ∪ E22 ∩ E2p = E21 ∩ E2p,p E2p,p

2p|d

2p|d p∈P1

2p|d p∈P2

and D

(35)

D

Ep α ∩

pα |d

E2p =

2p|d

D

D

S1 (p op ) ∪

p|d

S2 (p op ),

p|d

where Si (2o2 ) = E2i ∩

o2 D

E2 α ,

α=1

Si (p ) = op

⎧ op D ⎪ ⎪ ⎪ Ep α ⎪ ⎨E2p,p ∩ op ⎪ D ⎪ ⎪ ⎪ Ep α ⎩

if p ∈ Pi ,

α=1

if p ∈ / Pi , p > 2.

α=1

The family of sets {Si (p op )}p|d satisfies for i = 1, 2 the assumptions of Lemma 5 and by (33) the two summands in (35) are disjoint. Hence D D d −2 Ep α ∩ E2p d −2 |S1 (p op )| + d −2 |S2 (p op )|. = pα |d

c

2p|d

p|d

p|d

On the other hand, Si (2o2 ) =

o2 D

E2 α \ (D \ E2i ) ∩ E2 β ,

α=1

d −2 |Si (2o2 )| e2 − d −2 (D \ E2i ) ∩ E2 β = e2 − 2−2β (D \ E2i ) ∩ E2 β ∩ D2β = e2 − 2−β−1 , o2 D E2 α = e2 , d −2 |S1 (2o2 )| + d −2 |S2 (2o2 )| = d −2 S1 (2o2 ) ∪ S2 (2o2 ) = d −2 α=1

whence d −2 |Si (2o2 )| =

e2 1 + (−1)i ε 2

where

|ε| 2−β e2−1 − 1.

399


Further, for p > 2 op D d −2 |Si (p op )| d −2 Ep α − d −2 D \ E2p,p α=1 −2

(D \ E2p,p ) ∩ Dp = ep − 2p −2

= ep − p

d −2 |Si (p op )| = ep

if

if

p ∈ Pi ,

p∈ / Pi .

Hence

D D Ep α ∩ E2p ep ( 21 − 21 ε)Π1 + ( 21 + 21 ε)Π2 (36) d −2 pα |d

2p|d

p|d

ep

1

2 Π1

+ 21 Π2 − 21 (2−β e2−1 − 1)|Π1 − Π2 | ,

p|d

where Πi =

p∈Pi

c

1−

p∈Pi

It follows from Lemma 6 that ∞ Π1 Π2 1− p=3

c

(1 − 2p −2 cp−1 ) =

2(p 2 − 1) . p(p 3 − p 2 − 3p + 1)

2(p 2 − 1) = C > 0.3683 p(p 3 − p 2 − 3p + 1)

and, if 3 ∈ P1 ∪ P2 , since 1 − 2 · 3−2 c3−1 =

7 15

0.627.

On the other hand, in both cases |Π1 − Π2 | 1 − C < 0.632. It follows from (29), (31), (32) and (36) that D ep 0.627 − (2−β e2−1 − 1)0.316 − 1.003 + 2−β e2−1 · 0.38 d −2 El c

l |d

p|d

p|d

c

and the proof is complete.

ep 0.004 + (2−β e2−1 − 1)0.064 > 0.004 ep > 0 p|d

Lemma 8. If A = ±B then each rational factor of Ax c + B is of degree at least c |AB|−1 .

400


Proof. Each zero of Ax c + B has absolute value |BA−1 |1/c . Hence any monic factor of Ax c + B of degree γ has constant term with absolute value |BA−1 |γ /c . If this term is rational we have in the notation in Lemma 1 of [6] c e B γ A−γ , Q = γ e(BA−1 , Q). However, since either BA−1 or B −1 A is not an integer we get by that lemma e(BA−1 , Q) = e(B −1 A, Q)

log(A2 + B 2 ) |AB|, 2 log 2

γ c|AB|−1 .

Proof of Theorem. Let a, b, d be integers from Lemma 7 and set c (37) c = a − b + d + d d −1 (b − a + |f |∗ |AB|) , 2 2 (38) e = b + d + d −bd −1 + d −1 log(f + A2 + B 2 )120(4c2 + 8)f +A +B , where as in [6] |f |∗ =

' max{|f |2 , 2} + 2 .

It follows c > |f |∗ |AB| max(|f |, 2)|AB|,

(39) (40)

e > 120(4c2 + 8)f +A

2 +B 2

log(f + A2 + B 2 ) > |f |.

We note that (41) (42)

(Ax c + B)(A + Bx c ) = x c f (x)f (x −1 ), K(Ax c + B), Kf (x) = L(Ax c + B), Lf (x) = 1.

(41) follows from (39) by comparison of degrees of both sides, (42) is obvious if A = ±B. If A = ±B any rational factor of Ax c + B is by Lemma 8 and (39) of degree greater than |f |, which implies (42). Assume now n = dt + e

(43)

(t 0)

and set in Lemmata 12 and 13 of [6] F (x1 , x2 ) = (Ax2c + B)x1 + f (x2 ),

n1 = n, n2 = 1.

The assumption of Lemma 13 is satisfied since by (41), (42) Ax2c + B f (x2 ) F (x1 , x2 ) = , = Ax2c + B, f (x2 ) c KF (x1 , x2 ) K(Ax2 + B) Kf (x2 ) Ax2c + B f (x2 ) F (x1 , x2 ) = , = . c L(Ax2c + B) Lf (x2 ) LF (x1 , x2 ) In view of (39) |F | = c > 2;

|F |∗ =

c2 + 2 ,

F = A2 + B 2 .


401

In view of (40) and (43) the numbers n1 , n2 do not satisfy any relation γ1 n1 + γ2 n2 = 0 with 0 < max{|γ1 |, |γ2 |} 120(2|F |∗ )2F log F . c

Therefore, by Lemma 12 of [6] there is an integral matrix M = [μij ] of order 2 such that (44)

0 μ21 < μ11 ,

0 = μ12 < μ22 ,

[n, 1] = [v1 , v2 ]M

(45) and (46)

s cμ μ μ μ can L (Ay2 22 + B)y1 11 y2 21 + f (y2 22 ) = const Fσ (y1 , y2 )eσ σ =1

c

implies s LFσ (x v1 , x v2 )eσ , L Ax n+c + Bx n + f (x) = const σ =1

where polynomials LFσ (σ s) are either irreducible or constant. Now by (44) and (45) μ22 = 1 and the left hand side of (46) becomes μ μ L (Ay2c + B)y1 11 y2 21 + f (y2 ) which itself is not reducible. Indeed, since c > |f | and Ay2c + B has no multiple factors (x v1 , x v2 )

c

±

f (y2 ) μ21 y2 (Ay2c + B)

is not a power in the field Q(y2 ) and by Capelli’s theorem μ

y1 11 +

f (y2 ) μ y2 21 (Ay2c + B)

is irreducible in this field. It follows that μ

μ

(Ay2c + B)y1 11 y2 21 + f (y2 ) μ (Ay2c + B)y2 21 , f (y2 ) is irreducible. Since by (42) and f (0) = 0 μ L(Ay2c + B)y2 21 , Lf (y2 ) = 1, we have on the right hand side of (46) s = 0 or s = e1 = 1. We infer that L Ax n+c + Bx n + f (x) is not reducible. By Lemma 13 of [6] we have L Ax n+c + Bx n + f (x) = K Ax n+c + Bx n + f (x) . Finally by (37), (38) and (43) n + c ≡ a, n ≡ b mod d and by Lemma 7 K Ax n+c + Bx n + f (x) = Ax n+c + Bx n + f (x), thus Ax m + Bx n + f (x) is irreducible for any m = n + c, n = dt + e (t 0). By (40)

402


we have n > |f |. On the other hand, by (18) and (37) c |f |∗ |AB| + d |f |∗ |AB| + 3 exp 25 |f | 5 exp

5

2 |f | + log |AB|

and for t = 0 we get by (18) and (38) 2 2 m = c + e c + d + log f + A2 + B 2 120(4c2 + 8)f +A +B 2 2 2 2 8 exp 25 |f | + log |AB| + 6f +A +B 108 exp(5|f | + 2 log |AB|)f +A +B < exp (5|f | + 2 log |AB| + 7)(f + A2 + B 2 ) .


Proof of Corollary. If f (0) = 0 we set g(x) = Ax n + Bx m + f (x) and apply Theorem with A = B = 1 if f (1) = −2, with A = −B = 1 if f (1) = −2. The inequality for |g0 | follows, even with f + 3 replaced by f + 2. If f (0) = 0 we set g(x) = Ax n +Bx m +f (x)+1 and apply Theorem with A = B = 1 if f (1) = −3, with A = −B = 1 if f (1) = −3. If f (x) ≡ 0 we have |f (x) + 1| = |f |, f (x) + 1 = f + 1, which implies the c inequality for |g0 |. If f (x) ≡ 0, |f | = −∞, we set g0 (x) = x.

References [1] H. T. Davis, Tables of Higher Mathematical Functions, vol. II. Principia Press, Bloomington 1935. [2] H. B. Mann, On linear relations between roots of unity. Mathematika 12 (1965), 107–117. [3] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers. Illinois J. Math. 6 (1962), 64–94. [4] A. Schinzel, On the reducibility of polynomials and in particular of trinomials. Acta Arith. 11 (1965), 1–34; Errata, ibid., 491; this collection: D2, 301–332. [5] −−, Reducibility of polynomials and covering systems of congruences. Acta Arith. 13 (1967), 91–101; this collection: D3, 333–343. [6] −−, Reducibility of lacunary polynomials I. Acta Arith. 16 (1969), 123–159; Corrigenda: Acta Arith. 19 (1971), 201; ibid. 24 (1978), 265; this collection: D4, 344–380.

Originally published in Acta Arithmetica XIX (1971), 195–201


A note on the paper “Reducibility of lacunary polynomials I” with J. Wójcik (Warszawa)

In the paper [1] mentioned in the title the first writer has left a gap in the proof of Lemma 1. The aim of this note is to fill this gap by proving a property of normal number fields which may be of independent interest. Let Ω be a number field of degree |Ω|, α ∈ Ω, α = 0. We denote by ζq a primitive qth root of unity and set following [1] 0 if α = ζq for some q, e(α, Ω) = maximal e such that α = ζq β e with suitable q and β ∈ Ω, otherwise. It is asserted in Lemma 1 of [1] that if α = 0, f (α) = 0, where f (x) = polynomial with integral coefficients and f =

m i=0

m

ai x i is a

i=0

ai2 ,

then

e(α, Ω) 25 |Ω| log f .

(1)

The proof for α not being an integer is correct. The proof for α being an integer is based on the following refinement of a result of Cassels ([1], p. 159 (1 )). If an algebraic integer β of degree n is not conjugate to β −1 then β > 1 + 1 , (2) 5n − 1 where β is the maximal absolute value of the conjugates of β. If α is an integer and α = ζq β e then β is also an integer (e > 0). However it does not follow that if α is not conjugate to α −1 then β is not conjugate to β −1 . The example ' √ 2 √ α = −1 − 2 = ζ4 ζ8 1 + 2 = ζ4 β 2 shows that even for all i ζqi β may be conjugate to ζq−i β −1 . (1 )

Page 379 in this volume.

404


Therefore the inequality (2) does not follow in an obvious way (which is assumed although not asserted in [1]) from the assumption of the lemma in question and we are not able to decide whether it follows at all. However the inequality (1) is a simple consequence (see Corollary below) of the following Theorem. If α = ζq β e is not conjugate to α −1 , β ∈ K(α) where K is a normal field of degree |K| and (|K|, q, e) = 1 then for some i, ζqi β is not conjugate to ζq−i β −1 . c

Lemma 1. Let p be a prime not dividing |K|, ζp ∈ K(α), β ∈ K(α). If σ1 , σ2 are two automorphisms of the normal closure of K(α), σ1 (α) = σ2 (α), σ1 (ζp ) = σ2 (ζp ) and σ1 (β p ) = σ2 (β p ) then σ1 (β) = σ2 (β). Proof. Set σ = σ2−1 σ1 . Let α be of degree r over K(ζp ) and let β = a0 + a1 α + . . . + ar−1 α r−1 ,

c

ak ∈ K(ζp ).

If σ (β) = β we have σ (β) = ζp β = 0. Therefore, σ (β) = ζp a0 + ζp a1 α + . . . + ζp ar−1 α r−1 , where at least one coefficient ζp ai , say, is non-zero. On the other hand, σ (β) = σ (a0 ) + σ (a1 )α + . . . + σ (ar−1 )α r−1 . Since K is normal, σ (ai ) ∈ K(ζp ). It follows that σ (ai ) = ζp ai ,

p

p

σ (ai ) = ai .

p

ai belongs, therefore, to the subfield L of K(ζp ) invariant with respect to σ . We have also ζp ∈ L, ai ∈ / L and by Abel’s theorem ai is of degree p over L. Since L ⊂ L(ai ) ⊂ K(ζp ) it follows |K(ζp )| ≡ 0 mod p and |K| ≡ 0 mod p, contrary to the assumption. Lemma 2. The theorem holds for q = 2ν . Proof. Set ζq = ζ . If e ≡ 0 (mod 2) we have for suitable i α = ζβ e = (ζ i β)e , hence ζ i β is not conjugate to ζ −i β −1 . Assume that e ≡ 0 mod 2, |K| ≡ 0 mod 2 and that for each i there exists an automorphism σi of the normal closure of K(α) such that σi (ζ i β) = ζ −i β −1 . If σi (ζ ) = ζ si we have σit (β) = ζ −i(si +1) β −1 (t odd), t t ζ si −ie(si +1) β −e for t odd, t σi (α) = t t ζ si −ie(si −1) β e for t even. t

If si ≡ −1 mod q then setting t = 1 we get σi (α) = α −1 , contrary to the assumption. This remark implies the validity of the lemma for ν = 1, 2. Indeed, if ν = 1 then

D6. A note on the paper “Reducibility of lacunary polynomials I”

405

si ≡ −1 mod 2. If ν = 2 then either s0 ≡ −1 mod 4 or s1 ≡ −1 mod 4. Otherwise s0 ≡ s1 ≡ 1 mod 4, σ0 (α) = ζβ −e = σ1 (α),

σ0 (β 2 ) = β −2 = σ1 (β 2 )

and by Lemma 1 σ0 (β) = σ1 (β),

β −1 = −β −1 ,

which is impossible. In order to prove the lemma for ν 3 we prove first the three assertions: (i)

if ν 3, 2ν−3 i − j then either si ≡ −1 mod 4 or sj ≡ −1 mod 4,

(ii) if ν 4, 2ν−4 i − j , si ≡ 3 mod 8 then sj ≡ −1 mod 4, (iii) if 1 < l < ν, 2ν−1−l i − j , 2l si + 1 then 2l /|| sj + 1. (i) We have σi σj (α) = ζ si sj −j e(sj +1)si +ie(si +1) β e . Since 2(i − j )(si + 1)(sj + 1) ≡ 0 mod 2ν it follows that σi σj (α) = σj σi (α) and σi σj (β 2 ) = σj σi (β 2 ), thus by Lemma 1 σi σj (β) = σj σi (β). Hence (i−j )(si +1)(sj +1) ≡ 0 mod 2ν , (si +1)×(sj +1) ≡ 0 mod 8 and either si ≡ −1 mod 4 or sj ≡ −1 mod 4. (ii) If si ≡ 3 mod 8 then 2(i − j )(si + 1)(sj + 1) ≡ 0 mod 2ν , σi σj (α) = σj σi (α), σi σj (β 2 ) = σj σi (β 2 ) thus by Lemma 1 σi σj (β) = σj σi (β), (i − j )(si + 1)(sj + 1) ≡ 0 mod 2ν , (si + 1)(sj + 1) ≡ 0 mod 16, sj ≡ −1 mod 4. (iii) Let si ≡ −5αi , sj ≡ −5αj mod 2ν . If 2l si + 1, 2l sj + 1 then 5αi ≡ 2l + 1 mod 2l+1 , 5αj ≡ 2l + 1 mod 2l+1 , hence 2l−2 αi , 2l−2 αj , (αj , 2ν−2 ) | αi . It follows that the congruence tαj ≡ αi mod 2ν−2 is soluble. Its root t must be odd since otherwise l − 1 ν − 2 implies αi ≡ tαj ≡ 0 mod 2l−1 , which is impossible. Thus we have for an odd t sjt ≡ si mod 2ν . Since 2(i − j )(si + 1) ≡ 0 mod 2ν we get σjt (α) = σi (α), σjt (β 2 ) = σi (β 2 ), thus by Lemma 1 σjt (β) = σi (β), (i − j )(si + 1) ≡ 0 mod 2ν , which is impossible. Let l be the greatest integer not exceeding ν such that si ≡ −1 mod 2l for suitable i. Since si ≡ −1 mod 2ν for all i we have l < ν and by (i) l > 1. Consider first the case ν = 3. Then q = 8, l = 2, si ≡ 3 mod 8. Taking in (iii) l = 2, ν = 3, j = i − 1 or i + 1 we get si−1 ≡ si+1 ≡ 1 mod 4. If si−1 ≡ si+1 mod 8 then 2[(i + 1) − (i − 1)](si−1 + 1) ≡ 4(si−1 + 1) ≡ 0 mod 8, hence σi−1 (α) = σi+1 (α), σi−1 (β 2 ) = σi+1 (β 2 ) and by Lemma 1 σi−1 (β) = σi+1 (β), 2(si−1 + 1) ≡ 0 mod 8 which is impossible. In the remaining cases: si−1 ≡ 1, si ≡ 3, si+1 ≡ 5 mod 8 and si−1 ≡ 5, si ≡ 3, si+1 ≡ 1 we have si−1 si si+1 ≡ −1, si ≡ 3 mod 8.

406


It follows σi−1 σi σi+1 (α) = ζ si−1 si si+1 −e[(i+1)si−1 si si+1 +si−1 si −si−1 +i−1] β −e = ζ −1−e(−i−1+3si−1 −si−1 +i−1) β −e = ζ −1−2e(si−1 −1) β −e = ζ −1 β −e = α −1 , since 2(si−1 − 1) ≡ 0 mod 8. Consider next the case ν 4. Let 2ν−1−l i − j , 2k sj + 1. For suitably chosen j we have k > 1. Indeed, if l > 2 then ν − 1 − l < ν − 3, 2ν−1−l i − j − 2ν−3 and by (i) sj ≡ −1 mod 4 or sj +2ν−3 ≡ −1 mod 4. If l = 2 and sj ≡ 1 mod 4 then by (i) sj +2ν−3 ≡ −1 mod 4, by (ii) sj +2ν−4 ≡ −1 mod 4, because sj +2ν+3 ≡ −1 mod 8 and again by (ii) sj ≡ −1 mod 4, a contradiction. By the definition of l we have k l and by (iii) k = l. Thus we get 1 < k < l < ν. Let si ≡ −5αi mod 2ν ,

sj ≡ −5αj mod 2ν .

It follows 5αi ≡ 1 mod 2l , 2l−2 | αi ,

5αj ≡ 2k + 1 mod 2k+1 ;

2k−2 αj ;

(αj , 2ν−2 ) | αi

and the congruence tαj + αi ≡ 0 mod 2ν−2 is soluble. Since k < l its root t must be even. Thus we have for an even t si sjt ≡ −1 mod 2ν . Since 2ν−1−l | i − j , 2l | si + 1 we get j (si sjt − si ) + i(si + 1) ≡ (i − j )(si + 1) ≡ 0 mod 2ν−1 and σi σjt (α) = ζ si sj −e[j (si sj −si )+i(si +1)] β −e = α −1 , t

t

which is impossible. Lemma 3. The theorem holds for q = pν , where p is an odd prime. Proof. Set ζq = ζ . If e ≡ 0 mod p we have for suitable i α = ζβ e = (ζ i β)e ,

hence ζ i β is not conjugate to ζ −i β −1 . Assume that e ≡ 0 mod p, |K| ≡ 0 mod p and that for each i there exists an automorphism σi of the normal closure of K(α) such that σi (ζ i β) = ζ −i β −1 . If i ≡ 0 mod pν−1 , t is odd then σit (β) = ζ −i(si +1) β −1 , t

c

σit (α) = ζ si β −e . t

407

D6. A note on the paper “Reducibility of lacunary polynomials I”

We have σ0 σpν−1 (α) = ζ

s0 spν−1

σ0 σpν−1 (ζp ) = σpν−1 σ0 (ζp ),

β e = σpν−1 σ0 (α),

σ0 σpν−1 (β p ) = σpν−1 σ0 (β p ),

thus by Lemma 1 σ0 σpν−1 (β) = σpν−1 σ0 (β);

pν−1 (s0 + 1)(spν−1 + 1) ≡ 0 mod pν .

Hence either s0 ≡ −1 mod p or spν−1 ≡ −1 mod p and we assume without loss of pν−1

generality that the first congruence holds. Then s0 which is impossible.

pν−1

≡ −1 mod pν , σ0

(α) = α −1 ,

Proof of the theorem. We proceed by induction with respect to ω(q) the number of distinct prime factors of q. If ω(q) = 0 the theorem is trivial. If ω(q) = 1 the theorem holds in virtue of Lemmata 2 and 3. Suppose that the theorem holds for ω(q) < n and consider ω(q) = n > 1. Let p be the least prime factor of q, q = p ν q1 , e = pμ e1 , where p /| q1 e1 . If μ = 0 then for suitable ζpν , ζq1 we have α = ζq β e = ζq1 (ζpν β)e . Since ζq = αβ −e ∈ K(α) we have ζpν ∈ K(α), ζpν β ∈ K(α) and by the inductive assumption for some i −1 ζ −1 , ζqi1 ζpν β is not conjugate to ζq−i ν β 1 p

which was to be proved. If μ > 0, by the assumption |K| ≡ 0 mod p. We have for suitable ζpν , ζq1 μ

α = ζq β e = ζpν (ζq1 β e1 )p . Since ζq1 β e1 ∈ K(α) it follows by the inductive assumption that for some i α1 = ζpi ν ζq1 β e1 is not conjugate to αi−1 . However we have for suitable j j

α1 = ζq1 (ζpν β)e1 1−ipμ pμ

j

and ζpν β ∈ K(ζpν , α1 ). Indeed, β ∈ K(α) and α = ζpν a1 . Moreover since (|K|, q, e) = 1 and p is the least prime factor of q |K(ζpν )|, q1 , e1 pν−1 (p − 1)|K|, q1 , e1 = 1. By the inductive assumption we have for some k: −j

ζqk1 ζpν β is not conjugate to ζq−k ζpν β −1 , 1 j

c

which was to be proved.

Remark 1. An examination of the proof shows that if K is abelian the assumption (|K|, q, e) = 1 can be replaced by (|K|, q, e) ≡ 1 mod 2.

408


Corollary. If α ∈ Ω is an integer not conjugate to α −1 , α = 0 and f (α) = 0, where f is a polynomial with integer coefficients then e(α, Ω) 25 |Ω| log f . Proof. Suppose first that Ω = Q(α), set e(α, Ω) = e and let β be an integer of Q(α) such that α = ζq β e . It follows that (3) log α = e log β . By the inequality of Carmichael–Masson α f 1/2 we have (4) log α 21 log f . On the other hand, by the theorem ζqi β is not conjugate to ζq−i β −1 for some i. Since ζqi β ∈ Q(α) we have by the inequality (2) β = ζ i β > 1 + q

1 , 5|Q(α)| − 1

thus log β >

1 5|Q(α)|

and by (3) and (4) e 25 |Q(α)| log f . In the general case we use the following assertion of Lemma 1 of [1] independent of (1). If Ω 1 ⊃ Ω then e(α, Ω 1 )

|Ω 1 | e(α, Ω). |Ω|

Taking Ω 1 = Ω, Ω = Q(α) we get |Ω| · 5 |Q(α)| log f = 25 |Ω| log f . |Q(α)| 2 Remark 2. The recent unpublished work of C. J. Smyth on the product of conjugates of an algebraic integer lying outside the unit circle allows one to strenghten considerably the Corollary and the relevant results of [1]. This will form an object of another paper (see the paper D7 in this volume). e(α, Ω)

Reference [1] A. Schinzel, Reducibility of lacunary polynomials I. Acta Arith. 16 (1969), 123–159; Corrigenda: Acta Arith. 19 (1971), 201; ibid. 24 (1978), 265; this collection: D4, 344–380.

Originally published in Acta Arithmetica XXXIV (1978), 227–265


Reducibility of lacunary polynomials III

1. The present paper is a sequel to [11] and the notation of that paper is used throughout. All the polynomials considered are supposed to have integral coefficients unless stated to the contrary. Reducibility means reducibility over the rational field Q. If f (x1 , . . . , xk ) = 0 is a polynomial then can

f (x1 , . . . , xk ) = const

s

fσ (x1 , . . . , xk )eσ

σ =1

means that the polynomials fσ are irreducible and prime to each other. k xiαi , where f is a polynomial prime to x1 x2 · · · xk If Φ(x1 , . . . , xk ) = f (x1 , . . . , xk ) i=1

and αi are integers, then we set J Φ(x1 , . . . , xk ) = f (x1 , . . . , xk ). A polynomial g such that J g(x1−1 , . . . , xk−1 ) = ±g(x1 , . . . , xk ) is called reciprocal. Let can

J Φ(x1 , . . . , xk ) = const

s

fσ (x1 , . . . , xk )eσ .

σ =1

We set KΦ(x1 , . . . , xk ) = const LΦ(x1 , . . . , xk ) = const

1

2

fσ (x1 , . . . , xk )eσ , fσ (x1 , . . . , xk )eσ ,

where 1 is extended over all fσ that do not divide J (x1δ1 · · · xkδk − 1) for any [δ1 , . . . , δk ] = 0, 2 is extended over all fσ that are non-reciprocal. The leading coefficients of KΦ and LΦ are assumed equal to that of J Φ. In particular for k = 1 KΦ(x) equals J Φ(x) deprived of all its cyclotomic factors and is called the kernel of Φ. For a polynomial F (x1 , . . . , xk ), F is the sum of squares of the coefficients of F ; if F = 0, |F | is the maximum of the degrees of F with respect to xi (1 i k), Ω(F ) is the

410


number of irreducible factors of F counted with multiplicities, expk and logk denote the kth iteration of the exponential and the logarithmic function respectively. τ (n) is the number of divisors and Ω0 (n) the number of prime divisors of n counted with multiplicities. The main object of [11] has been to describe the canonical factorization of LF (x n1 , . . . , x nk ) for any fixed polynomial F and a variable integral vector [n1 , . . . , nk ]. The much more difficult problem of describing the factorization of KF (x n1 , . . . , x nk ) has been solved only for k = 1 and for k = 2 provided KF (x1 , x2 ) = LF (x1 , x2 ), in particular if F (x1 , x2 ) = a0 + a1 x1 + a2 x2 . For k > 2 even the simplest case 3 F (x1 , x2 , x3 ) = a0 + aj xj (n1 < n2 < n3 ) has been settled only under very rej =1

strictive assumption about the aj ’s (see [3]). The aim of the present paper is to improve and to extend the above results in several ways. First, due to the recent progress made by Blanksby and Montgomery [1] and by Smyth [20] in the problem of distribution of the conjugates of an algebraic integer on the plane it has been possible to improve the result on KF (x n ) mentioned above. We have

c

Theorem 1. For any polynomial F (x) = 0 such that KF (x) = const, for some positive integer c(F ) and for any positive integer n there exist positive integers ν and u such that (i) (ii)

ν | c(F ), n = uν, can

(iii) KF (x ν ) = const

s σ =1

can

Fσ (x)eσ implies KF (x n ) = const

s σ =1

Fσ (x u )eσ .

Moreover,

2/3 1/3 log 2|KF | + log2 F log c(F ) |KF | log(2|KF |) log F

and if KF (x) = LF (x) log c(F )

(

log F log2 F + O log F log3 F , 2 log ϑ0

where ϑ0 is the real zero of x 3 − x − 1. In any case s log 2 + o(1) Ω KF (x n ) = eσ min |KF |τ (n), |KF |1+o(1) exp log2 F . log3 F σ =1

Examples will be given to show that in the first of the estimates for log c(F ) the exponent 1/3 cannot be lowered, in the second the main term is best possible and the estimate for Ω(KF (x n ) is sharp with respect to all three parameters involved n, |KF | and F . Corollary 1. For any polynomial F (x) such that F (0) = 0 and any n we have Ω F (x n ) |F |τ (n).

411

D7. Reducibility of lacunary polynomials III

Corollary 2. For any binomial b(x) we have log 2 + o(1) Ω Kb(x) exp log2 b . log3 b Corollary 3. For any trinomial t (x) we have log t Ω Kt (x) . 2 log ϑ0 + o(1)

The corollaries are of interest because for a general f (x) only Ω Lf (x) polynomial is known to be O log f and the estimates for Ω Kf (x) depend upon |f | (see [15] and the Corollary to Lemma 1). Coming back to [11] it is possible to improve also the estimates given there for the case k > 1. The improvements are however not drastic and the new estimates are probably still far from best possible, thus we shall not go into the matter. On the other hand using the result of E. Gourin [4] it is possible to describe the canonical factorization of KF (x1n1 , . . . , xknk ) for any k. We have Theorem 2. For any polynomial F (x1 , . . . , xk ) = 0 and any positive integers n1 , . . . , nk there exist positive integers ν1 , . . . , νk and v1 , . . . , vk such that (iv) νj | c(F ) (1 j k), (v) nj = νj vj (1 j k), can

(vi) KF (x1ν1 , . . . , xkνk ) = const can KF (x1n1 , . . . , xknk ) =

const

s σ =1 s σ =1

Fσ (x1 , . . . , xk )eσ implies Fσ (x1v1 , . . . , xkvk )eσ .

The constant c(F ) = 0 is effectively computable. This theorem is clearly stronger than its analogue with L in place of K announced in [13]. In the latter case it follows by the method of [11] that log c(F ) 9 · 2F −5 ; it seems however that this estimate is far from the best possible. Turning again to polynomials in one variable we shall obtain Theorem 3. Let k 3, aj (0 j k) be non-zero integers and n1 < n2 < . . . < nk positive integers. Then either there exist integers γj (1 j k) such that k

(vii)

γj nj = 0

j =1

and (viii)

k k 2 0 < max |γj | < exp2k−4 k2 j =0 aj +2 log aj2 1j k

j =0

412


or all primitive irreducible factors of f (x) = a0 +

k j =1

aj x nj except a single simple one

are reciprocal and monic, moreover if |a0 | + |ak |

(ix)

k−1

|aj |

j =1

they are cyclotomic and if for some g, h k ag2 ≡ ah2 mod g.c.d. aj · g.c.d. aj

(x)

j =g,h

0j k

none whatever. Besides, (ix) and (x) imply Ω Kf (x)/Lf (x) Ω0 (a0 , ak ) and Ω f (x)/Lf (x) Ω0 (a0 , ak ) , respectively. This is a refinement of Theorem 4 of [11]. A refinement in a different direction has been given in [14]. The last part of the paper is concerned with quadrinomials. Improving the results of [3] we shall prove Theorem 4. Let aj (0 j 3) be non-zero integers and (xi) either |a0 | + |a3 | |a1 | + |a2 | or for some g, h 3 ag2 ≡ ah2 mod g.c.d. aj · g.c.d. aj 0j 3

or |a0 | = |a3 |, |a1 | = |a2 |. Then for any quadrinomial q(x) = a0 +

3 j =1

j =g,h

aj x nj (0 < n1 < n2 < n3 ) that is not

reciprocal we have one of the following four possibilities. (xii) Kq(x) is irreducible. (xiii) q(x) can be divided into two parts that have the highest common factor d(x) being a non-reciprocal binomial. K q(x)d −1 (x) is then irreducible unless q(x)d −1 (x) is a binomial. (xiv) q(x) can be represented in one of the forms k(T 2 − 4T U V W − U 2 V 4 − 4U 2 W 4 ) = k(T − U V 2 − 2U V W − 2U W 2 )(T + U V 2 − 2U V W + 2U W 2 ), (1)

k(U 3 + V 3 + W 3 − 3U V W ) = k(U + V + W )(U 2 + V 2 + W 2 − U V − U W − V W ), k(U 2 + 2U V + V 2 − W 2 ) = k(U + V + W )(U + V − W ), where k = ±(a0 , a1 , a2 , a3 ) and T , U , V , W are monomials in Z[x]. The factors on the right hand side of (1) have irreducible kernels.

413


(xv) nj = vνj (1 j 3); v and νj are positive integers, ν3 < exp2 12 · 2q log q 3 and K a0 + aj x νj is reducible. j =1

Moreover 3 s can aj x νj = const Fσ (x)eσ K a0 + σ =1

j =1

implies can

Kq(x) = const

s

Fσ (x v )eσ .

σ =1

Besides s Ω q(x) = eσ σ =1

1 1 log q. + 2 log ϑ0 2 log 2

The condition (xi) is fulfilled for about 82% of quadruples (a0 , a1 , a2 , a3 ) of height H → ∞. Since a rule for obtaining the canonical factorization of binomials is contained in Theorem 1 (and a more practical one in Lemma 5 below), Theorem 4 gives a satisfactory 3 description of the canonical factorization of the kernel of q(x) = a0 + aj x nj (0 < n1 < j =1

n2 < n3 ) for all those quadruples (a0 , a1 , a2 , a3 ) provided only q(x) is not reciprocal. The factorization of q(x)/Kq(x) can be obtained easily by means of the results of Mann [8]. We content ourselves with stating the following Corollary 4. A non-reciprocal quadrinomial q(x) = a0 +

3 j =1

aj x nj (0 < n1 < n2 < n3 )

satisfying (xi) is reducible if and only if we have one of the conditions (xii)–(xv) or q(x) can be divided into two parts with the highest common factor equal to x δ ± 1 or finally a0 +

3

aj ζ nj /(n1 ,n2 ,n3 ) = 0,

where ζ 6 = 1.

j =1

A real enigma is the reducibility of reciprocal quadrinomials. A new idea seems to be needed even to solve the following simple Problem. Given a, b with |a| = |b| do there exist infinitely many quotients r such that for suitable integers m, n: m/n = r and K(ax m+n + bx m + bx n + a) is reducible? The proofs of Theorems 1, 2, 3 and 4 are given in Sections 2, 3, 4, 5 respectively. Before proceeding to the proofs we call the attention of the reader to an error in [11] repeated also in [12]. At the bottom of p. 133 in [11] certain inequalities for determinants are said

414


to follow from Hadamard’s inequality. Now the inequalities in question are true but need not follow from Hadamard’s inequality(1 ).

2.

c

In addition to the notation introduced in §1 we shall use the following: ζq is a primitive root of unity of order q, Xq is the qth cyclotomic polynomial. If Ω is a field and α ∈ Ω, α = 0, then 0 if α = ζq for some q, e(α, Ω) = maximal e such that α = ζq β e with some q and β ∈ Ω otherwise; 0 if α = ζq for some q, E(α, Ω) = maximal n such that α = ϑ n , ϑ ∈ Ω(ζn ) otherwise. For a given polynomial f =

l(f ) =

k j =0

k

aj x j

|aj |,

C(f ) = (a0 , a1 , . . . , ak ).

j =0

c

Small bold face letters denote vectors, capital bold face letters denote matrices except Ω that is a field. NΩ2 /Ω1 is the norm from Ω2 to Ω1 or from Ω2 (x) to Ω1 (x). Lemma 1. Let αi (i = 1, . . . , ) be a system of pairwise not conjugate zeros of Kf , where f is a polynomial, and let εi be the multiplicity of αi . Then ' (2) εi e αi , Q(αi ) 26|Kf | log(7|Kf |) log f , i=1

(2 ) where the sum x 3 − x − 1.

log f εi e αi , Q(αi ) , 2 log ϑ0

is taken over all αi not conjugate to αi−1 and ϑ0 is the real zero of

Proof. Let us consider the product P = |a0 |

|α|,

f (α)=0, |α|>1

where a0 is the leading coefficient of f . By the inequality of Landau [6] (3) (1 )

P < f 1/2 . The explanation is omitted, since in this edition [11] has been corrected.


415

On the other hand, let can

Kf (x) = c

fiεi (x),

i=1

where fi (αi ) = 0 and fi is primitive. We have P =c

(4)

i=1

|ai |εi

(j )

|αi |εi ,

(j )

|αi |>1

(j )

where αi runs over the conjugates of αi and ai is the leading coefficient of fi . We shall show that ⎧ , Q(α ) e α i i ⎪ ⎨exp always, (j ) 52|fi | log 7|fi | |αi | > (5) |ai | ⎪ ⎩ e(αi ,Q(αi )) (j ) |αi |>1 if αi is not conjugate to αi−1 . ϑ0 Since αi is not a root of unity, we have by the definition of e αi , Q(αi ) (6) αi = ζq β e , β ∈ Q(αi ), e = e αi , Q(αi ) . If αi is not an integer we use an argument due to J. Wójcik and set αi = μ/ν, (μ) = dm, (ν) = dn, where d, m, n are ideals of Q(αi ) and (m, n) = 1. By Gauss’s Lemma the |f i | (j ) ν x − μ(j ) is primitive, N denoting the norm from Q(αi ) to Q. polynomial N(d)−1 j =1

Since it is also irreducible it coincides with fi up to a sign. It follows that ai = ±N d−1 N ν = ±N n. By (6) n = re and |ai | = N re 2e thus (5) holds. If αi is an integer, β is also. We have (j ) (7) |αi | = |β (j ) |e . c

(j )

|αi |>1

|β (j ) |>1

By the theorem of Blanksby and Montgomery [1] 1 1 |β (j ) | > 1 + > exp 52|fi | log 6|fi | 52|fi | log 6|fi | + 1 c |β (j ) |>1 1 > exp , 52|fi | log 7|fi | which together with (7) gives the first part of (5). If αi is not conjugate to αi−1 then by the result of [18] applied with K = Q, ζqr β is not conjugate to ζq−r β −1 for a suitable r. By Smyth’s theorem [20] |β (j ) | = |(ζqr β)(j ) | ϑ0 , |β (j ) |>1

|β (j ) |>1

which together with (7) gives the second part of (5).

416


Now (3), (4) and (5) give

e αi , Q(αi ) εi < 21 log f , 52|fi | log 7|fi | i=1 log f εi e αi , Q(αi ) < . 2 log ϑ0

(8)

The inequality (2 ) follows at once. In order to prove (2) let us notice that

52|fi | log 7|fi | 52|Kf | log 7|Kf |.

i=1

Since ' εi e αi , Q(αi ) =

( e αi , Q(αi ) εi · 52εi |fi | log 7|fi | 52|fi | log 7|fi |

(2) follows from (8) by the Schwarz inequality. Corollary. We have

26|Kf | log(7|Kf |) log f , log f Ω(Lf ) < . 2 log ϑ0

Ω(Kf )
0 is an integer, ν = n, E α, Q(α) then can

Φ(x ν ) = Φ1 (x) · · · Φr (x) implies can

Φ(x n ) = Φ1 (x n/ν ) · · · Φr (x n/ν ). Moreover r |Φ|τ (ν). Proof. Since Φ is irreducible, Φ(x) and hence also Φ(x ν ) has no multiple factors. Clearly Φj (x n/ν ) (1 j r) are prime to each other and to prove the first assertion of the lemma we have only to show that they are irreducible. Let fj (x) be an irreducible factor of Φj (x n/ν ). Clearly (25)

fj (x) | Φ(x n ).

We now use the following Lemma of Capelli (cf. [21], p. 289): if (26)

xn − α =

l

gi (x)

i=1

is the canonical factorization of x n − α in Ω = Q(α) then (27)

can

Φ(x n ) = const

l

NΩ/Q gi (x).

i=1

It follows from (25) and (27) that for some i l (28)

const fj (x) = NΩ/Q gi (x).


421

On the other hand, it follows from (26) and Lemma 3 that gi (x) = Gi (x n/ν ),

(29)

where Gi (x) ∈ Ω[x]. By (28), (29) and the choice of fj (30)

const fj (x) = NΩ/Q Gi (x n/ν ) | Φj (x n/ν ),

thus NΩ/Q Gi (x) | Φj (x). Since Φj is irreducible Φj (x) = const NΩ/Q Gi (x), thus by (30) Φj (x n/ν ) = const fj (x) and by the choice of fj (x), Φj (x n/ν ) is irreducible. To prove the second assertion of the lemma we first remark that by (27) r = l.

(31)

By the definition of E(α, Ω) = E we have E > 0 or α is a root of unity. In the former case α = ϑ(ζE )E ,

(32)

where

ϑ ∈ Ω[x].

Let the Galois group G of Ω(ζE )/Ω be represented as a subgroup J of the multiplicative group E of reduced residues mod E, so that j

J = {j ∈ E : ∃ g ∈ G ζE = g(ζE )}.

(33)

For any j ∈ J we have by (32) j

ϑ(ζE )E = α = ϑ(ζE )E , hence j

e(j )

ϑ(ζE ) = ζE ϑ(ζE )

(34) for a suitable integer e(j ). On the other hand, by (32)

xν − α =

ν

x − ζνi ϑ(ζE )E/ν

i=1

and taking norms from Ω(ζE , x) to Ω(x) (35)

(x − α) ν

|G |

=

ν

NΩ(ζE )/Ω x − ζνi ϑ(ζE )E/ν ,

i=1

where |G | is the order of G . c The ith factor on the right hand side is a power of a polynomial irreducible over Ω E/ν with the exponent equal to the number ni of those elements of G that leave x − ζνi ϑ(ζE )

422


invariant. By (33) we have j ni = j ∈ J : ζνij ϑ(ζE )E/ν = ζνi ϑ(ζE )E/ν and by (34)

ni = j ∈ J : ij + e(j ) ≡ i mod ν .

(36)

Comparing the number of factors irreducible over Ω on both sides of (35) we get by (26), (31) and (36) r|G | =

ν

ni =

i=1

ν j ∈ J : ij + e(j ) ≡ i mod ν i=1

1 i ν : ij + e(j ) ≡ i mod ν = (ν, j − 1) (ν, j − 1). c

j ∈J

j ∈J

(j,E)=1

Now |G | = [Ω(ζE ) : Ω] by Lemma 4

ϕ(E) , |Φ|

(ν, j − 1) = τ (ν)ϕ(E),

(j,E)=1

and it follows that r |Φ|τ (ν). It remains to consider the case where α is a root of unity. We have then, for a suitable q, Φ(x) = const Xq (x). Let now n = n1 n2 , where every prime factor of n1 divides q and (n2 , q) = 1. It follows from the identity Xq (x n ) = Xqn1 d (x) d |n2

and from the irreducibility of cyclotomic polynomials that r τ (n2 ) τ (n) = τ (ν). c

In the next three lemmata we use the notation m(x) =

l.c.m. (p − 1) for any

p|x, p prime

positive integer x. Lemma 6. For any integer x > 1 either there exist three positive integers x1 , x2 , x3 such that (37) xm(x) | x1 m(x1 ), x2 m(x2 ), x3 m(x3 ) √ √ √ √ c and x1 + x2 + x3 < x or x = q α r β s, where q, r are primes, s is an integer, r < 50, s < 50, α > 0, β 0. Proof. Let q be the greatest prime factor of x, x = q α y, q /| y. If y 50, but (q − 1, y) = 1

423


we set x1 = q α , x2 = y, x3 = 1 and get √ √ √ √ √ √ √ √ x − x1 − x2 − x3 q −1 y−1 −2> 3−1 50 − 1 − 2 > 0. If (q − 1, y) > 1 let r be a common prime factor of y and q − 1 y = r β s,

q − 1 = r γ t,

r /| st.

If either r 50 or s 50 we set x1 = q α ,

x2 = r β+γ ,

x3 = r β st,

easily verify (37) and get √ √ √ √ x − x1 − x2 − x3 q α/2 r β/2 s 1/2 − q α/2 − r (β+γ )/2 − r β/2 (st)1/2 √ 1 1 1 1 √ 2 > x 1 − β/2 1/2 − − γ 1/2 > x 1 − √ − √ > 0. 1/2 r s (st) (r ) 2 50 √ γ The case st = 1 is excluded since q − 1 = r implies r = 2 < 50. Lemma 7. If xi are positive integers and

j √

xi

√ x then

i=1

log l.c.m. xi m(xi ) x 1/3 (log x)2/3 . i=1,...,j

Proof. Let M = max l.c.m. xi m(xi ), where the maximum is taken over (finitely many) i=1,...,j

integral points (x1 , . . . , xj ) satisfying xi > 1,

j √

√ x. Let (x10 , . . . , xk0 ) be a point

xi

i=1

j √

in which the maximum is attained with the least value of

xi . By Lemma 6 we have

i=1 β

xi0 = qiαi ri i si (i = 1, . . . , k), where qi , ri are primes and αi > 0, ri < 50, si < 50. It follows that (38)

β

M = l.c.m. xi0 m(xi0 ) l.c.m. qiαi (qi − 1) l.c.m. ri i l.c.m.(ri − 1)si m(si ) 1ik

1ik

1ik

1ik

β

l.c.m. qiαi (qi − 1) l.c.m. ri i . 1ik

1ik

β

Since ri i x and ri < 50 we have (39)

β

log l.c.m. ri i π(50) log x = 15 log x. 1ik

Similarly (40)

log l.c.m. qiαi (qi − 1) 2n log x, i=1,...,k

where n is the number of distinct terms among qiαi (i = 1, . . . , k). Let n1 , n2 be the number k √ α /2 of distinct terms with αi = 1 and αi 2, respectively. The condition qi i x i=1

424 implies

D. Polynomials in one variable n1 i=1

1/2

pi

n2 √ √ 1/2 x, where pi is the ith prime and Pi x, where Pi is the ith

i=1

perfect power with an exponent 2. Using pi " i log i and Pi " i 2 we get √ √ 3/2 n1 (log n1 )1/2 x and n22 x. Hence n = n1 n2 x 1/3 (log x)−1/3 ,

(41)

and the lemma follows from (38)–(41). Lemma 8. If xi are positive integers and

j

xi x then

i=1

log l.c.m. xi m(xi ) i=1,...,j

x log x + O x log2 x .

Proof. Let M = max l.c.m. xi m(xi ), where the maximum is taken over all integral points 1ij

(x1 , . . . , xj ) satisfying xi > 1,

j i=1

xi x, and let (x10 , . . . , xk0 ) be a point in which the

maximum is attained with the least value of

k

xi . Since

√

√ √ √ x1 + x2 + x3 < x implies

i=1

x1 + x2 + x3 < x it follows from Lemma 6 that β

xi0 = qiαi ri i si ,

where qi , ri are primes and ri < 50, si < 50, αi > 0

and as in the proof of Lemma 7 we find (42)

log M < log l.c.m. qiαi (qi − 1) + 15 log x + O(1). 1ik

Now by the classical result of Landau ([7], §61) if

k

xi x then

i=1

log l.c.m. xi

√ x log x + O x

log l.c.m. qiαi

√ x log x + O x .

1ik

hence (43)

1ik

In order to estimate l.c.m. (qi − 1) we divide the primes qi into two classes C1 i=1,...,k

and C2 assigning qi to C1 if qi − 1 has a prime factor ti between a = log x/ log2 x and b = x log2 x and to C2 otherwise. Since qi − 1 x log2 x ti log x qi ∈C1

we have by the quoted Landau’s result log l.c.m. qi ∈C1

√ qi − 1 x log2 x + O x ti


425

and (44)

qi − 1 ti b √ √ b+O + x log2 x + O x = 2 x log2 x + O x . log b

log l.c.m.(qi − 1) log l.c.m. ti + log l.c.m. qi ∈C1

qi ∈C1

qi ∈C1

In order to estimate l.c.m.(qi − 1) we may assume without loss of generality that qi ∈C2

C2 = {q1 , . . . , qn } and q1 < q2 < . . . < qn . By the upper sieve theory (see [5], p. 134, Theorem 4.2) the number C2 (t) of qi ∈ C2 , qi t satisfies for t b t 1 t log2 x C2 (t) 1− . log t p log t log x a (L1 /L2 )1/5 or not, and adding the bounds obtained for log((3Dω Δ)Δ+4 ) in these cases, we obtain 8n log 2 Sω (T ) < T 8/9 exp 25(L1 )4/5 (L2 )1/5 + 1250m4 log(m + 1) + . 9(m + 1) Case 3. dω = 2. In this case, by Lemma 3, Dω 2m. We take 1/6 E = max 3m, 21 L1 and apply Theorem 5 of [1] to the polynomial Pω (t, t E + y), assumed irreducible over C (if it is reducible over C the opening argument in the proof of Lemma 1 applies). As in Case 2 we may take N = (m + 1)2n H T E > T E + (m + 1)2n H T m (note that the degree of Pω (t, t E + y) is 2E).

F3. Parameter in Hilbert’s Irreducibility Theorem

857

Observe that the condition N > exp(26 E 6 ) (an assumption of the theorem in question) is equivalent to (m + 1)2n H T E > exp max{(6m)6 , log H } and is satisfied provided T exp(2(6m)5 ), as we are assuming. The mentioned theorem gives Sω (T ) < N 1/(2E) exp 12 2E log N log log N . Now 2E (log N )1/6 , and log log N (log N )1/6 , since log N log T 366 by assumption. Hence ( log N 3/2 log log N Sω (T ) < exp 1 + 12(2E) 2E log N 1/6 1 log N 1 + 12 exp 2E log N 2 log N 2n log 2 5/6 exp T 2/3 exp 2L1 + . 3E 9m Observe now that since H > 20 we have 4/5 1/5

L1 L2

5/6

< 2L1 .

Using this inequality in the first two cases, comparing the three estimates and summing over ω, an operation which at most multiplies the bound by 2n , we obtain 5/6 2/5 S(T ) T 8/9 exp 50L1 + 250m4 log(m + 1) + 4(m + 1)2 n + 5nL1 . We have still to take into account the solutions of a0 (t ∗ )D(t ∗ ) = 0, but these are at most 2m(n + 1) < exp((m + 1)2 n) in number. This concludes the proof. Proof of Theorem. Let m, n, H be as in the statement of the Theorem, and let T satisfy the lower bound in the statement of Lemma 4. Then the total number R of positive integers t ∗ T such that at least one of the polynomials Fi (t ∗ , x) is reducible over Q satisfies R hT 8/9 exp 50(log H )5/6 + 250m4 log(m + 1) + 5(m + 1)2 n + 5n(log H )2/5 . To find a suitable value of t ∗ T it thus suffices that this quantity is less than T , which holds if T > h9 exp 450(log H )5/6 + 2250m5 + 45(m + 1)2 n + 45n(log H )2/5 . Combining this with the lower bound necessary for an application of Lemma 4, we obtain the Theorem. Remark. It is obviously possible by changing the splitting into cases to obtain a corresponding theorem, with different numerical values for the coefficients appearing in the final estimate.

858

F. Hilbert’s Irreducibility Theorem

References [1] E. Bombieri, J. Pila, The number of integral points on arcs and ovals. Duke Math. J. 59 (1989), 337–357. [2] S. D. Cohen, The distribution of Galois groups and Hilbert’s irreducibility theorem. Proc. London Math. Soc. (3) 43 (1981), 227–250. [3] P. Dèbes, Parties hilbertiennes et progressions géométriques. C. R. Acad. Sci. Paris Sér. I Math. 302 (1986), 87–90. [4] −−, Hilbert subsets and S-integral points. Manuscripta Math. 89 (1996), 107–137. [5] K. Dörge, Einfacher Beweis des Hilbertschen Irreduzibilitätssatzes. Math. Ann. 96 (1927), 176–182. [6] M. Eichler, Zum Hilbertschen Irreduzibilitätssatz. Math. Ann. 116 (1939), 742–748. [7] E. Fogels, On the abstract theory of primes III. Acta Arith. 11 (1966), 293–331. [8] M. Fried, On Hilbert’s irreducibility theorem. J. Number Theory 6 (1974), 211–231. [9] H. Hasse, Number Theory. Springer, Berlin 1980. [10] D. Hilbert, Über die Irreducibilität ganzer rationaler Functionen mit ganzzahligen Coefficienten. J. Reine Angew. Math. 110 (1892), 104–129; Ges. Abhandlungen II, Springer, Berlin 1970, 264–286. [11] E. Inaba, Über den Hilbertschen Irreduzibilitätssatz. Jap. J. Math. 19 (1944), 1–25. [12] E. Landau, Sur quelques théorèmes de M. Petrovitch relatifs aux zéros des fonctions analytiques. Bull. Soc. Math. France 33 (1905), 1–11. [13] F. Mertens, Über die Zerfällung einer ganzen Funktion einer Veränderlichen in zwei Faktoren. Sitzungsber. K. Akad. Wiss. Wien 120 (1911), Math. Naturwiss. Cl., 1485–1502. [14] P. Roquette, Nonstandard aspects of Hilbert’s Irreducibility Theorem. In: Model Theory and Algebra (A memorial tribute to Abraham Robinson), Lecture Notes in Math. 498, Springer, Berlin 1975, 231–275. [15] A. Schinzel, Selected Topics on Polynomials. University of Michigan Press, Ann Arbor 1982. [16] C. L. Siegel, Über einige Anwendungen diophantischer Approximationen. Abh. Preuss. Akad. Wiss. Phys.-Math. Kl. Nr. 1 (1929); Ges. Abhandlungen I, Springer, Berlin 1966, 209–266. [17] Th. Skolem, Untersuchungen über die möglichen Verteilungen ganzzahliger Lösungen gewisser Gleichungen. Kristiania Vid. Selskab. Skrifter I (1921), No. 17. [18] V. G. Sprindzhuk, Diophantine equations with prime unknowns. Trudy Mat. Inst. Steklov. 158 (1981), 180–196 (Russian). [19] M. Yasumoto, Algebraic extensions of nonstandard models and Hilbert’s irreducibility theorem. J. Symbolic Logic 53 (1988), 470–480.


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18:34 Uhr

Seite 1

Heritage of European Mathematics

Advisory Board Michèle Audin, Strasbourg Ciro Ciliberto, Roma Ildar A. Ibragimov, St. Petersburg Wladyslaw Narkiewicz, Wroclaw Peter M. Neumann, Oxford Samuel J. Patterson, Göttingen


5.3.2007

18:34 Uhr

Seite 2

Andrzej Schinzel in 2007


5.3.2007

18:34 Uhr

Seite 3

Andrzej Schinzel

Selecta Volume II Elementary, Analytic and Geometric Number Theory

Edited by Henryk Iwaniec Wladyslaw Narkiewicz Jerzy Urbanowicz


5.3.2007

18:34 Uhr

Seite 4

Author: Andrzej Schinzel Institute of Mathematics Polish Academy of Sciences ul. Śniadeckich 8, skr. poczt. 21 00-956 Warszawa 10 Poland

Editors: Henryk Iwaniec Department of Mathematics Rutgers University New Brunswick, NJ 08903 U.S.A. [email protected]

Władysław Narkiewicz Institute of Mathematics University of Wrocław pl. Grunwaldzki 2/4 50-384 Wrocław Poland [email protected]

Jerzy Urbanowicz Institute of Mathematics Polish Academy of Sciences ul. Śniadeckich 8, skr. poczt. 21 00-956 Warszawa 10 Poland [email protected]

2000 Mathematics Subject Classification: 11, 12

ISBN 978-3-03719-038-8 (Set Vol I & Vol II) The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.helveticat.ch. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2007 European Mathematical Society Contact address: European Mathematical Society Publishing House Seminar for Applied Mathematics ETH-Zentrum FLI C4 CH-8092 Zürich Switzerland Phone: + 41 (0)44 632 34 36 Email: [email protected] Homepage: www.ems-ph.org Printed in Germany 987654321

Contents

Volume 2 G. Arithmetic functions Commentary on G: Arithmetic functions by Kevin Ford . . . . . . . . . . . . . . . . . . . . . . G1 On functions ϕ(n) and σ (n) . . . . . . . . . . . . . . . . G2 Sur l’équation ϕ(x) = m . . . . . . . . . . . . . . . . . . G3 Sur un problème concernant la fonction ϕ(n) . . . . . . . G4 Distributions of the values of some arithmetical functions with P. Erd˝os . . . . . . . . . . . . . . . . . . . . . . G5 On the functions ϕ(n) and σ (n) with A. Makowski . . . . . . . . . . . . . . . . . . . . G6 On integers not of the form n − ϕ(n) with J. Browkin . . . . . . . . . . . . . . . . . . . . .

859 . . . .

. . . .

. . . .

. . . .

861 866 871 875

. . . . . . . . .

877

. . . . . . . . .

890

. . . . . . . . .

895

H. Divisibility and congruences Commentary on H: Divisibility and congruences by H. W. Lenstra jr. . . . . . . . . . . . . . . . . . . . . . . H1 Sur un problème de P. Erd˝os . . . . . . . . . . . . . . . . . . . H2 On the congruence a x ≡ b (mod p) . . . . . . . . . . . . . . . H3 On the composite integers of the form c(ak + b)! ± 1 . . . . . H4 On power residues and exponential congruences . . . . . . . . H5 Abelian binomials, power residues and exponential congruences H6 An extension of Wilson’s theorem with G. Baron . . . . . . . . . . . . . . . . . . . . . . . . . H7 Systems of exponential congruences . . . . . . . . . . . . . . H8 On a problem in elementary number theory with J. Wójcik . . . . . . . . . . . . . . . . . . . . . . . . . H9 On exponential congruences . . . . . . . . . . . . . . . . . . . H10 Une caractérisation arithmétique de suites récurrentes linéaires avec Daniel Barsky et Jean-Paul Bézivin . . . . . . . . . . . H11 On power residues with M. Skałba . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

899 . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

901 903 909 912 915 939

. . . . . . . . . . . .

971 975

. . . . . . . . . . . .

987 996

. . . . . . 1001 . . . . . . 1012

vi I.

I1 I2 I3 I4 I5 I6 J.

J1 J2 J3 J4 J5

Contents

Primitive divisors Commentary on I: Primitive divisors by C. L. Stewart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On primitive prime factors of a n − bn . . . . . . . . . . . . . . . . . . . . On primitive prime factors of Lehmer numbers I . . . . . . . . . . . . . . On primitive prime factors of Lehmer numbers II . . . . . . . . . . . . . . On primitive prime factors of Lehmer numbers III . . . . . . . . . . . . . Primitive divisors of the expression An − B n in algebraic number fields . . An extension of the theorem on primitive divisors in algebraic number fields Prime numbers Commentary on J: Prime numbers by Jerzy Kaczorowski . . . . . . . . . . . . . . . . . . . . . . . . . . Sur certaines hypothèses concernant les nombres premiers with W. Sierpiński . . . . . . . . . . . . . . . . . . . . . . . . . . . . Remarks on the paper “Sur certaines hypothèses concernant les nombres premiers” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A remark on a paper of Bateman and Horn . . . . . . . . . . . . . . . . On two theorems of Gelfond and some of their applications Section 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On the relation between two conjectures on polynomials . . . . . . . . .

K. Analytic number theory Commentary on K: Analytic number theory by Jerzy Kaczorowski . . . . . . . . . . . . . . . . . . . . . . K1 On Siegel’s zero with D. M. Goldfeld . . . . . . . . . . . . . . . . . . . . . . K2 Multiplicative properties of the partition function with E. Wirsing . . . . . . . . . . . . . . . . . . . . . . . . . K3 On an analytic problem considered by Sierpiński and Ramanujan K4 Class numbers and short sums of Kronecker symbols with J. Urbanowicz and P. Van Wamelen . . . . . . . . . . . . L.

L1 L2 L3 L4

Geometry of numbers Commentary on L: Geometry of numbers by Wolfgang M. Schmidt . . . . . . . . . . . . . . . . . . . A decomposition of integer vectors II with S. Chaładus . . . . . . . . . . . . . . . . . . . . . . . A decomposition of integer vectors IV . . . . . . . . . . . . . A property of polynomials with an application to Siegel’s lemma On vectors whose span contains a given linear subspace with I. Aliev and W. M. Schmidt . . . . . . . . . . . . . . .

1031 1033 1036 1046 1059 1066 1090 1098 1103

. 1105 . 1113 . 1134 . 1142 . 1145 . 1154 1193

. . . . . 1195 . . . . . 1199 . . . . . 1211 . . . . . 1217 . . . . . 1224 1245

. . . . . . 1247 . . . . . . 1249 . . . . . . 1259 . . . . . 1274 . . . . . . 1288

vii

Contents

M. Other papers Commentary on M: Other papers by Stanisław Kwapień . . . . . . . . . . . . . . . . . . . . The influence of the Davenport–Schinzel paper in discrete and computational geometry by Endre Szemerédi . . . . . . . . . . . . . . . . . . . . . . M1 Sur l’équation fonctionnelle f [x + y · f (x)] = f (x) · f (y) avec S. Gołab . . . . . . . . . . . . . . . . . . . . . . . . . M2 A combinatorial problem connected with differential equations with H. Davenport . . . . . . . . . . . . . . . . . . . . . . M3 An analogue of Harnack’s inequality for discrete superharmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . M4 An inequality for determinants with real entries . . . . . . . . . M5 Comparison of L1 - and L∞ -norms of squares of polynomials with W. M. Schmidt . . . . . . . . . . . . . . . . . . . . . .

1303 . . . . . . 1305

. . . . . . 1311 . . . . . . 1314 . . . . . . 1327 . . . . . . 1338 . . . . . . 1347 . . . . . . 1350

Unsolved problems and unproved conjectures Unsolved problems and unproved conjectures proposed by Andrzej Schinzel in the years 1956–2006 arranged chronologically Publication list of Andrzej Schinzel

1365 . . . . 1367 1375

viii

Contents

Volume 1 A.


1

Commentary on A: Diophantine equations and integral forms by R. Tijdeman . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Sur les nombres de Mersenne qui sont triangulaires avec Georges Browkin . . . . . . . . . . . . . . . . . . . . . . . . . .

11

A2

Sur quelques propriétés des nombres 3/n et 4/n, où n est un nombre impair

13

A3

Sur l’existence d’un cercle passant par un nombre donné de points aux coordonnées entières . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

A1

A4 A5 A6

Sur les sommes de trois carrés . . . . . . . . . . . . . . . . . . . . . . . . ϑ On the Diophantine equation nk=1 Ak xk k = 0 . . . . . . . . . . . . . . .

18 22

Polynomials of certain special types with H. Davenport and D. J. Lewis . . . . . . . . . . . . . . . . . . . .

27

A7

An improvement of Runge’s theorem on Diophantine equations . . . . . .

36

A8

On the equation y m = P (x) with R. Tijdeman . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

Zeta functions and the equivalence of integral forms with R. Perlis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

A10 Quadratic Diophantine equations with parameters with D. J. Lewis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

54

A11 Selmer’s conjecture and families of elliptic curves with J. W. S. Cassels . . . . . . . . . . . . . . . . . . . . . . . . . . .

62

A12 Families of curves having each an integer point . . . . . . . . . . . . . . .

67

A13 Hasse’s principle for systems of ternary quadratic forms and for one biquadratic form . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

A14 On Runge’s theorem about Diophantine equations with A. Grytczuk . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93

A15 On sums of three unit fractions with polynomial denominators . . . . . . .

116

= + k in a finite field A16 On equations with M. Skałba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

124

B.

127

A9

y2

xn

Continued fractions Commentary on B: Continued fractions by Eugène Dubois . . . . . . . . . . . . . . . . . . . . . . . . . . . .

129

B1

On some problems of the arithmetical theory of continued fractions . . . .

131

B2

On some problems of the arithmetical theory of continued fractions II . . .

149

B3

On two conjectures of P. Chowla and S. Chowla concerning continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161

ix

Contents

C. Algebraic number theory Commentary on C: Algebraic numbers by David W. Boyd and D. J. Lewis . . . . . . . . . . . . . . . . . . . . C1 A refinement of two theorems of Kronecker with H. Zassenhaus . . . . . . . . . . . . . . . . . . . . . . . . . . . . C2 On a theorem of Bauer and some of its applications . . . . . . . . . . . . C3 An extension of the theorem of Bauer and polynomials of certain special types with D. J. Lewis and H. Zassenhaus . . . . . . . . . . . . . . . . . . . C4 On sums of roots of unity. (Solution of two problems of R. M. Robinson) . C5 On a theorem of Bauer and some of its applications II . . . . . . . . . . . C6 On the product of the conjugates outside the unit circle of an algebraic number C7 On linear dependence of roots . . . . . . . . . . . . . . . . . . . . . . . . C8 On Sylow 2-subgroups of K2 OF for quadratic number fields F with J. Browkin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C9 A class of algebraic numbers . . . . . . . . . . . . . . . . . . . . . . . . C10 On values of the Mahler measure in a quadratic field (solution of a problem of Dixon and Dubickas) . . . . . . . . . . . . . . . . . . . . . . . . . D.

Polynomials in one variable Commentary on D: Polynomials in one variable by Michael Filaseta . . . . . . . . . . . . . . . . . . . . . . . . . . D1 Solution d’un problème de K. Zarankiewicz sur les suites de puissances consécutives de nombres irrationnels . . . . . . . . . . . . . . . . D2 On the reducibility of polynomials and in particular of trinomials . . . D3 Reducibility of polynomials and covering systems of congruences . . . D4 Reducibility of lacunary polynomials I . . . . . . . . . . . . . . . . . D5 Reducibility of lacunary polynomials II . . . . . . . . . . . . . . . . . D6 A note on the paper “Reducibility of lacunary polynomials I” with J. Wójcik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D7 Reducibility of lacunary polynomials III . . . . . . . . . . . . . . . . D8 Reducibility of lacunary polynomials IV . . . . . . . . . . . . . . . . D9 On the number of terms of a power of a polynomial . . . . . . . . . . D10 On reducible trinomials . . . . . . . . . . . . . . . . . . . . . . . . . D11 On a conjecture of Posner and Rumsey with K. Gy˝ory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D12 Reducibility of lacunary polynomials XII . . . . . . . . . . . . . . . . D13 On reducible trinomials II . . . . . . . . . . . . . . . . . . . . . . . . D14 On reducible trinomials III . . . . . . . . . . . . . . . . . . . . . . . . D15 On the greatest common divisor of two univariate polynomials I . . . . D16 On the greatest common divisor of two univariate polynomials II . . . D17 On the reduced length of a polynomial with real coefficients . . . . . .

167 169 175 179 190 197 210 221 238 253 264 272 281

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549 563 580 605 632 646 658

x E.

E1 E2 E3 E4 E5 E6 E7 E8 E9 E10

Contents

Polynomials in several variables Commentary on E: Polynomials in several variables by Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . . Some unsolved problems on polynomials . . . . . . . . . . . . . . Reducibility of polynomials in several variables . . . . . . . . . . Reducibility of polynomials of the form f (x) − g(y) . . . . . . . Reducibility of quadrinomials with M. Fried . . . . . . . . . . . . . . . . . . . . . . . . . . . A general irreducibility criterion . . . . . . . . . . . . . . . . . . . Some arithmetic properties of polynomials in several variables with H. L. Montgomery . . . . . . . . . . . . . . . . . . . . . . On difference polynomials and hereditarily irreducible polynomials with L. A. Rubel and H. Tverberg . . . . . . . . . . . . . . . . On a decomposition of polynomials in several variables . . . . . . On weak automorphs of binary forms over an arbitrary field . . . . Reducibility of symmetric polynomials . . . . . . . . . . . . . . .

693 . . . .

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695 703 709 715

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720 739

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Hilbert’s Irreducibility Theorem Commentary on F: Hilbert’s Irreducibility Theorem by Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . . . F1 On Hilbert’s Irreducibility Theorem . . . . . . . . . . . . . . . . . . F2 A class of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . F3 The least admissible value of the parameter in Hilbert’s Irreducibility Theorem with Umberto Zannier . . . . . . . . . . . . . . . . . . . . . . .

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849

Part G Arithmetic functions


Commentary on G: Arithmetic functions by Kevin Ford

Schinzel spent much of the early years of his career studying Euler’s totient function ϕ(n) and the sum of divisors function σ (n). His teacher Wacław Sierpiński and Pál Erd˝os corresponded about numerous problems concerning ϕ and σ , and Sierpiński encouraged Schinzel to work on some of these. Papers G1–G6 showcase Schinzel’s considerable skill with elementary methods. G1, G4. Arithmetic functions at consecutive integers Somayajulu [27] proved in 1950 that the ratio ϕ(n + 1)/ϕ(n) takes arbitrarily large and arbitrarily small values. In a series of four papers ([20], [21], [23], G1), Schinzel improved and generalized this result, finally proving in G1 for any positive integer h that the vectors (g(n + 1)/g(n), . . . , g(n + h)/g(n + h − 1)) are dense in [0, ∞)h , where g = ϕ or g = σ . A few years later, he teamed with Erd˝os in G4 to study analogous problems for a wide class of multiplicative functions. For a positive multiplicative function g(n), f (n) = log g(n) is additive, and the authors chose to state their results in terms of additive functions. To obtain results about ϕ and σ , one applies these theorems with f (n) = log(n/ϕ(n)) and with f (n) = log(σ (n)/n). There is a vast literature on the distribution of additive functions, and paper G4 is a major contribution to the topic. Erd˝os and Schinzel give necessary and sufficient conditions on f so that for any h 1, there is a ch so that for any a1 , . . . , ah ch and ε > 0, there are x integers n x satisfying |f (n + i) − ai | < ε (1 i h) (Theorem 2 and discussion at the end of §1). They also give sufficient (and conjecturally necessary) conditions on f in order to conclude that for any real a1 , . . . , ah and ε > 0, there are x integers n x with |f (n+i)−f (n+i −1)−ai | < ε (1 i h) (Theorem 1 and discussion at the end of §1). Finally, they give very general conditions under which (f (n+1), . . . , f (n+h)) and (f (n+1)−f (n), . . . , f (n+h)−f (n+h−1)) have continuous distribution functions (Theorems 3 and 4). There is no claim that these conditions are necessary, but likely they cannot be relaxed too much. Condition 1 of Theorem 1 is closely related to the classical Kolmogorov three series theorem of probability theory(1 ). (1 )

The editors thank Kevin Ford for correcting a serious mistake in the proof of Lemma and an inaccuracy in the proof of Theorem 1 in G4. The corrections have been incorporated in the text.

862

G. Arithmetic functions

Although the results of G4 are “best possible” (or nearly so), the theorems in G1 may be extended in other directions. For example, Erd˝os in [8] determined the maximum rate of growth of h(n) in order to have lim inf

max

n→∞ 1ih(n)

ϕ(n + i)/

min

1ih(n)

ϕ(n + i) = 1.

The answer (Theorems 1, 2 of [8]) involves the 6th iterate of the logarithm! In another direction, Alkan, Ford and Zaharescu [2] have proven, for a wide class of multiplicative functions g including ϕ and σ , that for every h 1 there is a Ch > 0 so that for any positive a1 , . . . , ah there are infinitely many n so that |g(n + i)/g(n + i − 1) − ai | < n−Ch

(1 i h).

G2, G3, parts of J1, J2. Multiplicity problems Let A(m) be the number of solutions x of ϕ(x) = m, and let B(m) be the number of solutions of σ (x) = m. The famous Carmichael Conjecture ([4], [5]) states that A(m) = 1 for all m. Around 1955, Sierpiński made related conjectures that for all k 0, there are infinitely many m with B(m) = k and for all k 0, k = 1, A(m) = k for infinitely many m. Sierpiński in 1956 gave an infinite sequence of numbers m with A(m) = 2 and in G2, Schinzel gives explicit infinite sequences of numbers m with A(m) = 3. Schinzel also provides explicit infinite sequences of m for which (i) A(m) = 0 (in G2) and (ii) A(m) is unbounded (in G3). In fact, in G2 Schinzel gives a construction, for any positive integer n, of an infinite sequence of integers k with (iii) A(kn) = 0. The existence of sequences satisfying (i), (ii) or (iii) (without giving them explicitly) had earlier been proved by Pillai [18] in 1929, as a corollary of his bound V (x) x/(log x)(log 2)/e , where V (x) is the number of m x with A(m) > 0. A couple of years later, Erd˝os [8] showed with sieve methods that if there is one integer m with A(m) = k, then there are infinitely many such integers (the same method works also for B(m)). Later, in J2, Schinzel deduced both conjectures of Sierpiński (labelled C14 and C15 in J2) from his Hypothesis H, using a clever construction requiring the values of certain polynomials to be simultaneously prime for some argument n. There has been much activity on these problems since Schinzel’s papers. The Sierpiński conjectures for A(m) and B(m) have now been proved (in [12] and [14], respectively). 10 Carmichael’s conjecture remains open, but any counterexample m must exceed 1010 and a single counterexample implies that a positive proportion of all m with A(m) > 0 are counterexamples [11]. Estimates for V (x) have been progressively refined by Erd˝os [6], Erd˝os and Hall, Pomerance, Maier and Pomerance, and Ford [11] (see [11] for more on the history of the problem and further references). Combining the results of [11], [12] and [14], it is now known that for any k 2, a positive proportion of numbers with A(m) > 0 have A(m) = k and for every k 1, a positive proportion of numbers with B(m) > 0 have B(m) = k. Erd˝os in [6] showed that for some c > √ 0 there are infinitely many m with A(m) > mc . This was proved for any c < 3 − 2 2 by Wooldridge in 1979, and larger c values were obtained successively by Pomerance, Balog, Fouvry and Grupp, and Friedlander. The current record is c = 0.7039, due to Baker and Harman [3]. The best value of c is closely tied to the problem of finding primes p for which p − 1 is composed only of small prime factors (see [3] and the references therein). Trivial modifications of

Arithmetic functions

863

the analysis yield infinitely many m with B(m) > mc with the same value of c. Erd˝os’ conjecture that one may take any c < 1 remains open. G5. Compositions of ϕ and σ and Schinzel examined in G5 the lim sup and lim inf of the functions Makowski ϕ(ϕ(n))/n, ϕ(σ (n))/n, σ (ϕ(n))/n and σ (σ (n))/n. Alaoglu and Erd˝os [1] had earlier shown that lim inf ϕ(σ (n))/n = 0 and that lim sup σ (ϕ(n)) = ∞. Five of the six remaining cases are resolved in G5. The proofs use a combination of elementary methods and a result of Rényi that implies that for some c > 0 and infinitely many primes p, p − 1 has no prime factor > pc . They could not determine lim inf σ (ϕ(n))/n, but showed that 1 1 + 2 234 − 4 and conjectured, but could not prove, that the left side is > 0. This last assertion was proved by Pomerance [19] in 1989 using sieve methods. The authors in G5 also pose a problem P486: Is the inequality lim inf σ (ϕ(n))/n

σ (ϕ(n))/n

(1)

1 2

1 and due true for all n ? This remains open, the best result to date being σ (ϕ(n))/n 39.4 to Ford [13]. Inequality (1) has been verified for integers of certain types, and Luca and Pomerance [17] showed that (1) holds for a set of integers of asymptotic density 1.

G6. Integers of the form n − ϕ(n) Sierpiński conjectured [26] that there are infinitely many integers which are not of the form n − ϕ(n). Erd˝os in [9] settled the analogous conjecture for numbers of the form σ (n) − n, but it was not until 1995 that Sierpiński’s conjecture was proved by Browkin and Schinzel in G6. The proof uses the fact that there are integers n such that n2k − 1 is composite for all natural numbers k. This fact is proved using covering congruences, a method introduced by Erd˝os [7]. The smallest known value of n is 509203 and was discovered by H. Riesel in 1956. This may be the smallest n with this property, but a few smaller candidates have not been eliminated yet. In G6, Browkin and Schinzel show that none of the numbers 509203 · 2k are of the form n − ϕ(n). Computer calculations are needed for the case k = 1, and then the proof proceeds by induction on k. Building upon these ideas, Flammenkamp and Luca [10] have discovered similar families of numbers not of the form n − ϕ(n). Based on computations performed by D. H. Lehmer and A. Odlyzko, Browkin and Schinzel conjecture that a positive proportion of numbers are not of the form n − ϕ(n), and this remains an open problem. Other problems For every k 1, it is unknown if the equation (2)

ϕ(n) = ϕ(n + k)

has infinitely many solutions n. Sierpiński [25] showed that for each k there is at least one solution of (2), and by the work of Schinzel [22] and Schinzel and Wakulicz [24], we know that there are at least two solutions for each k 2 · 1058 . When k is even, Schinzel and

864


Sierpiński deduced from Hypothesis H (J1, Conjecture C2,1,2 ) that (2) has infinitely many solutions. For discussions about the distribution of solutions of (2), see the paper of Graham, Holt and Pomerance [15] and the references therein. In particular, it is conjectured that most solutions of (2) when 2 | k are generated by certain pairs of generalized twin primes ([15], Theorems 1 and 2). Solutions when k ≡ 3 (mod 6) are particularly rare. The infinitude of solutions of the equation σ (m) = ϕ(n) is also unknown, although it follows easily if there are infinitely many twin primes or infinitely many Mersenne primes (it also follows from the Extended Riemann Hypothesis for Dirichlet L-functions by unpublished work of Pomerance). Schinzel and Sierpiński deduce from Hypothesis H (J1, C8 ) a stronger result: for every k, there are integers m for which simultaneously A(m) k and B(m) k. More information about the arithmetic function problems investigated in G1–G6, J1 and J2, including additional references to related work, may be found in Richard Guy’s book [16], especially sections B13, B36, B38, B39, B41 and B42.

References [1] L. Alaoglu, P. Erd˝os, A conjecture in elementary number theory. Bull. Amer. Math. Soc. 50 (1944), 881–882. [2] E. Alkan, K. Ford, A. Zaharescu, Diophantine approximation with arithmetic functions. Preprint. [3] R. C. Baker, G. Harman, Shifted primes without large prime factors. Acta Arith. 83 (1998), 331–361. [4] R. D. Carmichael, On Euler’s φ-function. Bull. Amer. Math. Soc. 13 (1907), 241–243. [5] −−, Note on Euler’s ϕ-function. Bull. Amer. Math. Soc. 28 (1922), 109–110. [6] P. Erd˝os, On the normal number of prime factors of p − 1 and some related problems concerning Euler’s φ-function. Quart. J. Math. Oxford Ser. 6 (1935), 205–213. [7] −−, On integers of the form 2k +p and some related problems. Summa Brasil. Math. 2 (1950), 113–123. [8] −−, Some remarks on Euler’s ϕ-function. Acta Arith. 4 (1958), 10–19. [9] −−, Über die Zahlen der Form σ (n) − n und n − ϕ(n). Elem. Math. 28 (1973), 83–86. [10] A. Flammenkamp, F. Luca, Infinite families of noncototients. Colloq. Math. 86 (2000), 37–41. [11] K. Ford, The distribution of totients. The Ramanujan J. 2 (1998), 67–151. [12] −−, The number of solutions of φ(x) = m. Ann. of Math. (2) 150 (1999), 283–311. [13] −−, An explicit sieve bound and small values of σ (φ(m)). Period. Math. Hungar. 43 (2001), 15–29. [14] K. Ford, S. Konyagin, On two conjectures of Sierpiński concerning the arithmetic functions σ and φ. In: Number Theory in Progress, Vol. 2 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 795–803. [15] S. W. Graham, J. J. Holt, C. Pomerance, On the solutions to φ(n) = φ(n + k). In: Number Theory in Progress, Vol. 2 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 867–882. [16] R. K. Guy, Unsolved Problems in Number Theory, third edition. Problem Books in Math., Springer, New York 2004.

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[17] F. Luca, C. Pomerance, On some problems of Makowski–Schinzel and Erd˝os concerning the arithmetical functions φ and σ . Colloq. Math. 92 (2002), 111–130. [18] S. S. Pillai, On some functions connected with φ(n). Bull. Amer. Math. Soc. 35 (1929), 832–836. [19] C. Pomerance, On the composition of the arithmetic functions σ and ϕ. Colloq. Math. 58 (1989), 11–15. [20] A. Schinzel, Quelques théorèmes sur les fonctions ϕ(n) et σ (n). Bull. Acad. Polon. Sci. Cl. III 2 (1954), 467–469. [21] −−, Generalization of a theorem of B. S. K. R. Somayajulu on the Euler’s function ϕ(n). Ganita 5 (1954), 123–128. [22] −−, Sur l’équation ϕ(x + k) = ϕ(x). Acta Arith. 4 (1958), 181–184. [23] A. Schinzel, W. Sierpiński, Sur quelques propriétés des fonctions ϕ(n) et σ (n). Bull. Acad. Polon. Sci. Cl. III 2 (1954), 463–466. [24] A. Schinzel, A. Wakulicz, Sur l’équation ϕ(x + k) = ϕ(x), II. Acta Arith. 5 (1959), 425–426. [25] W. Sierpiński, Sur une propriété de la fonction φ(n). Publ. Math. Debrecen 4 (1956), 184–185. [26] −−, Teoria liczb, cze´ sc´ 2 (Number Theory, Part II). Monografie Matematyczne 38, PWN, Warszawa 1959 (Polish). [27] B. S. K. R. Somayajulu, The Euler’s totient function ϕ(n). Math. Student 18 (1950), 31–32.

Originally published in Bulletin de l’Académie Polonaise des Sciences Cl. III 3 (1955), 415–419


On functions ϕ(n) and σ (n)

The object of the present note is to prove the following two theorems: Theorem 1. For every finite sequence of positive numbers a1 , a2 , . . . , ak there exists an increasing infinite sequence of natural numbers n1 , n2 , . . . such that lim

k→∞

ϕ(nk + i) = ai for i = 1, 2, . . . , h. ϕ(nk + i − 1)

Theorem 2 is obtained from Theorem 1 by replacing the letter ϕ by σ . Lemma 1. Let un (n = 1, 2, . . . ) denote an infinite sequence of real numbers such that un+1 lim un = +∞, lim = 1. n→∞ n→∞ un Then, for every real number C > 1 and for every increasing infinite sequence of natural numbers nk , there exists an infinite sequence of natural numbers lk such that ul lim k = C. k→∞ unk Proof. Let nk (k = 1, 2, . . . ) denote a given increasing infinite sequence of natural numbers. For every natural number k let us denote by lk the least natural number for which ulk > C; unk such a number exists, since lim ul = +∞.

l→∞

Thus we have ul −1 ulk >C k , unk unk whence 1>C:

ul −1 ulk k ulk unk

Presented by W. Sierpiński on June 21, 1955

G1. On functions ϕ(n) and σ (n)

867

and, in view of ulk −1 = 1, k→∞ ulk ul lim k = C. k→∞ unk lim

Lemma 2. If g is a natural number, and A0 , A1 , . . . , Ag are natural numbers > 1 relatively prime in pairs, then there exists a natural number m such that, for i = 0, 1, . . . , g, m+i Ai , Ai | m + i, = 1, (A0 A1 · · · Ag )2 > m + i. Ai Proof. In virtue of so-called Chinese remainder theorem there exists a natural number m1 such that m1 + i ≡ Ai (mod A2i ) for i = 0, 1, . . . , g. m1 + g . We shall have Ai | m + i and Let m = m1 − A20 A21 · · · A2g 2 2 2 A A · · · A c g 0 1 m+i Ai , = 1, since Ai m+i m1 + i Ai | m1 + i, ≡ ≡ 1 (mod Ai ) for i = 0, 1, . . . , g. Ai Ai c

c

We shall also have −g m < (A0 A1 · · · Ag )2 − g. If we had m 0, we should have for i = −m, 0 i g and A2i | m + i, which is impossible, since Ai > 1. Thus m is an integer such that for i = 0, 1, . . . , g (A0 A1 · · · Ag )2 > m + i > 0.

Lemma 3a. For every finite sequence of real numbers, b1 , b2 , . . . , bg , satisfying the inequality ϕ(i) > bi > 0 for i = 1, 2, . . . , g, i there exists a sequence mk such that ϕ(mk + i) lim = bi , i = 1, 2, . . . , g. k→∞ mk + i Proof. Assume in Lemma 1 that un =

n i=1

1 1 − 1/pi

(where pi is the i-th prime number) and that Ci =

1 ϕ(i) , bi i

and for k g, i = 1, 2, . . . , g, let A0,k = g! p1 · p2 · · · pk ,

Ai,k = pli−1,k +1 · · · pli,k ,

868


where l0,k = k and li,k is the number lk from Lemma 1 suitably chosen for the number Ci and the sequence li−1,k . In virtue of Lemma 2 there exists a number mk such that m + i k Ai,k | mk + i, , Ai,k = 1, (A0,k A1,k · · · Ag,k )2 > mk + i > 0 Ai,k for i = 0, 1, . . . , g. Moreover, for i, j > 0 (A0,k , mk + i) = (A0,k , i) = i mk + i

A0,k mk + i p1 p2 · · · pk , , =1

i i i (Aj,k , mk + i) = (Aj,k , i − j ) = 1

(1) c

(2) (3)

for i = j , because k g. Accordingly, for i = 1, 2, . . . , g, α

αi,s

α

mk + i = iAi,k qi,1i,1 qi,2i,2 · · · qi,si i , where plg,k < qi,1 < qi,2 < . . . < qi,si are prime numbers and, of course, si < 2lg,k . Hence, si ϕ(i) ϕ(Ai,k ) 1 ϕ(i) ϕ(Ai,k ) ϕ(mk + i) 1− · = · > i Ai,k mk + i i Ai,k qi,j j =1

>

ϕ(i) ϕ(Ai,k ) · · i Ai,k

s1

1−

j =1

ϕ(i) ϕ(A ) plg,k 1 i,k · > · plg,k + j i Ai,k plg,k + 2lg,k

and, in virtue of the formulas ul i , lim i−1,k = bi · k→∞ uli,k ϕ(i)

lim

k→∞

plg,k plg,k + 2lg,k

= 1,

we have ϕ(mk + i) = bi . k→∞ mk + i lim

Lemma 3b. For every finite sequence of real numbers, b1 , b2 , . . . , bg , satisfying the inequality σ (i) i there exists a sequence mk such that bi >

for i = 1, 2, . . . , g,

σ (mk + i) = bi , k→∞ mk + i lim

i = 1, 2, . . . , g.

869

G1. On functions ϕ(n) and σ (n)

Proof. Assume in Lemma 1 that un =

n

1 pi

1+

i=1

(where pi is the i-th prime number) and that Ci = bi · i/σ (i), and, for k g, i = 1, 2, . . . , g, let A0,k = g! p1 · p2 · · · pk ,

Ai,k = pli−1,k +1 · · · pli,k ,

where li,k are the numbers from Lemma 1 (on the whole different from those in Lemma 3a). Analogously to the proof of Lemma 3a, we prove the existence of a number mk such that, for i = 1, 2, . . . , g, α

αi,s

α

mk + i = iAi,k qi,1i,1 qi,2i,2 · · · qi,si i , where plg,k < qi,1 < qi,2 < . . . < qi,si are prime numbers (on the whole different from those in Lemma 3a) and si < 2lg,k . Hence, si qi,j σ (mk + i) σ (i) σ (Ai,k ) σ (i) σ (Ai,k ) · < < · i Ai,k mk + i i Ai,k qi,j − 1 j =1

1, on aurait donc p1 | p − 1 et p | p1 − 1, ce qui est impossible. On a donc α1 = 1, d’où p1 − 1 = p6k+1 (p − 1) et p1 = p6k+1 (p − 1) + 1 > p 2 + 1 > p2 − p + 1, et comme, d’autre part p1 = p6k+2 − p 6k+1 + 1 = p6k (p 2 − p + 1) − (p 6k − 1), p6 − 1 | p 6k − 1,

p 6 − 1 = (p3 − 1)(p + 1)(p 2 − p + 1),

on a 1 < (p 2 − p + 1) | p1 , ce qui est impossible, vu que le nombre p1 est premier. Le théorème 2 se trouve ainsi démontré.

G2. Sur l’équation ϕ(x) = m

873

Il en résulte immédiatement ce Corollaire 2. L’équation ϕ(x) = 6 · 712k+1 , où k est un nombre naturel, a précisément deux solutions : x = 712k+2 et x = 2 · 712k+2 . Théorème 3. Il existe une infinité de nombres naturels m pour lesquels l’équation ϕ(x) = m a précisément trois solutions. Tels sont, par exemple, les nombres m = 12 · 712k+1 où k = 1, 2, . . . . Démonstration. Soit k un nombre naturel et m = 12 · 712k+1 . On vérifie sans peine que m = ϕ(3 · 712k+2 ) = ϕ(4 · 712k+2 ) = ϕ(6 · 712k+2 ). Supposons maintenant que (3)

ϕ(x) = m,

x = 3 · 712k+2 ,

x = 4 · 712k+2

et

x = 6 · 712k+2 .

D’après ϕ(x) = m il ne peut pas être x = 2α , où α est un entier 0. S’il était x = 2α y, où α 2 et (y, 2) = 1, on aurait ϕ(x) = 2α−1 ϕ(y) = 12 · 712k+1 ,

donc

α=2

et

ϕ(y) = 6 · 712k+1 ,

et, d’après le corollaire 2 on aurait y = 712k+2 [puisque (y, 2) = 1], d’où x = 4y = 4 · 712k+2 , contrairement à (3). Donc, le nombre x n’est pas divisible par 4 et on a x = 2α p1α1 p2α2 · · · prαr où r est un nombre naturel, p1 , p2 , . . . , pr sont des nombres premiers impairs distincts, α 1 et αi 1 (i = 1, 2, . . . , r). On a donc ϕ(x) = ϕ(p1α1 )ϕ(p2α2 ) · · · ϕ(prαr ) = 12 · 712k+1 . S’il était r 3, on aurait 8 | ϕ(x) = 12 · 712k+1 , ce qui est impossible. On a donc r 2. S’il était r = 2 alors, les nombres ϕ(p1α1 ) et ϕ(p2α2 ) étant pairs, un d’eux, soit ϕ(p1α1 ) serait égal à 2 · 7l , où l est un entier 0, d’où, d’après le corollaire 1, l = 0 et p1α1 = 3, donc ϕ(p1α1 ) = 2 et ϕ(p2α2 ) = 6 · 712k+1 et, d’après le corollaire 2 on aurait p2α2 = 712k+2 , d’où x = 2α · 3 · 712k+2 , contrairement à (3). On a donc r = 1 et ϕ(x) = ϕ(p1α1 ) = 12 · 712k+1 et évidemment on a p1 = 3 et p1 = 7, donc α1 = 1 et p1 − 1 = 12 · 712k+1 , d’où p1 = 12 · 712k+1 + 1 > 5, ce qui est impossible, vu que le nombre 12 · 712k+1 + 1 est divisible par 5 (puisque 74 = 5t + 1 et 12 · 7 = 5u − 1). Nous avons ainsi démontré que l’équation ϕ(x) = m a précisément trois solutions. Le théorème 3 est ainsi démontré.

M. W. Sierpiński a exprimé l’hypothèse que, quel que soit le nombre naturel s > 1, il existe une infinité de nombres naturels m pour lesquels l’équation ϕ(x) = m a précisément s solutions. Or, nous ne savons pas démontrer même que pour tout nombre naturel s > 1 il existe au moins un nombre naturel m tel que l’équation ϕ(x) = m a précisément s solutions.

874


Il est encore à remarquer que dans une communication présentée au Congrès des mathématiciens tchécoslovaques à Prague en 1955(3 ) j’ai démontré d’une façon tout-àfait élémentaire que, quel que soit le nombre naturel m tel que l’équation ϕ(x) = m a plus que s solutions. Tel est, par exemple, le nombre m = (p1 − 1)(p2 − 1) · · · (ps − 1), où pi désigne le i-ème nombre premier. (L’équation ϕ(x) = m est ici vérifiée par les nombres x0 = p1 p2 · · · ps

et

xi = x0

pi − 1 . pi

où i = 1, 2, . . . , s.)

Bibliographie [1] V. L. Klee, jr., On a conjecture of Carmichael. Bull. Amer. Math. Soc. 53 (1947), 1183–1186.

(3 ) Voir G3, p. 875–876.

Originally published in Czechoslovak Mathematical Journal 6 (1956), 164–165


Sur un problème concernant la fonction ϕ(n)

Récemment M. W. Sierpiński m’a posé le problème suivant : k étant un nombre naturel quelconque, existe-t-il toujours un nombre naturel m tel que l’équation ϕ(x) = m ait plus que k solutions (en nombres naturels x) ? (ϕ(n) désigne ici le nombre de nombres naturels n et premiers avec n). Le but de cette communication est de démontrer que la réponse à ce problème est positive. Soit k un nombre naturel donné, pi — le i-ème nombre premier. Posons m = (p1 − 1)(p2 − 1) · · · (pk − 1),

(1) (2)

xi = p1 p2 · · · pi−1 (pi − 1)pi+1 · · · pk

pour

i = 1, 2, . . . , k,

xk+1 = p1 p2 · · · pk .

(3)

Les nombres x1 , x2 , . . . , xk , xk+1 sont évidemment naturels et distincts deux à deux. Soit maintenant i un des nombres 1, 2, . . . , k. Le nombre pi − 1 évidemment n’est pas divisible par aucun nombre premier > pi−1 : γ

γ

γ

i−1 pi − 1 = p1 1 p2 2 · · · pi−1

(4)

où γ1 , γ2 , . . . , γi−1 sont des entiers 0. D’après (2) on a donc γi−1 +1 γ +1 γ2 +1 p2 · · · pi−1 pi+1 pi+2 · · · pk ,

xi = p1 1 d’où γ

c

γ

γ

i−1 (p1 − 1) · · · (pi−1 − 1)(pi+1 − 1) · · · (pk − 1) ϕ(xi ) = p1 1 p2 2 · · · pi−1

et, d’après (4) et (1) on trouve ϕ(xi ) = m. Or, d’après (3) et (1) on a évidemment ϕ(xk+1 ) = m. Les k + 1 nombres naturels distincts x1 , x2 , . . . , xk+1 satisfont donc à l’équation ϕ(x) = m et notre assertion se trouve démontrée. En ce qui concerne l’équation ϕ(x) = m il est encore à remarquer que R. D. Carmichael suppose qu’il n’existe aucun nombre naturel m pour lequel elle ait précisément une solution [1] ; comme l’a démontré V. L. Klee jr., cela est vrai pour m 10400 [2]. Or, M. W. Sierpiński a récemment démontré qu’il existe une infinité de nombres naturels m pour lesquels l’équation ϕ(x) = m a précisément deux solutions : tels sont, par exemple, Communication présentée le 2 septembre 1955 au Congrès des mathématiciens tchécoslovaques à Prague par M. W. Sierpiński.

876


les nombres m = 2 · 36k+1 , où k = 1, 2, . . . . (Ces deux solutions sont ici x = 36k+2 et x = 2 · 36k+2 ). Après avoir pris connaissance avec la communication de A. Schinzel, M. P. Erd˝os a remarqué que S. Pillai a prouvé que le nombre des entiers m x pour lesquels l’équation ϕ(y) = m a des solutions est d’ordre o(x). En 1935 P. Erd˝os a démontré (dans le Quarterly Journal of Mathematics) l’existence d’une suite infinie croissante d’entiers nk (k = 1, 2, . . . ) telle que le nombre de solutions de l’équation ϕ(y) = nk est plus grand que nck où c est un nombre fixe positif, et il énonce l’hypothèse que pour tout ε > 0 il existe un tel c > 1 − ε. Quant à la démonstration de A. Schinzel, M. P. Erd˝os la considére comme la plus simple de toutes qui lui sont connues.

Travaux cités [1] R. D. Carmichael, Note on Euler’s ϕ-function. Bull. Amer. Math. Soc. 28 (1922), 109–110. [2] V. L. Klee, jr., On a conjecture of Carmichael. Bull. Amer. Math. Soc. 53 (1947), 1183–1186.

Originally published in Acta Arithmetica VI (1961), 473–485


Distributions of the values of some arithmetical functions with P. Erd˝os (Budapest)

1. Y. Wang and A. Schinzel proved, by Brun’s method, the following theorem ([3]): For any given sequence of h non-negative numbers a1 , a2 , . . . , ah and ε > 0, there exist positive constants c = c(a, ε) and x0 = x0 (a, ε) such that the number of positive integers n x satisfying

ϕ(n + i)

− ai < ε (1 i h)

ϕ(n + i − 1) is greater than cx/ logh+1 x, whenever x > x0 . They also proved the analogous theorem for the function σ . Shao Pin Tsung, also using Brun’s method, extended this result to all multiplicative positive functions fs (n) satisfying the following conditions ([4]): I. For any positive integer l and prime number p:

lim fs (p l )/p ls = 1 (p denotes primes). p→∞

There exists an interval a, b , a = 0 or b = ∞, such that for any integer M > 0 the set of numbers fs (N )/N s , where (N, M) = 1, is dense in a, b . (This formulation is not the same but equivalent to the original one.) In this paper we shall show without using Brun’s method that if we replace the condition I by the condition

fs (p) − p s 2 1. 2. There exists a number c1 such that, for any integer M > 0, the set of numbers f (N ), where (N, M) = 1, is dense in (c1 , ∞). Then, for any given sequence of h real numbers a1 , a2 , . . . , ah and ε > 0, there exist more than C(a, ε)x positive integers n x for which

f (n + i) − f (n + i − 1) − ai < ε (i = 1, 2, . . . , h); (1) C(a, ε) is a positive constant, depending on ε and ai . Lemma. There exists an absolute constant c such that the number of integers of the form pq > x for which one can find n x satisfying n ≡ b (mod a), n ≡ 0 (mod p) and n + 1 ≡ 0 (mod q) is for x > x0 (a) less than cx/a. Proof. Let c1 , c2 , . . . denote absolute constants. Assume p > x 1/2 (q > x 1/2 can be dealt similarly). Denote by Al (x) the number of integers of the form pq satisfying pq > x,

l

x 1−1/2 p < x 1−1/2

l+1

n ≡ b (mod a),

,

p | n, q | n + 1, 1 n x,

for some n,

c

and by Al (x) the number of integers pq for which l

x 1−1/2 p < x 1−1/2

l+1

,

q > x 1/2

l+1

,

n ≡ b (mod a),

p | n, q | n + 1,

for some n, Clearly Al (x) Al (x) and it will suffice to prove that for x > x0 (a),

∞ l=1

1 n x,

Al (x) < cx/a.

Define positive integer lx by the inequality 2lx log log x > 2lx −1 .

c

The number k of integers n satisfying (2)

n x,

n ≡ b (mod a),

for an l lx does not exceed

l

xp>x

c

Chebyshev k
x1 (a).

879

G4. Values of some arithmetical functions c

Denote the numbers satisfying (2) for an l lx and satisfying ν(n + 1) 10 log log x by a1 < a2 < . . . < ah x and denote the numbers satisfying (2) for an l lx and satisfying ν(n + 1) > 10 log log x by b1 < b2 < . . . < bj x. Since

2ν(n+1)

nx

d(n + 1) = O(x log x)

nx

we have j
x 1/2 . l+1

c

with Let g = [2l / l 2 ]. Given distinct primes q1 , q2 , . . . , qg greater than x 1/2 q1 q2 · · · qg x + 1, the number of integers n x satisfying n ≡ a (mod b) and x + 1. Thus, by theorems of Chebyshev and Mertens, q1 · · · qg | n + 1 is at most aq1 · · · qg

880


the number of integers of the second class is less than x 1 g x +O ag! q gq1 · · · qg−1 log(x/q1 · · · qg−1 ) l+1 q ,...,q x 1/2

1

c10 , x > x4 (a). By definition, νl (ai + 1) < 2l+1 . Thus, for l > c10 , the contribution of the numbers of the second class to (5) is < x/a · 2l−1 ; for l c10 the c +1 x. Thus, for l < l , x > x (a), c contribution is clearly < 2 10 x 4 c

Al (x) < c8 x/al 2 and in view of (3) we have for x > x0 (a) ∞

Al (x)
k2

1 1 1 + . < 2 p p 3(h + 1) p>k2

881

G4. Values of some arithmetical functions

Finally by condition 1 there exists a k3 such that

(9)

|f (p)|k3

f (p)2 ε2 < . p 48h

Let us put k = max(k1 , k2 , k3 ),

N = N1 N2 · · · Nh ,

P =

p,

Q = (h + 1)! N 2 P

pk, p/| N

and let us consider the following system of congruences n ≡ 1 (mod (h + 1)! P ),

n ≡ −i + Ni (mod Ni2 ),

0 i h.

By (6) and the Chinese Remainder Theorem there exists a number n0 satisfying these congruences. It is easy to see that (10) for every integer t the numbers (Qt + n0 + i)/(i + 1)Ni (i = 1, 2, . . . , h) are integers which are not divisible by any prime k; (11) the number of terms not exceeding x of the arithmetical progression Qt + n0 is x/Q + O(1). In order to prove Theorem 1 we shall estimate the number of integers n of the progression Qt + n0 which satisfy the inequalities (12) n x,

h

2

f (n + i) − f (n + i − 1) − f (i + 1)Ni + f (iNi−1 ) > 41 ε 2 .

i=1

We divide the set of integers n ≡ n0 (mod Q) for which the inequalities (12) hold into two classes. Integers n such that n(n + 1) · · · (n + h) is divisible by a prime p > k with |f (p)| μ, or by p2 , p > k, are in the first class and all other integers are in the second class. (13) The number of integers n x, n ≡ r (mod Q) which are divisible by a given integer d > 0 is equal to x/dQ + O(1) for (d, Q) = 1, hence the number of integers n x, n ≡ n0 (mod Q) of the first class is less than x 1 1 (h + 1) +O 1+ 1 . + Q p p2 2 p>k |f (p)|μ

p>k

px+h

p x+h

By the inequality (8) and the definition of k this number is less than 13 x/Q + o(x).

882


For the integers of the second class, by remark (10) we have 2

f (n + i) − f (n + i − 1) − f (i + 1)Ni + f (iNi−1 )

h n

i=1

=S=

h n

f (p) −

i=1 p|n+i p>k

2 f (p) ,

p|n+i−1 p>k

where means that the summation runs through the integers of the second class. In view of remark (13), since (Q, p) = 1 we have 2 h S f (p) − f (p) p|n+i n≡n0 (mod Q) i=1 nx p>k, |f (p)|k |f (p)|k, |f (p)|p>k |f (p)|k |f (p)|k, |f (p)| 2y 2. p log y c c

ypk2

1 1 1 + . < 2 p p 3h p>k2

By condition 1 there exists also a k3 such that ε2 f (p)2 < . (19) p 24h p>k3 , |f (p)| max{k1 , k2 , k3 } such that 1 2 η, νη

pk, |f (p)| k, pq x + h, |f (p)| < μ, |f (q)| < μ. From (20) we get (21)

2

f (p)f (q) pq

x+hp>k |f (p)|(x+h)1−1/2 p(x+h)/4

ε2 + 2μ2 C 2 96h

Let us put N = N1 N2 · · · Nh , P = (22)

l=2

pk, p/| N

Q = h! N 2 P h! N 2

x+hq>(x+h)/p

x+hq>(x+h)/p

μ q

μ q

x+hq

∞

l−1

μ p

μ p

μ μ p q

x+hp>(x+h)/4

c

√ x+hp> x+h

+2

2 l log 2 ε2 μ log log x + O < . l 2 −2 log x 24h p,

p=Q

pk, p/| N

and let us consider the following system of congruences: n ≡ 0 (mod h! P ),

n ≡ −i + Ni (mod Ni2 ).

By (16) and the Chinese Remainder Theorem there exists a number n0 satisfying these congruences. It is easy to see that Qt + n0 + i (i = 1, 2, . . . , h) are integers which are (23) for every integer t the numbers iNi not divisible by any prime k. Analogously, as in the proof of Theorem 1, we shall estimate the number of integers n of

885


the progression Qt + n0 which satisfy the inequalities n x,

(24)

h

2

f (n + i) − f (iNi )

> 41 ε 2 .

i=1

We divide the set of integers n ≡ n0 (mod Q) for which the inequalities (24) hold into two classes. Integers n such that (n + 1)(n + 2) · · · (n + h) is divisible by a prime p > k with |f (p)| μ or by p2 , p > k, are in the first class and all other integers are in the second class. By remark (13) the number of integers n x, n ≡ n0 (mod Q) of the first class is less than x 1 1 h +O 1+ 1 . + Q p p2 2 p>k, |f (p)|μ

p>k

p x+h

px+h

By the inequality (18) and the definition of k this number is less than 13 x/Q + o(x). For the integers of the second class, by remark (23), we have 2 h h 2

f (p) f (n + i) − f (iNi ) = i=1

i=1

c

p|n+i, p>k |f (p)|k |f (p)|k |f (p)|k |f (p)|k, |f (p)| 0 such that the number of integers n x satisfying |f (n + 1) − f (n)| < a is > cx, then g(p) 2 f (n) = α log n + g(n) with < ∞. p

2. The proof of Theorem 2 is very similar to the proof of Lemma 1 of P. Erd˝os’ paper [1]. Using ideas and results from that paper we can prove the following theorem. Theorem additive function

satisfying condition 1 of Theorem 1 and 3. Let f (n) be an (1/p) be divergent,

f (p) /p convergent, then the distribution function of let f (p)=0 h-tuples f (m + 1), f (m + 2), . . . , f (m + h) exists, and it is a continuous function. Proof. We denote by N (f ; c1 , c2 , . . . , ch ) the number of positive integers m not exceeding n for which f (m + i) ci ,

i = 1, 2, . . . , h,

where ci are given constants. It is sufficient to consider, as in [1], the special case in which, for any α, f (p α ) = f (p), so that f (m) = f (p). p|m

888


Let us also consider the function fk (m) =

f (p). We are going to show that

p|m, pk

the sequence N (fk ; c1 , c2 , . . . , ch )/n is convergent. Since fk (m + A) = fk (m), where A= p, we can see that the integers m for which pk

c

fk (m + i) ci c

(i = 1, 2, . . . , h)

are distributed periodically with the period A. Hence N (fk ; c1 , c2 , . . . , ch )/n has a limit. To prove the existence of a limit of N (f ; c1 , c2 , . . . , ch )/n it is sufficient to show that for arbitrary ε > 0 there exists k0 such that for every k > k0 and n > n(ε)

N (f ; c1 , c2 , . . . , ch ) − N (fk ; c1 , c2 , . . . , ch ) /n < ε. To show this, it is enough to prove that the number of integers m n for which there exists i h such that fk (m + i) < ci and f (m + i) ci or fk (m + i) ci and f (m + i) < ci is less that εhn. But it is an immediate consequence of the analogous theorem for h = 1 proved in [1], p. 123. In order to prove that the distribution function is continuous we must show that for every ε > 0, there exists a δ > 0 such that Δ = N(f ; c1 − δ, c2 − δ, . . . , ch − δ) − N (f ; c1 + δ, c2 + δ, . . . , ch + δ) < ε. Now Δ=

h

N (f ; c1 + δ, . . . , ci−1 + δ, ci − δ, . . . , ch − δ)

i=1

− N (f ; c1 + δ, . . . , ci + δ, ci+1 − δ, . . . , ch − δ)

and by Lemma 2 of [1] each term of this sum is less than ε/ h for suitably chosen δ. This completes the proof.

We conclude 3 that function f satisfies condi if an additive from Theorems 2 and tions 1, 2, (1/p) is divergent and

f (p) /p convergent, then the distribution f (p)=0

function of {f (m + 1), . . . , f (m + h)} exists, is continuous and strictly decreasing on some half straight-line, thus the sequence of integers n for which inequality (15) holds has a positive density. Similarly we can prove the following: 1 f (p) 2 Theorem 4. Assume that = ∞ and that < ∞ then p f (p)=0 p f (n + 1) − f (n), f (n + 2) − f (n + 1), . . . , f (n + k) − f (n + k − 1) has a continuous distribution function. It is easy to see that condition 2 can be replaced by the conditions lim f (p) = 0 and |f (p)| = ∞. p→∞

p


889

3. Y. Wang proved in [6] that the number N of primes p < x satisfying

ϕ(p + ν + 1)

− aν < ε, 1 ν k,

ϕ(p + ν) is greater than x c(a, ε) . k+2 log log x (log x) By our methods we can obtain in that case x . log x

After having passed to the additive function log

ϕ(n)/n the proof is similar to the proof of Theorem 1. We use the fact that log ϕ(n)/n is always negative, and apply the asymptotic formula for the number of primes in arithmetical progression instead of (11) and the Brun–Titchmarsh theorem instead of (13). We can also prove that there exists distribution function N (c1 , c2 , . . . , ck ) defined as 1 ϕ(p + ν) lim N p < x; cν , ν = 1, 2, . . . , k . x→∞ π(x) p+ν N > c1 (a, ε)

References [1] P. Erd˝os, On the density of some sequences of numbers III. J. London Math. Soc. 13 (1938), 119–127. [2] −−, On the distribution function of additive functions. Ann. of Math. (2) 47 (1946), 1–20. [3] A. Schinzel,Y. Wang, A note on some properties of the functions ϕ(n), σ (n) and θ(n). Bull.Acad. Polon. Sci. Cl. III 4 (1956), 207–209; Ann. Polon. Math. 4 (1958), 201–213; Corrigendum, ibid. 19 (1967), 115. [4] P. T. Shao, On the distribution of the values of a class of arithmetical functions. Bull. Acad. Polon. Sci. Cl. III 4 (1956), 569–572. [5] P. Turán, On a theorem of Hardy and Ramajunan. J. London Math. Soc. 9 (1934), 274–276. [6] Y. Wang, A note on some properties of the arithmetical functions ϕ(n), σ (n) and d(n). Acta Math. Sinica 8 (1958), 1–11.

Originally published in Colloquium Mathematicum XIII (1964), 95–99


On the functions ϕ(n) and σ (n) with A. Makowski (Warsaw)

In this paper ϕ(n) and σ (n) denote the Euler function and the sum of the divisors of n, respectively, p denotes odd primes, pi the i-th prime. It has been asked in [5] whether the inequality k times

σ . . . σ (n) lim inf 1 and N (a, p) = (a p − 1)/(a − 1), then

ϕ N (a, p) σ N (a, p) = lim = 1. lim p→∞ N (a, p) p→∞ N (a, p)

891

G5. On the functions ϕ(n) and σ (n)

Proof. Put N(a, p) = N = q1α1 q2α2 · · · qsαs , where qi (1 i s) are different primes, αi 1. Clearly s

1−

(5)

i=1

s σ (N) 1 −1 1 ϕ(N ) 1− . 1 qi N N qi i=1

For p > a + 1, we have p /| a − 1, thus qi ≡ 1 (mod p) (cf. [3], p. 381) and qi > p (i = 1, 2, . . . , s). It follows that N pα1 +...+αs ps and

ap − 1

s

log N = log p

log

a−1 log p

1− → 1. 1− qi p p

The lemma follows from (5) and (6). Lemma 2. The following formula holds:

ϕ 21 (p − 1) σ 21 (p − 1) lim sup 1 = lim inf 1 = 1. 2 (p − 1) 2 (p − 1) Proof. Clearly (7)

ϕ

1

2 (p − 1) 1 2 (p − 1)

1

σ

1

2 (p − 1) 1 2 (p − 1)

.

On the other hand, it has been proved by Wang ([6], Appendix, formulae (7) and (8)) that c x cq (x) q + O . Pω (x, q, x 1/6.5q ) > 12.9η 2 ϕ(q) log x log3 x Here, Pω (x, q, ξ ) is the number of primes p satisfying p x, p ≡ a (mod q), p ≡ ai (mod pi ) (i = 1, . . . , r), where ω = a, q, ai (1 i r) is a sequence of integers such that q x, (a, q) = 1, ai ≡ 0 (mod pi ) and pi are all primes ξ not dividing 2q; cq is a certain positive constant (cf. [6], formula (6)), η = δ/(δ − 1), where as stated on p. 1054 r times

one can take δ = 1.5. It follows after the substitution ω = 3, 4, 1, . . . , 1 that there exist infinitely many primes p such that every prime factor of (p − 1)/2 is greater than p1/20 . Let ε be any number > 0 and take p of the above kind greater than 2020 ε −20 . Let 1 2 (p

− 1) = q1α1 q2α2 · · · qsαs ,

892


where qi (1 i s) are different primes and αi 1. Clearly s < 20, and s

1−

i=1

1 1 20 20 > 1 − 1/20 > 1 − 1/20 > 1 − ε. qi p p

On the other hand,

s σ 21 (p − 1) ϕ 21 (p − 1) 1 −1 1 1 1− 1− . 1 qi qi 2 (p − 1) 2 (p − 1) i=1

s i=1

It follows that (1 − ε)

−1

>

σ

1

2 (p − 1) 1 2 (p − 1)

ϕ

1

2 (p − 1) 1 2 (p − 1)

> 1 − ε.

In view of (7), this completes the proof.

Proof of the Theorem. We begin with formula (1). For any ε > 0 we take a prime r > 1 + ε −1 and put a = r in Lemma 1. We have N (r, p) = σ (r p−1 ). Hence σ σ (r p−1 ) σ σ (r p−1 ) σ (r p−1 ) · = lim p−1 p→∞ p→∞ σ (r p−1 ) r r p−1

σ N (r, p) σ (r p−1 ) r · lim = lim = < 1 + ε. p→∞ N (r, p) p→∞ r p−1 r −1 lim

Since σ σ (n)/n 1 for all n, formula (1) is proved. Proof of formula (2) is similar. For any M we take a number t such that t i=1

pi >M pi − 1

and put successively a = p1 , p2 , . . . , pt in Lemma 1. We have t t p−1 σ = pi N (pi , p). i=1

i=1

Hence ϕσ lim sup p→∞

t i=1 t i=1

p−1

pi

ϕ = lim sup

p−1 pi

=

t i=1

p→∞

t

N (pi , p)

i=1 t i=1

lim sup p−1 pi

p→∞

t ϕ N (pi , p) p−1

pi

i=1

t t ϕ N (pi , p) N (pi , p) pi lim lim = · > M. p→∞ N (pi , p) p→∞ p p−1 pi − 1

This completes the proof of (2).

i=1

i

i=1

G5. On the functions ϕ(n) and σ (n)

893

Formula (3) follows at once from Lemma 2, since

ϕ 21 (p − 1) p − 1 ϕϕ(p) ϕ(p − 1) = lim sup lim sup 1 , lim sup · p p 2p p→∞ p→∞ p→∞ 2 (p − 1)

and, on the other hand, ϕϕ(n)/n 21 for all n > 1. In order to prove formula (4) assume that m is any positive integer divisible by 4. By Lemma 2

σ ϕ 21 mp σ 2ϕ( 21 m) σ 21 (p − 1) lim inf lim inf 1 1 p→∞ p→∞ 2 mp 2 mp

σ 21 (p − 1) σ ϕ(m) σ ϕ(m) = · lim inf . = 1 p→∞ m m 2p Since 233 − 1 1 1 σ ϕ(234 − 4) = = + 34 , 34 34 2 −4 2 −4 2 2 −4 the proof of the theorem is complete.

The following equalities supplement the theorem: σ σ (n) = ∞, n ϕσ (n) (9) = 0, lim inf n ϕϕ(n) lim inf (10) = 0, n σ ϕ(n) lim sup (11) = ∞. n Equalities (8) and (10) are trivial, equalities (9) and (11) have been proved by Alaoglu and Erd˝os [1]. In that paper the following conjecture has been announced: for sufficiently large n the sequence

(8)

lim sup

σ (n), σ σ (n), ϕσ σ (n), σ ϕσ σ (n), . . . tends to infinity. We remark that this conjecture implies the finiteness of the set of Mersenne primes. Indeed, if 2p − 1 is a prime, then ϕσ σ (2p−1 ) = ϕσ (2p − 1) = ϕ(2p ) = 2p−1 , and the sequence in question is periodical. It seems a natural question to ask whether formula (4) can be improved. Mrs. K. Kuhn has investigated the quotient σ ϕ(n)/n for n having at most 6 prime factors and has found i that σ ϕ(n)/n 21 for such n’s, the equality being realized only if n = 22 +1 − 2 (0 i 5). This suggests a problem P486. Is the inequality σ ϕ(n)/n

1 2

true for all n ?

894


Remark. Even the weaker inequality inf σ ϕ(n)/n > 0 remains still unproved.

References [1] L. Alaoglu, P. Erd˝os, A conjecture in elementary number theory. Bull. Amer. Math. Soc. 50 (1944), 881–882. [2] R. Bojanić, Solution to problem 4590. Amer. Math. Monthly 62 (1955), 498–499. [3] L. E. Dickson, History of the Theory of Numbers, vol. I. Chelsea, New York 1952. [4] A. Rényi, On the representation of an even number as the sum of a prime and of an almost prime. Izv. Akad. Nauk SSSR Ser. Mat. 12 (1948), 57–78 (Russian); English transl.: Amer. Math. Soc. Transl. (2), 19 (1962), 299–321. [5] A. Schinzel, Ungelöste Probleme, Nr. 30. Elem. Math. 14 (1959), 60–61. [6] Y. Wang, On the representation of large integer as a sum of prime and an almost prime. Sci. Sinica 11 (1962), 1033–1054.

Originally published in Colloquium Mathematicum LXVIII (1995), 55–58


On integers not of the form n − ϕ(n) with J. Browkin (Warszawa)

W. Sierpiński asked in 1959 (see [4], pp. 200–201, cf. [2]) whether there exist infinitely many positive integers not of the form n − ϕ(n), where ϕ is the Euler function. We answer this question in the affirmative by proving Theorem. None of the numbers 2k · 509203 (k = 1, 2, . . . ) is of the form n − ϕ(n). Lemma 1. The number 1018406 is not of the form n − ϕ(n). Proof. Suppose that 10108406 = n − ϕ(n)

(1) and let (2)

n=

j

qiαi

(q1 < q2 < . . . < qj primes).

i=1

If for any i j we have αi > 1 it follows that qi | 2 · 509203, and since 509203 is a prime, either qi = 2 or qi = 509203. In the former case n − ϕ(n) ≡ 0 ≡ 10108406 (mod 4), in the latter case n − ϕ(n) > 1018406, hence (3)

αi = 1

(1 i j ).

Since n > 2 we have ϕ(n) ≡ 0 (mod 2), hence n ≡ 0 (mod 2). However, n/2 cannot be a prime since 1018405 is composite. Hence ϕ(n) ≡ 0 (mod 4) and n ≡ 2 (mod 4). Moreover, n ≡ 1 (mod 3) would imply ϕ(n) ≡ n − 1018406 ≡ 2 (mod 3), which is impossible, since 0 (mod 3) if at least one qi ≡ 1 (mod 3), ϕ(n) ≡ 1 (mod 3) otherwise. Hence n ≡ 2 (mod 12) or n ≡ 6 (mod 12) and (4)

n − ϕ(n) >

1 n. 2

896


Let pi denote the ith prime and consider first the case n = 12k + 2. We have q1 = 2, qi pi+1 (i 2). Since 7

(5)

pi+1 > 1018406,

i=2

it follows from (1)–(4) that j 6 and

6 1 ϕ(n) 1 2/5 if n ≡ 0 (mod 5), 1− 2 pi+1 n 1/2 otherwise. i=2

Hence if n = 12k + 2 satisfies (1) we have either 116381 < k < 141446 or 141446 k < 169735 and k ≡ 4 (mod 5). Consider now n = 12k + 6. Here q1 = 2, q2 = 3, qi pi . By (1)–(5), j 7 and 7

1−

i=1

1 ϕ(n) 1 . pi n 3

Hence if n = 12k + 6 satisfies (1) we have 103561 < k < 127301. The computation performed on the computer SUN/SPARC of the Institute of Applied Mathematics and Mechanics of the University of Warsaw using the program GP/PARI has shown that no n corresponding to k in the indicated range satisfies (1).

Lemma 2. All the numbers 2k · 509203 − 1 (k = 1, 2, . . . ) are composite. Proof. We have 509203 ≡ 2ai (mod qi ), where qi , ai is given by 3, 0 , 5, 3 , 7, 1 , 13, 5 , 17, 1 and 241, 21 for i = 1, 2, . . . , 6, respectively. Now, 2 belongs mod qi to the exponent mi , where mi = 2, 4, 3, 12, 8 and 24 for i = 1, 2, . . . , 6, respectively. It is easy to verify that every integer n satisfies one of the congruences n ≡ −ai (mod mi )

(1 i 6).

If k ≡ −aj (mod mj ) we have 2k · 509203 ≡ 2aj −aj ≡ 1 (mod qj ), and since 2k · 509203 − 1 > qj the number 2k · 509203 − 1 is composite.

Remark 1. Lemma 2 was proved by H. Riesel, already in 1956 (see [3], Anhang). The following problem, implicit in [1], suggests itself. Problem 1. What is the least positive integer n such that all integers 2k n−1 (k = 1, 2, . . . ) are composite?

G6. On integers not of the form n − ϕ(n)

897

Proof of the theorem. We shall prove that n − ϕ(n) = 2k · 509203 by induction on k. For k = 1 this holds by virtue of Lemma 1. Assume that this holds with k replaced by k − 1 (k 2) and that n − ϕ(n) = 2k · 509203.

(6)

If ϕ(n) ≡ 0 (mod 4) we have n ≡ 0 (mod 4) and n n −ϕ = 2k−1 · 509203, 2 2 contrary to the inductive assumption. Thus ϕ(n) ≡ 2 (mod 4) and n = 2p α , where p is an odd prime. From (6) we obtain p α−1 (p + 1) = 2k · 509203. By Lemma 2, α = 1 is impossible. If α > 1 we have p | 2k · 509203, and since 509203 is a prime, p = 509203, α = 2 and 509204 = 2k , which is impossible. The inductive proof is complete.

Remark 2. D. H. Lehmer on the request of one of us has kindly computed the table of all numbers not of the form n − ϕ(n) up to 50 000. This table and its prolongation up to 100 000 seems to indicate that the numbers not of the form n − ϕ(n) have a positive density, about 1/10. This suggests Problem 2. Have the integers not of the form n − ϕ(n) a positive lower density? Added in proof (November 1994). A computation performed by A. Odlyzko has shown that there are 561 850 positive integers less than 5 000 000 not of the form n − ϕ(n).

References [1] A. Aigner, Folgen der Art ar n + b, welche nur teilbare Zahlen liefern. Math. Nachr. 23 (1961), 259–264. [2] P. Erd˝os, Über die Zahlen der Form σ (n) − n und n − ϕ(n). Elem. Math. 28 (1973), 83–86. [3] W. Keller, Woher kommen die größten derzeit bekannten Primzahlen? Mitt. Math. Ges. Hamburg 12 (1991), 211–229. [4] W. Sierpiński, Teoria liczb, cze´ sc´ 2 (Number Theory, Part II). Monografie Matematyczne 38, PWN, Warszawa 1959.

Part H Divisibility and congruences


Commentary on H: Divisibility and congruences by H. W. Lenstra jr.

The eleven papers in Section H are naturally divided in two categories: four papers in elementary number theory, and seven papers on local-global results concerning exponential equations. The four papers H1, H3, H6, H8 in the first category show Schinzel’s resourcefulness in elementary arithmetic. Problem P217, formulated in H1, has been studied by E. Burkacka and J. Piekarczyk in their master dissertations at the University of Warsaw; see Colloq. Math. 10 (1963), 365. The joint paper H6 with G. Baron, which has a combinatorial flavour, has applications in ring theory. An algebraic proof of the main result of this paper was obtained by M. Van Den Bergh and M. Van Gastel [5]. Especially noteworthy is the joint paper H8 with J. Wójcik, which was suggested by work in group theory. The theme of the paper is reminiscent of the local-global results discussed below, but the tools used are quite different and more elementary. A negative answer to the question posed at the end of the paper was given by J. Wójcik [7]. The seven papers H2, H4, H5, H7, H9, H10, H11 in the second category address fundamental issues related to exponential diophantine equations, the emphasis being on local-global results such as the following Theorem 2 from H4: if H is a finitely generated subgroup of the multiplicative group of an algebraic number field K, and a non-zero element b ∈ K has the property that for almost all primes p of K the element b mod p belongs to H mod p, then b belongs to H . The result admits numerous variations and applications, and Schinzel investigated many of them. His work helped inspire a development in which the role of the multiplicative group is played by general algebraic groups. A good discussion and a substantial bibliography can be found in a paper by E. Kowalski [1]. The technique used by Schinzel in most papers in the second category, consists of combining density theorems from algebraic number theory with information on Galois groups of field extensions obtained by adjoining radicals. Several of Schinzel’s auxiliary results on such Galois groups are new. The most notable one (H5, Theorem 2) asserts the following. Let K be a field, n a positive integer not divisible by the characteristic of K, and w the number of nth roots of unity in K. Then, for a ∈ K, the Galois group of X n − a over K is abelian if and only if there exists b ∈ K with a w = bn . This basic result has been finding its way into the field-theoretic literature as Schinzel’s theorem. It is important in the context of Stark’s conjectures (J. Tate, [4]; see Chapitre IV, Exercice 1.4).

902

H. Divisibility and congruences

One also encounters Schinzel’s theorem when one attempts to describe, for any field K ¯ the Galois group of the “maximal radical extension” of K, which with algebraic closure K, one obtains by adjoining all α ∈ K¯ for which there exists an integer n not divisible by the characteristic of K with α n ∈ K. P. Stevenhagen ([3], cf. [6]) gave a proof of Schinzel’s theorem that is so elegant that it deserves to be included here. We may assume a = 0. Denote by L the splitting field of Xn − a over K, and generally by ζm a primitive mth root of unity. First suppose a w = bn with b ∈ K. Then L is contained in the composite of the Kummer extension K(b1/w ) and the cyclotomic extension K(ζwn ) of K. Both of these extensions are abelian, and therefore so is L. For the converse, suppose L has an abelian Galois group G. We fix an nth root α of a in L. For each σ ∈ G, one has σ (α)/α = ζσ ∈ ζn , and σ (ζn ) = ζnc(σ ) with c(σ ) ∈ Z. For any σ , τ ∈ G one has τ (α c(σ ) )/α c(σ ) = ζτc(σ ) = σ (τ (α)/α) = τ σ (α)/σ (α); hence α c(σ ) /σ (α) is fixed by all τ so belongs to K, and taking the nth power one sees that a c(σ )−1 ∈ K ∗n . Thus, if v denotes the gcd of n and all numbers c(σ ) − 1 (for σ ∈ G), then one has a v ∈ K ∗n . To finish the proof it suffices to show v = w. A divisor d of n divides v if and only if d divides all numbers c(σ )−1, if and only if all elements σ (ζd )/ζd = ζdc(σ )−1 are 1, if and only if ζd ∈ K, and if and only if d divides w. Therefore we have v = w, as required. The Corollary to Theorem 5 in H5 has the condition a = d 3 + 3d. Schinzel himself proved that this condition can be omitted [2].

References [1] E. Kowalski, Some local-global applications of Kummer theory. Manuscripta Math. 111 (2003), 105–139. [2] A. Schinzel, On the congruence un ≡ c (mod p), where un is a recurring sequence of the second order. Acta Acad. Paedagog. Agriensis Sect. Mat. (N.S.) 30 (2003), 147–165. [3] P. Stevenhagen, Ray class groups and governing fields. Thesis, Universiteit van Amsterdam, 1989. [4] J. Tate, Les conjectures de Stark sur les fonctions L d’Artin en s = 0. Progr. Math. 47, Birkhäuser, Boston 1984. [5] M. Van Den Bergh, M. Van Gastel, On the structure of non-commutative regular local rings of dimension two. Comm. Algebra 30 (2002), 4575–4588. [6] J. Wójcik, Criterion for a field to be abelian. Colloq. Math. 68 (1995), 187–191. [7] −−, On a problem in algebraic number theory. Math. Proc. Cambridge Philos. Soc. 119 (1996), 191–200.

Originally published in Colloquium Mathematicum V (1958), 198–204


Sur un problème de P. Erd˝os

Désignons pour k naturels et n entiers par Hk,n la proposition suivante :

Hk,n : il existe un entier i tel que 0 i < k et n − i nk . P. Erd˝os a posé le problème, si Hk,n subsiste pour tous les nombres naturels k et n 2k. La réponse à ce problème est négative, comme le prouve l’exemple k = 15, n = 99215. En effet, les décompositions en facteurs premiers 99215 = 5 · 19843,

99214 = 2 · 113 · 439,

99213 = 3 · 33071,

2

99212 = 2 · 17 · 1459,

99211 = 7 · 14173,

99210 = 2 · 3 · 5 · 3307,

99209 = 11 · 29 · 311,

99208 = 2 · 12401,

99207 = 32 · 73 · 151,

99206 = 2 · 49603,

99205 = 5 · 19841,

99204 = 22 · 3 · 7 · 1181,

99203 = 132 · 587,

99202 = 2 · 193 · 257,

99201 = 3 · 43 · 769

3

entraînent que

n 1◦ k = 13P , où (P , 15!) = 1, 2◦ chacun des nombres n − i (0 i < k) a un diviseur premier pi < 15.

Donc, si l’on avait n − i nk pour un i, où 0 i < k, on aurait pi nk = 13P , donc

n

pi = 13 | n − i, n − i = 99203, 132 | n − i

= 13P , 13 | P , k ce qui implique une contradiction. On peut démontrer que la proposition Hk,n est vraie pour k < 15, n 2k et pour k = 15, 30 n < 99215. En omettant la démonstration assez pénible de ce dernier fait, j’indiquerai plus loin certaines méthodes d’examiner si pour un k fixé, il existe un n pour lequel Hk,n est un défaut. Je commencerai par les remarques suivantes. 1. Soit n1 ≡ n2 (mod k!). Alors on l’équivalence Hn1 ,k ≡ Hn2 ,k . En effet, on a

n

1 n1 − i

k dans ce et seulement dans ce cas n1 · · · (n1 − i + 1)(n1 − i − 1) · · · (n1 − k + 1) ≡ 0 (mod k!), c’est-à-dire, si n2 · · · (n2 − i + 1)(n2 − i − 1) · · · (n2 − k + 1) ≡ 0 (mod k!), ce

904 qui équivaut à


n

2 n2 − i

. k

Il en résulte que si pour un nombre naturel k donné il existe un nombre naturel n 2k tel que la proposition Hk,n est fausse, il existe une infinité de tels nombres naturels n. 2. On a l’équivalence Hk,n ≡ Hk,−n+k−1 . En effet, vu l’identité n −n + k − 1 = (−1)k , k k

on a la formule n − i nk , où 0 i < k, dans ce et seulement dans ce cas, si

−n + k − 1

(−n + k − 1) − j

, où 0 j = k − i − 1 < k. k La proposition Hk,n étant vraie pour 0 n < 2k (ce qu’on déduit sans peine du théorème de Tchebycheff), il résulte de la remarque 2 que si Hk,n est vraie pour le nombre naturel k et pour les entiers n 2k, elle est vraie pour n entier quelconque. Désignons par Hk la proposition affirmant que Hk,n subsiste pour tout entier n. c

Lemme 1. S’il existe pour k et n donnés un i tel que 0 i 1, on a p | n donc (n + 1, p) = 1 et

k n − (k − 1), k − (k − 1) = (n − k + 1, k) = (n + 1, k) = (n + 1, pα ) = 1. c k−1 Dans tous les deux cas Hk,n est vraie d’après le Lemme 1.

Théorème 2. La proposition Hk est vraie pour k = 6, 10, 12, 14, 18, 20, 24, 26, 28, 30. Démonstration. À titre d’exemple nous donnerons la démonstration pour k = 30. Pour les autres k la démonstration est analogue.

905

H1. Sur un problème de P. Erd˝os

Supposons d’abord que n ≡ 0 (mod 2). La vérité de Hk,n résulte alors du Lemme 1 dans lequel selon les restes de n de la division par 3, 5, 29, 7, 13, 23 et 11 les on substitue valeurs envisagées dans la table suivante, pour lesquels on a n − i, 30 (30 − i) = 1 : i le reste de la division de n par 3 5 29 7 13 n ≡ 2 n ≡ 4 0 4 n ≡ 1 0 4 1 n ≡ 4 n ≡ 12 4 1 n ≡ 5 12 0 0 4 1 5 12 4 1 4 n ≡ 5 0 4 1 4 5 0 1 4 n ≡ 27 n ≡ 6 4 n ≡ 3 6 1 1 4 3 6 n ≡ 5 1 4 3 6 5 4 1 27 n ≡ 3 4 1 27 3 n ≡ 5 4 27 3 5 1 2 n ≡ 1 n ≡ 1 2 n ≡ 2 1 n ≡ 6 1 6 2 n ≡ 3 1 n ≡ 27 n ≡ 6 2 2 1 n ≡ 3 6 1 2 3 6 n ≡ 12

i 29 1 25 5 23 5 23 27 3 5 23 3 5 23 1 27 3 27 3 25

3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

5 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3

le reste de la division de n par 29 7 13 23 3 6 12 n ≡ 7 3 6 12 7 3 6 12 7 27 n ≡ 3 27 3 n ≡ 12 27 3 12 n ≡ 7 27 3 12 7 27 3 12 7 1 n ≡ 3 1 3 n ≡ 12 1 3 12 n ≡ 21 1 3 12 n ≡ 9 1 3 12 9 1 3 12 21 1 3 12 21 1 6 n ≡ 12 1 6 12 n ≡ 7 1 6 12 7 1 6 12 7

11 n ≡ 9 9

n ≡ 9 9 n ≡ 10 10 10 n ≡ 9 9 n ≡ 10 10

i 7 9 19 3 25 7 9 19 3 25 21 9 19 9 19 25 7 21 9

Le cas n ≡ 1 (mod 2) qui reste à examiner se réduit au précédant d’après la remarque 2 et la congruence −n + k − 1 ≡ −n + 29 ≡ 0 (mod 2).

Lemme 2. Soient 2 = p1 < p2 < . . . < pl k tous les nombres premiers k. S’il existe un système des entiers a(p1 ), a(p2 ), . . . , a(pl ) tel que 1◦ 2◦

tout entier i, 0 i < k, satisfait à une des congruences i ≡ a(pj ) (mod pj ) (1 j l), quel que soit le nombre naturel j l, on a k a(p ) − k a(p ) j j + = , pj pj pj

alors Hk est en défaut. Démonstration. Supposons que le système a(pj ) (1 j l) satisfait aux conditions 1◦ –2◦ . Posons ln k a(p ) j αj = + 1, a(p ¯ j ) = a(pj ) − pj . ln pj pj

906


D’après 2◦ on a pour j l k a(p ¯ j ) − k a(p ¯ j) (1) + = = 0. pj pj pj D’après le théorème chinois sur les restes il existe un entier n tel que α

n ≡ a(p ¯ j ) − pj (mod pj j )

(2)

(1 j l).

Admettons que, pour un certain i0 , 0 i0 < k et

n

(3) n − i0

. k D’après la condition 1◦ il existe un j0 l tel que i0 ≡ a(p ¯ j0 ) (mod pj0 ), d’où, d’après (2), n − i0 ≡ 0 (mod pj0 ) et d’après (3)

n

. (4) pj0

k D’autre part, parmi les nombres n − i (0 i < k) les seuls nombres divisibles par pj0 sont, d’après (2), les nombres n − i(t), où a(p ¯ j0 ) − k i(t) = a(p ¯ j0 ) + tpj0 , 0 t − − 1. pj0 Or, d’après (1), a(p k ¯ j0 ) − k =− , pj0 pj0 αj

¯ j0 ) + pj0 + i(t) k < pj0 0 . donc i(t) a(p ¯ j0 ) + [k/pj0 ]pj0 − pj0 , d’où 0 < −a(p αj

D’autre part, d’après (2), −a(p ¯ j0 ) + pj0 + i(t) ≡ −[n − i(t)] (mod pj0 0 ). Donc pj0 figure dans le développement de −a(p ¯ j0 ) + pj0 + i(t) en facteurs premiers avec le même exposant que dans le développement de n − i(t).

On a donc pj0 , nk = 1, contrairement à la formule (4), ce qui achève la démonstration.

Théorème 3. S’il existe un système de nombres premiers k, Q(k) = {q1 , q2 , . . . , qm } β

c

β

β

tel que pour tout nombre s = q1 1 q2 2 · · · qmm k, où β1 , β2 , . . . , βm 0, on a (k + 1 − s, q1 q2 · · · qm ) > 1, alors Hk est en défaut. Démonstration. Soient r1 , r2 , . . . , rl−m tous les nombres premiers k qui n’appartiennent pas à Q(k) et posons a(qj ) = −1, a(rj ) = k. Nous prouverons que le système des nombres a ainsi défini remplit les conditions 1◦ et 2◦ du Lemme 2. 1◦ Pour tout nombre i, 0 i < k, on a β

β

γ

γ

γ

l−m βm k − i = q1 1 q2 2 · · · qm · r1 1 r2 2 · · · rl−m ,

où β

β1 , . . . , βm , γ1 , . . . , γl−m 0. β

Donc (k − i, r1 r2 · · · rl−m ) > 1 ou bien k − i = q1 1 · · · qmm .

H1. Sur un problème de P. Erd˝os

907

Dans le premier cas il existe un j l − m tel que i ≡ k = a(rj ) (mod rj ), et dans le second cas, en posant s = k−i, nous obtenons d’après l’hypothèse (i+1, q1 q2 · · · qm ) > 1, d’où, pour un certain j m, i ≡ −1 = a(qj ) (mod qj ). 2◦ Il faut vérifier que −1 − k k + = −1 (1 j m) qj qj et que k − k rj

+

k rj

=

k rj

(1 j l − m).

La deuxième de ces égalités est évidente et on obtient la première en changeant le signe dans l’inégalité k 1+k k > +1 .

qj qj qj Corollaire 1. La proposition Hk est fausse pour k = 15, 21, 33, 35, 45, 55, 63, 65, 69, 75, 77, 85, 87, 91, 93, 95, 99. La démonstration résulte du Théorème 3 dans lequel il faut prendre comme Q(k) les systèmes suivants des nombres premiers : Q(15) = {5, 11},

Q(21) = {5, 7, 17},

Q(33) = {11, 23},

Q(35) = {7, 29},

Q(45) = {3, 19, 37, 43},

Q(55) = {5, 13, 17, 31, 43},

Q(63) = {3, 11, 31, 37, 53, 61},

Q(65) = {13, 53},

Q(69) = {23, 47},

Q(75) = {5, 17, 59, 71},

Q(77) = {11, 67},

Q(85) = {7, 17, 23, 37, 79},

Q(87) = {29, 59},

Q(91) = {13, 79},

Q(93) = {5, 7, 23, 29, 31, 59, 71, 89},

Q(95) = {11, 17, 19, 79},

Q(99) = {11, 89}. Comme on voit, pour k = 15, 33, 35, 65, 69, 77, 87, 91, 99 le système Q(k) est formé de deux nombres. Comme on le vérifie aisément, cela a lieu dans ce et seulement dans ce cas si k = aq1 , où les nombres q1 > a > 1 et q2 = (a − 1)q1 + 1 sont tous les deux premiers, donc, par exemple, si k = 3q1 , où les nombres q1 > 3 et 2q1 + 1 sont tous les deux premiers. Grâce à ce fait, la réponse positive au problème suivant est très probable : P216. Existe-t-il une infinité de nombres k pour lesquels Hk est fausse ? Théorème 4. La proposition Hk est fausse pour k = 22. La démonstration résulte du Lemme 2 où l’on pose a(2) = 0, a(3) = 1, a(5) = 4, a(7) = 3, a(11) = 10, a(13) = 11, a(17) = 5, a(19) = 15.

908


Des théorèmes 1, 2 et 4 et du corollaire 1 il résulte le Corollaire 2. La proposition Hk est vraie pour tous les nombres k 33 sauf pour les nombres k = 15, 21, 22 et 33. Il se pose ici le problème suivant : P217. Est-ce que la proposition Hk est vraie pour une infinité de nombres k = pα , où p est un nombre premier et α un entier 0 ? Je suppose que la réponse est négative. Ajouté pendant la correction des épreuves. P. Erd˝os a demontré que la réponse au Problème P216 est positive. Voici l’esquisse de sa démonstration — l’extrait de sa lettre à l’auteur du 5 février 1957 : “Let q be a large prime. I want to find√an n = aq so that Hn is false. Let p ≡ 1 (mod q), p = (a1 − 1)q + 1, n1 = a1 q, p < e q but p so large that the number of primes ≡ 1 (mod q) which are p is

p 1 + o(1) q log p (such a p exists by Page–Walfisz–Siegel theorem). Choose a(q) = 0, a(r) = −1 (r prime) except for a “few” primes pi which I define now, a(p) = q − 1. All the primes pi for which a(pi ) = −1 will satisfy pi ≡ 1 (mod q), pi > n/2. Thus the only numbers n which are not yet eliminated by our congruences are q 2 − 1, q 3 − 1, …, q k − 1 (q k − 1 < n1 q k+1 − 1). If there are at least l primes pi ≡ 1 (mod q) in (n1 , n1 − q l + 1) for every l 2, we successively eliminate these integers by the primes pi and avoid contradiction against Lemma 2. If not, then for some l there are fewer than l primes pi ≡ 1 (mod q) in (n1 − q l + 1, n1 ). Consider then the greatest prime p2 ≡ 1 (mod q), p2 < n1 − q l + 1, put p2 = (a2 − 1)q + 1, n2 = a2 q and repeat the same argument until nk < n1 /2. But then the number of primes ≡ 1 (mod q) in (n1 /2, n1 ) is < n1 /q 2 , but since q was “small” compared with n1 , by Page–Walfisz, the number of these primes is

n1 n1 1 + o(1) > 2 2q log n1 q √

if n1 < e q . This contradiction shows that before nk becomes < n1 /2, we have that for every l 2, (nk − q l + 1, nk ), where nk = ak q, (ak − 1)q + 1 is a prime, contains at least l primes p ≡ 1 (mod q) and Hnk is false.”

Originally published in Bulletin de l’Académie Polonaise des Sciences Série des sci. math., astr. et phys. VIII (1960), 307–309


On the congruence a x ≡ b (mod p)

The aim of this paper is to prove the following theorem signalled in [1]. Theorem. If a, b are rational integers, a > 0 and b = a k (k—rational integer), then there exist an infinite number of rational primes, p, for which the congruence a x ≡ b (mod p) has no solutions in rational integers x. Lemma. Let l be an arbitrary rational prime, ζl —a primitive root of unity of order l, k = Γ (ζl )—the field obtained by adjoining ζl to the field Γ of rational numbers. If a system of rational integers γ1 , γ2 , . . . , γt has the property that γ1m1 γ2m2 · · · γtmt is an l-th c power of a rational integer only when l | mi (i = 1, . . . , t), then for arbitrary rational integers c1 , c2 , . . . , ct there exists an infinite number of prime ideals p of the field k whose degree is 1, and for which γ i = ζlci (i = 1, 2, . . . , t). c p Proof. If the number γ1m1 γ2m2 · · · γtmt is not an l-th power of a rational integer, then the mt m1 m2 l c polynomial x − γ is irreducible over Γ . On the basis of a well known 1 γ2 · · · γt theorem ([2], p. 298, Th. 16) the polynomial remains irreducible over the field Γ (ζl ), mt m1 m2 c therefore γ is not an l-th power of the integer of the field k. The thesis of 1 γ2 · · · γt the lemma follows from this directly, in view of Chebotarev’s improvement of a Hilbert’s theorem ([3], cf. [4], p. 276, Th. 152).

c

Proof of the Theorem. The cases a = 1 and b = 0 are trivial. Assume that a > 1, b = 0, hence |ab| > 1 and let q1 , q2 , . . . , qs be all the prime factors of ab. Let further c

c

a = q1α1 q2α2 · · · qsαs ,

β

β

b = ±q1 1 q2 2 · · · qsβs

(αi , βi 0).

If b < 0, we observe that the numbers l = 2, t = s + 1, γi = qi (i = 1, 2, . . . , s), γs+1 = −1 satisfy the conditions of our Lemma. Then there exists an infinite number of prime ideals p of the field Γ (−1) (i.e., simply rational primes) for which we have q −1 i = 1, = −1, p p Presented by W. Sierpiński on March 17, 1960

910


whence

a p

= 1,

b p

= −1.

Assume now that b > 0 and that for some indices i, j s we have αi βj − βi αj = 0. Choose a rational prime l > |αi βj − βi αj |. The numbers γi = qi (i = 1, 2, . . . , s), as different rational primes and the number l satisfy the conditions of the Lemma. Then there exists an infinite number of prime ideals p of the field Γ (ζl ), the degree of which is 1 and for which we have q q q −α j ν i = 1 (ν = i, j ), = ζl j , = ζlαi , c p p p c

whence c

a p

= 1,

b p

αi βj −βi αj

= ζl

= 1.

In both considered cases, therefore, there exists such a rational prime l that the field b = 1, but Γ (ζl ) contains infinitely many prime ideals p, of the degree 1, for which p ax a = 1, whence = 1, then the congruence a x ≡ b (mod p) is insoluble. c p p The same property has, of course, the congruence a x ≡ b (mod p), where the rational c prime p is the norm of the ideal p. As a prime p can be a norm of only a finite l number of prime ideals, there exists in the considered cases an infinite number of rational primes for which the congruence a x ≡ b (mod p) is insoluble. We have still to examine the case, when b > 0 and when for all i, j s: αi βj − αj βi = 0. c As αi + βi > 0 (i = 1, 2, . . . , s) and not all αi are = 0, it follows from the last formula βi β1 that all αi are = 0 and that, for i s, = holds. c αi α1 Let α1 β1 = α, = β. c (α1 , β1 ) (α1 , β1 ) c

βi β As (α, β) = 1, = (i s), we have αi = αδi , βi = βδi , where δi are positive αi α integers. Putting c = γ1δ1 γ2δ2 · · · γsδs we get a = cα , b = cβ . c If α = 1, one obtains b = a β , in spite of the conditions assumed. Hence, α > 1 and h | 2(δ , δ , . . . , δ ). c there exists a rational prime l | α. Choose a positive integer h so that l / 1 2 s As the numbers qi are primes, it follows from the last formula that c is neither of the form h h/2 h nl nor of the form 2l nl , where n is a rational integer. By Trost’s theorem ([5]) there exists, therefore, an infinite set P of rational primes p, for which c is not a residue of l h -th degree. c As l | α, l /| β for any rational integer x: l /| αx − β. Hence, for all p ∈ P , for all x: αx−β c c is not a residue of l h -th degree, then the congruence cαx ≡ cβ (mod p), i.e. the congruence a x ≡ b (mod p) is impossible. This completes the proof.

c

H2. On the congruence a x ≡ b (mod p) c

911

Remark. The Theorem remains true, if a > 0 is not assumed, but the proof is longer.

References [1] A. Schinzel, W. Sierpiński, Sur les congruences x x ≡ c (mod m) et a x ≡ b (mod p). Collect. Math. 11 (1959), 153–164. [2] N. Tschebotaröw [Chebotarev], Grundzüge der Galoisschen Theorie. Übersetzt und bearbeitet von H. Schwerdtfeger, Noordhoff, Groningen–Djakarta 1950. [3] N. Chebotarev, Über einen Satz von Hilbert. Vestnik Ukr. Akad. Nauk, 1923, 3–7; Sobranie Sochineni˘ı I, Izd. Akad. Nauk SSSR, Moscow–Leningrad 1949–50, 14–17. [4] D. Hilbert, Die Theorie der algebraischen Zahlkörper. In: Ges. Abhandlungen I, Berlin 1932, 63–363. [5] E. Trost, Zur Theorie von Potenzresten. Nieuw Arch. Wisk. 18 (1934), 58–61.

Originally published in Nordisk Matematisk Tidskrift 10 (1962), 8–10


On the composite integers of the form c(ak + b)! ± 1

Summary. The author raises the problem whether there exist infinitely many composite integers of the form c(ak + b)! ± 1. An affirmative answer in many cases when c = 1 follows immediately from Wilson’s theorem; other cases are answered in the Theorem.

It follows immediately from the theorems of Wilson (1 ) and Leibniz (2 ) that there exist infinitely many composite integers of the forms (ak + b)! + 1 and (ak + b)! − 1 if a > 0 and (a, b + 1) = 1 or (a, b + 2) = 1 respectively. This suggests the problem whether for arbitrary integers a > b 0 and rational c > 0 there exist infinitely many composite integers of the form c(ak + b)! ± 1. All the cases, except the two mentioned above, for which I have found a positive answer to this problem are given by the following Theorem. There exist infinitely many composite integers of each of the forms 1) c(4k)! + 1, c(4k + 2)! + 1, c(6k)! + 1, c(6k + 2)! + 1, c(6k + 4)! + 1; 2) c(2k)! − 1, c(2k + 1)! + 1, c(2k + 1)! − 1; 3) [b(2k + 1)]! + 1. Here b is a positive odd integer and c a positive rational number. Proof. An immediate generalisation of the theorem of Wilson gives (p − i − 1)! i! ≡ (−1)i+1 (mod p), c

p prime, 0 i p − 1.

Hence for arbitrary c = d/n (d, n positive integers): (1)

p | ni! + ε(−1)i+1 d

implies

p | d(p − i − 1)! + εn

(0 i p − 1, ε = ±1). (al − b − 2)! ≡ 0 (mod a), Let now a = 4 or 6, b even. For sufficiently large l we have c therefore (al − b − 2)! − 1 ≡ −1 (mod a), c (1 ) (2 )

(p − 1)! ≡ −1 (mod p). (p − 2)! ≡ +1 (mod p).

H3. Composite integers of the form c(ak + b)! ± 1

913

and since positive integers of the form 4t − 1 resp. 6t − 1 have a prime factor of the same (al − b − 2)! form, − 1 has a prime factor pl ≡ −1 (mod a). For sufficiently large l, pl c must be > al − b − 2 > n, and in view of (1): pl | d(pl − al + b + 1)! + n.

(2)

It follows that d(pl − al + b + 1)! + n pl > al − b − 2, whence lim (pl − al + b + 1) = ∞,

l→∞

and for l large enough, c(pl − al + b + 1)! ≡ 0 (mod a). Since pl ≡ −1 (mod a) we have pl = c(pl − al + b + 1)! + 1, and the number c(pl − al + b + 1)! + 1 is composite, because by (2) pl | c(pl − al + b + 1)! + 1. Since pl − al + b + 1 ≡ b (mod a), the proof of part 1 of the theorem is complete. To prove part 2, let us assume a = 2, b = 0 or 1, ε = ±1, and if 2l − b d denote (2l − b)! by pl the greatest prime factor of + ε(−1)b+1 . For l large enough, each prime c factor p of the above number is > 2l − b > n, thus in view of (1): p | d(p − 2l + b − 1)! + εn. It follows hence that d(p − 2l + b − 1)! + εn p > 2l − b. For sufficiently large l we have n | (p − 2l + b − 1)! and thus p | c(p − 2l + b − 1)! + ε.

(3) In particular

pl | c(pl − 2l + b − 1)! + ε.

(4) Suppose that (5) If

pl = c(pl − 2l + b − 1)! + ε.

(2l − b)! + ε(−1)b+1 has any prime factor p < pl , we have c p c(p − 2l + b − 1)! + ε

in view of (3), and therefore for sufficiently large l: pl − p c(pl − 2l + b − 1)! + ε − c(p − 2l + b − 1)! − ε = c(p − 2l + b − 1)! (pl − 2l + b − 1) · · · (p − 2l + b) − 1 (p − ε)(pl − p − 1) (2l − b)(pl − p − 1) > pl − p, which is impossible. Equality (5) implies therefore (6)

(2l − b)! + ε(−1)b+1 = plα . c

914


For sufficiently large l we have further 6 | c(pl − 2l + b − 1)! = pl − ε, pl − ε pl − ε 2l − b > > 6d, 2 3 thus d(pl − ε)2 | (2l − b)! and in view of (6): (pl − ε)2 | plα − ε(−1)b+1 . Hence (pl − ε)2 | α(pl − ε)ε α−1 + ε α − ε(−1)b+1 ,

(pl − ε) | α,

and we get (2l − b)! p −ε + ε(−1)b+1 = plα pl l > (2l − b)2l−b , c which for l large enough gives a contradiction. We must therefore have pl = c(pl − 2l + b − 1)! + ε, and in view of (4) the number c(pl − 2l + b − 1)! + ε is composite. Since pl ≡ 1 (mod 2), pl − 2l + b − 1 ≡ b (mod 2), which proves part 2 of the theorem. In order to prove part 3, we shallshow that if b is odd andb(2l + 1) > 3, at least one of the numbers [b(2l + 1)]! + 1 and [b(2l + 1)]! − b(2l + 1) ! + 1 is composite. In fact, suppose that [b(2l + 1)]! + 1 is a prime p. Then, in view of (1): p | [b(2l + 1)]! − b(2l + 1) ! + 1, and if [b(2l + 1)]! − b(2l + 1) ! + 1 is not composite, we have [b(2l + 1)]! − b(2l + 1) ! + 1 = p = [b(2l + 1)]! + 1, [b(2l + 1)]! − b(2l + 1) = b(2l + 1), [b(2l + 1) − 1]! = 2, b(2l + 1) = 3, against the assumption. On the other hand, both numbers b(2l + 1) and [b(2l + 1)]! − b(2l + 1) are of the form b(2k + 1), which completes the proof.

Originally published in Acta Arithmetica XXVII (1975), 397–420


On power residues and exponential congruences

In memory of Yu. V. Linnik

The main aim of this paper is to extend the results of [6] to algebraic number fields. We shall prove Theorem 1. Let K be an algebraic number field, ζq a primitive qth root of unity and ∈ K. Let n1 , . . . , nk , n be positive inteτ the greatest integer such that ζ2τ + ζ2−1 τ gers, ni | n; α1 , . . . , αk , β be non-zero elements of K. The solubility of the k congruences x ni ≡ αi mod p (1 i k) implies the solubility of the congruence x n ≡ β mod p for almost all prime ideals p of K if and only if at least one of the following four conditions is satisfied for suitable rational integers l1 , . . . , lk , m1 , . . . , mk and suitable γ , δ ∈ K: k nm /n (i) β αi i i = γ n ; i=1

(ii) n ≡ 0 mod 2τ ,

2|ni

(iii) n ≡ 2τ mod 2τ +1 ,

αili = −δ 2 and β 2|ni

k i=1

nmi /ni

αi

= −γ n ;

αili = −δ 2 and

β

k

nmi /ni

n/2 n = − ζ2τ + ζ2−1 γ ; τ +2

nmi /ni

n/2 n = ζ2τ + ζ2−1 γ . τ +2

αi

i=1

(iv) n ≡ 0 mod 2τ +1 and β

k i=1

αi

If ζ4 ∈ K, the conditions (ii), (iii), (iv) imply (i); if τ = 2, (ii) implies (i) for not necessarily the same m1 , . . . , mk and γ . Almost all prime ideals means all but for a set of Dirichlet density zero or all but finitely many. In this context it comes to the same in virtue of Frobenius density theorem. Corollary 1. If each of the fields K(ξ1 , ξ2 , . . . , ξk ), where ξini = αi , contains at least one η satisfying ηn = β then at least one of the conditions (i)–(iv) holds. This corollary may be regarded as a generalization of the well known result concerning Kummer fields (see [3], p. 42). As one can see from Lemmata 6 and 7 below it holds for arbitrary fields K of characteristic not dividing n (with τ = ∞, if necessary).

916


Corollary 2. The congruences x n ≡ α mod p and x n ≡ β mod p are simultaneously soluble or insoluble for almost all prime ideals p of K if and only if either βα u = γ n , or n ≡ 0 mod 2τ +1 and

n/2 n βα u = ζ2τ + ζ2−1 γ τ +2

where (u, n) = 1 and γ ∈ K. This is a simultaneous refinement of the theorems of Flanders [1] and Gerst [2] concerning α = 1 and K = Q, respectively. We shall prove further Theorem 2. If α1 , . . . , αk , β are non-zero elements of K and the congruence α1x1 α2x2 · · · αkxk ≡ β mod p c

is soluble for almost all prime ideals p of K then the corresponding equation is soluble in rational integers. This is a refinement of a theorem of Skolem [7], in which he assumes that the congruence is soluble for all moduli (also composite). Skolem’s proof is defective but it can be amended. On the lines indicated by Skolem we prove Theorem 3. Let αij , βi (i = 1, . . . , h, j = 1, . . . , k) be non-zero elements of K, D a positive integer. If the system of congruences k

x

αijj ≡ βi mod m (i = 1, . . . , h)

j =1

is soluble for all moduli m prime to D then the corresponding system of equations is soluble in integers. We show on an example that already for h = 2, k = 3 one cannot replace here “all moduli prime to D” by “all prime moduli”. On the other hand, the present approach gives no clue to Skolem’s very interesting conjecture: If the congruence h

xk x1 αi0 αi1 · · · αik ≡ 0 mod m

i=1

is soluble for all moduli m then the corresponding equation is soluble in rational integers. The proof of Theorem 1 is based on nine lemmata. In formulating and proving them we use as much as possible the matrix notation. Integral matrices are denoted by bold face capital letters, integral vectors are treated as matrices with one row and denoted by

H4. On power residues and exponential congruences

917

bold face lower case letters. AT is the transpose of A. A congruence a ≡ b mod M or a T ≡ bT mod M means that for a certain x, a−b = xM, a congruence a ≡ b mod (M, N) means that a −b = xM +yN. Instead of mod nI or mod (nI , N ), where I is the identity matrix, we write mod n or mod (n, N), respectively. The congruence a ≡ b mod (n, N ) implies aR ≡ bR mod (n, N R) for any R and a ≡ b mod (n, RN ) for any unimodular R. Lemma 1. For every integral matrix A there exist two unimodular matrices P and Q such that all elements of P AQ outside the diagonal are zero.

Proof. See [8], p. 13.

c

Lemma 2. Let A be an integral matrix, b an integral vector. If for all integral vectors x the congruence xA ≡ 0 mod n implies xbT ≡ 0 mod n then bT ≡ AcT mod n for an integral vector c. Proof. Let A = [aij ] ir , b = [b1 , . . . , br ]. If aij = 0 for i = j then the congruence j s

c

xA ≡ 0 mod n is satisfied by ⎧ n ⎪ , 0, . . . , 0 (1 i q = min(r, s)), 0, . . . , 0, ⎪ ⎪ (n, aii ) ⎪ ⎨ xi = i ⎪ ⎪ (q < i r). . . . , 0, 1, 0, . . . , 0) ⎪(0, ⎪ ⎩ i

It follows that x i bT ≡ 0 mod n (1 i r) and hence 0 mod (n, aii ) (1 i q), bi ≡ 0 mod n (q < i r). Thus bi ≡ aii ci mod n for suitable ci (1 i q) and setting c = [c1 , . . . , cq , 0, . . . , 0] we get bT ≡ AcT mod n. In the general case let P , Q have the property asserted in Lemma 1. If xP AQ ≡ 0 mod n then xP A ≡ 0 mod n hence xP bT ≡ 0 mod n. By the already proved case of our lemma P bT ≡ P AQd T mod n for a suitable integral d and since P is unimodular bT ≡ AQd T mod n. Thus we can take c = dQT .

Lemma 3. Let A and b satisfy the assumptions of Lemma 2, let besides a ≡ 0 mod np−1 and b ≡ 0 mod np −1 , where p is a prime and p n. If for all integral vectors x the congruence xA ≡ a mod n implies xbT ≡ b mod n then bT ≡ Ad T mod n and b ≡ ad T mod n for an integral vector d. Proof. Let A = [aij ]ir, , a = (a01 , . . . , a0s ). As in the proof of Lemma 2 it is enough j s

to consider the case where aij = 0 for i = 0, j . In virtue of that lemma we have bT ≡ AcT mod n, for a certain c.

918


If the congruence xA ≡ a mod n is soluble then we take d = c. Indeed, we have for a suitable x 0 b ≡ x 0 bT ≡ x 0 AcT ≡ acT mod n. If the congruence xA ≡ a mod n is insoluble we have, for a certain j min(r, s), (n, ajj ) /| a0j , hence in view of a0j ≡ 0 mod np−1 p | ajj

(1)

and

p /| a0j .

We determine d from the system of congruences (2)

d ≡ 0 mod np−1 ,

(3)

a0j d ≡ (b − acT ) mod p

and set d = c + (0, . . . , 0, d , 0, . . . , 0). j

It follows from (1) and (2) that Ad T ≡ AcT ≡ bT mod n and by (3) ad T ≡ b mod n. Lemma 4. Let An be a subgroup of the multiplicative group of residues mod n and B the set of all integers b ≡ 1 mod (4, n) the residues of which belong to An . Let d be the greatest common factor of all numbers b − 1, where b ∈ B; n = n1 n2 , where each prime factor of n1 divides d and (n2 , d) = 1. If an integer valued function h on B satisfies the congruences (4)

h(ab) ≡ ah(b) + h(a) mod n,

(5)

h(b) ≡ 0 mod n1 if b ≡ 1 mod n1

then h(b) ≡ c(b − 1) mod n for a suitable c and all b ∈ B. Proof. Let n = p1ν1 p2ν2 · · · psνs be the factorization of n into primes. Assume that pi | n1 for i r, pi | n2 for i > r. Let bi be an element of B such that ordpi (bi − 1) is minimal, equal to, say μi . We have μ

μ

d = p1 1 p2 2 · · · prμr ,

1 μi νi (i r),

μi = 0 (i > r).

For i r let gi be a primitive root mod piνi +1 or if pi = 2, νi 2 then gi = 5. Let, for c a ≡ 0 mod pi , a ≡ 1 mod 4, if pi = 2, νi 2, ind i a be defined by the congruence giindi a ≡ a mod piνi +1 and set (6)

ϕ (p ν ) =

2ν−2 pν−1 (p − 1)

if p = 2, ν 2, otherwise.

919


indi a is determined mod ϕ (piνi +1 ), moreover for μ νi + 1 c

(7)

μ

a ≡ 1 mod pi

if and only if

indi a ≡ 0 mod ϕ (pi ). μ

Since indi a ν ≡ ν indi a mod ϕ (piνi +1 ) it follows from (6) and (7) that

min νi + 1, ordpi (a ν − 1) = min νi + 1, ordpi ν + ordpi (a − 1) and since νi can be arbitrarily large ordpi (a ν − 1) = ordpi ν + ordpi (a − 1),

(8) c

provided pi > 2, a ≡ 1 mod pi or pi = 2, a ≡ 1 mod 4 or pi = 2, aν odd. Since for all a ∈ B ordpj (a − 1) μj ,

(9) we have in particular

ordpj (bin1 d and hence bin1 d

−1

− 1) νj

(1 j r)

≡ 1 mod n1 . By (5) −1

h(bin1 d ) ≡ 0 mod n1 .

(10) c

−1

On the other hand, by (4) h(be ) ≡

(11)

be − 1 h(b) mod n. b−1

−1

μ

The formula (8) gives ordpi (bini d −1) = νi and we infer from (10) and (11) that pi i | h(bi ) for all i r. The same holds clearly for i > r. We now choose c to satisfy the system of congruences −μi

(12) c

c≡

h(bi )pi

−μi

(bi − 1)pi

mod piνi

(1 i s).

For every b ∈ B and i r we have by (6), (7) and (9)

indi bi , ϕ (piνi ) | indi b. Choosing xi so that xi indi bi + indi b ≡ 0 mod ϕ (piνi ) we get (13)

bixi b ≡ 1 mod piνi .

It follows from (8) and (9) with a = bixi b that −νi

ordpj (bixi b)n1 pi − 1 νj

(1 j r)

920


and thus −νi

(bixi b)n1 pi c

≡ 1 mod n1 .

Hence by (5) and (11)

h

−νi

(bixi b)n1 pi

−νi

≡

(bixi b)n1 pi bixi b

−1

−1

h(bixi b) ≡ 0 mod n1 .

However by (8) the cofactor of h(bixi b) above is prime to pi , thus h(bixi b) ≡ b

bixi − 1 h(bi ) + h(b) ≡ 0 mod piνi bi − 1

and by (12) and (13) h(b) ≡ c(b − 1) mod piνi

(14)

(1 i r).

On the other hand, for i > r we have by (4) h(bbi ) ≡ bh(bi ) + h(b) ≡ bi h(b) + h(bi ) mod piνi , hence by (12) (15)

h(b) ≡

h(bi ) (b − 1) ≡ c(b − 1) mod piνi bi − 1

(r < i s),

and the lemma follows from (14) and (15).

Lemma 5. Let An be a subgroup of the multiplicative group of residues mod n and A the set of all integers the residues of which belong to An . Let M be a non-singular square matrix such that nM −1 is integral. Let f and g be functions on A into set of integral vectors or integers respectively, satisfying the conditions (16)

f (a) ≡ f (b),

g(a) ≡ g(b) mod n if a ≡ b mod n,

(17)

f (ab) ≡ af (b) + f (a) mod M,

(18)

g(ab) ≡ ag(b) + g(a) mod n.

If for all a ∈ A the congruence f (a) ≡ 0 mod (a − 1, M) implies the congruence g(a) ≡ 0 mod (a − 1, n) c

then there exist vectors u1 and u2 and an integer c such that for all a ∈ A, a ≡ 1 mod (4, n) g(a) ≡ c(a − 1) + f (a)nM −1 uT1 mod n and for all a ∈ A g(a) ≡ f (a)uT2 mod (2, n),

MuT2 ≡ 0 mod (2, n).


921

Proof. By Lemma 1 there exist unimodular matrices P and Q such that ⎤ ⎡ e1 0 . . . 0 ⎢ 0 e2 . . . 0 ⎥ ⎥ ⎢ . (19) P MQ = ⎢ . . . . . ... ⎥ ⎦ ⎣ .. .. 0

0 . . . ek

Since M is non-singular the entries ei are non-zero and since nM −1 is integral we have ei | n (1 i k). Any congruence x ≡ 0 mod (m, P MQ) where x = [x1 , . . . , xk ] is equivalent to the system of congruences xi ≡ 0 mod (m, ei ) (1 i k), which will be frequently used in the sequel. Let n1 , n2 have the meaning defined in Lemma 4. For each prime pi | n2 there exists bi ∈ A such that bi ≡ 1 mod pi . If piνi n2 we get by (17) for all a ∈ A f (a)(bi − 1) ≡ f (bi )(a − 1) mod M,

(20)

f (a)(bi − 1)Q ≡ 0 mod (a − 1, P MQ),

f (a)Q ≡ 0 mod (a − 1, piνi ), P MQ ,

f (a)Q ≡ 0 mod (a − 1, n2 ), P MQ .

Let a1 , . . . , ar represent all residue classes of An congruent to 1 mod n1 . If x1 , . . . , xr are integers not necessarily positive and a ≡ a1x1 · · · arxr mod n we have by (16), (17) and (18) (21)

f (a) ≡ x1 f (a1 ) + . . . + xr f (ar ) mod (n1 , M),

(22)

g(a) ≡ x1 g(a1 ) + . . . + xr g(ar ) mod n1 .

Let us set

⎡

⎤ f (a1 ) ⎢f (a2 )⎥ ⎢ ⎥ F = ⎢ . ⎥, ⎣ .. ⎦

g = [g(a1 ), . . . , g(ar )].

f (ar ) By (21) f (a) ≡ xF mod (n1 , P M) and (23)

f (a)Q ≡ xF Q mod (n1 , P MQ).

Now suppose that for a vector x we have (24)

xF nM −1 ≡ 0 mod n1 .

Then n2 xF Q ≡ 0 mod P MQ

922


and in view of (19) xF Q ≡ 0 mod (n1 , P MQ). By (23) we can write the above congruence in the form f (a)Q ≡ 0 mod (n1 , P MQ). This together with (20) gives

f (a)Q ≡ 0 mod (a − 1, n), P MQ

and since ei | n we infer that f (a)Q ≡ 0 mod (a − 1, P MQ), f (a) ≡ 0 mod (a − 1, M). By the assumption g(a) ≡ 0 mod (a − 1, n) and by (22) xg T ≡ 0 mod n1 .

(25)

Thus (24) implies (25) and by Lemma 2 we get g T ≡ F nM −1 uT1 mod n1 for a suitable u1 . On comparing the components it follows g(ai ) ≡ f (ai )nM −1 uT1 mod n1

(1 i r).

However every a ≡ 1 mod n1 satisfies a ≡ ai mod n for a suitable i r, thus by (16) the function h(a) = g(a) − f (a)nM −1 uT1 c

satisfies h(a) ≡ 0 mod n1 for all a ≡ 1 mod n1 . By (17) and (18) it satisfies also h(ab) ≡ ah(b)+h(a) mod n and by Lemma 4 we infer that for all a ∈ A, a ≡ 1 mod (4, n), h(a) ≡ c(a − 1) mod n for suitable c. This gives the first assertion of the lemma. In order to prove the second one it is enough to consider the case where 4 | n and A contains an integer a¯ 0 ≡ −1 mod 4. Let n0 be the greatest odd factor of n1 and a0 = a¯ 0n0 . Clearly a0 ≡ −1 mod 4 and by (8) a0 ≡ 1 mod n0 (a¯ 0 − 1). Hence by (17) and (18) (26) (27)

a0 − 1 f (a¯ 0 ) ≡ 0 mod (n0 , M), a¯ 0 − 1 a0 − 1 g(a¯ 0 ) ≡ 0 mod n0 . g(a0 ) ≡ a¯ 0 − 1

f (a0 ) ≡


923

Let a1 , . . . , as represent all residue classes of A congruent to 1 mod 4n0 . If a ≡ a0 a1x1 · · · asxs mod n

(28) we have by (16), (17) and (18)

(29) f (a) ≡ f (a0 ) + f (a1x1 · · · asxs ) ≡ f (a0 ) + x1 f (a1 ) + . . . + xs f (as ) mod (4n0 , M), (30)

g(a) ≡

Let us set

g(a0 ) + g(a1x1

· · · asxs ) ≡ g(a0 ) + x1 g(a1 ) + . . . + xs g(as ) mod 4n0 .

⎡

⎤ f (a1 ) ⎢ ⎥ F 0 = ⎣ ... ⎦ ,

g 0 = [g(a1 ), . . . , g(as )].

f (as ) By (29) f (a) ≡ f (a0 ) + xF 0 mod (4n0 , P M) and c

f (a)Q ≡ f (a0 )Q + xF 0 Q mod (4n0 , P MQ).

(∗)

Now suppose that for a vector x we have (31)

xF 0 R + f (a0 )R ≡ 0 mod 2n0 ,

where

⎡ ⎢ ⎢ R = Q⎢ ⎢ ⎣

2n0 ... (2n0 , e1 ) .. .. . . 0

...

⎤ 0 .. . 2n0 (2n0 , ek )

⎥ ⎥ ⎥. ⎥ ⎦

Then xF 0 Q + f (a0 )Q ≡ 0 mod (2n0 , P MQ) c

and since by (28) (2n0 , ei ) = (a − 1, n1 , ei ) we have by (∗)

f (a)Q ≡ 0 mod (a − 1, n1 ), P MQ . This together with (20) gives f (a)Q ≡ 0 mod (a − 1, P MQ), f (a) ≡ 0 mod (a − 1, M). By the assumption g(a) ≡ 0 mod (a − 1, n) and by (30) (32)

xg T0 + g(a0 ) ≡ 0 mod 2n0 .

924


Thus (31) implies (32). On the other hand, by the already proved part of the lemma and since a − 1 ≡ 0 mod 2n0 , g T0 ≡ F 0 nM −1 uT1 mod 2n0 .

c

Also

nM −1

⎡ n(2n , e ) 0 1 ... 0 ⎢ 2n0 e1 ⎢ .. .. .. = R⎢ . . . ⎢ ⎣ n(2n0 , ek ) 0 ... 2n0 ek

⎤ ⎥ ⎥ ⎥P ⎥ ⎦

and finally by (26) and (27) f (a0 )R ≡ 0 mod n0 ,

g(a0 ) ≡ 0 mod n0 .

The assumptions of Lemma 3 are satisfied with p = 2 and we infer that for a suitable vector d g T0 ≡ F 0 Rd T mod 2n0 ,

g(a0 ) ≡ f (a0 )Rd T mod 2n0 .

Setting u2 = dR T we get

(33)

⎡ 2n e 0 1 ... ⎢ (2n0 , e1 ) ⎢ .. .. MuT2 = MRd T = P −1 ⎢ . . ⎢ ⎣ 0 ...

⎤ 0 .. . 2n0 ek (2n0 , ek )

⎥ ⎥ T ⎥ d ≡ 0 mod 2. ⎥ ⎦

On the other hand, for each i s g(ai ) ≡ f (ai )uT2 mod 2n0 c

and since every a ∈ A, a ≡ 1 mod 2n0 is congruent to ai or to a0 ai mod n we infer from (16), (29) and (30) that g(a) ≡ f (a)uT2 mod 2n0

c

for all a ∈ A, a ≡ 1 mod 2n0 . By (8) for any a ∈ A, a n0 ≡ 1 mod n0 and hence g(a n0 ) − f (a n0 )uT2 ≡ 0 mod 2n0 . On the other hand by (17), (18) and (33) g(a n0 ) − f (a n0 )uT2 ≡ and since

a n0 − 1 g(a) − f (a)uT2 mod 2 a−1

a n0 − 1 is odd a−1 g(a) ≡ f (a)uT2 mod 2.

Lemma 6. Let K be an arbitrary field, n a positive integer not divisible by the characteristic of K, ni divisors of n and α1 , . . . , αk , β non-zero elements of K. Let G be the Galois


925

√ √ group of the field K(ζn , n1 a1 , . . . , nk ak ) and assume that every element of G which fixes one of the fields K(ξ1 , . . . , ξk ), where ξini = αi , fixes at least one η with ηn = β. Then for any choice of numbers ξi and η and for suitable exponents m0 , m1 , . . . , mk , q1 , . . . , qk ζnm0 ηξ1m1 · · · ξkmk ∈ K(ζ4 ), and if n ≡ 0 mod 2, q

q

ηn/2 ξ1 1 · · · ξk k ∈ K,

2qi ≡ 0 mod ni (1 i k).

Proof. Let us choose some ξi and η. It is clear that η ∈ K(ζm , ξ1 , . . . , ξk ) = L. The elements σ of G act on L in the following way σ (ζn ) = ζnα ,

σ (ξi ) = ζntii ξi .

G contains a normal subgroup H = {σ : σ (ζn ) = ζn }. The vectors [t1 , . . . , tk ] such that for a σ ∈ H σ (ξi ) = ζntii ξi

(1 i k)

constitute a lattice Λ. The fundamental vectors of Λ written horizontally form a matrix, say M. Since the vectors [n1 , 0, . . . , 0], [0, n2 , 0, . . . , 0], …, [0, . . . , 0, nk ] belong to Λ, M is non-singular and ⎤ ⎡ 0 n/n1 . . . ⎢ .. ⎥ .. (34) nM −1 = SN for S = ⎣ ... . . ⎦ 0

. . . n/nk

and a certain integral matrix N . Let A be the set of all integers a such that for a σ ∈ G : σ (ζn ) = ζna . The residues of a ∈ A mod n form a subgroup An of the multiplicative group of residues mod n, isomorphic to G /H , and every integer the residue of which belongs to An is in A. Let f (1) = 0, for an a ∈ A, 1 < a < n, f (a) = [f1 (a), . . . , fk (a)] be any vector such that for a σ ∈ G : f (a)

σ (ζn ) = ζna ,

σ (ξi ) = ζnii ξi (1 i k) a and for all the other a let f (a) = f a − n . Thus f (a) = f (b) for a ≡ b mod n. n On the other hand, for every σ ∈ G we have (35)

σ (ζn ) = ζna ,

f (a)+ti

σ (ξi ) = ζnii

ξi

for a suitable a ∈ A and a suitable [t1 , . . . , tk ] ≡ 0 mod M. Since G is a group with respect to superposition we get for all a, b ∈ A f (ab) ≡ af (b) + f (a) mod M. Now for every pair a, t where a ∈ A, t ≡ 0 mod M we define σ by (35) and ϕ(a, t) by the condition (36)

σ (η) = ζnϕ(a,t) η,

0 ϕ(a, t) < n.

926


Since σ2 σ1 (η) = σ2 σ1 (η) we get

(37) ϕ a1 a2 , a2 t 1 + f (a1 ) + t 2 + f (a2 ) − f (a1 a2 ) ≡ a2 ϕ(a1 , t 1 ) + ϕ(a2 , t 2 ) mod n and in particular ϕ(1, t 1 + t 2 ) ≡ ϕ(1, t 1 ) + ϕ(1, t 2 ) mod n c

It follows that ϕ(1, 0) = 0 and if t = xM,

for

t 1 ≡ t 2 ≡ 0 mod M.

⎤ m1 ⎢m2 ⎥ ⎢ ⎥ M=⎢ . ⎥ ⎣ .. ⎦ ⎡

mk then ϕ(1, t) ≡ x1 ϕ(1, m1 ) + . . . + xk ϕ(1, mk ) mod n.

c

c

Since tS ≡ 0 mod n implies ϕ(1, t) = ϕ(1, 0) = 0 we infer by Lemma 2 that for an integral vector c ⎡ ⎤ ϕ(1, m1 ) ⎢ ⎥ .. T ⎣ ⎦ ≡ MSc mod n . ϕ(1, mk ) and thus ϕ(1, t) ≡ tScT mod n. Hence by (37) with a2 = 1, t 1 = 0 ϕ(a, t) ≡ ϕ(a, 0) + tScT mod n.

(38)

The condition (37) takes the form

ϕ(a1 a2 , 0) + (a2 t 1 + t 2 )ScT + a2 f (a1 ) + f (a2 ) − f (a1 a2 ) ScT ≡ a2 ϕ(a1 , 0) + a2 t 1 ScT + ϕ(a2 , 0) + t 2 ScT mod n. It follows that the function g(a) = ϕ(a, 0) − f (a)ScT

(39)

satisfies the conditions g(a) ≡ g(b) mod n for a ≡ b mod n and g(ab) ≡ ag(b) + g(a) mod n. Now suppose that for an a ∈ A we have f (a) ≡ 0 mod (a − 1, M).

(40)

It follows that for a suitable v = [v1 , . . . , vk ] f (a) − (a − 1)v ≡ 0 mod M and G contains σ such that c

σ (ζn ) = ζna ,

iξ σ (ξi ) = ζn(a−1)v i i

(1 i k).


927

We have i ξ ) = ζ −vi ξ σ (ζn−v i i ni i

(1 i k),

thus by the assumption σ (ζn−v0 η) = ζn−v0 η for a suitable v0 . We obtain from (36), (38) and (39)

−v0 a + ϕ a, (a − 1)v − f (a) ≡ −v0 mod n,

ϕ(a, 0) + (a − 1)v − f (a) ScT ≡ (a − 1)v0 mod n, g(a) ≡ 0 mod (a − 1, n).

(41) c

Thus (40) implies (41) and we infer by Lemma 5 that for all a ∈ A, a ≡ 1 mod (4, n) g(a) ≡ −m0 (a − 1) + f (a)nM −1 uT1 mod n

(42) c

and for all a ∈ A (43)

g(a) ≡ f (a)uT2 mod (2, n),

MuT2 ≡ 0 mod (2, n).

Set m = [m1 , . . . , mk ] = −c − u1 N T , where N is defined by (34). If σ is defined by (35) and a ≡ 1 mod (4, n) we get σ (ζnm0 ηξ1m1 · · · ξkmk ) = ζne1 ηξ1m1 · · · ξkmk , where by (36), (38), (39) and (42)

e1 = am0 + ϕ(a, t) + f (a) + t SmT

≡ am0 + g(a) + f (a) + t ScT + f (a) + t SmT

≡ am0 − m0 (a − 1) + f (a)nM −1 uT1 − f (a) + t SN uT1 ≡ m0 − tnM −1 uT1 ≡ m0 mod n. Thus σ (ζ4 ) = ζ4 implies σ (ζnm0 ηξ1m1 · · · ξkmk ) = ζnm0 ηξ1m1 · · · ξkmk and the first assertion of the lemma follows. In order to prove the second one assume 2 | n and set n n (44) q = [q1 , . . . , qk ] = c + u2 S −1 . 2 2 q is integral since by (34) and (43) (nu2 S −1 )T = nS −1 uT2 = N MuT2 ≡ 0 mod 2. If σ is defined by (35) we get q

q

q

q

σ (ηn/2 ξ1 1 · · · ξk k ) = ζne2 ηn/2 ξ1 1 · · · ξk k ,

928


where by (36), (38), (39) and (43)

n ϕ(a, t) + f (a) + t Sq T 2

n

n n n ≡ g(a) + f (a) + t)ScT + f (a) + t ScT + f (a) + t uT2 ≡ 0 mod n. 2 2 2 2 It follows that e2 =

q

q

ηn/2 ξ1 1 · · · ξk k ∈ K. Also, by (44), 2qi ≡ 0 mod ni .

Lemma 7. Let K be an arbitrary field of characteristic different from 2 and τ the greatest integer such that ζ2τ + ζ2−1 τ ∈ K if there are only finitely many of them, otherwise τ = ∞. If ϑ ∈ K(ζ4 ), ϑ n ∈ K, then at least one of the following four conditions is satisfied for a suitable γ ∈ K: (i) ϑ n = γ n , (ii) n ≡ 0 mod 2τ , ϑ n = −γ n ,

n/2 n (iii) n ≡ 2τ mod 2τ +1 , ϑ n = − ζ2τ + ζ2−1 γ , τ +2

n/2 n (iv) n ≡ 0 mod 2τ +1 , ϑ n = ζ2τ + ζ2−1 + 2 γ . τ Remark. If n is a power of 2 the lemma is contained in Satz 2 of [4]. Proof. Set ζ4 = i, ϑ = α + βi, α, β ∈ K. If i ∈ K we have (i); if i ∈ / K then (α + βi)n = κ ∈ K implies (α − βi)n = κ hence α + βi = ζnν (α − βi),

(45) ζnν + ζn−ν =

α − βi 2(α 2 − β 2 ) α + βi + = ∈ K. α − βi α + βi α2 + β 2

It follows that the only conjugate of ζnν over K is ζn−ν and the only possible conjugates ν are of ζ2n ε2 ν ε1 ζ2n

(ε1 = ±1, ε2 = ±1).

2 = ζ .) (ζ2n is chosen so that ζ2n n Let

(46)

μ = ord2 2n/(2n, ν).

Then (47) If σ is an automorphism of (48)

ν

ζ2μ = ζ2n , ν ) K(ζ2n

≡ 1 mod 2.

and ε2 ν ν ) = ε1 ζ2n σ (ζ2n

we get (49)

σ (ζ2μ ) = ε1 ζ2εμ2 .


929

If μ = 2 we have ζnν = 1, by (45) ζν + 1 , α = βi nν ζn − 1 n n 2i 2i n n n ν = β (−1) ϑ =β ν − ζ −ν ζnν − 1 ζ2n 2n

(50)

ν ) over K and by (48) and (49) for all automorphisms σ of K(ζ2n 2i 2i = ν σ ν −ν −ν . ζ2n − ζ2n ζ2n − ζ2n

2i −ν ∈ K and by (46) and (50) we get (i) if ν is even and (ii) if ν is odd. − ζ2n If μ = 2, ζnν = −1 and by (45)

Thus

(51)

ν ζ2n

ζν − 1 , βi = α nν ζn + 1 n n 2 2 = α n (−1)ν ν . ϑ n = αn ν −ν ζn + 1 ζ2n + ζ2n

ν ), If σ is an automorphism of K(ζ2n −ν ν + ζ2n )(ζ2μ + ζ2−1 δ = (ζ2n μ )

we have by (48) and (49) σ (δ) = δ. Thus δ ∈ K and since ζ2μ + ζ2−1 μ = 0 we get from (51) n n 2α (52) ϑ n = (−1)ν (ζ2μ + ζ2−1 . μ ) δ On the other hand, μ τ + 1. This is clear if μ = 0 and if μ > 0 it follows from (47) that α + βi α − βi ν −ν ζ2μ−1 + ζ2−1 = + ∈K μ−1 = ζn + ζn α − βi α + βi thus μ − 1 τ . Denoting by γ a suitable element of K we can draw from (46) and (52) the following conclusions: If μ τ , ν ≡ 0 mod 2 then ϑ n = γ n ; if μ τ , ν ≡ 1 mod 2 then ϑ n = −γ n ; n ≡ 0 mod 2τ ;

n/2 n γ and n ≡ 2τ mod 2τ +1 ; if μ = τ + 1, ν ≡ 1 mod 2, then ϑ n = − ζ2τ + ζ2−1 τ +2

n/2 γ n, if μ = τ + 1, ν ≡ 0 mod 2, then ϑ n = ζ2τ + ζ2−1 τ +2 which correspond to the conditions (i), (ii), (iii), (iv), respectively.

Lemma 8. Let K be an algebraic number field, fi (x) polynomials over K with integral coefficients and discriminants Di and p a prime ideal of K not dividing D1 · · · Dk . The k

930


congruences fi (x) ≡ 0 mod p (1 i k) are soluble mod p if and only if p has a prime factor of degree one in at least one field K(ξ1 , . . . , ξk ), where fi (ξi ) = 0. Proof. The sufficiency of the condition is obvious. In order to prove the necessity we proceed by induction. For k = 1 the condition follows from Dedekind’s theorem applied to a suitable irreducible factor of f . Suppose that the condition holds for less than k polynomials and that the k congruences fi (x) ≡ 0 mod p are soluble. Then p has a prime factor P of degree one in K(ξ1 , . . . , ξk−1 ), where ξi is a certain zero of fi (x). The congruence fk (x) ≡ 0 mod P being soluble it follows by Dedekind’s theorem that P has a prime factor of relative degree one in at least one field K(ξ1 , . . . , ξk ) where fk (ξk ) = 0. This factor is of degree one over K, which completes the proof.

Lemma 9. If K is an algebraic number field, τ is defined as in Theorem 1 and ν > τ then ν 2ν−1 mod p is soluble for all prime ideals p of K. the congruence x 2 ≡ (ζ2τ + ζ2−1 τ + 2)

Proof. See [5], p. 156.

Proof of Theorem 1. Necessity. Suppose that the Galois group G of the extension √ √ L = K ζn , n1 α1 , . . . , nk αk of K contains an element σ which fixes one of the fields K(ξ1 , . . . , ξk ), where ζini = αi , but does not fix any η with ηn = β. By Frobenius density theorem prime ideals p of K belonging to the division of σ in G have a positive density. Every such prime ideal p has a prime factor of degree one in K(ξ1 , . . . , ξk ) (1 i k) where ξ1 , . . . , ξk are suitably chosen roots of ξini = αi , but it has no prime factor of degree one in any of the fields K(η), where ηn = β. By Lemma 8, for almost all p’s the congruences x ni ≡ αi mod p are soluble and the congruence x n ≡ β mod p is insoluble. The obtained contradiction shows that the assumptions of Lemma 6 are satisfied. Let us choose some values of ξ1 , . . . , ξk and η. By Lemma 6 there exist integers m0 , m1 , . . . , mk , q1 , . . . , qk such that ϑ = ζnm0 ηξ1m1 · · · ξkmk ∈ K(ζ4 ) and if n ≡ 0 mod 2 (53)

q

q

κ = ηn/2 ξ1 1 · · · ξk k ∈ K,

2qi ≡ 0 mod ni

Since ϑn = β

k

nmi /ni

αi

∈K

i=1

we have by Lemma 7 for a suitable γ ∈ K either (54)

β

k

nmi /ni

= γ n,

nmi /ni

= −γ n

αi

i=1

or n ≡ 0 mod 2τ (55)

β

k i=1

αi

(1 i k).


931

or n ≡ 2τ mod 2τ +1 (56)

β

k

n/2 n = − ζ2τ + ζ2−1 γ τ +2

nmi /ni

αi

i=1

or n ≡ 0 mod 2τ +1 (57)

β

k

nmi /ni

αi

n/2 n = ζ2τ + ζ2−1 γ . τ +2

i=1

(54) and (57) correspond to the conditions (i) and (iv), respectively. If n ≡ 0 mod 2, (55) reduces to (54). If n ≡ 0 mod 2 we get from (53) k

β

2qi /ni

αi

= κ 2,

κ ∈ K.

i=1

This together with (55) and (56) gives on division k

αili = −λ2 ,

where li =

i=1

However if ni is odd, li is even, thus

nmi − 2qi , λ ∈ K. ni

αili = −δ 2 ,

δ ∈ K.

ni even

Sufficiency. The sufficiency of the condition (i) is obvious. To show that (ii) and (iii) are sufficient we argue as follows. The equality l αii = −δ 2 ni even

implies that for any choice of ξi satisfying ξini = αi ζ4 ∈ K(ξ1 , . . . , ξk ). Since

1−2τ −2 τ −2 2ζ2τ = ζ2τ + ζ2−1 + ζ2−1+2 τ + ζ 4 ζ2 τ τ

ζ2τ + ζ2−1 τ ∈ K,

we have K(ζ4 ) = K(ζ2τ ). Hence ζ2τ ∈ K(ξ1 , . . . , ξk ). Let ν = ord2 n. The conditions β

k

nmi /ni

αi

= −γ n ,

ν < τ,

i=1

and β

k i=1

nmi /ni

αi

n/2 n = − ζ2τ + ζ2−1 γ , τ +2

ν = τ,

932


can be rewritten for a suitable η and as η

k

ξimi = ζ2τ γ

and η

i=1

k

ξimi = (ζ2τ + 1)γ ,

i=1

respectively. It follows that η ∈ K(ξ1 , . . . , ξk ) and any ideal p which has a prime factor of degree one in K(ξ1 , . . . , ξk ) has a prime factor of degree one in K(η). Since this is valid for any choice of ξi and a suitable η, we infer by Lemma 8 that the solubility of x n ≡ αi (1 i k) implies the solubility of x n ≡ β mod p. The sufficiency of condition (iv) follows from Lemma 9, since the solubility of the congruence

2ν−1 ν x 2 ≡ ζ2τ + ζ2−1 mod p τ +2 clearly implies the solubility of the congruence

n/2 x n ≡ ζ2τ + ζ2−1 mod p. τ +2 If ζ4 ∈ K then ζ2τ ∈ K and the equalities τ (−1)n/2 ζ2τ

−1 = ζ2nν+1 if ν < τ, n/2 n + ζ2−1 = ζ2τ + 1 , τ +2

if

ντ

show that the conditions (ii), (iii), (iv) imply (i). If τ = 2 and n ≡ 0 mod 2τ we have either n ≡ 1 mod 2 in which case −γ n = (−γ )n or n ≡ 2 mod 4. In the latter case we get from (ii) β

k

nmi /ni

αi

l n/2

αii

= (γ δ)n ,

ni even

i=1

which leads to (i). The proof is complete.

Proof of Corollary 1 follows at once from Lemma 8.

Proof of Corollary 2. If the congruences x n ≡ α mod p and x n ≡ β mod p are for almost all p simultaneously soluble or insoluble, we have by Theorem 1 the following seven possibilities: n

α = βt ,

(58) (59) (60) (61) (62)

n

n ≡ 0 mod 2τ ,

n

α = β t = −δ 2 ,

n ≡ 2τ mod 2τ +1 ,

n

n

β = −α s ;

α = β t = −δ 2 ,

n ≡ 0 mod 2τ +1 , n ≡ 0 mod 2τ ,

n

β = αs ;

n

α = βt ,

α = −β t = −δ12 ,

n

β = −ωα s ; n

β = ωα s ; n

β = −α s = −δ22 ;

933


(63)

n ≡ 2τ mod 2τ +1 ,

n

n ≡ 0 mod 2τ +1 ,

(64)

n

α = −ωβ t = −δ12 ,

β = −ωα s = −δ22 ;

n

n

α = ωβ t ,

β = ωα s

and three other possibilities obtained by the permutation of α and β in (59), (60) and (61).

n/2 n . Moreover, in Here γ = δ means that γ /δ is an nth power in K and ω = ζ2τ +ζ2−1 τ +2 / K. Let us choose an integer x such that u = s +(st −1)x (59) to (64) it is assumed that ζ4 ∈

is prime to n. If s is even or t is odd x will be chosen odd, which is possible because then s + st − 1, 2(st − 1) = 1. n

n

n

Now, (58) gives α = α st , α st−1 = 1, β = α u ; n

n

n

(59) gives t ≡ 1 mod 2, α = −α st , α st−1 = −1, β = α u ; n

n

n

(60) gives t ≡ 1 mod 2, α = −ωα st , α st−1 = −ω, β = α u ; n

n

n

(61) gives α = ωt α st , α st−1 = ωt , β = ωt+1 α u ; (62) gives s ≡ t ≡ 0 mod 2. Indeed, if for instance t ≡ 1 mod 2 then −δ12 = −β t = δ22t n n n and ζ4 ∈ K. If s ≡ t ≡ 0 mod 2 then α = −α st , α st−1 = −1, β = α u . n

n

(63) gives like (62) that s ≡ t ≡ 0 mod 2. In that case α = −ωα st , α st−1 = −ω, β = αu. n

n

n

n

Finally (64) gives α = ωt+1 α st , α st−1 = ωt+1 , β = ωt+x+1 α u . n

n

On the other hand, if β = α u or n ≡ 0 mod 2τ +1 and β = ωα u , where (u, n) = 1 then n n also α = β ν or α = ωβ ν , respectively and by Theorem 1 the congruences x n ≡ α mod p n and x ≡ β mod p are simultaneously soluble or insoluble for almost all prime ideals p of K.

To prove Theorem 2 we need two lemmata both due to Skolem. Lemma 10. In every algebraic number field K there exists an infinite sequence of elements l d πj such that every element of K is represented uniquely in the form ζ πj j , where ζ is j =1

a root of unity and dj are rational integers.

Proof. See [9].

Lemma 11. If a system of linear congruences is soluble for all moduli, then the corresponding system of equations is soluble in rational integers.

Proof. See [7]. Proof of Theorem 2. Let αi = ζwai0

l j =1

a

πj ij ,

β = ζwb0

l j =1

b

πj j ,

934


where w is the number of roots of unity contained in K, πj have the property asserted in Lemma 10 and aij , bj are rational integers. If the congruence α1x1 · · · αkxk ≡ β mod p is soluble for almost all p then for every positive integer n the solubility of the k congruences x n ≡ αi mod p (1 i k) implies the solubility of x n ≡ β mod p. It follows hence by Theorem 1 with n = 2τ +1 m that for every positive integer n there exist γ ∈ K and rational integers m1 , . . . , mk such that βα1m1 · · · αkmk = γ m . By Lemma 10 the last equality implies for a suitable m0 b0 +

k

ai0 mi + wm0 ≡ 0 mod m,

i=1

bj +

k

aij mi ≡ 0 mod m

(1 j l).

i=1

By Lemma 11 there exist rational integers m0 , . . . , mk such that b0 +

k

ai0 mi + wm0 = 0,

i=1

bj +

k

aij mi = 0

(1 j l)

i=1

and this gives β=

k

αimi .

i=1

The above proof is modelled on Skolem’s proof ([7]) of his theorem that the solubility of the congruence α1x1 · · · αkxk ≡ β mod m for all moduli implies the solubility of the corresponding equation. That proof uses instead of Theorem 1 the case D = 1 of the following Lemma 12. Let ξ0 = ζw , ξ1 , . . . , ξt be any t distinct terms of the sequence πj . For any positive integer m there exists μ ∈ K prime to D such that the congruence y

y

y

ξ0 0 ξ1 1 · · · ξt t ≡ 1 mod μ implies y0 ≡ 0 mod w, y1 ≡ . . . ≡ yt ≡ 0 mod m. Skolem’s proof of the above lemma given only in the case of fields with class number one is defective because he claims the existence of prime ideals p0 , . . . , pl of K such that j x m ≡ ξr mod ps is soluble for r = s and x m ≡ ξr mod pr is insoluble for j ≡ 0 mod m,

H4. On power residues and exponential congruences c

935

r = 0 and j ≡ 0 mod (m, w), r = 0. The assertion is false for K = Q, t = 1, ξ1 = 2, m = 4. Proof of Lemma 12. We can assume without loss of generality that m ≡ 0 mod 2τ +1 w. For every p | m set n = m(p, 2). Suppose that the solubility of (65)

x n ≡ ξi mod p

(i = r = 0)

implies the solubility of (66)

m/p

x n ≡ ξr

for almost all p. Then by Theorem 1 m/p

ξr

mod p

ξimi = γ n/2

i=r

for suitable γ ∈ K and suitable exponents mi . We get m n ≡ 0 mod , p 2

m m(p, 2) ≡ 0 mod , p 2

which is impossible. The obtained contradiction shows that for a certain prime ideal p prime to D the congruences (65) are soluble, but (66) is insoluble. Denoting this prime ideal by pp,r we infer from ξ0x0 ξ1x1 · · · ξtxt ≡ 1 mod pp,r that

m m(p, 2), xr /| , p

hence ordp xr ordp m. If p | w, suppose that the solubility of the congruences (67)

x n ≡ ξi mod p

(1 i t)

implies the solubility of the congruence x n ≡ ζp mod p

(68) for almost all p. Then by Theorem 1

ζp

t

ξimi = γ n/2

i=1

for suitable γ ∈ K and suitable exponents mi . We get n w w w(p, 2) ≡ 0 mod ,w , ≡ 0 mod . p 2 p 2

936


The obtained contradiction shows that for a certain prime ideal p prime to D the congruences (67) are soluble, but (68) is insoluble. Denoting this prime ideal by pp,0 we infer from ξ0x0 ξ1x1 · · · ξtxt ≡ 1 mod pp,0 that (x0 , w) /|

w p

hence ordp x0 ordp w. For μ we can choose any number prime to D divisible by k

pp,r

p|m r=1

pp,0 .

p|w

Proof of Theorem 3. Let for i h, j k αij =

t

a

βi =

ξs ij s ,

s=0

t

ξsai0s

s=0

in the notation of Lemma 12 and let m, D be positive integers. Let μ be a modulus with the property asserted in Lemma 12. Then the congruences k

x

αijj ≡ βi mod μ

(i = 1, . . . , h)

j =1

imply k

aij 0 xj ≡ ai00 mod w

(i = 1, . . . , h),

j =1 k

aij s xj ≡ ai0s mod m

(i = 1, . . . , h; s = 1, . . . , t)

j =1

and by Lemma 11 there exist rational integers xj (j = 1, . . . , k), yi (i = 1, . . . , h) satisfying the system of equations k

aij 0 xj = ai00 + wyi

(i = 1, . . . , h),

j =1 k j =1

aij s xj = ai0s

(i = 1, . . . , h; s = 1, . . . , t).


937

Hence k

x

αijj = βi

(i = 1, . . . , h).

j =1


We proceed to the example showing that Theorem 3 is no longer valid if the solubility for all moduli prime to D is replaced by the solubility for all prime moduli. Let us consider the system 2x 3y ≡ 1 mod p, 2y 3z ≡ 4 mod p.

(69)

For p = 2, 3 it has the solution (x, y, z) = (0, 1, 0), (0, 0, 0), respectively. For other p it is equivalent to the system (70)

x ind 2 + y ind 3 ≡ 0 mod p − 1, y ind 2 + z ind 3 ≡ 2 ind 2 mod p − 1,

where indices are taken with respect to a fixed primitive root mod p. Now

(ind 2)2 , (ind 3)2 | ind 2 ind 3. Hence

(ind 2)2

, ind 3 ind 2 (ind 2, ind 3)

and the equation t

(ind 2)2 + z ind 3 = 2 ind 2 (ind 2, ind 3)

is soluble in integers. The numbers x =

t ind 2 −t ind 3 ,y = and z satisfy (ind 2, ind 3) (ind 2, ind 3)

the system (70) and hence also (69).

References [1] H. Flanders, Generalization of a theorem of Ankeny and Rogers. Ann. of Math. (2) 57 (1953), 392–400. [2] I. Gerst, On the theory of n-th power residues and a conjecture of Kronecker. Acta Arith. 17 (1970), 121–139. [3] H. Hasse, Bericht über neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper II. Jahresber. der Deutschen Mathematiker-Vereinigung 6 (1930); reprint: PhysicaVerlag, Würzburg–Wien 1965. [4] −−, Zum Existenzsatz von Grunwald in der Klassenkörpertheorie. J. Reine Angew. Math. 188 (1950), 40–64.

938


[5] H. B. Mann, Introduction to Algebraic Number Theory. The Ohio Univ. Press, Columbus 1955. [6] A. Schinzel, A refinement of a theorem of Gerst on power residues. Acta Arith. 17 (1970), 161–168. [7] Th. Skolem, Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen. Vid. Akad. Avh. Oslo I 1937 nr. 12. [8] −−, Diophantische Gleichungen. Berlin, 1938. [9] −−, On the existence of a multiplicative basis for an arbitrary algebraic field. Norske Vid. Selsk. Forh. (Trondheim) 20 (1947), no. 2, 4–7.

Originally published in Acta Arithmetica XXXII (1977), 245–274


Abelian binomials, power residues and exponential congruences* In memory of Marceli Stark

This paper supplements the results of [6] concerning power residues and extends those pertaining to exponential congruences. We begin however with the study of binomials. G. Darbi [1] and E. Bessel-Hagen (cf. [10], p. 302) have found all binomials x n − a normal over the rational field Q. (Their argument extends to fields K such that a primitive n-th root of unity ζn is of degree ϕ(n) over K.) We shall do the same for an arbitrary field and n equal to a prime power. In fact, we shall prove Theorem 1. Let K be a field, p a prime different from the characteristic of K. A binomial ν x p − α is the product of factors normal over K if and only if at least one of the following conditions is satisfied for a suitable integer λ and a suitable γ ∈ K: min(ω,ν)

ν

(i) α p = γp ; (ii) p = 2, ω = 1, ν τ , α = −γ 2 ;

( (iii) p = 2, ω = 1, ν = τ + 1, α = −γ 2 , − ζ2τ + ζ2−1 τ + 2 ∈ K; 2λ 2λ+1

γ , 1 λ τ − 2; (iv) p = 2, ω = 1, ν = τ + 1, α = − ζ2τ + ζ2−1 τ +2 2ν−2 2ν−1

−1 γ . (v) p = 2, ω = 1, ν τ + 2, α = − ζ2τ + ζ2τ + 2 ω Here ω is the greatest integer such that ζp ∈ K if there are only finitely many of them, ω = ∞ otherwise; τ is the greatest integer such that ζ2τ + ζ2−1 τ ∈ K if there are only finitely many of them, τ = ∞ otherwise. ( / K, λ = 1, If the binomial in question is irreducible (iv) implies − ζ2τ + ζ2−1 τ +2 ∈ τ 3; (v) implies τ = 2. Theorem 2. Let n be a positive integer not divisible by the characteristic of K. A binomial x n − α has over K an abelian Galois group if and only if α wn = γ n , where γ ∈ K and wn is the number of n-th roots of unity contained in K. When a binomial satisfying this condition is irreducible then its group is cyclic if n ≡ 0 mod 4 or ζ4 ∈ K and the product of cyclic groups of order 2 and n/2 otherwise. From Theorem 2 and a result of Hasse [2] concerning the case n = pν we shall deduce *

Addendum and corrigendum, Acta Arith. 36 (1980), 101–104. Written within the Research Program I.1.

940


Theorem 3. Let n be a positive integer not divisible by the characteristic of K. If α = ϑ n,

ϑ ∈ K(ζn )

then ασ = γ n, where (vi)

σ = wn ,

l.c.m.

γ ∈ K,

q |n q prime or q=4

[K(ζq ) : K] .

Moreover if for a certain m prime to n ⎧ ⎪ either ζ(4,n) ∈ K and nm ≡ 0 mod wn l.c.m. [K(ζq ) : K] ⎪ ⎪ q |n ⎨ q prime (vii) ⎪ / K, τ < ∞ and nm ≡ 0 mod 2τ wn l.c.m. [K(ζq ) : K] ⎪ ⎪or ζ(4,n) ∈ q |n ⎩ q prime

then α = γ n/σ ,

γ ∈ K.

Next we shall assume that K is an algebraic number field and prove the following extension of Kummer’s theorem (see [3], Satz 152) on power residues.

c

Theorem 4. Let K be an algebraic number field, w the number of roots of unity contained in K, σ given by (vi). If α1 , . . . , αk ∈ K ∗ are such that (viii)

α1σ x1 · · · αkσ xk = γ n ,

γ ∈ K implies x1 ≡ x2 ≡ . . . ≡ xk mod n/σ

then for any integers c1 , . . . , ck ≡ 0 mod σ there exist infinitely many prime ideals p of K(ζn ) such that α i = ζnci . p n If α1 , . . . , αk satisfy the stronger condition that (ix)

ζwx0 α1x1 · · · αkxk = γ n/σ implies x1 ≡ x2 ≡ . . . ≡ xk ≡ 0 mod n/σ

and n satisfies the condition (vii) of Theorem 3 then for any integers c1 , . . . , ck ≡ 0 mod σ and any c0 there exist infinitely many prime ideals p of K(ζn ) such that ζ α w i c0 = ζ(w,n) , = ζnci . p n p n If n = p ν , p prime and p > 2 or ν = 1 or w ≡ 0 mod 4 then the assertion holds without any restriction on ci . Thus, for n = p, ν = 1 we obtain Chebotarev’s refinement [9] of Kummer’s theorem. For K = Q and n arbitrary a more precise result has been obtained by Mills [5]. We shall use Theorem 4 to prove two theorems on exponential congruences.

941

H5. Abelian binomials, power residues and exponential congruences

Theorem 5. Let f (x) be a polynomial of degree g over K, α1 , . . . , αk ∈ K ∗ . If the congruence f (α1x1 · · · αkxk ) ≡ 0 mod p is soluble for almost all prime ideals p of K then the equation f (α1x1 · · · αkxk ) = 0 is soluble in rational numbers x1 , . . . , xk with the least common denominator not exceeding max{1, g − 1}. This is a generalization of Theorem 2 of [6] and the examples which we give further show that it is essentially best possible. Corollary. Let a sequence un of rational integers satisfy the recurrence relation un+1 = aun + bun−1 , where a 2 + 4b = 0. If the congruence un ≡ c mod p is soluble for almost all primes p and either b = 0, −1 or b = 1, a = d 3 + 3d (d integer), then c = um for an integer m. Here as in Theorem 5 almost all means all except a set of density zero. It is conjectured that the Corollary holds for all recurring sequences of the second order satisfying a 2 + 4b = 0. Theorem 6. Let αhij , βhi be non-zero elements of K, D a positive integer. If the system of congruences gi k xj αhij − βhi ≡ 0 mod m (i = 1, 2, . . . , l) h=1

j =1

is soluble for all moduli prime to D then the corresponding system of equations is soluble in integers. This is a generalization of Theorem 3 of [6]. According to Skolem’s conjecture Theorem 6 with D = 1 remains valid if gi k xj αhij − βhi h=1

j =1

is replaced by gi h=1

βhi

k

x

j αhij ,

j =1

but that we cannot prove. Lemma 1. If p is a prime different from the characteristic of K, ζp ∈ K, ξ p ∈ K ∗ , ν ηp ∈ K ∗ ξ and η ∈ K(ξ ) then either η ∈ K ∗ ξ or p = 2, ζ4 ∈ / K and ζ4 ∈ K ∗ ξ . ∗ ∗ Here K ξ is the multiplicative group generated by K and ξ . μ

942


Proof. For p > 2 this is an easy consequence of a theorem of Kneser [4], since however for p = 2 we have to go through Kneser’s proof all over again, we can cover at once the general case. The proof is by induction with respect to μ and ν. If μ = 0 or ν = 0 the lemma is obvious. Assume it is true for μ = m − 1 and all ν. We prove it first for μ = m, ν = 1. Suppose that ξ ∈ K(ξ p ). Then using the inductive assumption with ξ1 = ξ p , η1 = ξ , ∗ p c ν1 = 1 we get either ξ ∈ K ξ or p = 2, ζ4 ∈ / K and ζ4 ∈ K ∗ ξ 2 . The former ∗ c possibility gives ξ ∈ K, the latter ζ4 ∈ K ξ , thus in this case lemma holds. Suppose now that ξ ∈ / K(ξ p ). Then also ξ ζp ∈ / K(ξ p ), ξ satisfies over K(ξ p ) the p p irreducible equation x − ξ = 0 and denoting by N the norm from K(ξ ) to K(ξ p ) we have N ξ = (−1)p−1 ξ p . c

c

On the other hand, ηp ∈ K ∗ ξ , hence ηp = aξ pk+q , where 0 q < p, a ∈ K ∗ . Consider first the case q > 0. Taking the norms of both sides we get q

2 (−1)p−1 ξ p = (N η)p a −p ξ −p k . For p > 2 it follows that ξ p ∈ K(ξ p )p and ξ ∈ K(ξ p ) which has been excluded. For p = 2 we get −ξ 2 = (N η)2 a −2 ξ −4k ,

ζ4 ξ ∈ K(ξ 2 ),

η2 = ±ζ4 N (η),

hence ζ4 ∈ K(ξ ), ζ4 ∈ / K(ξ 2 ). Writing η = g + ζ4 h with g, h ∈ K(ξ 2 ) we obtain 2 2 c g = h , η = (1 ± ζ4 )g. Hence g 4 = −η4 /4 ∈ K ∗ ξ 2 and by the inductive assumption with ξ1 = ξ 2 , η1 = g, ν1 = 2 we infer that g ∈ K ∗ ξ 2 , ζ4 = ± 21 η2 g −2 ∈ K ∗ ξ . Consider now the case q = 0. Let S be an automorphism of the normal closure of K(ξ ) p p ∈ K(ξ p ), Sηp = ηp , c over K(ξ ) such that Sξ = ξ ζp . From q = 0 we infer that η r η. It follows that S(ηξ −r ) = ηξ −r , ηξ −r ∈ K(ξ p ). Since ηp ξ −rp ∈ K ∗ ξ p , Sη = ζ c p we apply the inductive assumption with ξ1 = ξ p , η1 = ηξ −r , ν1 = 1 and obtain that −r ∈ K ∗ ξ p or p = 2, ζ ∈ ∗ 2 ∗ c ηξ 4 / K, ζ4 ∈ K ξ . The former possibility gives η ∈ K ξ and the proof for μ = m, ν = 1 is complete. Assume now that n 2, the lemma holds pn ∈ K ∗ ξ . Using the inductive assumption with η = ηp , c for μ = m, ν < n and that η 1 p ∈ K ∗ ξ , or p = 2, ζ ∈ ∗ c ν1 = n − 1, we get η 4 / K and ζ4 ∈ K ξ . In the former case ∗ c we use the inductive assumption with ν1 = 1 and obtain η ∈ K ξ , which completes the inductive proof.

c

ν

Lemma 2. Let K be a field of characteristic different from 2. If ϑ ∈ K(ζ4 ), ϑ 2 ∈ K then at least one of the following four conditions is satisfied for a suitable γ ∈ K: ν

(2) (3) (4)

ν

ϑ2 = γ 2 ;

(1)

ν < τ,

ν

ν

ϑ 2 = −γ 2 ;

2ν−1 2ν ν γ ; ϑ 2 = − ζ2τ + ζ2−1 τ +2

2ν−1 2ν −1 2ν γ . ν > τ, ϑ = ζ2τ + ζ2τ + 2

ν = τ,


943

Proof. This is a special case, n = 2ν of Lemma 7 of [6]. Let us remark that the conditions (3) and (4) do not depend upon the choice of ζ2τ . Indeed, for any odd j

j −1 −j ζ2τ +1 + ζ2τ +1 ζ2τ +1 + ζ2−1 ∈ K, τ +1 hence

−j

j

ζ2 τ + ζ2 τ + 2

c

2ν−1

ζ2τ + ζ2−1 τ +2

−2ν−1

ν

∈ K2 .

(The same remark applies to the general case.)

Lemma 3. Let τ1 be the greatest integer such that ζ2τ1 ∈ K(ζ4 ), if there are only finitely many of them, τ1 = ∞ otherwise. Then ( τ + 1 if τ < ∞ and − ζ2τ + ζ2−1 τ + 2 ∈ K, τ1 = τ otherwise. Proof. We have for all σ 2

σ −2 σ −2 2ζ2σ = ζ2σ + ζ2−1 + ζ4 ζ21−2 + ζ2−1+2 σ σ σ which implies τ1 τ . If we had τ < ∞ and ζ2τ +2 ∈ K(ζ4 ) it would follow by Lemma 2 that

2τ 2τ +1 τ +1 τ +1 −1 = ζ22τ +2 = γ 2 or ζ2τ + ζ2−1 γ , γ ∈ K, τ +2 hence ζ4 ∈ K and ζ2τ +2 ∈ K, ζ2τ +2 + ζ2−1 τ +2 ∈ K contrary to the definition of τ . This proves τ1 < τ + 2. If ζ2τ +1 ∈ K(ζ4 ), then ζ4 ∈ / K and ζ2τ is conjugate over K to ζ2−1 τ . Hence ζ4 ζ2τ +1 is −1 −1 conjugate over K either to ζ4 ζ2τ +1 or to −ζ4 ζ2τ +1 . However the latter possibility gives ζ21+2 τ +1

τ −1

+ζ2−1−2 τ +1

τ −1

ζ4 ζ2τ +1 +ζ4 ζ2−1 τ +1

∈ K contrary to the definition of τ . Thus the former possibility holds and ( ( ∈ K, − ζ2τ + ζ2−1 − ζ2τ + ζ2−1 τ + 2 ∈ K. Conversely if τ +2 ∈ K

then ζ2τ +1 + ζ2−1 τ +1 ∈ K(ζ4 )

and ζ2τ +1 =

ζ2τ + 1 ζ2τ +1 + ζ2−1 τ +1

∈ K(ζ4 ).

This proves the lemma. μ

Lemma 4. If ξ p = β ∈ K, ζpμ ∈ K(ξ ) and either p > 2, ζp ∈ K or p = 2, ζ4 ∈ K then j

β = ζpκ γ p

μ−κ

,

0 κ min(μ, ω),

(j, p) = 1,

Proof. By Lemma 1 we have in any case ζpμ ∈ K ∗ ξ , c

(5)

ζpμ = δξ i ,

δ ∈ K ∗,

1 i pμ .

γ ∈ K.

944


Let i = pκ h,

(h, p) = 1,

hj ≡ 1 mod p μ−κ .

Raising both sides of (5) to the power pμ−κ j we get j

ζp κ = δ p

c

c

μ−κ j

β hj ,

hence κ ω and the lemma holds with γ = β (1−hj )p

κ−μ

δ −j .

ν

Proof of Theorem 1. Necessity. Assume that x p − α is the product of normal factors. Let μ be the least nonnegative integer such that α = βp

ν−μ

β ∈ K.

,

If μ = 0 then the theorem holds with γ = β min(ν,ω) . If μ > 0 then j

β = ζpν−μ δ p ,

(6)

δ ∈ K.

pμ

Hence if p > 2 or p = 2, ζ4 ∈ K, then x − β is irreducible and by the assumption normal. Denoting any of its zeros by ξ we get ζpμ ∈ K(ξ ),

(7) and since [K(ξ ) : K] = have

pμ , [K(ζp ) j

: K] | p − 1, it follows that ζp ∈ K. By Lemma 4 we

β = ζpκ γ p

(8) pκ

ζp ∈ K(ξ )

μ−κ

;

0 κ min(μ, ω)

pν

and α = γ , which proves (i). μ Assume now that p = 2, ζ4 ∈ / K. Then either x 2 − β is irreducible or μ 2, μ β = −4δ 4 , δ ∈ K. In the former case we get again (7) for any zero ξ of x 2 − β, in the latter case let be the least nonnegative integer such that (1 + ζ4 )δ = η2 c

The binomial x

2

μ−2−

η ∈ K(ζ4 ).

,

− η is irreducible over K(ζ4 ), hence

f (x) = NK(ζ4 )/K (x 2 − η) is irreducible over K. The polynomial f (x) is a factor of

μ−2 μ−2− μ−1 μ−2 μ c − η2 + 2δx 2 + 2δ 2 | x 2 + 4δ 4 , = x2 NK(ζ4 )/K x 2 hence it is normal. Let ξ be a zero of x 2 − η, ξ a zero of x 2 − η , where η is conjugate to η over K. We have

ξ ∈ K(ξ ), ξ

(9) on the other hand

ξ 2μ−2 ξ

=

η 2μ−2− η

=

(1 − ζ4 )δ = −ζ4 , (1 + ζ4 )δ


945

hence ξ /ξ = ζ2μ , (j, 2) = 1 and from (9) we get again (7). Using now Lemma 1 we get ζ4 ∈ K ∗ ξ . Hence j

ζ4 = δξ i ,

(10)

δ ∈ K,

2 = [K ∗ ζ4 : K ∗ ] = [K ∗ ξ i : K ∗ ] = 2μ−ord2 i ,

i = 2μ−1 j,

(j, 2) = 1

and on squaring both sides of (10) we get −1 = δ 2 β j ,

β = −γ 2 .

It follows from (6) that μ = ν α = β = −γ 2 .

(11)

On the other hand, applying Lemma 4 to the field K(ζ4 ) we get α = ζ2σ ϑ 2

ν−σ

0 σ min(ν, τ1 ),

,

ϑ ∈ K(ζ4 ).

If ν τ1 , then by (11) and Lemma 3 we have (ii) or (iii). If ν > τ1 then, since ζ4 ∈ /K by (11) and Lemma 2, σ = 0 is impossible. We get α2

σ −1

= −ϑ 2

ν−1

and by Lemma 2 either α2

(12)

σ −1

= −γ 2

ν−1

γ ∈K

,

or c

(13)

ν − 1 = τ = τ1 ,

α2

σ −1

2τ −1 2τ = ζ2τ + ζ2−1 γ τ +2

or c

(14)

ν − 1 > τ,

α2

2ν−2 2ν−1 = − ζ2τ + ζ2−1 γ τ +2

σ −1

Since ζ4 ∈ / K, (12) and (14) imply σ = 1 and then we get (i) or (v) respectively. Finally (13) in view of (11) implies σ > 1 2τ −σ 2τ −σ +1

γ , α = ± ζ2τ + ζ2−1 τ +2 again by (11) and Lemma 3, σ < τ and the upper sign is excluded. This gives (iv). Sufficiency. To prove the sufficiency of (i) we proceed by induction with respect to ν. The case ν ω is trivial. If ν > ω (i) gives α = ζpκ γ p

(15)

ν−ω

,

0 κ ω.

If κ < ω we have ν

xp − α =

p−1

j =0

xp

ν−1

j

− ζp ζpκ+1 γ p

ν−ω−1

.

946


Each of the factors on the right hand side is by the inductive assumption the product of ν normal factors, hence the same holds for x p − α. If κ = ω = 0 we have ν x ν Xp μ , xp − α = α γ μ=0

where Xn (x) is the nth cyclotomic polynomial. Every zero of Xn

x γ

generates over K

all the other zeros, hence the desired result. ν Finally if κ = ω > 0 let ξ denote as in the sequel any zero of x p − α. We have by (15)

pω −1 pν−ω ξ γ = ζpω c

hence for an integer j

ω j ζpν = ξ p γ −1 . If (ii) or (iii) holds then ζ4 = ±ξ 2

ν−1

γ −1 .

Since, by Lemma 3, ν τ1 and by definition ζ2τ1 ∈ K(ζ4 ), it follows that ζ2ν ∈ K(ξ ). If (iv) holds then ξ2

τ −λ

j = ζ2λ+2 γ ζ2τ +1 + ζ2−1 τ +1 ,

(j, 2) = 1,

thus ξ2

τ −λ

ζ2τ +1 ∈ K(ζ2τ ),

τ

ζ4 ∈ K(ξ 2 ).

Since, by Lemma 3, K(ζ2τ ) = K(ζ4 ) it follows that ξ2

τ −λ

τ

ζ2τ +1 ∈ K(ξ 2 )

and ζ2τ +1 ∈ K(ξ ). If (v) holds then

j ξ 2 = ζ2ν ζ2τ +1 + ζ2−1 τ +1 γ ,

(j, 2) = 1,

thus ξ 2 ∈ K(ζ2ν ). We shall show that ξ 2 has as many distinct conjugates over K as ζ2ν . Indeed, if S is an automorphism of K(ζ2ν ) over K then

−1 S ζ2τ +1 + ζ2−1 τ +1 = ± ζ2τ +1 + ζ2τ +1 . j

j

Hence Sξ 2 = ξ 2 implies Sζ2ν = ζ2ν , Sζ2ν = ζ2ν or

j j −1 Sζ2ν = −ζ2ν , S ζ2τ +1 + ζ2−1 τ +1 = − ζ2τ +1 + ζ2τ +1 .

947


The latter case is however impossible, since Sζ2τ +1 = S(ζ2ν )2

ν−τ −1

. It follows that

ζ2ν ∈ K(ξ 2 ). ν

If the binomial x p − α is irreducible, then for p = 2, ν ( 2 we have α = −4γ 4 ,

/ K, hence for τ 3, α = −γ 4 , γ ∈ K. Thus (iv) implies τ 3, − ζ2τ + ζ2−1 τ +2 ∈ λ = 1; (v) implies τ = 2.

ν

c

Remark. Note that in case (i) if κ = ω > 0 and in cases (ii)–(v) every root of x p − α generates all the others. ν

Lemma 5. If a binomial x p − α satisfies condition (i) then its Galois group G over K is abelian. If it is irreducible then G is cyclic unless p = 2, ν 2, ω = 1, in which case G is of type (2, 2ν−1 ). If λ is the least nonnegative integer such that α = ζpκ γ p

ν−λ

0 κ λ ω,

,

γ ∈ K,

ν

then the Galois group of each irreducible factor of x p − α contains an element of order pλ and besides an element of order p ν−ω+κ if p

ν−τ +1

if

p, ω = 2, 1 and κ > max{0, ω − ν + λ}, p, ω = 2, 1 and κ = λ = 1 > τ − ν + 1.

Proof. We start by proving that an irreducible binomial satisfying (i) has a cyclic group G unless p = 2, ν 2, ω = 1. Since it is irreducible we have either ν −λ = 0 or κ = ω = λ. ν In the former case ν ω; if ξ is any zero of x p − α and S the substitution ξ → ζpν ξ we j have S j (ξ ) = ζpν ξ hence G is cyclic, generated by S. In the latter case let ξ be any zero ν of x p − α satisfying ω

ξ p = ζpν γ and consider the substitution S : ξ → ζpν ξ . We have pω +1

S(ζpν ) = ζpν

j −1

,

S j (ξ ) = ζpν i=0

(pω +1)i

ξ.

The order of S is the least j such that (16)

j −1

(p ω + 1)i =

i=0

(pω + 1)j − 1 ≡ 0 mod pν . pω

However if p > 2, a ≡ 1 mod p or p = 2, a ≡ 1 mod 4 we have (17)

ordp (a j − 1) = ordp j + ordp (a − 1)

(see [6], p. 401(1 ), formula (8)). Hence if p > 2 or p = 2, ω 2, (16) implies ordp j ν (1 )

Page 919 in this volume.

948


and S is of order p ν . The same is clearly true for p ν = 2. The remaining assertions of the lemma are trivial for λ = 0. If λ > 0 we consider first the case p > 2 or p = 2, ω 2. If κ > max{0, ω − ν + λ} we have the factorization x

pν

pω−κ −1

−α =

xp

ν−ω+κ

j

− ζpω−κ ζpω γ p

j =0

ν−λ−ω+κ

and the factors are irreducible since ζpω+1 ∈ / K. By the fact already established the Galois ν−ω+κ and since ν − ω + κ > λ contain also an element of groups are cyclic of order p order p λ . If κ ω − ν + λ then pν−λ

α = γ1

γ1 = ζpκ+ν−λ γ ∈ K.

,

We have the factorization x

pν

−α =

pν−λ −1 j =0

λ

j

x p − ζpν−λ γ1

j

p

and the factors are irreducible, since ζpν−λ γ1 = γ2 , γ2 ∈ K would imply αp

λ−1

pν

= γ2

contrary to the choice of λ. The Galois groups are cyclic of order p λ . If κ = 0 > ω − ν + λ we have the factorization x

pν

−α =

ω −1 p

x

pλ

j − ζp ω γ

j =0

ν−ω

ω −1 p

μ=λ+1

j =0 (j,p)=1

μ

j

x p − ζp ω γ p

μ−λ

.

The factors of the first product are irreducible for the same reason as before, the other factors are irreducible since ζpω+1 ∈ / K. The Galois groups are cyclic of order pμ (λ μ ν − ω). Consider now the case p = 2, ω = λ = 1. Let τ1 have the meaning of Lemma 3. If κ = 1 > τ − ν + 1 we have ν τ + 1 τ1 and the factorization x

2ν

−α =

τ1 −1 2

ν−τ1 +1 ν−τ1 j . NK(ζ4 )/K x 2 − ζ2τ1 γ 2

j =1 j ≡1 mod 4 ν−τ +1

j

ν−τ

/ K(ζ4 ) If fj (x) = x 2 1 − ζ2τ1 γ 2 1 were reducible over K(ζ4 ), then since ζ2τ1 +1 ∈ we should have ν = τ1 = τ + 1 and j

ζ2τ1 γ = ϑ 2 , 2τ

c

2τ +1

ϑ ∈ K(ζ4 ),

contrary to Lemma 2. Thus fj (x) is irreducible over K(ζ4 ) and whence −γ = ϑ NK(ζ4 )/K fj (x) = fj (x)fj (x) is irreducible over K.

949


In order to determine the Galois group of fj fj it is necessary to distinguish between the cases τ1 = τ + 1 and τ1 = τ . Let ξ be a zero of fj (x) satisfying c

j

ξ 2 = ζ2ν γ .

(18) −j

−j (1+2τ −1 )

j

If τ1 = τ + 1 then −ζ2τ1 is conjugate over K to ζ2τ1 hence ζ2ν fj (x). Let S be the substitution −j (1+2τ −1 )

ξ → ζ2ν

ξ is a zero of

ξ.

We have by (18) −j (1+2τ )

j

S(ζ2ν ) = ζ2ν hence

−j (1+2τ −1 )

S r (ξ ) = ζ2ν

r−1

i=0 (−1−2

τ )i

ξ.

The order of S is the least r such that −j (1 + 2τ −1 )

r−1

(−1 − 2τ )i = j

i=0

(−1 − 2τ )r − 1 ≡ 0 mod 2ν . 2

Clearly r must be even and since by (17)

ord2 (1 + 2τ )r − 1 = ord2 r + τ c we get r ≡ 0 mod 2ν−τ +1 . The order of S is thus equal to the degree of fj fj and since the latter polynomial is normal, its group is cyclic of order 2ν−τ +1 . −j j −j If τ1 = τ then ζ2τ1 is conjugate over K to ζ2τ1 hence ζ2ν ξ is a zero of fj (x). Let S be −j

j

the substitution ξ → ζ2ν ξ and T the substitution ξ → ζ2ν−τ +1 ξ . We have by (18) −j

j

S(ζ2ν ) = ζ2ν , j (1+2τ )

j

T (ζ2ν ) = ζ2ν

S 2 (ξ ) = ξ ; j

r−1

T r (ξ ) = ζ2ν−τ i=0 +1

,

(1+2τ )i

ξ.

Using (17) we infer that T is of order 2ν−τ +1 , moreover −j ≡ 0 mod 2τ −1 , hence S = T r . However −j

j

−j

−j (1+2τ −1 )

ST (ξ ) = S(ζ2ν−τ +1 )S(ξ ) = ζ2ν−τ +1 ζ2ν ξ = ζ2ν −j

−j (1+2τ ) 2ν

T S(ξ ) = T (ζ2ν )T (ξ ) = ζ

j

ξ,

−j (1+2τ −1 ) 2ν

ζ2ν−τ +1 ξ = ζ

ξ,

thus ST (ξ ) = T S(ξ ) and the group of fj fj being of order 2ν−τ +2 must be abelian of type (2, 2ν−τ +1 ). ν In particular, if x 2 − α is irreducible, we have τ1 = τ = 2 and the group is abelian of ν−1 type (2, 2 ).

950


Consider now the case κ = 1 τ − ν + 1. We have the factorization ν −1 2

ν

x2 − α =

j

NK(ζ4 )/K (x 2 − ζ2ν γ ).

j =1 j ≡1 mod 4

The factors that are not irreducible are products of two quadratic factors and hence satisfy the condition of the lemma. The irreducible factors have groups abelian of type (2, 2) −j j generated by the substitutions (ξ → −ξ ) and (ξ → ζ2ν ξ ), where ξ is a zero of x 2 − ζ2ν γ . 4 2 In particular this applies to the case of an irreducible binomial x + γ . It remains to consider the case κ = 0. Then the assertions of the lemma follow by induction with respect to ν. They are true for ν = 1. For ν > 1 we have the factorization

ν−1 ν ν−1 ν−2 2ν−1 ν−2 x2 − γ 2 = x2 − γ 2 x . + γ2 The first factor on the right hand side has an abelian Galois group and all its irreducible factors are of even degree by the inductive assumption, the second factor has this property by the already considered case κ = 1 of the lemma.

Proof of Theorem 2. Necessity. Assume that the splitting field of x n − α is abelian over K. ν Then also the splitting field of x p − α is abelian over K for any pν | n and since every ν subgroup of an abelian group is normal x p − α is the product of normal factors. Thus we have one of the conditions (i)–(v) listed in Theorem 1. We shall show that under our assumption (ii)–(v) lead to (i). Consider first (ii), (iii) or (iv). Let μ be the least nonnegative integer such that α = −γ12 c

,

γ1 ∈ K.

Clearly μ < ν. If μ 1 we have (i). If μ > 1 then ν − μ + 2 τ , unless ν = τ + 1, μ = 2, in which case by (iii) or (iv) and, by Lemma 3, ν − μ + 2 τ1 . Thus ζ2ν−μ+2 ∈ K(ζ4 ).

(19) c

ν−μ

x

2ν

− α has over K(ζ4 ) the factor μ

f (x) = x 2 − ζ2ν−μ+1 γ1 .

c

Now c

ζ2ν−μ+1 γ1 = ϑ 2 ,

ϑ ∈ K(ζ4 ),

since otherwise, by (19) γ1 = ϑ12 , γ1 ∈ K(ζ4 ) and by Lemma 2 γ1 = ±γ22 , c c

γ2 ∈ K;

α = −γ22

ν−μ+1

contrary to the choice of μ. Thus f (x) is irreducible over K(ζ4 ) and NK(ζ4 )/K f (x) = f (x)f (x) is irreducible over K. By the assumption the latter polynomial is normal over K. Let ξ be any zero of f (x), ξ1 = ζ2−1 ν ξ,

ξ2 = ζ2−1−2 ν

ν−μ

ξ.


951

ξ1 , ξ2 are zeros of f (x). Let Si be the automorphism of the Galois group of ff over K such that Si ξ = ξi

(i = 1, 2).

−1 We have Si ζ2ν−μ+1 = ζ2−1 ν−μ+1 hence Si ζ2ν = εi ζ2ν (εi = ±1). It follows that

S1 S2 ξ = S1 ξ2 = S1 (ζ2−1−2 ν

ν−μ

)S1 ξ = ε1 ζ2μ ξ,

= ε2 ζ2−1 μ ξ. ( By Lemma 3 we have ε1 = ε2 unless ν = τ + 1 and − ζ2τ + ζ2−1 / K. In the τ +2 ∈ c c latter case by (iv) μ = ν − λ 3, thus in both cases S1 S2 ξ = S2 S1 ξ and the group in question is not abelian. ( Consider now the case (v). If − ζ2τ + ζ2−1 ∈ K then we get (i). If τ +2 ( − ζ2τ + ζ2−1 / K then by Lemma 3 τ +2 ∈ S2 S1 ξ = S2 ξ1 =

ζ2τ +1 ∈ / K(ζ4 ).

(20) c

x

2ν

− α has over K(ζ4 ) the factor f (x) = x 2

c c

S2 (ζ2−1 ν )S2 ξ

ν−τ +1

2ν−τ −1 2ν−τ − ζ2τ ζ2τ + ζ2−1 γ . τ +2

By (20) and the inequality ν τ + 2, f (x) is irreducible over K(ζ4 ). Hence NK(ζ4 )/K f (x) = f (x)f (x) is irreducible over K and by the assumption normal. Let ξ be a zero of f (x) satisfying

(21) ξ 2 = ζ2ν ζ2τ +1 + ζ2−1 τ +1 γ and let ξ1 = ζ2−1 ν ξ,

ξ2 = ζ2−1−2 ν

τ −1

ξ.

ξ1 and ξ2 are zeros of f (x). Let Si (i = 1, 2) be the automorphism of the Galois group of ff over K such that Si ξ = ξi

(i = 1, 2).

We have for a suitable εi = ±1

−1 Si ζ2τ +1 + ζ2−1 τ +1 = ε1 ζ2τ +1 + ζ2τ +1 ; then by (21) (22)

S1 ζ2ν = ε1 ζ2−1 ν ,

S2 ζ2ν = ε2 ζ2−1−2 ν

τ

hence

−1 2ν−τ −1 2ν−τ −1 + ε 1 ζ2 ν = ζ2τ +1 + ζ2−1 S1 ζ2τ +1 + ζ2−1 τ +1 = ε1 ζ2ν τ +1 , ν−τ −1

−1−2τ 2

τ 2ν−τ −1 + ε2 ζ21+2 ) = −ζ2τ +1 − ζ2−1 S2 ζ2τ +1 + ζ2−1 ν τ +1 = ε2 ζ2ν τ +1 .

952


Thus ε1 = 1, ε2 = −1 and (22) implies

τ −1 S1 ξ = ζ2ν−τ +1 ξ, S1 S2 ξ = S1 ξ2 = S1 ζ2−1−2 ν

−1 S2 S1 ξ = S2 ξ1 = S2 ζ2ν S2 ξ = −ζ2ν−τ +1 ξ.

k piνi Hence S1 S2 ξ = S2 S1 ξ and the group in question is not abelian. Therefore, if n = i=1 is the canonical factorization of n we get for each i k ν

pi i

i

α w = γi

γi ∈ K,

,

where we have put for abbreviation wi = wpνi . It follows that i

α

ν nwn /pi i

nwn /wi

= γi

.

If now 1 ri = n piνi k

(23)

i=1

we obtain α wn =

k

r wn /wi

n

γi i

i=1

and the proof is complete. Sufficiency. Assume that α wn = γ n ,

γ ∈ K,

k and let again n = piνi , w i = wpνi . i i=1 w n νi Since , pi = 1 we have wi ν

pi i

i

α w = γi

(24)

.

Thus α satisfies the assumptions of Lemma 5 for all pi and by that lemma the Galois groups over K of all binomials (25)

νi

x pi − α

(1 i k) νi

are abelian. If ξ is any zero of x n − α then ξ n/pi is a zero of (25) and defining ri by (23) we get ξ=

k

νi

ξ n/pi

ri

.

i=1

Hence the splitting field of x n − α as the composite of the splitting fields of (25) is abelian. Moreover if x n − α is irreducible then also the binomials (25) are irreducible and their groups are cyclic of order piνi unless piνi ≡ 0 mod 4 and ζ4 ∈ / K in which case the group


953

of (25) has a cyclic factor of order 2νi −1 . Since the direct product of cyclic groups of orders prime in pairs is again cyclic we get all assertions of the theorem.

ν

Lemma 6 (Hasse). If α = ηp , η ∈ K(ζpν ), then at least one of the following conditions is satisfied for a suitable γ ∈ K: ν

α = γp ;

(26)

ν

p = 2, ω = 1, 1 < ν < τ, α = −γ 2 ; 2τ −1 2τ

γ ; p = 2, ω = 1, ν = τ, α = − ζ2τ + ζ2−1 τ +2 ν−1

ν 2 γ2 , p = 2, ω = 1, ν > τ, α = ζ2τ + ζ2−1 τ +2 c

where ω and τ have the meaning of Theorem 1, p = char K. Proof which we give is based on the previous results and therefore much shorter than the original Hasse’s proof ([2], for char K = 0). ν Since all the subextensions of K(ζpν ) are normal over K the binomial x p − α satisfies the conditions of Theorem 1. Hence we have either (26) or ω 1. In the latter case Lemma 1 applies with ξ = ζpν , and we get either

c

η ∈ K ∗ ζpν ;

(27)

j

η = γ ζpν ,

γ ∈ K∗

or p = 2, ζ4 ∈ / K. (27) gives at once (26). To settle the case p = 2 we apply the already proved case of our lemma for the field K(ζ4 ) and get ν

α = ϑ2 ,

ϑ ∈ K(ζ4 ).

Now Lemma 6 follows immediately from Lemma 2.

ν Proof of Theorem 3. We start by estimating for each

p n the greatest exponent μp μ p such that p divides the order of an element in Gal K(ζnp−ν )/K . Since K(ζnp−ν ) is the composite of K(ζq s ), where q = p is a prime and q s n, we have

ordp [K(ζq s ) : K]. μp max s

(#)

q n q=p

Let r be the largest integer such that ζq r ∈ K(ζq ). Then for q s > 2 q max(0,s−r) [K(ζq ) : K] if (q, r) = (2, 1), [K(ζq s ) : K] = max(0,s−τ ) 1 [K(ζ ) : K] 2 if (q, r) = (2, 1). 4 This gives μp max ordp [K(ζq ) : K].

(28)

q |n q prime

(Actually we have here the equality(2 ).) (2 )

See Addendum, p. 967.

954


Assume now that α = ϑ n,

(29)

ϑ ∈ K(ζn ).

ν

Then for each p ν n, the binomial x p − α is abelian over K and by Theorem 2 c

(30)

pν−λ

α = ζ p κ γp

;

κ λ min(ν, ω),

γp ∈ K.

where ω has the meaning of Theorem 1. Suppose first pν ≡ 0 mod 4 or ζ4 ∈ K. Then by Lemma 6 it follows from (29) that pν

α = ϑ1 , ϑ1 ∈ K(ζnp−ν ).

By Lemma 5 Gal K(ϑ1 )/K contains an element of order p λ hence Gal K(ζnp−ν )/K contains such an element and by (28) we have

(31)

λ μp max ordp [K(ζq ) : K]. q |n q prime

By (30) we have also λ ordp wn

(32)

hence λ ordp σ . The same inequality follows directly from (32) if pν ≡ 0 mod 4, ζ4 ∈ c / K. Thus, by (30), for each p we have pν

α σ = δp ,

δp ∈ K,

whence by the standard argument (see the proof of Theorem 2) ασ = γ n,

γ ∈ K.

Assume now in addition to (29) that for a certain m prime to n ⎧ ⎪ either ζ(4,n) ∈ K and nm ≡ 0 mod wn l.c.m. [K(ζq ) : K] ⎪ ⎪ q |n ⎨ q prime (33) ⎪ or ζ(4,n) ∈ / K and nm ≡ 0 mod 2τ wn l.c.m. [K(ζq ) : K] ⎪ ⎪ q |n ⎩ q prime

and consider again (30) for any

n. If ν ω then from (30) we get immediately pν

(34) c

pν−λ

α = δp

,

δp ∈ K.

If ν > ω and either p > 2 or ζ4 ∈ K we get from (28) and (33), (35)

ν ω + μp ,

hence in particular (##)

ω − ν + λ −μp + ordp σ 0.


955

Thus if (30) holds with κ > 0 we get by Lemma 5 and (31) ν − ω + 1 μp , which contradicts (35). If ν > ω = 1 and p = 2 we get from (33) ν τ + 1 + μ2 > τ.

(36)

Let τ2 be the greatest integer such that

ζ2τ2 + ζ2−1 τ2 ∈ K ζn2−ν .

τ2 −τ we have Since K ζ2τ2 + ζ2−1 τ2 is over K cyclic of degree 2 τ2 − τ μ2 and by (36) ν τ2 + 1. Hence by (29) and Lemma 6 α = ϑ12 c

ν−1

,

ϑ1 ∈ K(ζn2−ν .

Thus if (30) holds with κ > 0 we get by Lemma 5 that Gal(K(ϑ1 )/K) contains an element of order 2ν−τ , hence ν − τ μ2 contrary to (36). Therefore (34) holds in any case and by the standard argument α = γ n/σ ,

γ ∈ K.

Remark (3 ). If (wn , n/wn ) = 1 the number σ occurring in Theorem 3 is the least integer with the required property. Indeed, by the definition

of σ there exists a character χ mod n belonging to exponent σ on the group G = Gal K(ζn )/K . Let τ (χ, ζn ) be the corresponding Gauss sum. Clearly τ (χ ) ∈ K(ζn ). Suppose that n = γ n , γ ∈ K, 0 < < σ . Then τ (χ ) = ζ α γ and applying an automorphism c τ (χ) n j

ζn → ζn

(∗) a(j −1)

from G we get χ (j ) = ζn a(j 2 −1)

ζn

. It follows that 2a(j −1)

= χ (j 2 ) = χ (j )2 = ζn

;

a(j −1)2

ζn

= 1,

a(j − 1)2 ≡ 0 mod n and since this holds for all automorphisms (∗) from G, awn2 ≡ 0 mod n. Since (wn , n/wn ) = 1 we get a ≡ 0 mod

n , wn

contrary to the choice of χ . (3 ) See Addendum, p. 967.

a(j − 1) ≡ 0 mod n

and χ (j ) = 1

956


Lemma 7. Under the assumption (viii) of Theorem 4 the group ) )

G0 = Gal K(ζn , n α1 , . . . , n αk /K(ζn ) contains the substitution

) n

αi → ζnci

) n

αi (1 i k),

under the assumptions (vii) and (ix) the group ) ) )

c G1 = Gal K(ζn , n ζw , n α1 , . . . , n αk /K(ζn ) contains the substitution ) ) c0 n n ζw → ζ(w,n) ζw ,

) n

αi → ζnci

) n

αi (1 i k)

for any c0 and any ci ≡ 0 mod σ (1 i k). √ √ Proof. Let us denote any value of n αi by ξi (1 i k) and of n ζw by ξ0 . To prove the first part of the lemma it is clearly sufficient to prove that G0 contains each of the substitutions (1 i k) (37)

ξj → ξj (1 j k, j = i),

ξi → ζnσ ξi .

If (37) were not contained in G0 , we should have

n (38) di = Gal K(ζn , ξ1 , . . . , ξk )/K(ζn , ξ1 , . . . , ξi−1 , ξi+1 , . . . , ξk ) ≡ 0 mod . σ Now by Kneser’s theorem c di = K(ζn )∗ ξ1 , . . . , ξk : K(ζn )∗ ξ1 , . . . , ξi−1 , ξi+1 , . . . , ξk hence di is the least exponent such that c

(39)

x

x

i−1 i+1 ξidi = ϑξ1x1 · · · ξi−1 ξi+1 · · · ξkxk ,

ϑ ∈ K(ζn ).

By raising (39) to nth power we get that x

x

i−1 i+1 αidi = ϑ n α1x1 · · · αi−1 αi+1 · · · αkxk ,

ϑ n ∈ K and, by Theorem 3, ϑ nσ = γ n ; γ ∈ K, α1−σ x1 · · · αidi σ · · · αk−σ xk = γ n . By the assumption di σ ≡ 0 mod n, contrary to (38). To prove the second part of the lemma we have similarly to prove that G1 contains each of the substitutions (1 i k) ξ0 → ζ(n,w) ξ0 , ξj → ξj (1 j k); ξ0 → ξ0 , ξj → ξj (1 j k, j = i), ξi → ζnσ ξi . This reduces to proving that the least exponents ei (0 i k) such that (40)

ξiei ∈ K(ζn )ξ0 , . . . , ξi−1 , ξi+1 , . . . , ξk


957

satisfy e0 ≡ 0 mod (n, w), ei ≡ 0 mod n/σ . Now (40) implies for i = 0 ζwe0 = ϑ n α1x1 · · · αkxk , c

ϑ ∈ K(ζn );

hence by (vii) and (ix) ζwe0 α1−x1 · · · αk−xk = γ n/σ , ζwe0

=

γ ∈ K;

y n/σ γ n/σ α1 1

xi ≡ 0 mod n/σ ;

y n/σ · · · αk k

n/σ

= γ1

.

j

γ1 must be a root of unity contained in K; γ1 = ζw and so we get e0 ≡ 0 mod (w, n/σ ). c However by the condition (vii) n/σ ≡ 0 mod (w, n) and thus e0 ≡ 0 mod (w, n). For i > 0, (40) implies x

x

i−1 i+1 αi+1 · · · αkxk , αiei = ϑ n ζwx0 α1x1 · · · αi−1

ζw−x0 α1−x1 · · · αiei · · · αk−xk = γ n/σ ,

ϑ ∈ K(ζn );

γ ∈ K;

ei ≡ 0 mod n/σ.

Proof of Theorem 4. We use Chebotarev’s density theorem and get the existence of infinitely many prime ideals P of L0 = K(ζn , ξ1 , . . . , ξk ) or L1 = K(ζn , ξ0 , . . . , ξk ) dividing p in K(ζn ) such that for all η ∈ L0 or L1 respectively ηNp ≡ Sη mod P, where S is the automorphism described in Lemma 7, ξ0n = ζw , ξin = αi . Setting η = ξi we get Np

≡ ξi ζnci mod P (i > 0),

ξi

Np

ξ0

c0 ≡ ξ0 ζ(w,n) mod P,

consequently (Np−1)/n

αi

≡ ζnci mod P,

c0 ζw(Np−1)/n ≡ ζ(w,n) mod P

and the same mod p. One has only to remark that N p ≡ 1 mod n.

Remark. If (wn , n/wn ) = 1 the number σ occurring in the first part of Theorem 4 is the least integer with the required property. This follows from the Remark after Theorem 3 on taking k = 1, α = τ (χ )σ . If (wn , n/wn ) > 1 σ need not be best possible. In particular, if / K, n ≡ 0 mod 2τ +1 , σ can be replaced by wn , l.c.m. [K(ζq ) : K] . ζ4 ∈ q |n, q prime

Lemma 8. If every integral vector [t0 , t1 , . . . , tr ] satisfies at least one of the congruences (41)

r

ahs ts ≡ 0 mod m (1 h g)

s=0

then for at least one h we have ah0 ≡ 0 mod (ah1 , . . . , ahr , m)

958


and (42)

m max(g − 1, 1). (ah1 , . . . , ahr , m)

Proof. Let us choose in {1, 2, . . . , g} a minimal subset M with the property that every integral vector [1, t1 , . . . , tr ] satisfies at least one congruence (41) with h ∈ M. Put dh = (ah1 , . . . , ahr , m). For h ∈ M we have ah0 ≡ 0 mod dh , since otherwise the congruence (43)

ah0 +

c

r

ahs ts ≡ 0 mod m

s=1

would not be satisfied by any [t1 , . . . , tr ]. Hence, by a theorem of Frobenius the congruence (43) has dh mr−1 solutions mod m. If for a certain h ∈ M we have m/dh < g (42) follows. If for all h ∈ M, m/dh g then either dh r, hence r k n yj (w, cr+1 , . . . , ck )tj − yj cj t0 w j =1 j =r+1 r t (w, cr+1 , . . . , ck ) n cs n dhs ts + t0 + dhs ηs mod n. ≡ − bh0 t0 + w w es es

c

s=1

s=r+1

It follows that t

dhs ηs ≡ 0 mod m,

s=r+1

t

dhs ηs = 0

s=r+1

and by the choice of ηr+1 , . . . , ηs : dhs = 0 (r < s t). Hence all integer vectors [t0 , . . . , tr ] satisfy at least one congruence r r ndhs ncs dhs n n (w, cr+1 , . . . , ck )ts + t0 − bh0 ≡ 0 mod (w, cr+1 , . . . , ck ), wes wes w w c c

s=1

c

s=1

such that dhs = 0 (r < s t). It follows by Lemma 8 that for a certain h (44)

n/w q= max(g − 1, 1), n dhs n , g.c.d. w 1sr w es

dhs = 0

(r < s t)

and (45)

r n cs dhs n n (w, cr+1 , . . . , ck ). − bh0 ≡ 0 mod w es w wq s=1

For h satisfying (44) we have c

ps dhs , = es q

ps integer (1 s r),

(p1 , . . . , pr , q) = 1


961

and by (45) there exist integers u1 , . . . , uk such that (46) c

r r k n cs dhs n n dhs − bh0 + n us + cs us ≡ 0 mod n. w es w es wq s=1

s=1

s=r+1

Let us fix any values of log πs (1 s t) and set log ζw = many valued and

k

2π i . The function αjx is w

x

j =1

αj j can take any value

k k t k 2πi V = exp aj 0 xj + log πs aj s xj + 2π i vj xj , w j =1

j =1

s=1

j =1

where [v1 , . . . , vk ] is an integral vector. Taking [v1 , . . . , vk ] = [u1 , . . . , ur , 0, . . . , 0](P −1 )T , uk dh1 dhr ur+1 [x1 , . . . , xk ] = ,..., P ,..., , e1 er q q we get r k t r uj dhj 2πi dhj V = exp cj + cj + bhs log πs + 2π i uj . w ej q ej j =1

j =r+1

s=1

j =1

By (46) V = βh , hence f (V ) = 0.

Remark. Theorem 5 is essentially best possible, as the following example shows: f (t) = (t − β1 )

q−1

j t − β1 β2 ,

j =0

where q is a prime and β1 , β2 are integers of K multiplicatively independent. The congruence

q q f α1x1 α2x2 ≡ 0 mod p, where α1 = β1 , α2 = β2 , is soluble for every prime ideal p. Indeed, let γ be a primitive root mod p. If ordq indγ β1 > ordq indγ β2 then the equation qx1 indγ β1 + qx2 indγ β2 = indγ β1 is soluble and so is the congruence α1x1 α2x2 ≡ β1 mod p. If on the other hand, ordq indγ β1 ordq indγ β2 then there is a j < q such that

ordq j indγ β1 + indγ β2 > ordq indγ β1 .

962


This implies the solubility of the equation qx1 indγ β1 + qx2 indγ β2 = j indγ β1 + indγ β2 and of the congruence c

j

α1x1 α2x2 ≡ β1 β2 mod p.

The solubility of f α1x1 α2x2 ≡ 0 mod p if p | α1 α2 is trivial. On the other hand, the equation 1 j 1

f α1x1 α2x2 = 0 has only the solutions (x1 , x2 ) = ,0 , , (0 j q). q q q For β1 = ζq , β2 different from a root of unity we get an example with k = 1. Let us note further that Theorem 5 does not extend to all exponential congruences even in one variable, e.g. the congruence

(α x + α) (−α)x − α ≡ 0 mod p is soluble for all prime ideals p, but the corresponding equation has no rational solutions if α is not a root of unity. Proof of the Corollary. For b = 0 it suffices to put in Theorem 5 f (t) = u1 t − c,

k = 1,

α1 = a.

− α −1 )

For b = −1, we have − at − b = (t − α)(t with α = ±1 since a 2 − 4 = 0. It n −n c is well known that un = λ1 α + λ2 α and it suffices to put in Theorem 5 t2

f (t) = λ1 t 2 − ct + λ2 ,

k = 1,

α1 = α.

For b = 1 we have t 2 − at − b = (t − α)(t + α −1 ) with α = ±1 since a = 0. Now un = λ1 α n + λ2 (−α)−n , where λ1 , λ2 are conjugate in the field Q(α). If c = 0, we set in Theorem 5 f (t) = λ1 t + λ2 ,

k = 1,

α1 = −α 2 .

If c = 0, we set f (t) = (λ1 t 2 − ct + λ2 )(λ1 t 2 − ct − λ2 ),

k = 1,

α1 = α,

where α is chosen negative (one of the numbers α, −α −1 is always negative). We infer that the equation f (α x ) = 0 has a solution x = m/q, where (m, q) = 1, q 3. If q = 1 and λ1 α 2m − cα m + (−1)m λ2 = 0 we get c = um . If q = 1 and (47)

λ1 α 2m − cα m − (−1)m λ2 = 0

we get a contradiction. Indeed, since λ1 α 2m − um α m + (−1)m λ2 = 0 we obtain 2λ1 α 2m − (c + um )α m = 0, and from (47) c = 0.

λ1 = 21 (c + um )α −m ,

λ2 = 21 (c + um )(−α)−m


963

q = 2 is impossible since then both numbers λ1 α 2x − cα x ± λ2 have a non-zero imaginary part. Finally q = 3 is impossible for the following reason. If α = β 3 , β ∈ Q(α) then α m/3 2 c is of degree 3 over Q(α) and cannot satisfy the equation λ1 t − ct ± λ2 = 0 for any choice of sign. If α = β 3 , β ∈ Q(α) then β satisfies an equation t 2 − dt − 1 = 0, d integer, and we get a = Tr β 3 = d 3 + 3d, contrary to the assumption.

Lemma 10. If every integral vector [t1 , . . . , tk ] satisfies at least one congruence of the set S: (48)

ah0 +

k

ahj tj ≡ 0 mod m (1 h g)

j =1

and no proper subset of S has the same property then for all h, j M(g)ahj ≡ 0 mod m, where

M(g) =

p [(g−1)/(p−1)] .

pg p prime

Proof. Let dh = (ah1 , ah2 , . . . , ahk , m). If dh /| ah0 the congruence k

ahj tj + ah0 ≡ 0 mod m

j =1

is never satisfied contrary to the minimal property of S. Hence for all h ah0 ≡ 0 mod dh

(49)

and the congruences (48) take the form (50)

k ahj j =1

dh

tj +

ah0 m ≡ 0 mod dh dh

(1 h g).

For a given prime p let nr be the number of indices h g such that p r be the largest r with nr = 0. We have nr ns nr+1 (51) + 2 + . . . + s−r+1 1 p p p

m and let s dh

(1 r s).

In order to prove this assume that for a certain r s nr nr+1 ns (52) + 2 + . . . + s−r+1 < 1. p p p m The congruences (48) with p r /| form a proper subset of the set S and by the assumption dh 0 there is a vector t which does not satisfy any of them.

964


m (q r) is in virtue of Frobenius’s dh theorem (used in proof of Lemma 8) satisfied by at most a ahk h1 ,..., , pq−r+1 p (q−r+1)(k−1) = p(q−r+1)(k−1) dh dh On the other hand, a congruence (50) with p q

integral vectors t mod p q satisfying t ≡ t 0 mod p r−1 .

(53)

The alternative of all congruences in question is satisfied by at most s q=r

nq p (s−r+1)k p q−r+1

integral vectors t mod p s satisfying (53). Since the number of all integral vectors t mod ps satisfying (53) is p(s−r+1)k , (52) implies the existence of a vector t 1 ≡ t 0 mod p r−1 which m satisfies no congruence (50) and consequently no congruence (48) with pr | . By the dh Chinese remainder theorem there exists a vector t such that m t ≡ t 0 mod l.c.m. , r p /| m/dh dh t ≡ t 1 mod p s . This vector satisfies no congruence (48). The obtained contradiction proves (51). Consider the lower bound of the function n1 + n2 + . . . + ns = f (n1 , . . . , ns ) under the condition (51), where now n1 , . . . , ns are nonnegative real numbers. Since f (n1 , n2 , . . . , ns ) max nr the lower bound is attained. 1rs (0) (0) (n1 , . . . , ns ) be a

Let respect to s − r

point in which it is attained. We shall show by induction with

n(0) r =p−1

(54)

(0)

(1 r < s), (0)

n(0) s = p. (1)

(0)

(1)

Indeed, (51) for r = s gives ns p. If ns > p, we set nr = nr for r < s −1, ns−1 = 1 (0) (0) (1) (1) (1) (0) (0) ns−1 + (ns − p), ns = p, verify (51) and find f (n1 , . . . , ns ) < f (n1 , . . . , ns ) p which is impossible. Assume now that (54) holds for s − r < s − q, i.e. r > q. The condition (51) for r = q gives s−1 (0) nq p−1 p−1 p − s−r+1 1− ; p p q−r+1 p p

n(0) q p − 1.

q=r+1

(0)

If nq > p − 1, we set

(1)

nq−1

(0) n(1) for r = q − 1, q; r = nr 1 (0) = nq−1 + (n(0) − p + 1), n(1) q = p − 1, p q


965

verify (51) and find again (0) (1) (0) f (n(1) 1 , . . . , ns ) < f (n1 , . . . , ns ),

which is impossible. (0) Since n(0) 1 + . . . + ns g it follows from (54) that g − 1 s(p − 1) + 1 g, s p−1 m and thus for all h g, | M(g). dh This together with (49) gives the lemma.

Lemma 11. If every integral vector [t1 , . . . , tr ] satisfies at least one of the congruences ah0 +

(55)

r

ahs ts ≡ 0 mod m

s=1

(1 h g) then for at least one h (56)

ah0 ≡ 0 mod m and M(g)ahs ≡ 0 mod m (1 s r),

where M(g) has the meaning of Lemma 10. Proof. Choose in {1, 2, . . . , g} a minimal subset M with the property that every integral vector satisfies at least one congruence (55) with h ∈ M. To the set of these congruences Lemma 10 applies. The congruence satisfied by the vector [0, 0, . . . , 0] satisfies also the conditions (56).

Remark. M(g) is the least number with the property formulated in Lemmata 10 and 11, as the following example shows already in dimension one: m = p[(g−1)/(p−1)] (p prime), a11 = 1, a10 = 0 and for h = (p − 1)q + r + 1, 1 r p − 1, 2 h g, ah1 = p q , m ah0 = r. p For k = 1 Lemma 10 is contained in a stronger result of S. Znám [11], however his c proof does not extend to k > 1. Lemma 12. Let H, I be two finite sets and let Mhi (h ∈ H , i ∈ I ) be inhomogeneous linear forms with integral coefficients. If for every positive integer m and a suitable h ∈ H the system of congruences (57)

Mhi (x) ≡ 0 mod m (i ∈ I )

is soluble then for a suitable h ∈ H the system of equations (58)

Mhi (x) = 0 (i ∈ I )

is soluble in integers. Proof. Suppose that no system (58) is soluble in integers. Then by Lemma 9 of [6] for each h ∈ H there exists an mh such that the system (57) is insoluble for m = mh . Taking

966 m =


mh we infer that the system (57) is insoluble for any h ∈ H contrary to the

h∈H

assumption. Proof of Theorem 6. Let us set (59)

a

αhij = ζwhij 0

r

a

βhi = ζwbhi0

πs hij s ,

s=1

r

πsbhis ,

s=1

where w is the number of roots of unity contained in K and πs are elements of the multiplicative basis described in Lemma 9. Consider the linear forms Lhi0 = wx0 +

k

ahij 0 xj − bhi0 ,

j =1

(60) Lhis =

k

ahij s xj − bhis

(1 s r)

j =1

and let H be the set of all vectors h = [h1 , h2 , . . . , hl ] with 1 hi gi (1 i l), I be the set of all vectors i = [i, s] with 1 i l, 0 s r. For any h ∈ H , i = [i, s] ∈ I we put Mhi = Lhi is .

(61)

We assert that for any positive integer m there exists an h ∈ H such that the system of congruences Mhi (x0 , x1 , . . . , xk ) ≡ 0 mod m

(62)

is soluble. Let us take n = 2τ wM(max gi )m

l.c.m.

qm+max gi q prime

(i ∈ I )

(q − 1), where τ is the relevant parameter

of K. By Theorem 4 for any choice of t1 , . . . , tr mod n/w there exists a prime ideal p of K(ζn ) prime to D such that π ζ w s = ζw , = ζnwts (1 s r). (63) p n p n Let m be the product of all these prime ideals p. The solubility of the system of congruences gi k xj αhij − βhi ≡ 0 mod m

(i = 1, . . . , l)

h=1 j =1

implies that for any vector [t1 , . . . , tr ] and any i l there is an h gi such that k k αhij xj βhi xj αhij ≡ βhi mod p, = p n p n j =1

j =1


967

for some p satisfying (63). This implies by (59) that k r r n n ahij 0 + wts ahij s xj ≡ bhi0 + wts bhis mod n, w w j =1

s=1

s=1

whence k r k n ahij 0 xj − bhi0 + w ts ahij s xj − bhis ≡ 0 mod n. w j =1

j =1

s=1

Using now Lemma 11 we get that for any i l and a certain hi gi k

k

ahi ij 0 xj − bhi i0 ≡ 0 mod w,

j =1

ahi ij s xj − bhi is ≡ 0 mod m

(1 s r).

j =1

c

In virtue of (60) and (61) this is equivalent for a suitable x0 to the system (62) in which h = [h1 , . . . , hl ]. Therefore, by Lemma 12 there exists a vector h0 = [h01 , . . . , h0l ] such that the system of equations Mh0 i (x0 , x1 , . . . , xk ) = 0

(i ∈ I )

is soluble in integers. Denoting a solution by [x00 , x10 , . . . , xk0 ] we get from (60) and (61) for all i l wx00 k j =1

+

k j =1

ah0 ij 0 xj0 − bh0 i0 = 0, i

i

ah0 ij s xj0 − bh0 is = 0 i

i

(1 s r)

hence by (59) gi h

x0

j αhij − βhi

=0

(1 i l).

h=1 j =1

Addendum 1. Dr. J. Wójcik has pointed out that the equality in formula (28) which is only said to hold but not proved is actually used in the formula (##) on p. 954. The equality in question follows from the formula (#) on p. 953, where also can be replaced by =. The latter is a consequence of the fact that for q > 2 the extension K(ζq s )/K is cyclic and for q = 2, p = 2 we have ordp [K(ζq s ) : K] = 0. 2. The remark made on p. 955 has not been proved rigorously, since it is not clear why τ (χ ) = 0. Therefore, we return to the question and we shall prove more than was asserted

968


namely that the number σ occurring in Theorem 3 is the least integer with the required property, provided (σ, n/wn ) = 1. By the definition of σ there exists a character χ belonging to the exponent σ on the

group G = Gal K(ζn )/K represented as a multiplicative group of residue classes mod n. Let xy τy = χ (x)ζn . x∈G xy

Since χ (x) are non-zero and the Vandermonde determinant |ζn |

x∈G y=1,2,...,|G|

is non-zero

there exists a y such that τy = 0. Let us fix such a y and denote the corresponding τy by τ (χ) = 0. Since χ (x) ∈ K, χ (x)σ = 1 we have τ (χ ) ∈ K(ζn ), τ (χ )σ ∈ K, τ (χ)nσ ∈ K n . Suppose that τ (χ )nσ = γ n , γ ∈ K. Then τ (χ )(n/wn ) = ζwαn γ ∈ K j

and applying an automorphism ζn → ζn with j ∈ G we get τ (χ )(n/wn ) χ (j )(n/wn ) = τ (χ )(n/wn ) . Since τ (χ) = 0 it follows that χ (j )(n/wn ) = 1

c

and by the choice of χ

n

. σ

wn

Hence if (σ, n/wn ) = 1 we get σ | . If (σ, n/wn ) = 1 σ need not be the least integer with in Theorem 3. In particular if ζ4 ∈ / K, n ≡ 0 mod 2τ +1 , ( the property asserted −1 − ζ2τ + ζ2τ + 2 ∈ K, σ can be replaced by (wn , l.c.m. [K(ζq ) : K]). q |n, q prime

The remark on p. 957 remains valid on replacing (wn , n/wn ) by (σ, n/wn ) which makes it stronger. 3. Theorem 6 has the following equivalent form much more useful in applications. Theorem 7. Let fr (z1 , . . . , zp ) (1 r s) be polynomials with coefficients in an algebraic number field K and αij (1 i p, 1 j q) non-zero elements of K, M a positive integer. If the system of equations fr (z1 , . . . , zp ) = 0 (1 r s)

(A1)

has only finitely many solutions in the complex field and the system of congruences q q xj xj (A2) fr α1j , . . . , αpj ≡ 0 mod m (1 r s) j =1

j =1


969

is soluble for all moduli m prime to M then the system of equations (A3)

fr

q

q xj xj α1j , . . . , αpj j =1 j =1

= 0 (1 r s)

is soluble in rational integers xj . Proof. Since the system (A1) has only finitely many solutions they all lie in a finite extension K1 of K. Let them be (βh1 , . . . , βhp ) (1 h g). Thus we have the equivalence + * * fr (z1 , . . . , zp ) = 0 ≡ zi = βhi rs

hg ip

and by the distributive property of alternative with respect to conjunction * * * * + fr (z1 , . . . , zp ) = 0 ≡ ··· zih = βhih i1 p i2 p

rs

(A4) ≡

* * i1 p i2 p

ig p hg g *

···

(zih − βhih ) = 0.

ig p h=1

By the Hilbert theorem on zeros it follows that for every integral vector i = [i1 , . . . , ig ] ∈ {1, 2, . . . , p}g = I and a suitable exponent ei we have (A5)

g

zih − βhih

ei

=

s

fr (z1 , . . . , zp )Fri (z1 , . . . , zp ),

r=1

h=1

where Fri ∈ K1 [z1 , . . . , zp ]. If m is prime to the denominators of Fri and to the numerators as well as the denominators of αij , the system of congruences (A2) with m = me , e = max ei , and the identity (A5) imply i∈I

(A6)

g q h=1 j =1 βhih =0

x αihjj

− βhih

≡ 0 mod m

(i ∈ I ).

Therefore, the system (A6) is soluble for all moduli prime to D = M times a certain finite product. Applying Theorem 6 we infer that the system of equations g q h=1 j =1

x

αihjj − βhih

=0

(i ∈ I )

is soluble in integers. By the equivalence (A4) this system is equivalent to (A3) and the proof is complete.

970


References [1] G. Darbi, Sulla reducibilità delle equazioni algebriche. Ann. Mat. Pura Appl. 4 (1925), 185–208. [2] H. Hasse, Zum Existenzsatz von Grunwald in der Klassenkörpertheorie. J. Reine Angew. Math. 188 (1950), 40–64. [3] D. Hilbert, Die Theorie der algebraischen Zahlkörper. In: Ges. Abhandlungen I, Chelsea, New York 1965. [4] M. Kneser, Lineare Abhängigkeit von Wurzeln. Acta Arith. 26 (1975), 307–308. [5] W. H. Mills, Characters with preassigned values. Canad. J. Math. 15 (1963), 169–171. [6] A. Schinzel, On power residues and exponential congruences, Acta Arith. 27 (1975), 397–420; this collection: H4, 915–938. [7] Th. Skolem, Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen. Vid. Akad. Avh. Oslo I 1937, no. 12. [8] −−, On the existence of a multiplicative basis for an arbitrary algebraic field. Norske Vid. Selsk. Forh. (Trondheim) 20 (1947), no. 2, 4–7. [9] N. Tschebotaröw [Chebotarev], Über einen Satz von Hilbert (Russian). Vestnik Ukr. Akad. Nauk, 1923, 3–7; Sobranie Sochineni˘ı I, Moscow–Leningrad 1949–50, 14–17. [10] −−, Grundzüge der Galoisschen Theorie. Übersetzt und bearbeitet von H. Schwerdtfeger, Noordhoff, Groningen–Djakarta 1950. [11] S. Znám, On properties of systems of arithmetic sequences. Acta Arith. 26 (1975), 279–283.

Originally published in Comptes Rendus Mathématiques de l’Académie des Sciences. La Société Royale du Canada Mathematical Reports of the Academy of Sciences. The Royal Society of Canada 1 (1979), 115–118


An extension of Wilson’s theorem with G. Baron (Wien)

The aim of this note is to prove the following extension of Wilson’s theorem conjectured by W. Snyder in his Ph.D. thesis A concept of Bernoulli numbers in algebraic function fields, Univ. of Maryland 1977. Snyder has found interesting applications of his conjecture to differentials in rings of characteristic p. Theorem. For any prime p and any residues xi mod p we have

(1) xσ (1) xσ (1) + xσ (2) · · · xσ (1) + . . . + xσ (p−1) σ ∈Sp−1

p−1

mod p, ≡ x1 + . . . + xp−1

where the summation is taken over all permutations σ of {1, 2, . . . , p − 1}. r Let for positive integers a1 , . . . , ar C(a1 , . . . , ar ) denote the coefficient of X = xiai i=1 in the sum Pσ , where σ ∈Sn

Pσ = xσ (1) xσ (1) + xσ (2) · · · xσ (1) + . . . + xσ (n)

Lemma 1. Let aj > 1 for j s, aj = 1 for s < j r,

r

(σ ∈ Sn ).

ai = n. Then

i=1

C(a1 , . . . , ar ) = (n − r)

s

C(a1 , . . . , ai−1 , ai − 1, ai+1 , . . . , ar )

i=1

+ (n − r + 1)(r − s)C(a1 , . . . , ar−1 ). Proof. We have σ ∈Sn

Pσ =

n

j =1 σ ∈Sn ,σ (n)=j

Presented by P. Ribenboim, F.R.S.C.

Pσ =

n j =1

j .

972


The coefficient of X in Pσ is the same as in Pτ σ where τ ∈ Sn is any permutation stable on {1, 2, . . . , s} and fixing the set {s + 1, . . . , r}. Hence the coefficient of X in j is 0 if j s, is equal to the coefficient C1 of X in r if s < j r and equal to the coefficient C2 of X in n = (x1 + . . . + xn ) Pσ if j > r. If r > s, C1 is equal to C(a1 , . . . , ar−1 ). σ ∈Sn−1

On the other hand C2 =

s

C(a1 , . . . , ai−1 , ai − 1, ai+1 , . . . , ar ) +

i=1

r

C(a1 , . . . , ar−1 ).

i=s+1

Hence C(a1 , . . . , ar ) =

r

C1 +

j =s+1

= (n−r)

s

n

C2

j =r+1

C(a1 , . . . , ai−1 , ai −1, ai+1 , . . . , ar )+(n−r+1)(r−s)C(a1 , . . . , ar−1 ).

i=1

In order to evaluate C(a1 , . . . , ar ) we introduce the following notation valid for all systems of r b a 0 positive real numbers ai : R = {1, 2, . . . , r}, S1 (a, b, r, q) =

b i=a+1

S2 (a, b, r, q) =

b i=a+1

q ∗ (|Tk | + 1)! ai i A(Tk ) + ai k=1

∗ i

q j =1

q j =1, j =k

|Tj |! , A(Tj ) + 1

|Tj |! , A(Tj ) + 1

where 1 q < r and the inner summation ∗i in both sums is taken over all partitions (the order of summands neglected) of R − {i} into q non-empty sets Tj of cardinality |Tj | and A(Tj ) = al . Moreover we set l∈Tj

S1 (a, b, r, r) = 0, S2 (a, b, r, 0) = 0 (r 2), S2 (0, 0, 1, 0) = 0, S2 (0, 1, 1, 0) = 1. Lemma 2. For any positive q r the following identity holds S1 (0, r, r, q) + S2 (0, r, r, q − 1) = (A(R) + q) where

∗∗

q ∗∗ j =1

|Rj |! , A(Rj ) + 1

is taken over all partitions of R into q non-empty sets Rj .

Proof. For q = r = 1 the identity holds trivially. For q = 1, r 2 we have S1 (0, r, r, q) + S2 (0, r, r, q − 1) =

r i=1

ai

r! r! = r! = (A(R) + 1) . A(R) A(R) + 1

973

H6. An extension of Wilson’s theorem

For q 2 we group together all terms in S1 (0, r, r, q) in which Tk ∪ {i} = T . For any i, k we get |T | 2. On the other hand for any T ⊂ R with |T | 2 we have r

ai

∗ i

i=1

q

(|Tk | + 1)! A(Tk ) + ai

k=1, Tk ∪{i}=T

=

i∈T

where

∗

T

q

|Tj |! A(Tj ) + 1

j =1, j =k

q−1 ∗ |T |! q−1 ∗ |Tj |! |Tj |! = , ai |T | T A(T ) T A(Tj ) + 1 A(Tj ) + 1 j =1

j =1

is taken over all partitions of R − T into q − 1 non-empty sets Tj .

Hence

S1 (0, r, r, q) =

∗ T

T ⊂R, |T |2

|T |!

q−1 j =1

|Tj |! . A(Tj ) + 1

Now setting in S2 (0, r, r, q − 1), {i} = T we get ∗

S2 (0, r, r, q − 1) =

T

T ⊂R, |T |=1

|T |!

q−1 j =1

|Tj |! , A(Tj ) + 1

thus S1 (0, r, r, q) + S2 (0, r, r, q − 1) =

q ∗∗

|Rj |! (A(Rj ) + 1) A(Rj ) + 1 q

j =1

j =1

= (A(R) + q)

q ∗∗ j =1

|Rj |! .

A(Rj ) + 1

Lemma 3. For any positive integers r, a1 , . . . , ar with a1 + . . . + ar = n we have (2)

C(a1 , . . . , ar ) =

∗∗ |Rj |! (n − r)! , (−1)r−q (n + q) a1 ! · · · ar ! A(Rj ) + 1 r

q

q=1

j =1

the inner sum being taken over all partitions of R into q non-empty subsets Rj .

Proof by induction on n. For n = 1 the lemma holds trivially. Assume that it is true for all r r ai = n − 1 and consider a sequence ai with ai = n 2. In sequences ai satisfying i=1

i=1

view of symmetry we may assume that aj > 1 for j s, aj = 1 for j > s. Let us denote the right hand side of (2) by D(a1 , . . . , ar ). By Lemma 1 and the inductive assumption

974


we have

(n − r)! (−1)r−q (n + q − 1)! S1 (0, s, r, q) + S2 (0, s, r, q − 1) a1 ! · · · ar ! r

C(a1 , . . . , ar ) =

q=1

(n − r + 1)! (−1)r−q−1 (n + q − 1)! S2 (s, r, r, q). a1 ! · · · ar ! r−1

+

q=1

On the other hand, by Lemma 2 D(a1 , . . . , ar ) =

(n − r)! (−1)r−q (n + q − 1)! S1 (0, r, r, q) + S2 (0, r, r, q − 1) a1 ! · · · ar ! r

q=1

hence a1 ! · · · ar ! D(a1 , . . . , ar ) − C(a1 , . . . , ar ) = (n + r − 1)! S1 (s, r, r, r) (n − r)! r−1

+ (−1)r−q (n − q + 1)! S1 (s, r, r, q) − (r + q − 1)S2 (s, r, r, q) q=1

+ (−1)r−1 n! S2 (s, r, r, 0).

However S1 (s, r, r, r) = 0, q r ∗ (|Tk | + 1)!

S1 (s, r, r, q) =

i=s+1

=

r i=s+1

A(Tk ) + 1

i

k=1 q ∗

i

|Tk | + 1

k=1

q j =1, j =k q

j =1

|Tj |! A(Tj ) + 1

|Tj |! = (r + q − 1)S2 (s, r, r, q) A(Tj ) + 1

and since r 2 or s 1, S2 (s, r, r, 0) = 0. This gives D(a1 , . . . , ar ) = C(a1 , . . . , ar ).

Proof of the theorem. Since both sides of the congruence (1) are symmetric it is enough to show that a1 + . . . + ar = n = p − 1 implies C(a1 , . . . , ar ) ≡

(p − 1)! mod p. a1 ! · · · ar !

Now in formula (2) terms corresponding to q > 1 are divisible by p since (n + q)! ≡ 0 mod p and A(Rj ) + 1 < A(R) + 1 = p. Hence (n + 1)! r! (n − r)! (−1)r−1 a1 ! · · · ar ! n+1 (p − 1)! (p − 1)! (−1)r−1 (p − r − 1)! r! ≡ (p − 2)! mod p ≡ a1 ! · · · ar ! a1 ! · · · ar !

C(a1 , . . . , ar ) ≡

c

c

and (1) follows from Wilson’s theorem.

Originally published in Demonstratio Mathematica XVIII (1985), 377–394


Systems of exponential congruences

Dedicated to the memory of Professor Roman Sikorski

Some years ago I proved the following theorem ([1], Theorem 2). Let K be an algebraic number field, α1 , . . . , αk , β non-zero elements of K. If for almost all prime ideals p of K the congruence k

x

αj j ≡ β (mod p)

j =1

is soluble in integers xj then the equation k

x

αj j = β

j =1

is soluble in integers. I have shown by an example that this theorem does not extend to systems of congruences of the form k

(1)

x

αijj ≡ βi (mod p)

(i = 1, 2, . . . , h)

j =1

even for h = 2, k = 3. Recently L. Somer [4] has considered systems of the form (1) for k = 1. The study of his work has suggested to me that the connection between the local and the global solubility of (1) may hold if for some i h the numbers αij are multiplicatively independent. The aim of this paper is to prove this assertion in the form of the following theorem. Theorem 1. Let K be an algebraic number field, αij , βi (i = 1, 2, . . . , h; j = 1, 2, . . . , k) non-zero elements of K and assume that for some i h k

x

αijj = 1, xj ∈ Z implies xj = 0 for all j k.

j =1

If for almost all prime ideals p of K in the sense of the Dirichlet density the system (1) is

976


soluble in integers xj then the system of equations k

(2)

x

αijj = βi (i = 1, 2, . . . , h)

j =1

is soluble in integers. The following corollary is almost immediate. Corollary. If the system of congruences αix ≡ βi (mod p) (i = 1, 2, . . . , h) is soluble in integers x for almost all prime ideals p of K then the system of equations αix = βi (i = 1, 2, . . . , h) is soluble in integers. Somer [4] has proved the above corollary under the assumption that either none of the αi ’s is a root of unity or all the αi ’s are roots of unity. The next theorem shows that Theorem 1 cannot be extended further. Theorem 2. For every k 2 there exist non-zero rational integers αij , βi (i = 1, 2; j = 1, 2, . . . , k) such that α12 , . . . , α1k are multiplicatively independent, the system (1) with h = 2 is soluble for all rational primes p, but the system (2) is insoluble in integers. In the sequel ζq denotes a primitive q-th root of unity. For a rational matrix M den M denotes the least common denominator of the elements of M and M T the transpose of M. The proofs are based on eight lemmata. Lemma 1. For every rational square matrix A there exists a non-singular matrix U whose elements are integers in the splitting field of the characteristic polynomial of A such that ⎤ ⎡ A1 ⎥ ⎢ A2 ⎥ ⎢ (3) U −1 AU = ⎢ ⎥ . . ⎦ ⎣ . An with Aν a square matrix of degree ν : ⎤ ⎡ λν 1 ⎥ ⎢ λν 1 ⎥ ⎢ ⎢ . . .. .. ⎥ (4) Aν = ⎢ ⎥ ⎥ ⎢ ⎣ λν 1 ⎦ λν where the empty places (not the dots) are zeros.

(ν = 1, 2, . . . , n)

H7. Systems of exponential congruences

977

Proof (see [5], §88). The elements of U can be made algebraic integers, since the left hand side of (3) is invariant with respect to the multiplication of U by a number.

Lemma 2. Let L0 , Lj , Mj ∈ Z[t1 , . . . , tr ] (j = 1, 2, . . . , k) be homogeneous linear forms and Mj (j = 1, 2, . . . , k) linearly independent. If the system of congruences k

xj Lj (t1 , . . . , tr ) ≡ L0 (t1 , . . . , tr ) (mod m)

j =1

(5)

k

xj Mj (t1 , . . . , tr ) ≡ 0 (mod m)

j =1 c

is soluble in xj for all moduli m and all integer vectors [t1 , . . . , tr ], then L0 = 0. Proof. Let Lj =

r

lj s ts (0 j k), Mj =

s=1

r

mj s ts (1 j k). Taking if necessary

s=1

lj s = mj s = 0 for s > k we can assume that r > k. Since Mj ’s are linearly independent we can assume also that the matrix M = [mj s ]j,sk is non-singular. Put M ∗ = [mj s ] L = [lj s ]1j,sk , l 0 = [l01 , . . . , l0k ],

j k , k<sr L∗ = [lj s ]1j k , k<sr l ∗0 = [l0,k+1 , . . . , l0r ].

Let K0 be the splitting field of the characteristic polynomial of LM −1 . In virtue of Lemma 1 there exists a matrix U whose elements are integers of K0 such that ⎤ ⎡ A1 ⎥ ⎢ A2 ⎥ ⎢ (6) U −1 LM −1 U = ⎢ ⎥ . .. ⎦ ⎣ An c

where Aν of order ν is given by (4) (ν = 1, 2, . . . , n). We proceed to show that l 0 = 0 and l ∗0 = 0. Let us write (7)

l 0 M −1 U = [l1 , . . . , lk ].

Suppose that l 0 = 0 hence l 0 M −1 U = 0 and let the least κ k for which lκ = 0 satisfy μ < κ μ . (8) σν = μ 0,

(12)

t ∗ = [0, . . . , 0, den(L − λM)−1 ]T . With this choice of t ∗ we can find a t ∈ Zk such that (L − λM)t = λM ∗ t ∗ − L∗ t ∗ and then the system (5) gives for x = [x1 , . . . , xk ] xλ(Mt + M ∗ t ∗ ) ≡ l0r den(L − λM)−1 (mod m), x(Mt + M ∗ t ∗ ) ≡ 0 (mod m), hence l0r den(L − λM)−1 ≡ 0 (mod m). The obtained contradiction with (12) completes the proof.

Lemma 3. For every rational square matrix A there exists a non-singular matrix U such that (3) holds with Aν a square matrix of order ν (in general not the same as in Lemma 2), ⎤ ⎡ −αν1 1 ⎥ ⎢ −αν2 1 ⎥ ⎢ ⎥ ⎢ . . .. .. (13) Aν = ⎢ ⎥ ⎥ ⎢ ⎣−αν,ν −1 1⎦ c −ανν c

where ανj ∈ Q and x ν +

ν j =1

ανj x ν −j is a power of a polynomial irreducible over Q.

Proof (see [5], §88). The form of the matrix A has been changed by applying central symmetry (matrices symmetric to each other with respect to the common centre are similar). U can be made integral via multiplication by a suitable integer.

Lemma 4. Let L0 , Lj , Mj ∈ Z[t1 , . . . , tr ] (j = 1, 2, . . . , k) be homogeneous linear forms, Mj ’s linearly independent. Let a0 , aj , bj ∈ Z (j = 1, 2, . . . , k) and w be a fixed positive integer. If for all moduli m ≡ 0 (mod w) and for all integer vectors [t1 , . . . , tr ] the system of congruences k

(14)

j =1

m m xj Lj (t1 , . . . , tr ) + aj ≡ L0 (t1 , . . . , tr ) + a0 (mod m), w w k j =1

m xj Mj (t1 , . . . , tr ) + bj ≡ 0 (mod m) w

is soluble in integers xj then L0 = 0 and a0 ≡ 0 (mod w).

980


Proof. When m runs through all positive integers divisible by w, m/w runs through all positive integers, hence applying Lemma 2 we infer that L0 = 0. In order to show a0 ≡ 0 (mod w) we adopt the meaning of L, L∗ , M, M ∗ from the proof of Lemma 2. In virtue of Lemma 3 there exists a non-singular integral matrix U such that ⎤ ⎡ A1 ⎥ ⎢ A2 ⎥ ⎢ (15) U −1 LM −1 U = ⎢ ⎥, .. ⎦ ⎣ . An c

where Aν of order ν is given by (13). We can assume without loss of generality that ανν = 0, 1 ν for ν n0 and ανν = 0 for ν > n0 (n0 may be 0). It follows from the ν ανj x ν −j that condition on x ν + j =1

c

⎡ ⎢ ⎢ Aν = ⎢ ⎣

(16)

⎤

1

⎥ ⎥ ⎥ ⎦

1 ..

.

(1 ν n0 ),

1 where the empty places are zeros as before. Now put ⎡ ⎤ ⎡ ⎤ a1 b1 −1 ⎢ .. ⎥ −1 ⎢ .. ⎥ (17) U ⎣ . ⎦ = con(a 1 , . . . , a n ), U ⎣ . ⎦ = con(b1 , . . . , bn ), ak

bk

where con denotes concatenation and for ν = 1, 2, . . . , n ⎤ ⎤ ⎡ ⎡ aν1 bν1 ⎥ ⎥ ⎢ ⎢ (18) a ν = ⎣ ... ⎦ , bν = ⎣ ... ⎦ . aνν

c

bνν

Take (19)

m0 = w den M −1 den U −1 l.c.m. den A−1 ν n0 k and that σ is the least index > k such that cσ = 0 we take m = 2wek |cσ |, ⎧ bs m ⎪ ⎪ ⎨− e w for s k, s ts = 1 for s = σ, ⎪ ⎪ ⎩ 0 for s > k, s = σ c and get from (31) cσ ≡ 0 (mod 2|cσ |), bj m for a contradiction. Therefore cs = 0 for all s > k and taking m = 2wek , tj = − ej w j k we get from (31) m bj cj m − ≡ 0 (mod m), w ej w k

l10

j =1

hence l10 ≡

(32)

k bj cj j =1

ej

(mod w + ).

Finally taking m = wek and for a fixed j k ⎧ m bs ek ⎪ ⎪ − + ⎪ ⎪ ej ⎨ w es ts = − m bs ⎪ ⎪ w es ⎪ ⎪ ⎩ 0

if s = j, if s = j, s k if s > k,

we get from (31) and (32) yj ek ≡ cj ek /ej (mod ek ),

cj /ej ∈ Z.

Integers ξj defined by [ξ1 , . . . , ξk ] = [c1 /e1 , . . . , ck /ek ]B −1 satisfy (27) and (28) for i = 1 in virtue of (25), (29), (30) and (32). Take now i 1 and consider the system of two congruences: k j =1

m xj Lij (t1 , . . . , tr ) + lij w ≡ Li0 (t1 , . . . , tr ) + li0

k m m − ξj Lij (t1 , . . . , tr ) + lij (mod m) w w j =1

984


and k j =1

m xj L1j (t1 , . . . , tr ) + l1j ≡ 0 (mod m). w

0 ] is a solution of the system (26), the above system has the solution If [x10 , . . . , xm 0 0 − ξ ], hence it is soluble for all moduli m and all integer vectors [x1 − ξ1 , . . . , xm m [t1 , . . . , tr ]. Since L1j are linearly independent we have in virtue of Lemma 4

Li0 −

k

ξj Lij = 0

and li0 −

j =1

k

ξj lij ≡ 0 (mod w),

j =1

thus (27) and (28) hold for all i h.

Lemma 7. In any algebraic number field K there exists a multiplicative basis, i.e., such a r πsxs , sequence π1 , π2 , . . . that any non-zero element of K is represented uniquely as ζ s=1

where xs are rational integers and ζ is a root of unity.

Proof. See [3].

Lemma 8. Let K be an algebraic number field, w the number of roots of unity contained in K, w ≡ 0 (mod 4), n a positive integer,

σ = w, n, l.c.m. [K(ζq ) : K] . q |n, q prime

If

n ≡ 0 mod (w, n)

(33)

l.c.m. [K(ζq ) : K]

q |n, q prime

and α1 , . . . , αr ∈ K have the property that (34)

ζwx0

r

αsxs = γ n/σ , γ ∈ K implies x1 ≡ x2 ≡ . . . ≡ xr ≡ 0 (mod n/σ )

s=1

then for any integers c1 , . . . , cr ≡ 0 (mod σ ) and any c0 there exists a set of prime ideals q of K(ζn ) of a positive Dirichlet density such that ζ α w s c0 (35) = ζ(w,n) , = ζncs (1 s r). q n q n Proof. This is a special case (ζ4 ∈ K) of Theorem 4 of [2]. In this theorem only the existence of infinitely many prime ideals q with property (35) is asserted, but the existence of a set of a positive Dirichlet density is immediately clear from the proof based on the Chebotarev density theorem.

985

H7. Systems of exponential congruences

Proof of Theorem 1. Without loss of generality we may assume that ζ4 ∈ K and that α1j (j = 1, 2, . . . , k) are multiplicatively independent. Let us set r

a

αij = ζwij 0

(36)

a

πs ij s ,

βi = ζwbi0

s=1

r

πsbis ,

s=1

where w is the number of roots of unity contained in K and πs are elements of the multiplicative basis described in Lemma 7. Take an arbitrary modulus m ≡ 0 (mod w) and set in Lemma 8 n = mm1 , where m1 = l.c.m. (p − 1) and P is the greatest pP , p prime

prime factor of m. Since every prime factor q of n satisfies q P the number n satisfies (33). The condition (34) is clearly satisfied by αs = πs (1 s r). Hence for any integers c1 , . . . , cr ≡ 0 (mod w) there exists a set S of prime ideals q of K(ζn ) of positive Dirichlet density such that ζ π w s (37) = ζw , = ζncs (1 s r). q n q n The ideals p of K divisible by at least one q ∈ S form a set of positive Dirichlet density, hence by the assumption there exist integers xj satisfying k

x

αijj ≡ βi (mod q)

(i = 1, 2, . . . , h)

j =1

for at least one q ∈ S. It follows from (36) and (37) that k r r n n xj aij s cs + aij 0 bis cs + bi0 (mod n) ≡ w w j =1

s=1

(1 i h).

s=1

Now take cs = wm1 ts (1 s r), ⎧ r ⎪ ⎪ aij s ts ⎨Lij = w s=1 (38) r ⎪ ⎪ bis ts ⎩Li0 = w

(1 i h, 1 j k), (1 i h).

s=1

It follows that for all moduli m ≡ 0 (mod w) and all integer vectors [t1 , . . . , tr ] the system of congruences k

xj Lij (t1 , . . . , tr ) + aij 0

j =1

m m ≡ Li0 (t1 , . . . , tr ) + bi0 (mod m) w w

is soluble in integers xj . Since the numbers α1j are multiplicatively independent the linear forms L1j are linearly independent (1 j k). Hence by Lemma 6 there exist integers ξ1 , . . . , ξk such that k j =1

ξj Lij = Li0

and

k

ξj aij 0 ≡ bi0 (mod w)

(1 i h).

j =1

It follows from (36) and (38) that ξ1 , . . . , ξk satisfy the system (2).

986


Proof of Corollary. In view of Theorem 1 it remains to consider the case when for each i h the number αi is a root of unity. But then either there exists a positive integer x w such that αix = βi

(1 i h)

or the system of congruences αix ≡ βi (mod p)

(1 i h)

is soluble only for prime ideals p dividing w

g.c.d. (αix − βi ).

x=1 1ih

Proof of Theorem 2. Since here K = Q we write p instead of p and denote by pj the j -th prime. We take c

α11 = −1, α21 = 2,

α1j = pj −1 (2 j k), α2j = 1 (2 j k),

β1 = −1, β2 = 1.

For p = 2 (1) has the solution xj = 0 (1 j k). For p > 2 we consider the index p−1 of 2, ind 2 with respect to a fixed primitive root of p. If is odd, (1) has a (ind 2, p − 1) solution determined by ⎧ ⎨ 1 (mod 2) x1 ≡ xj = 0 (2 j k). p−1 ⎩ 0 mod , (ind 2, p − 1) p−1 is even, (1) has a solution determined by If (ind 2, p − 1) p−1 x1 = 0, x2 ind 2 ≡ (mod p − 1), xj = 0 (3 j k). 2 On the other hand, (2) is clearly insoluble.

References [1] A. Schinzel, On power residues and exponential congruences. Acta Arith. 27 (1975), 397–420; this collection: H4, 915–938. [2] −−, Abelian binomials, power residues and exponential congruences. Acta Arith. 32 (1977), 245–274; Addendum, ibid. 36 (1980), 101–104; this collection: H5, 939–970. [3] Th. Skolem, On the existence of a multiplicative basis for an arbitrary algebraic field. Norske Vid. Selsk. Forh. (Trondheim) 20 (1947), no. 2, 4–7. [4] L. Somer, Linear recurrences having almost all primes as maximal divisors. In: Fibonacci Numbers and their Applications (Patras, 1984), Math. Appl. 28, Reidel, Dordrecht 1986, 257–272. [5] B. L. van der Waerden, Algebra II Teil. Springer, Berlin 1967.

Originally published in Mathematical Proceedings of the Cambridge Philosophical Society 112 (1992), 225–232


On a problem in elementary number theory with J. Wójcik (Warsaw)

Let ordq (a) be the exponent with which a prime q occurs in the factorization of a rational number a = 0 and, if ordq (a) = 0, let Mq (a) be the multiplicative group generated by a modulo q. In the course of a group-theoretical investigation J. S. Wilson found he needed some results about integers a, b such that Mq (a) = Mq (b), indeed also for algebraic integers, and he proved some of what he needed. J. W. S. Cassels observed that Wilson’s argument naturally proved the existence of infinitely many primes q with Mq (a) = Mq (b) for rational integers a, b with ab > 0, |a| > 1, |b| > 1. J. G. Thompson found a proof for the case of integers a, b with ab < 0, |a| > 1, |b| > 1. He also posed the problem for rational a, b (all this is unpublished). The aim of this paper is to prove that the answer to Thompson’s question is affirmative. We also include the case ab > 0 settled by Thompson himself. We use the same technique devised by Wilson which has been elaborated by Cassels and Thompson. We thank Professor Cassels for the simplification of our original exposition and the referee for his suggestions. Theorem. For all a, b ∈ Q \ {0, 1, −1} there exist infinitely many primes q such that Mq (a) = Mq (b). Without loss of generality a = b. The strategy is first to find one prime q with ordq a = ordq b = ordq (a − b) and

Mq (a) = Mq (b).

This requires a fairly elaborate subdivision into cases. At the end, a uniform argument deduces the existence of infinitely many such q. It is convenient to enunciate one key idea in the following trivial Proposition. Let q be a prime. Suppose that ordq a = ordq b = 0 and that a n ≡ b mod q where n is prime to q − 1. Then Mq (a) = Mq (b). In the argument, we shall most often take n to be prime. Given a, b we consider primes l in an appropriate arithmetic progression. A study of the factorization of a l − b then shows that it must be divisible by some prime q ≡ 1 mod l.

988


Notation. For all primes p put αp = ordp a, and further put P =

βp = ordp b,

γp = ordp (a − b)

P1 =

p γp ,

αp =βp =0

(p − 1).

p|P

Lemma 1. If either ab > 0, |a| = 1, |b| = 1 or for some prime p0 we have αp0 βp0 > 0 and αp0 , βp0 , γp0 not all equal, then there exists a prime q such that αq = βq = γq = 0 and Mq (a) = Mq (b). Proof. Replacing if necessary a by a −1 and b by b−1 , we may assume without loss of generality (1)

1 < |a| < |b|

if ab > 0,

(2)

0 < αp0 βp0

if ab < 0.

Let l be a sufficiently large prime such that l ≡ 1 mod pγp (p − 1)

for all

p |P.

Then a l ≡ a mod pγp +1 , so that ordp (a l − b) = γp

for all

p |P.

On the other hand for all primes p with αp βp > 0 we have since l is large lαp if αp < 0, l (3) ordp (a − b) = βp if αp > 0 and for all primes p with αp βp 0 the same is true in an obvious way. Finally if αp = 0, βp = 0 we have (4)

ordp (a l − b) = min{0, βp }.

Hence a l − b = sgn a · P

αp 0

where s is an integer prime to the numerators and the denominators of a, b, a − b and positive by (1). We have p lαp ≡ p αp mod l (5) a l − b ≡ a − b mod l, αp 0 prime to the γp +1 exponent λ(a) or λ(b) to which a or b, respectively belongs mod p0 0 and such that ordp0 (a r − b) = γp0 or

ordp0 (br − a) = γp0 ,

respectively, then there exists a prime q such that αq = βq = γq = 0 and Mq (a) = Mq (b). Proof. Assume without loss of generality that (r, λ(a)) = 1 and ordp0 (a r − b) = γp0 . We choose a positive integer r0 prime to P P1 in the arithmetic progression λ(a)x + r. Let ordp (a r0 − b) = ep

for all

p |P.

Let l be a sufficiently large prime in the arithmetic progression

p ep P1 x + r0 . We have

p|P

a l ≡ a r0 mod p ep +1

for all

p |P,

ordp (a l − b) = ep

for all

p |P.

and hence

Since l is large we again have (3) and (4). Hence a l − b = sgn a p ep p lαp p min{0,βp } p βp s, p|P

αp 0

where s is an integer prime to the numerators and the denominators of a, b, a − b and is positive since ab < 0. We again have (5) and if s ≡ 1 mod l we would have sgn a p ep p αp p min{0,βp } p βp ≡ a − b mod l. p|P

αp 0

990


For l large enough both sides of the congruence are equal, so γp0 = ep0 = ordp0 (a r0 − b). γp0 +1

However r0 ≡ r mod λ(a), a r0 ≡ a r mod p0

and so

γp0 = ordp0 (a r − b), contrary to the assumption. Hence s ≡ 1 mod l and s has a prime factor q ≡ 1 mod l, αq = βq = γq = 0. By the Proposition with n = l we have Mq (a) = Mq (b).

Lemma 3. If for a prime p0 | P we have γp0 > min ordp0 (a 2 − 1), ordp0 (b2 − 1) , then p0 satisfies the assumptions of Lemma 2. Proof. Without loss of generality we may assume that γp0 > ordp0 (a 2 − 1). γp

Let λ0 be the exponent to which a belongs mod p0 0 . Clearly λ0 > 2, hence there exists an r0 ≡ 1 mod λ0 such that (r0 , λ0 ) = 1. The arithmetic progression λ0 x + r0 contains a positive integer r prime to λ(a) and we have ordp0 (a r − a r0 ) = ordp0 (a r−r0 − 1) γp0 , ordp0 (a r0 − a) = ordp0 (a r0 −1 − 1) < γp0 , ordp0 (a − b) = ordp0 (a r − a r0 + a r0 − a + a − b) < γp0 . r

Lemma 4. If for a prime p0 | P we have γp0 = ordp0 (a 2 − 1) = ordp0 (b2 − 1)

(7)

then p0 satisfies the assumptions of Lemma 2. Proof. The condition (7) implies (8) (9) (10)

μ

a = ε + p 0 a1 ,

μ

b = ε + p 0 b1 ,

where

ordp0 a1 = ordp0 b1 = 0, γp0 − 1 if p0 = 2, μ= γp0 if p0 > 2.

We choose a positive integer r such that a1 r ≡ b1 mod 4 (11) a1 r ≡ b1 mod p0 and (12)

ε = ±1, a1 , b1 ∈ Q,

if p0 = 2, if p0 > 2

r, p0 (p0 − 1) = 1.

991

H8. On a problem in elementary number theory

This is possible in view of (9). Now (8), (10) and (11) imply

ordp0 (a r − b) γp0 + 1,

while (12) implies r, λ(a) = 1.

Lemma 5. For all a, b ∈ Q\{0, 1, −1}, there exists a prime q such that αq = βq = γq = 0 and Mq (a) = Mq (b). Proof. In view of Lemmas 1–4 it suffices to consider the case where ab < 0 and for all primes p either αp βp 0

(13)

or

αp = βp = γp ,

for all p | P , γp min{ordp (a 2 − 1), ordp (b2 − 1)}

(14) and

γp < max{ordp (a 2 − 1), ordp (b2 − 1)}.

(15)

The last two conditions can be reformulated. If p = 2,

γ2 < ord2 (a 2 − b2 ) = ord2 (a 2 − 1) − (b2 − 1) and so (15) implies (16)

γ2 < min ord2 (a 2 − 1), ord2 (b2 − 1) ,

which is stronger than (14). If p > 2

γp = ordp (a 2 − b2 ) min ordp (a 2 − 1) , ordp (b2 − 1)}

hence (14) and (15) are equivalent to (16) and (17) min ordp (a 2 − 1), ordp (b2 − 1) = γp < max ordp (a 2 − 1), ordp (b2 − 1)

(p = 2).

Note that (16) and (17) are invariant under the replacement of a, b by a −1 , b−1 . Assume first that for a prime l /| P αl = βl = γl = 0. Replacing if necessary a, b by a −1 , b−1 we may suppose that αl = βl = γl > 0. Hence l > 2. For every p with |αp | + |βp | > 0, αp βp 0 we have ⎧ ⎪ if αp < 0, ⎨lαp ordp (a l − b) = min{0, βp } if αp = 0, ⎪ ⎩ if αp > 0. βp = γp The same is true by virtue of (13) if αp βp > 0.

992


For p | P if p = 2 or p > 2, ordp (a 2 − 1) > γp we have ordp (a l − a) > γp ,

ordp (a l − b) = ordp (a l − a + a − b) = γp ;

if p > 2, ordp (a 2 − 1) = γp < ordp (b2 − 1) we have ordp (a 2l − 1) = γp , ordp (a l − b) = ordp (a 2l − b2 ) = ordp (a 2l − 1 − (b2 − 1)) = γp . Therefore al − b =

p (l−1)αp (a − b)s,

αp 0,

(A1 , A2 ) = (B1 , B2 ) = 1. By (18), (Ai , Bi ) = 1 (i = 1, 2) and we have P = |A1 B2 − A2 B1 | 2. If P is odd, it has an odd prime factor l. If P is even we have by (16) a ≡ b ≡ ε mod 2γ2 ,

ε = ±1,

hence (19)

A1 = εA2 + 2γ2 A3 ,

B1 = εB2 + 2γ2 B3 ,

A3 , B3 ∈ Z \ {0}

and P = 2γ2 |A3 B2 − A2 B3 | = 2γ2 |A3 B1 − A1 B3 |. Since A1 B1 < 0 < A2 B2 it follows that P > 2γ2 , so P has an odd prime factor l. By (17) and since |a| = 1, |b| = 1 we may assume without loss of generality that a = ε + l μ a1 ,

b = ε + l ν b1 ,

993

H8. On a problem in elementary number theory

where ε = ±1,

a1 , b1 ∈ Q,

ordl a1 = ordl b1 = 0,

μ = γl < ν.

Clearly a − b ≡ l γl a1 mod l γl +1 .

(20) n−μ

Consider now the number a l − b (where n μ). For every prime p with |αp | + |βp | > 0 we have by (18) ⎧ n−μ α ⎪ if αp < 0, p ⎨l l n−μ ordp (a − b) = min{0, βp } if αp = 0, ⎪ ⎩ if αp > 0. βp For every p | P with p = l, if p = 2 or p > 2, ordp (a 2 − 1) > γp , we have ordp (a l

n−μ

if p > 2, ordp

− a) > γp ,

(a 2

ordp (a l

− 1) = γp < ordp

n−μ

(b2

ordp (a

− b) = ordp (a

2l n−μ

n−μ

− a + a − b) = γp ;

− 1) we have n−μ

− 1) = γp ,

n−μ − b ) = ordp a 2l − 1 − (b2 − 1) = γp .

ordp (a 2l l n−μ

− b) = ordp (a l

2

Finally we obtain by induction on n μ al

n−μ

≡ ε + l n a1 mod l n+1 ,

hence al

(21) (22)

ν−μ

a

− b ≡ l ν (a1 − b1 ) mod l ν+1 ,

l ν−μ+1

If a1 ≡ b1 mod l we obtain al

ν−μ

−b =

− b ≡ −l ν b1 mod l ν+1 .

p (l

ν−μ −1)α p

l ν−μ (a − b)s,

αp d such that

Since a F (n) (8)

bG(n) for some n

G(n2 ) ≡ 0 mod q2 ,

q2 /| D.

Take in Lemma 3 q = q2 , A = a, B = b. We cannot have (3) with Γ ∈ Q since it would imply q2 | D, hence Γ satisfying (3) would be of degree q2 and thus by (6) d q2 , contrary to the choice of q2 . Therefore, by Lemma 3 there exists a prime ideal p2 of degree 1 in K dividing neither Cabq2 nor the denominator of ϑ and such that x q2 ≡ a mod p is solvable

H9. On exponential congruences

999

in K, but x q2 ≡ b mod p is not. We have ϑ ≡ m2 mod p2 , m2 ∈ Z. Consider the arithmetic progression P2 = {n ∈ Z : n ≡ n2 mod q2 , n ≡ m2 mod p2 }, where p2 is the prime divisible by p2 . If n ∈ P2 and (4) does not hold we have h(n) ≡ h(ϑ) ≡ 0 mod p2 , hence a F (n) ≡ bG(n) mod p2 , and by (8) G(n) ≡ 0 mod q2 . It follows that for some integers ξ, η 1 = q2 ξ − G(n)η. Since a ≡

q a2 2

c

mod p2 , a2 ∈ K we infer that

−F (n)η q2 mod p2 , b = bq2 ξ −G(n)η ≡ bξ a2

contrary to the choice of p2 .

Lemma 5. Let [K : Q] < ∞; α1 , α2 , β ∈ K ∗ . If the congruence α1x1 α2x2 ≡ β mod p is solvable for almost all prime ideals p of K then the corresponding equation is solvable in rational integers. Proof. This is a special case of Theorem 2 of [2].

Proof of Theorem 2. It suffices to consider the case where h is irreducible over Q and primitive. The case h | f g is trivial, hence assume that h /| f g. Let us again define ϑ, K and d by (6). We shall consider two cases. Case 1. g(ϑ) = a k b−l , k, l ∈ Z; f (ϑ) Case 2. g(ϑ) = a k b−l , k, l ∈ Z. f (ϑ) 1. In this case we have h(t) | f (t)a k − g(t)bl and since h is primitive, also for all n ∈ Z h(n) | a − min{0,k} b− min{0,l} (f (n)a k − g(n)bl ). If (2) does not hold it follows that

h(n) | f (n), h(n) a − min{0,k} b− min{0,l} a F (n) bl − a k bG(n) . Now (f (n), h(n)) | R, where R is the resultant of f and h and since h /| f , R = 0. The assertion follows on taking in Lemma 4 C = a k−min{0,k} bl−min{0,l} |R|

1000


and on replacing there F by F − k, G by G − l . 2. In this case by Lemma 5 there exists a prime ideal p0 of degree 1 in K not dividing the denominator of ϑ such that for all x, y ∈ Z : f (ϑ)a x − g(ϑ)by ≡ 0 mod p0 . We have c ϑ ≡ m0 mod p0 , m0 ∈ Z. Let P0 = {n ∈ Z : n ≡ m0 mod p0 }, where p0 is the prime divisible by p0 . If n ∈ P0 and (2) does not hold we have h(n) ≡ h(ϑ) ≡ 0 mod p0 , hence f (ϑ)a F (n) − g(ϑ)bG(n) ≡ f (n)a F (n) − g(n)bG(n) ≡ 0 mod p0 contrary to the choice

of p0 . Remark. By a slightly more complicated argument one can prove the following extension of Theorem 2. Let [K : Q] < ∞, f, g, h ∈ K[t]; (f, g, h) = 1, h ∈ K; F, G ∈ Q[t] be integer valued, α, β ∈ K ∗ , α F (n) β −G(n) be not constant for n ∈ Z. Then there exists an arithmetic progression P such that for n ∈ P f (n)α F (n) − g(n)β G(n) h(n)

is not an integer of K.

References [1] C. Corralez Rodrigáñez, R. Schoof, The support problem and its elliptic analogue. J. Number Theory 64 (1997), 276–290. [2] A. Schinzel, On power residues and exponential congruences. Acta Arith. 27 (1975), 397–420; this collection: H4, 915–938. [3] −−, Selected Topics on Polynomials. University of Michigan Press, Ann Arbor 1982. [4] −−, Systems of exponential congruences. Demonstratio Math. 18 (1985), 377–394; this collection: H7, 975–986.


Originally published in Journal für die reine und angewandte Mathematik 494 (1998), 73–84

Une caractérisation arithmétique de suites récurrentes linéaires avec Daniel Barsky (Villetaneuse) et Jean-Paul Bézivin (Caen)

A Monsieur le Professeur Martin Kneser pour le soixante-dixième anniversaire de sa naissance Résumé. On étudie la relation entre deux suites récurrentes un et vn de nombres algébriques, quand tout diviseur premier de un divise vn pour chaque n.

I. Introduction Soit u = (un )n∈N une suite récurrente linéaire d’éléments de Z, c’est-à-dire une suite vérifiant une relation de la forme : c

un+s + as−1 un+s−1 + . . . + a0 un = 0 avec les ai dans Z fixés. On considère l’ensemble A(u) = {p premier : ∃ n ∈ N, un = 0 et p | un }, que nous appelerons l’ensemble des diviseurs premiers de la suite u = (un )n∈N . On sait qu’en général l’ensemble A(u) est infini, [10]. On peut demander quelle est la proportion des nombres premiers qui sont des diviseurs d’une suite récurrente donnée. Le premier pas dans cette direction a été accompli par H. Hasse, [6], qui montre par exemple que la densité au sens de Dirichlet de l’ensemble des diviseurs premiers de la suite un = 2n + 1 est 7/24. D’autres résultats de ce type, utilisant la méthode de démonstration de Hasse, ont été obtenus par J. Lagarias, [7], et C. Ballot, [1]. On peut demander aussi dans quelle mesure cet ensemble caractérise la suite récurrente linéaire. On voit aisément que si la suite possède un zéro entier, i.e. s’il existe m ∈ N tel que um = 0, alors l’ensemble A(u) est l’ensemble des nombres premiers, à un nombre fini d’exception près. Il nous faut donc préciser un peu les conditons à imposer. Nous allons nous intéresser à un résultat démontré par C. Corrales et R. Schoof, [4], et généralisé par A. Schinzel, [12], dont un cas particulier est le suivant :

1002


Soient a et b deux éléments de Z, tels que |a| > 1 et |b| > 1. Alors si {p premiers : p | a n − 1} = {p premiers : p | bn − 1} pour tout n ∈ N, on a a = b. Soit K un corps de nombres. Nous généralisons ce résultat à des suites récurrentes linéaires liées à des polynômes de K[x1 , . . . , xs ] dont l’ensemble Ω des solutions dans les racines de l’unité est non vide et fini. Typiquement soit P (x, y) = x + y + 1 et soit j une racine primitive cubique de l’unité alors Ω = {(j, j 2 ), (j 2 , j )} et si l’on a pour tout n : {p, idéaux premiers de K : p | a n + bn + 1} = {p, idéaux premiers de K : p | α n + β n + 1} où a, b, α, β ∈ K∗ alors, avec quelques conditions techniques sur a, b, α, β (voir plus loin) et quitte à permuter, a = α, b = β.

II. Résultats Soit T un polynôme à s variables, à coefficients dans K. On note Γ∞ le groupe des racines de l’unité et Γm le groupe des racines m-ièmes de l’unité (éventuellement plongé dans un corps suffisamment grand). On note : Ω = Ω(T ) = {γ = (γ1 , . . . , γs ) : γi ∈ Γ∞ , 1 i s, tels que T (γ1 , . . . , γs ) = 0}, Ωm (T ) = Γms ∩ Ω(T ). Soient α1 , . . . , αr , β1 , . . . , βs des éléments non nuls de K. Nous allons démontrer aux paragraphes III et IV les résultats suivants : Théorème 1. Soient T (x1 , . . . , xr ) ∈ K[x1 , . . . , xr ] et T (x1 , . . . , xs ) ∈ K[x1 , . . . , xs ]

c

deux polynômes tels que Ω(T ) soit non vide et que Ω(T ) soit fini. Soient α = (α1 , . . . , αr ) ∈ (K∗ )r et β = (β1 , . . . , βs ) ∈ (K∗ )s . On suppose que les αi sont multiplicativement indépendants, c’est-à-dire que l’égalité r αixi = 1 avec x1 , . . . , xr ∈ Z implique x1 = . . . = xr = 0.

i=1

Enfin on suppose que pour tout entier n > 0 et que pour presque tout idéal premier p de K (i.e. tous sauf un nombre fini) on a : p | T (α1n , . . . , αrn ) =⇒ p | T (β1n , . . . , βsn ). Alors Ω(T ) est non vide et il existe un entier naturel d > 0 tel que, pour tout j ∈ {1, . . . , s}, il existe des ei,j ∈ Z tels que : e

e

βjd = α1j,1 · · · αr j,r . Théorème 2. On suppose que dans les hypothèses du théorème 1, on a r = s = 1, de sorte que α = (α1 ) et n’est pas une racine de l’unité, et que β = (β1 ). On pose α1 = α et

H10. Suites récurrentes linéaires

1003

β1 = β. On suppose en outre que T (x) est séparable et T (0) = 0. Soit w le nombre de racines de l’unité contenues dans le corps K, et ζw une racine primitive w-ième de 1. L’implication p | T (α n ) =⇒ p | T (β n ) est vraie pour presque tout idéal premier p de K et pour tout entier n > 0, si et seulement s’il existe des entiers a0 , a1 > 0, b0 , b1 et γ ∈ K tels que (a1 , b1 ) = 1 et α = ζwa0 γ a1 , β = ζwb0 γ b1 et si, pour tout μ ∈ {0, . . . , w − 1} T (ζ a0 μ x a1 ) | T (ζ b0 μ x b1 )x − min{0,b1 } deg(T ) . Théorème 3. Soit Φn le polynôme cyclotomique d’ordre n, et k, l deux entiers positifs, avec l sans facteur carré. Soient α, β ∈ K∗ , avec α non racine de l’unité. L’implication p | Φk (α n ) =⇒ p | Φl (β n ) est vraie pour tout entier n > 0 et presque tout idéal premier de K, si et seulement si l divise k et β = α kλ/ l , où (λ, l) = 1. Corollaire 1. On suppose que dans les hypothèses du théorème 1, on a r = s, et que de plus les αi et les βi sont des entiers de K, et qu’aucun d’eux n’est une unité de K ni égal, à une racine de l’unité près, à une puissance parfaite dans K. On suppose de plus que les αi sont deux à deux premiers entre eux, ainsi que les βi . Alors l’implication p | T (α1n , . . . , αsn ) =⇒ p | T (β1n , . . . , βsn ) est vraie pour tout entier n > 0 et presque tout idéal premier p seulement s’il existe une permutation ji de {1, . . . , s}, et des entiers bi tels que : βi = ζwbi αji . Corollaire 2. On suppose que les éléments α1 , α2 , β1 , β2 satisfont aux hypothèses du corollaire 1, avec s = r = 2 ; on suppose de plus que w = 2. Si p | α1n + α2n + 1 =⇒ p | β1n + β2n + 1 pour tout entier n > 0 et presque tout idéal premier p de K, alors on a {α1 , α2 } = {β1 , β2 }.

III. Lemmes préliminaires Lemme 1. Dans le corps de nombres K, il existe une base multiplicative, c’est-à-dire une suite πi (i = 1, . . . ) d’éléments non nuls de K telle que tout élément α ∈ K∗ possède une unique représentation sous la forme : α = ζwa0

h i=1

où a0 ∈ {0, . . . , w − 1}, et ai ∈ Z, 1 i h.

πiai

1004


Preuve. Voir Skolem [15].

Lemme 2. Avec les mêmes notations que dans le lemme précédent, pour tout entier positif m et tout vecteur (t0 , . . . , th ) de Zh+1 , il existe une infinité d’idéaux premiers p de K(ζwm ) ζ π α w i wti tels que = ζwt0 et = ζwm , 1 i h, où l’on a noté le symbole p wm p wm p wm de Legendre pour les puissances wm-ièmes.

Preuve. C’est un cas particulier du théorème 4 de Schinzel, [13].

Lemme 3. Soient k, l deux entiers positifs, et a ∈ Zr , A ∈ Mr,h (Z). On suppose que A est de rang r, et que la matrice B ∈ Ms,h (Z) possède la propriété que pour tout t ∈ Zh et tout entier positif m divisible par k on ait : (1)

At =

m a =⇒ lBt ≡ 0 (mod m). k

Alors il existe une matrice C ∈ Ms,r (Q) telle que B = CA.

Preuve. On peut supposer sans nuire à la généralité, que la matrice A1 constituée des r premières colonnes de A est inversible. Soit B1 la matrice constituée des r premières colonnes des B. Supposons que B = (B1 A−1 1 )A. Alors r < h et on peut supposer que si a r+1 et br+1 sont les r + 1-ième colonnes de A et B respectivement, on a : ⎛ ⎞ c1 ⎜ .. ⎟ −1 ⎝ . ⎠ = br+1 − (B1 A1 )a r+1 = 0. ch Soit (2)

c = max {|ci |} 1ih

et posons (3)

m = 2ckl (det(A1 ))2 .

Comme c det(A1 ) ∈ Z, m est un entier positif divisible par k det(A1 ). On définit maintenant un vecteur entier (t1 , . . . , tr ) par ⎛ ⎞ t1 ⎜ .. ⎟ m −1 −1 (4) ⎝ . ⎠ = A1 a − det(A1 )A1 a r+1 k tr

1005


et posons

⎛

t1 .. .

⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ tr ⎟ ⎜ ⎟ ⎟ t =⎜ ⎜det(A1 )⎟ . ⎜ 0 ⎟ ⎜ ⎟ ⎜ . ⎟ ⎝ .. ⎠ 0

(5)

m a, et par suite on a k lBt ≡ 0 (mod m).

Les formules précédentes impliquent que At = (6)

On a de plus, en tenant compte de (4) et de (5) : ⎛ ⎞ t1 ⎜ .. ⎟ lBt = lB1 ⎝ . ⎠ + l det(A1 )br+1 tr =

lm −1 B1 A−1 1 a − l det(A1 )B1 A1 a r+1 + l det(A1 )br+1 . k

Avec la relation (6), ceci donne :

l det(A1 ) br+1 − (B1 A−1 )a ≡ 0 mod r+1 1

m . k det(A1 )

En vertu de (3), on obtient alors br+1 − (B1 A−1 1 )a r+1 ≡ 0 (mod 2c) ce qui est une contradiction avec la relation (2). On a donc bien B = (B1 A−1 1 )A.

Lemme 4. Si A, B, C, l sont des entiers tels que l | C, l sans facteur carré, et pour tout C t ∈ Z on a (At + B, C) = , alors C | A. l C C a, B = b, avec a, b ∈ Z. l l En outre, pour tout t ∈ Z, on a (at + b, l) = 1. Comme l est sans facteur carré, si l /| a, c il existe un nombre premier p divisant l et ne divisant pas a. En résolvant la congruence at + b ≡ 0 (mod p), on trouve (at + b, l) ≡ 0 (mod p), ce qui est impossible.

Preuve. En prenant t = 0 et t = 1 on trouve que A =

Lemme 5. Soient et m entiers positifs. On a Φd (x), si (, m) = 1, Φ (x m ) = d |m

Φ (x ) = Φm (x), m

si tout facteur premier de m divise .

1006


Preuve. Pour m premier le lemme est bien connu. La cas général en résulte par récurrence sur le nombre de facteurs premiers de m.

IV. Démonstrations des résultats Preuve du théorème 1. On pose : a

αi = ζwi,0

(7)

h

a

1 i r,

b

1 i s,

πj i,j ,

j =1 b

βi = ζwi,0

(8)

h

πj i,j ,

j =1

A = (ai,j ),

B = (bi,j ).

Comme les αi sont multiplicativement indépendants, le rang de la matrice A est r. Par hypothèse, il existe deux entiers k, l tels que (9)

Ωk (T ) = ∅,

Ω(T ) ⊂ Ωl (T ).

⎛ ⎞ a1 ⎜ .. ⎟ a1 ar Soit ζk une racine primitive k-ième de 1, et (ζk , . . . , ζk ) ∈ Ωk (T ) ; on pose a = ⎝ . ⎠.

ar Nous allons démontrer que, pour tout vecteur t ∈ Zh , et tout entier positif m divisible par k, l’égalité m (10) At = a k implique que (11)

lBt ≡ 0 (mod m).

En vertu du lemme 2, il existe une infinité d’idéaux premiers p de K(ζwm ) tels que ζ π wt j w (12) = 1, = ζwmj , 1 j h, p wm p wm wm/k

où l’on a choisi la racine primitive ζwm telle que ζwm = ζk . Donc en vertu de (7) et de (10), α i = ζkak p wm et comme α α1 r T =0 ,..., wm p p wm il en résulte que

(N(p)−1)/(wm) (N(p)−1)/(wm) , . . . , αr ≡ 0 (mod P) T α1


1007

où N est la norme de K(ζwm ) sur K, et P désigne l’idéal premier de K divisible par p. Si la norme de p est assez grande, on a par hypothèse que

(N(p)−1)/(wm) (N(p)−1)/(wm) c , . . . , βs ≡ 0 (mod P) T β1 donc T ce qui implique

β 1

p

wm

,...,

β s

p

wm

≡ 0 (mod p)

β β1 s ≡ 0 (mod N (p)). ,..., N T wm p p wm

Le côté gauche de cette congruence est borné par une constante ne dépendant que de T et de m. Si N(p) est suffisamment grande, cette congruence implique donc que β β1 s T ,..., =0 p wm p wm β l i =1 et par suite Ω(T ) = ∅. Par l’inclusion (9) Ω(T ) ⊂ Ωl (T ), il en résulte que p wm pour tout i = 1, . . . , s. Il suffit alors d’utiliser (8) et (12), qui donne la relation (11). Le lemme 3 implique alors l’existence d’une matrice C ∈ Ms,r (Q) telle que On pose C = et où les ei,j

e

i,j

B = CA.

, 1 i s, 1 j r, où d est un entier positif divisible par w d sont dans Z, et on obtient ei,j αj .

βid =

Preuve du théorème 2. Nous montrons tout d’abord que la condition est nécessaire. Supposons donc que pour tout entier n et presque tout idéal premier p de K on ait p | T (α n ) =⇒ p | T (β n ). En vertu du théorème 1, il existe d ∈ N∗ , et e ∈ Z tels que β d = α e . b1 e Posons = , avec a1 , b1 ∈ Z, a1 > 0 et (a1 , b1 ) = 1. d a1 a1 (d,e) β = 1, on a β a1 = ζwc1 α b1 . Comme α b1 En prenant des entiers u, v tels que ua1 − vb1 = 1, et en posant a0 = c1 u, b0 = c1 v, et γ = α u β −v , il vient α = ζwa0 γ a1 ,

β = ζwb0 γ b1 .

Comme α n’est pas une racine de l’unité, il en est de même de γ . Supposons maintenant que pour une valeur de μ ∈ {0, . . . , w − 1}, on ait (13)

T (ζwa0 μ x a1 ) /| T (ζwb0 μ x b1 )x − min{0,b1 } deg(T ) .

1008


Comme T est séparable, et T (0) = 0, le polynôme T (ζw0 x a1 ) est aussi séparable, et si on pose

D(x) = T (ζwa0 μ x a1 ), T (ζwb0 μ x b1 )x − min{0,b1 } deg(T ) a μ

on a 1=

T (ζ a0 μ x a1 ) w

D(x)

, T (ζwb0 μ x b1 )x − min{0,b1 } deg(T ) .

Il existe donc des polynômes U, V ∈ K[x] tels que T (ζw0 x a1 ) U (x) + T (ζwb0 μ x b1 )x − min{0,b1 } deg(T ) V (x) = 1. D(x) a μ

(14)

Puisque T (0) = 0, on a que m ∈ N.

a μ

T (ζw0 x a1 ) n’est pas de la forme cx m , avec c ∈ K et D(x) T (ζw0 γ a1 (wν+μ) ) D(γ wν+μ ) a μ

Par un résultat d’Evertse ([5]), il en résulte que la suite récurrente

(ν = 1, 2, . . . ) a une infinité de diviseurs p de K. En choisissant un diviseur p tel que γ soit une unité p-adique et les coefficients de U, V , D des entiers p-adiques, il en résulte du fait que p | T (α wν+μ ) =⇒ p | T (β wν+μ ) une contradiction avec la relation (14). On a donc bien T (ζwa0 μ x a1 ) | T (ζwb0 μ x b1 )x − min{0,b1 } deg(T )

c

∀μ ∈ {0, . . . , w − 1}.

Montrons maintenant que la condition est suffisante. On suppose donc que α = ζwa0 γ a1 ,

β = ζwb0 γ b1

et T (ζwa0 μ x a1 ) | T (ζwb0 μ x b1 )x − min{0,b1 } deg(T )

∀μ ∈ {0, . . . , w − 1}.

Soit p un idéal premier tel que γ soit une unité p-adique, et que les coefficients des polynômes T (ζw0 x b1 )x − min{0,b1 } deg(T ) a μ T (ζw0 x a1 ) b μ

soient des entiers p-adiques. Alors : c

p | T (ζwa0 μ γ (wν+μ)a1 ) =⇒ p | T (ζwb0 μ γ (wν+μ)b1 ). Donc, pour tout n ≡ μ (mod w), on a : p | T (α n ) =⇒ p | T (β n )

1009


et comme cette propriété est vraie pour tout μ ∈ {0, . . . , w − 1}, ceci achève la démonstration.

Preuve du théorème 3. Montrons d’abord que la condition est nécessaire. On applique le théorème 2 avec T = Φk , T = Φl , et on obtient (15)

α = ζwa0 γ a1 ,

β = ζwb0 γ b1 ,

(a1 , b1 ) = 1, a1 > 0

et Φk (ζwa0 μ x a1 ) | Φl (ζwb0 μ x b1 )

∀μ ∈ {0, . . . , w − 1}.

Comme les zéros de Φk sont les ζkκ , κ premier à k, et les zéros de Φl les ζlλ , λ premier à l, on obtient que pour μ ∈ Z, tout κ premier à k et tout t ∈ Z on a : b μka1 −b1 a0 μk+b1 w(κ+tk)

−a0 μ κ+tk 0 ζka1 = ζkwa ζwb0 μ ζwa 1 1

= ζlλ

avec λ premier à l. Donc l divise kwa1 , et

kwa1 (16) k(b0 a1 − b1 a0 )μ + wκb1 + wkb1 t, kwa1 = . l En appliquant le lemme 4 deux fois, on obtient (17)

kwa1 | wkb1 ,

kwa1 | k(b0 a1 − b1 a0 )

comme (a1 , b1 ) = 1, la première relation donne a1 = 1, et la seconde w | (b0 − a0 b1 ) ; donc en vertu de (15), il vient β = α b1 . Il résulte de (16) et de (17) que (wκb1 , kw) =

kw l

k λ, (λ, l) = 1, donc l | k et β = α kλ/ l . l Montrons maintenant que la condition est suffisante. Si l divise k et si β = α kλ/ l , désignons par τ le plus grand diviseur de k premier à l et soit n ∈ N. Alors d’après le lemme 5 Φld (α kn/(lτ ) ) = Φkd/τ (α n ). Φl (β n ) = Φl (α kλn/ l ) = et comme (k, κ) = 1, on trouve b1 =

d |λτ

c

Il en résulte que Φl (β n )Φk (α n )−1 =

d |λτ

Φkd/τ (α n )

d |λτ, d=τ

et donc, pour tout idéal premier p de K tel que α soit une unité p-adique on a : p | Φk (α n ) =⇒ p | Φl (β n ).

Preuve du corollaire 1. La condition que les αi sont premiers deux à deux et ne sont pas des unités de K implique qu’ils sont multiplicativement indépendants. En appliquant le

1010


théorème 1, il vient que βid =

r

e

αj i,j

j =1

avec d entier > 0. Comme les βi sont des entiers et les αi ne sont pas des unités algébriques, il vient que les ei,j sont des entiers 0. Comme les βi sont premiers entre eux deux à deux, il vient que pour chaque j , au plus un des ei,j est non nul. Mais comme βi n’est pas une unité de K, il en résulte que pour chaque i au moins un des ei,j est non nul. Il existe donc une permutation ji de {1, . . . , r} telle que ei,j

βid = αji i . Comme ni αi , ni βi n’est, aux racines de l’unité près, une puissance parfaite dans K, il vient finalement que βi = ζwbi αj , avec bi ∈ Z.

Preuve du corollaire 2. On applique le corollaire 1 avec T = T = x1 + x2 + 1. Il en résulte, que quitte à renuméroter, on peut écrire βi = εi αi , où εi ∈ {±1}, pour i = 1, 2. • Si ε1 ε2 = −1, on obtient une contradiction pour n impair en utilisant un idéal premier de norme assez grande divisant α1n + α2n + 1 et donc 2α1n ou 2α2n . • Si ε1 = ε2 = −1, on obtient de même que p divise la somme (α1n + α2n + 1) + (β1n + β2n + 1) = 2.

D’où le résultat.

V. Exemples de polynômes Proposition 1. Soit F (x) =

h

ah1 ,...,ht x1h1 · · · xtht et G(y) =

l

bl1 ,...,ls y1l1 · · · ysls des

polynômes à coefficients dans Z tels que Ω(F ) et Ω(G) soient finis et non vides.

ah ,...,h . Posons H (x, y) = μF (x) + λG(y) avec λ, μ ∈ Z et Soit F = t 1 h

|λ| > |μ| · F . Alors Ω(H ) = Ω(F ) × Ω(G) et est donc fini et non vide. Preuve. Soit z = (x, y) ∈ Ω(H ), et K un corps de nombres de degré N , contenant tous les xi , yj et leur conjugués. Soit T = Gal(K/Q), alors :

F (x σ ) = (−λ)N G(y σ ) =⇒ (−λ)N G(y σ ) F N |μ|N . μN σ ∈T

σ ∈T

σ ∈T

Si G(y) = 0, il vient |λ| |μ| · F , contradiction. Donc G(y) = 0 et par conséquent F (x) = 0.

Cette proposition permet de construire des exemples à volonté : (a) 1 + x1 + λ(1 + y1 ) avec |λ| 3.


1011

(b) 1 + x1 + x2 + λ(1 + y1 + y2 ) avec |λ| 4. (c) On peut aussi mélanger les variables 1 + x1 + x2 + λ(1 + x1 + y2 ) avec |λ| 4 (il faut vérifier que Ω est non vide). (d) Le polynôme x + y + z − 3xyz, où Ω = {(1, 1, 1), (−1, −1, −1)}, permet aussi de construire des exemples de suites récurrentes linéaires caractérisées par l’ensemble de leurs diviseurs premiers. (e) Enfin, on trouvera dans [11] une étude de ce type de polynômes ; par exemple, le polynôme T (x, y) = x 2 y −2xy 2 +2x −y

où

Ω = {(−i, j i), (i, −j i), (−i, j 2 i), (i, −j 2 i)}

(cf. [11], page 132), est donc un polynôme possédant les propriétés utilisées dans les lignes qui précèdent.

Bibliographie [1] Ch. Ballot, Density of primes divisors of linear recurrence. Mem.Amer. Math. Soc. 115 (1995), no. 551. [2] Z. I. Borevitch, I. R. Chafarevitch, Théorie des nombres. Gauthier–Villars, Paris 1967. [3] J. W. S. Cassels, A. Fröhlich (ed.), Algebraic Number Theory. Academic Press, London 1967. [4] C. Corralez Rodrigáñez, R. Schoof, The support problem and its elliptic analogue. J. Number Theory 64 (1997), 276–290. [5] J.-H. Evertse, On sums of S-units and linear recurrences. Compositio Math. 53 (1984), 225–244. [6] H. Hasse, Über die Dichte der Primzahlen p, für die eine vorgegebene ganzrationale Zahl a = 0 von gerader bzw. ungerader Ordnung mod p ist. Math. Ann. 166 (1966), 19–23. [7] J. C. Lagarias, The set of primes dividing the Lucas numbers has density 2/3. Pacific J. Math. 118 (1985), 449–461; Errata, ibid. 162 (1994), 393–396. [8] S. Lang, Algebraic Number Theory. Addison Wesley, Reading 1970. [9] W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, 2nd edition. Springer and PWN, Berlin and Warsaw, 1990. [10] G. Pólya, Arithmetische Eigenschaften Reihenentwicklungen rationaler Funktionen. J. Reine Angew. Math. 151 (1921), 1–31. [11] W. M. Ruppert, Solving algebraic equations in roots of unity. J. Reine Angew. Math. 435 (1993), 119–156. [12] A. Schinzel, On exponential congruences. In: Diofantovy Priblizheniya, Matematicheskie Zapiski 2, Moskva 1996, 121–126 (Russian); this collection: H9, 996–1000. [13] −−, Abelian binomials, power residues and exponential congruences. Acta Arith. 32 (1977), 245–274; Addendum, ibid. 36 (1980), 101–104; this collection: H5, 939–970. [14] I. Schur, Über die Existenz unendlich vieler Primzahlen in einigen speziellen arithmetischen Progressionen. Sitzungsber. Berlin. Math. Ges. II (1912), 40–50. [15] T. Skolem, On the existence of a multiplicative basis for an arbitrary algebraic field. Norske Vid. Selsk. Forh. (Trondheim) 20 (1947), no. 2, 4–7.

Originally published in Acta Arithmetica 108 (2003), 77–94


On power residues with M. Skałba (Warszawa)

Let n be a positive integer, K a number field, αi ∈ K (1 i k), β ∈ K. A simple necessary and sufficient condition was given in [7] in order that, for almost all prime ideals p of K, solubility of the k congruences x ni ≡ αi (mod p) should imply solubility of the congruence x n ≡ β (mod p), where ni | n. The aim of this paper is to extend that result to the case where the congruence x n ≡ β (mod p) is replaced by the alternative of l congruences x n ≡ βj (mod p). The general result is quite complicated, but it simplifies if n or K satisfy some restrictions. Here are precise statements, in which ζn denotes a primitive nth root of unity, |A| is the cardinality of a set A, K n = {x n : x ∈ K} and F is the family of all subsets of {1, . . . , l}. Theorem 1. Let n and ni be positive integers, ni | n (1 i k), K be a number field and αi , βj ∈ K ∗ (1 i k, 1 j l). Consider the implication (i) solubility in K of the k congruences x ni ≡ αi (mod p) implies solubility in K of at least one of the l congruences x n ≡ βj (mod p). Then (i) holds for almost all prime ideals p of K if and only if (ii) for every unitary divisor m > 1 of n and, if n ≡ 0 (mod 4), for every m = 2m∗ , where m∗ is a unitary divisor of the odd part of n, there exists an involution σm of F such that for all A ⊂ {1, . . . , l} |σm (A)| ≡ |A| + 1 (mod 2),

(1) (2)

j ∈σm (A)

βj =

j ∈A

βj

k

a m/(m,ni )

αi i

Γ m,

i=1

where ai ∈ Z, Γ ∈ K(ζm )∗ . Corollary 1. Let wn (K) be the number of n-th roots of unity contained in K and assume that (3)

(wn (K), lcm[K(ζq ) : K]) = 1,

where the least common multiple is over all prime divisors q of n and additionally q = 4 if 4 | n. The implication (i) holds for almost all prime ideals p of K if and only if there exists

1013

H11. On power residues

an involution σ of F such that for all A ⊂ {1, . . . , l} |σ (A)| ≡ |A| + 1 (mod 2)

(4) and

(5)

βj =

j ∈A

j ∈σ (A)

βj

k

a n/ni

αi i

γ n,

i=1

where ai ∈ Z, γ ∈ K ∗ . The condition (3) holds for every K if n = 2 or n = l e , where l is an odd prime, and for K = Q if n is odd. Corollary 2. For n = ni = 2 (1 i k), (i) holds for almost all prime ideals p of K if and only if (iii) there exists a subset A0 of {1, . . . , l} such that |A0 | ≡ 1 (mod 2)

(6) and

(7)

βj =

j ∈A0

where ai ∈ Z, γ0 ∈

k

αiai γ02 ,

i=1

K ∗.

Corollary 2 contains as a special case (K = Q, k = 0) a theorem of Fried [3], rediscovered by Filaseta and Richman [2]. The case n = 2e (e 2) is covered by the following corollary, in which τ denotes the greatest integer such that ζ2τ + ζ2−1 / K, τ ∈ K. This corollary is of interest only if ζ4 ∈ otherwise (3) holds. Corollary 3. For n = 2e (e 2) and ni > 1 (1 i k), (i) holds for almost all prime ideals p of K if and only if simultaneously (iii) holds and (iv) there exists an involution σ of F such that for all A ⊂ {1, . . . , l} we have (4) and

(8)

j ∈σ (A)

where ai ∈ Z, γ ∈ (9)

βj = ε

j ∈A

βj ·

k

a n/ni

αi i

γ n,

i=1

K∗

and {1, −1} if e < τ , ε∈ τ −1 n/2 n/2 (ζ2τ + ζ2τ + 2) } if e τ . {1, (−1)

The case K = Q, n odd is covered by Corollary 1. The case K = Q, n even is covered by the following

1014


Theorem 2. Let n = 2ν n∗ , ν > 0, n∗ odd, ni | n (1 i k), K = Q. The implication (i) holds for almost all prime ideals p of K if and only if (v) for every m = 2ν m∗ and, if ν = 2, for every m = 2m∗ , where m∗ is a unitary divisor of n∗ , there exists an involution σm of F such that for all A ⊂ {1, . . . , l} we have (1) and

βj = ε

where ai ∈ Z, γ ∈

βj

j ∈A

j ∈σm (A)

Q∗ ,

k

a m/(m,ni ) m/2 m

αi i

δ

γ ,

i=1

δ is a fundamental discriminant dividing m and {1, −2m/2 } if m ≡ 4 (mod 8), ε∈ {1} otherwise.

Corollary 4. Let n = 2ν n∗ , ν 0, n∗ odd, β1 , β2 ∈ Q∗ . The alternative of congruences x n ≡ βj (mod p) (1 j 2) is soluble for almost all primes p, if and only if either βi ∈ Qn

(10)

for some i 2, or there is a j 2, a prime q | n∗ with q e n∗ and some γ1 , γ2 ∈ Q such that one of the following holds: • ν = 1 and

n/2 n n/q e (11) βj = (−1)(q−1)/2 q γ1 , β3−j = γ2 , •

ν = 2 and either βj = −2n/2 γ1n ,

(12)

n/2

β3−j = γ2

or βj = q n/2 γ1n ,

(13) •

n/q e

β3−j ∈ {γ2

e

n/q e

, −2n/2q γ2

},

ν 3 and either βj = 2n/2 γ1n

(14) or (15)

βj ∈ {q n/2 γ1n , 2n/2 q n/2 γ1n },

n/q e

β3−j ∈ {γ2

e

n/q e

, 2n/2q γ2

}.

The proofs are based on eight lemmas and use the n-th power residue symbol, which is defined as follows. If a number field K contains ζn , then for every prime ideal p of K j prime to n and every p-adic unit α of K, (α|p)n is a unique number ζn that satisfies the congruence j

α (Np−1)/n ≡ ζn (mod p), where N p is the absolute norm of p. Moreover, ind α is the index of α with respect to a fixed primitive root modulo the relevant prime ideal.

1015


We give two proofs of Corollary 2, one short using Theorem 1 and the other longer, but using neither Theorem 1 nor the lemmas below, except the classical Lemma 3. At the end of the paper we give a deduction of the more difficult necessity part of Theorem 1 of [7] from Theorem 1 above. We thank Professor J. Browkin for some helpful suggestions. 5 its group of characters and gj ∈ G Lemma 1. Let G be a finite abelian group, G (1 j l). If l

(16)

(χ (gj ) − 1) = 0

j =1

5 then there exists an involution σ of F such that for all A ⊂ {1, . . . , l} we for every χ ∈ G have (4) and gj = gj . j ∈A

j ∈σ (A)

Proof. For g ∈ G let c(g) =

(−1)|A| .

A⊂{1,...,l} j ∈A gj =g

The equality (16) can be written in the form c(g)χ (g) = 0 g∈G

or, if h is any fixed element of G,

c(g)χ (gh−1 ) = 0.

g∈G

Summing over all characters χ gives |G|c(h) = 0, hence c(h) = 0, and h being arbitrary, c(g) = 0 for all g ∈ G. It follows that for all g ∈ G the number of subsets A of {1, . . . , l} gj = g and |A| odd equals the corresponding number with |A| even, hence there with j ∈A gj = g such that is an involution σg of the family of subsets A of {1, . . . , l} with j ∈A

|σg (A)| ≡ |A| + 1 (mod 2). The involution σ is obtained by putting together all involutions σg , i.e., σ (A) = σg (A) for gj .

g= j ∈A

Lemma 2. Let n be a positive integer, K and L be number fields, K(ζn ) ⊂ L, βj ∈ K ∗ (1 j l). Let H be the multiplicative group generated by β1 , . . . , βl , and H1 the

1016


intersection of H with Ln . For every χ ∈ H /H1 there exists a set P , with positive Dirichlet density, of prime ideals P of L such that χ ([x]) = (x|P)n .

(17)

where [x] is the coset of H1 in H containing x. Proof. By a theorem of Skolem [9] the field L has a multiplicative basis ζw , π1 , π2 , . . . , where ζw is a root of unity and π1 , π2 , . . . are generators of infinite order. Let πs be the c last generator that occurs in the representation of β1 , . . . , βl . We have H /H1 < J /J n , where J is the group generated by ζw , π1 , . . . , πs . Indeed, H < J and the relations h1 ∈ H , −1 n h2 ∈ H and h1 h−1 2 ∈ J together imply h1 h2 ∈ H1 . Hence for every χ ∈ H /H1 there n exists χ1 ∈ J /J such that χ (y) = χ1 (y) for

(18)

y ∈ H /H1 .

Clearly χ1 (y)n = 1 for all y ∈ J /J n . On the other hand, by Theorem 4 of [8] with σ = 1, for any integers c0 , . . . , cs there exist infinitely many prime ideals P of L such that (ζw |P)n = ζnc0 ,

(πr |P)n = ζncr (1 r s).

Since the proof is via the Chebotarev density theorem (see [8], p. 263), the infinite set of /J n there prime ideals in question has a positive Dirichlet density. Hence for every χ1 ∈ J exists a set P of positive Dirichlet density such that for P ∈ P , (19)

χ1 (x) = (x|P)n

for

x ∈ J,

where x is the coset of J n in J containing x. Since by (18), χ ([x]) = χ1 (x)

for

x ∈ H,

(17) follows from (19).

Lemma 3. Let n ∈ N, K be a number field, ζn ∈ K, and α1 , . . . , αk , β elements of K ∗ . If ) )

) n β ∈ K n α1 , . . . , n αk , then β=

k

αiai γ n ,

i=1

where ai ∈ Z, γ ∈ K ∗ . Proof. See [5], p. 222, formula (2).


1017

Lemma 4. The condition (i) for almost all prime ideals p of K implies the existence of an involution σ of F such that, for all A ⊂ {1, . . . , l}, (4) holds and (20)

βj =

j ∈σ (A)

βj ·

j ∈A

k

a n/ni

αi i

Γ n for some ai ∈ Z, Γ ∈ K(ζn )∗ .

i=1

Proof. Let χ be a character of the group H /H1 described in Lemma 2 with L = K(ζn , ξ1 , . . . , ξk ), where ξini = αi (1 i k). By Lemma 2 there exists a set P , with positive Dirichlet density, of prime ideals P of L such that (x|P)n = χ ([x]) for x ∈ H,

(21)

where [x] is the coset of H1 in H containing x. Since prime ideals of degree greater than 1 have Dirichlet density 0 and relative norms of prime ideals from P have positive Dirichlet density, there is P ∈ P such that p = NL/K P has the property that solubility in K of the k congruences x ni ≡ αi (mod p) implies solubility of at least one of the l congruences x n ≡ βj (mod p). Moreover, the congruence x ni ≡ αi (mod P) has the solution x = ξi in L, hence, P being of relative degree 1, the congruence x ni ≡ αi (mod p) has a solution in K and, by (i), l

(βj |P)n − 1 = 0.

j =1

By (21) we have l

χ ([βj ]) − 1 = 0

j =1

and, χ being arbitrary, it follows by Lemma 1 that there exists an involution σ of F such that (4) holds and [βj ] = [βj ]. j ∈A

j ∈σ (A)

The last formula means that (22) βj βj−1 = Γ1n

for some Γ1 ∈ L.

j ∈A

j ∈σ (A)

Since Γ1n ∈ K(ζn ), by Lemma 3 we have Γ1n =

k

a n/ni

αi i

Γn

for some ai ∈ Z, Γ ∈ K(ζn ),

i=1

which together with (22) gives (20).

1018


Lemma 5. If there exists an involution σ of F such that, for all A ⊂ {1, . . . , l}, (4) holds and

(23)

βj =

βj ·

j ∈A

j ∈σ (A)

k

a m/(m,ni )

αi i

Γm

i=1

for some ai ∈ Z and Γ ∈ K(ζm ), then the implication (i) holds for all prime ideals p of K such that all αi , βj are p-adic units and (N p − 1, n) = m. Proof. Let p satisfy the assumptions of the lemma and assume that the k congruences x ni ≡ αi (mod p), hence also x (m,ni ) ≡ αi (mod p), are soluble in K. Let g be a primitive root mod p and Φm the m-th cyclotomic polynomial. We have Φm (x) ≡ (x − g (Np−1)k/m ) (mod p), (k,m)=1

hence, by Dedekind’s theorem, p has a prime ideal factor P in K(ζm ) of relative degree 1. Solubility in K of the congruences in question implies

ai m/(m,ni ) αi |P m = 1 (1 i k) and, since (Γ m |P)m = 1, by (23) we have

βj P = βj P , m

j ∈σ (A)

j ∈A

m

hence 2

l

1 − (βj |P)m

j =1

=

(−1)|A|

A⊂{1,...,l}

=

j ∈A

A⊂{1,...,l}

βj P + (−1)|σ (A)| βj P m

(−1)|A| + (−1)|σ (A)|

j ∈σ (A)

m

βj P = 0.

j ∈A

m

Thus (βj |P)m = 1 for at least one j l. Since P is of relative degree 1, this means that the congruence x m ≡ βj (mod p) is soluble in K. Choosing an integer t such that (N p − 1)t ≡ m (mod n) we have for every p-adic unit x of K, x (Np−1)t ≡ 1 (mod p), hence the congruence x n ≡ βj (mod p) is soluble in K.


1019

Lemma 6. Let m, ni ∈ N (1 i k) and ni = ni ni , where (ni , m) = 1. Let αi , βj ∈ K ∗ (1 i k, 1 j l). If there exists a prime ideal p0 of K such that m, ni , αi , βj are p0 -adic units, the congruences (24)

x ni ≡ αi (mod p0 ) (1 i k)

are soluble in K and the congruences (25)

x m ≡ βj (mod p0 ) (1 j l)

are insoluble in K, then there exists a set P , with positive Dirichlet density, of prime ideals of K such that for p ∈ P the congruences (26)

x ni ≡ αi (mod p) (1 i k)

are soluble in K and the congruences (27)

x m ≡ βj (mod p) (1 j l)

are insoluble in K. Proof. Assume first that all ni are prime powers, ni = liνi , where li are primes and let I0 = {1 i k : li | m}, I1 = {1 i k : li | N p0 − 1} \ I0 , I2 = {1 i k} \ I0 \ I1 . Let further (N p0 − 1, m) = m . We set ) )

L = K ζni , ni αi (1 i k), ζm , m βj (1 j l) , take P0 to be a prime ideal factor of p0 in L, and let S be the element of the Galois group of L/K such that ϑ S ≡ ϑ Np0 (mod P0 ) for all P0 -adic units ϑ of L. By the assumption about congruences (24) the congruence x ni ≡ αi (mod p0 ) has a solution xi ∈ K for i ∈ I0 , hence there exists a zero Ai of x ni − αi such that Ai ≡ xi (mod P0 ) and then ASi = Ai .

(28)

For i ∈ I1 ∪ I2 and 1 j l, we choose Ai and Bj to be arbitrary zeros of x ni − αi and x m − βj , respectively. By the assumption about congruences (25) also the congruences (29)

x m ≡ βj (mod p0 )

(1 j l)

1020


are insoluble in K. We have Np0

ζmS = ζm

(30)

ASi = ζnaii Ai

= ζm ,

0 ζnSi = ζnNp i

(i ∈ I1 ∪ I2 ),

BjS =

(1 i k),

b ζmj Bj

(1 j l),

where ai , bj ∈ Z. Since the congruences (25) are insoluble in K we have bj ≡ 0 (mod m )

(31)

(1 j l).

Put now n0 = lcm{ni : i ∈ I1 }. We have 1 + Np0 + . . . + N pn0 0 −1 = (Npn0 0 − 1)/(N p0 − 1) ≡ 0 (mod ni ) 1 + N p0 + . . . + N pn0 0 −1

(i ∈ I1 ),

≡ n0 (mod m ).

It follows from (28) that ASi

(32)

n0

= Ai

(i ∈ I0 )

and from (30) and (31) that ASi

(33) (34)

BjS

n0

n0

=ζ

n −1

ai (1+Np0 +...+N p0 0

= ζni

n −1 bj (1+Np0 +...+N p0 0 ) m

Ai = Ai

(i ∈ I1 ∪ I2 ),

b n

Bj = ζmj 0 Bj = Bj

S n0 m

(35)

)

ζ

(1 j l),

= ζm .

If now P is a prime ideal of L such that the Frobenius symbol L/K = S n0 P and p is the prime ideal of K divisible by P we infer from (32)–(35) that the congruences (26) are soluble in K and the congruences

x m ≡ βj (mod p)

(1 j l),

hence also the congruences (27), are insoluble in K. However, by Chebotarev’s density theorem the set of relevant prime ideals p has a positive Dirichlet density. Consider now the general case. Let (36)

ni =

hi

qij

j =1

where for each i, qij (1 j hi ) are powers of distinct primes. Since the congruences (24) are soluble in K, for each i k and each j such that (qij , m) = 1 the congruence x qij ≡ αi (mod p0 )

1021


is soluble in K. Now, by the already proved case of the lemma, there exists a set P , with positive Dirichlet density, of prime ideals of K such that for each p ∈ P the congruences x qij ≡ αi (mod p) (1 i k, 1 j hi ) are soluble, but the congruences (27) insoluble. Thus for all i, j we have ind αi ≡ 0 (mod (N p − 1, qij )). It now follows from (36) that for all i, ind αi ≡ 0 (mod (N p − 1, ni )),

hence the congruences (26) are soluble. Lemma 7. Suppose that (i) holds for almost all prime ideals p of K.

(vi) If m is a unitary divisor of n, then for almost all prime ideals p of K, solubility in K of the k congruences x (m,ni ) ≡ αi (mod p)

(37)

implies solubility in K of at least one congruence (38)

x m ≡ βj (mod p) (1 j l).

(vii) If n ≡ 0 (mod 4) and m = 2m∗ , where m∗ is a unitary divisor of the odd part of n, then for almost all prime ideals p of K, solubility in K of the k congruences x (m,ni ) ≡ αi (mod p) implies solubility in K of at least one congruence x m ≡ −1 (mod p),

x m ≡ βj (mod p) (1 j l).

Proof. In order to prove statement (vi) assume to the contrary that there exists a prime ideal p0 of K such that m, ni , αi and βj are p0 -adic units, the congruences (37) are soluble and the congruences (38) are insoluble. We apply Lemma 6 with ni ni = (m, ni ), ni = . (m, ni ) The assumptions of the lemma are satisfied, since with our choice of m (m, ni ) =

(m2 , ni ) =1 (m, ni )

and the assertion of the lemma contradicts the assumptions of Lemma 7. A similar argument shows that if statement (vii) were false, there would exist a set P , with positive Dirichlet density, of prime ideals of K such that for p ∈ P the congruences (39)

∗

x ni ≡ αi (mod p)

(1 i k)

would be soluble and the congruences (40)

x m ≡ −1 (mod p),

x m ≡ βj (mod p)

(1 j l)

1022


insoluble, where n∗i is the greatest divisor of ni not divisible by 4. However, insolubility of x m ≡ −1 (mod p) implies Np − 1 = ind(−1) ≡ 0 (mod (N p − 1, m)), 2 hence for m ≡ 2 (mod 4), N p ≡ 3 (mod 4) and then solubility of (39) implies solubility of (26), while (40) is insoluble, contrary to the assumption of the lemma.

Proof of Theorem 1. Necessity. The existence of an involution σm satisfying (1) and (2) for m being a unitary divisor of n follows at once from Lemma 4 and (vi). In order to prove the same for m of the form 2m∗ , where m∗ is a unitary divisor of the odd part of n, denote by m the least unitary divisor of n divisible by m. Let Gm , resp. Gm , be the multiplicative m/(m,ni ) m/(m,ni ) (1 i k) and K(ζm )∗m , resp. by αi subgroup of K ∗ generated by αi (1 i k) and K(ζm )∗m . If Gm ⊂ Gm , then it suffices to take σm = σm . If Gm ⊂ Gm , let δ ∈ Gm \ Gm . We have k

δ=

(41)

a m/(m,ni )

αi i

Γ m,

i=1

)∗ .

where ai ∈ Z, Γ ∈ K(ζm By Theorem 3 of [8] we have Γ m = Γ0m for some Γ0 ∈ K(ζ4m∗ ). Taking norms of both sides of (41) with respect to K(ζm ) and denoting the norm of Γ0 by Γ1 we obtain k

δ = 2

2ai m/(m,ni )

αi

Γ1m ,

i=1

hence δ=±

k

a m/(m,ni )

αi i

m/2

Γ1

,

i=1

and, since

m

m ,

(m, ni ) (m, ni )

m

m , 2

Γ1 ∈ K(ζm ),

δ∈ / Gm ,

the plus sign is excluded and we have −1 ∈ / Gm

and δ ≡ −1 (mod × Gm ).

Since δ ≡ 1 (mod × Gm ) it follows that Gm : Gm ∩ Gm = 2, Gm = (Gm ∩ Gm ) ∪ δ(Gm ∩ Gm ). From the existence of σm satisfying (1) and (2) it follows that for each B ∈ K ∗ , (−1)|A| + (−1)|A| = 0, (42) A∈V (B)

A∈V (δB)

1023


where

Let S =

βj ≡ B (mod × Gm ∩ Gm ) . V (B) = A ∈ F : j ∈A

j ∈A

βj : A ∈ F and let {B1 , . . . , Br } be a subset of S maximal with respect to

the property that Bi ≡ B (mod × Gm ), Set

Bj ≡ Bi (mod × Gm ∩ Gm )

for

j = i.

βj ≡ B (mod × Gm ) . U (B) = A ∈ F : j ∈A

Replacing B by Bi in (42) and summing with respect to i we obtain (−1)|A| + (−1)|A| = 0. A∈U (B)

A∈U (−B)

However, from (vii) and Lemma 4 it follows that (−1)|A| + (−1)|A|+1 = 0. A∈U (B)

A∈U (−B)

Adding the last two inequalities we obtain 2 (−1)|A| = 0, A∈U (B)

hence there exists an involution B of the family of all subsets A of {1, . . . , l} with βj = B, such that

j ∈A

|B (A)| ≡ |A| + 1 (mod 2). The involution σm is obtained by combining all involutions B . Sufficiency. Consider a prime ideal p of K such that αi , βj are all p-adic units and let (43)

(Np − 1, n) = m1 .

If m1 = 1 the implication (i) is obvious. If m1 > 1, m1 ≡ 0 (mod 2) or m1 ≡ 0 (mod 4), let m be the least unitary divisor of n divisible by m1 . By condition (ii) we have (1) and (2) where Γ ∈ K(ζm ). However, Γ m ∈ K, hence also Γ m ∈ K(ζq : q | m, q prime or q = 4) =: K0 . It follows now from Theorem 3 of [8] that Γ m = Γ0m , where Γ0 ∈ K0 . However, by the definition of m, we have K0 ⊂ K(ζm1 ) and also

m m1

.

(m1 , ni ) (m, ni ) The implication (i) follows now from Lemma 5.

1024


If m1 ≡ 2 (mod 4), we take m = 2m∗ , where m∗ is the least unitary divisor of n

divisible by m1 /2, and argue as before. Proof of Corollary 1. Under the assumption (3) the conditions Γ n ∈ K, Γ ∈ K(ζn ) imply, by Theorem 3 of [8], that Γ n = γ n , γ ∈ K, hence for σ = σn , (1) implies (4) and (2) implies (5).

First proof of Corollary 2. The necessity of condition (iii) follows from Corollary 1 on taking A0 = σ (∅). Conversely, if (iii) holds, then we define the involution σ in Corollary 1 by σ (A) = A ÷ A0 (÷ denotes the symmetric difference) and notice that

βj =

βj

j ∈A

j ∈σ (A)

k i=1

αiai γ0

2 βj

,

j ∈A∩A0

hence (4) and (5) are satisfied and, by Corollary 1, (i) holds for almost all prime ideals p of K.

Second (direct) proof of Corollary 2. In order to prove the necessity of the condition, choose a maximal subset {i1 , . . . , is } of {1, . . . , l} such that s

√ √ βierr ∈ L2 , where L = K( α1 , . . . , αk ),

r=1

implies er ≡ 0 (mod 2) (1 r s). By the theorem of Chebotarev [1] there exists a set P , with positive Dirichlet density, of prime ideals P of L of degree 1 such that (βir |P)2 = −1

(44)

(1 r s).

Let p be the prime ideal of K divisible by P. Since P is of degree 1 and the k congruences x 2 ≡ αi (mod P) are soluble in L, they are soluble in K and, by the implication, (45)

(βj |p)2 = 1

for at least one j k.

On the other hand, for each j l, by the maximality of {i1 , . . . , is } we have (46)

βj =

s

e

βirj r γj2 ,

ej r ∈ {0, 1}, γj ∈ L.

r=1

If for each j we have s

ej r ≡ 1 (mod 2),

r=1

then the formulae (44) and (46) imply (βj |P)2 = −1, contrary to (45). If for a certain j0 we have s r=1

ej0 r ≡ 0 (mod 2),


1025

then taking A0 = {ir : ej0 r = 1} ÷ {j0 } we get (6) and if j0 ∈ A0 , βj20 γj−2 0 (47) βj = −2 γj0 if j0 ∈ A0 . j ∈A 0

However, since

γj−2 0

∈ K, it follows by Lemma 3 that = γj−2 0

k

αiai γ 2 ,

for some ai ∈ Z, γ ∈ K,

i=1

which together with (47) implies (7). In order to prove the sufficiency of the condition, let p be a prime ideal of K such that αi and βj are p-adic units and the k congruences x 2 ≡ αi (mod p) are soluble in K. Then (7) gives (βj |p)2 = 1 = (−1)|A0 | , j ∈A0

hence (βj |p)2 = 1 for at least one j ∈ A0 .

Proof of Corollary 3. Necessity. For n = 2e , by a theorem of Hasse [4] (see also Lemma 6 in [8]), Γ n ∈ K with Γ ∈ K(ζn ) implies Γ n = εγ n , where ε is given by (9) and γ ∈ K, hence (iv) follows from (ii) for σ = σn . Also (iii) follows from (ii), on taking m = 2 and A0 = σ2 (∅). Sufficiency. There is only one unitary divisor m > 1 of n = 2e , namely m = n, and for this m, (ii) follows from (iv) by the theorem of Hasse quoted above, used in the opposite direction. For m = 2, (ii) follows from (iii) on taking σ2 (A) = A ÷ A0 .

Lemma 8. Let m be even and α ∈ Q∗ . Then α ∈ Q(ζm )m if and only if α = εδ m/2 γ m , where γ ∈ Q∗ , δ is a fundamental discriminant dividing m and {1, −2m/2 } if m ≡ 4 (mod 8), ε∈ {1} otherwise. Proof. This is a reformulation of a lemma of Mills [6].

Proof of Theorem 2. The necessity of the conditions follows at once from Theorem 1 and Lemma 8. In order to prove the sufficiency we consider the cases ν 2 and ν 3 separately. If ν 2, then (ii) follows from (v) and Lemma 8 for every even unitary divisor m of n. For an odd unitary divisor m of n it suffices to take σm = σ2m . For ν 3 and m ≡ 2 (mod 4), (ii) follows as before, while for m ≡ 2 (mod 4) it suffices to take σm = σn . Indeed, for ν 3 we have ε = 1 and for every number of the form εδ n/2 γ n with δ, γ ∈ Q belongs to Qm .

1026


Proof of Corollary 4. Necessity. In the case ν = 0 the assertion follows at once from Corollary 1. We shall consider in detail only the case ν = 1; the proof in other cases is similar and will be only indicated briefly. Applying Theorem 2 for ν = 1 and m = n we infer that for {j } = σn (∅), n/2

βj = δn γnn

(48)

for some γn ∈ Q,

where δn is a fundamental discriminant dividing n. If δn = 1 we have βj ∈ Qn , hence (10) with i = j . If δn = (−1)(q−1)/2 q, where q is an odd prime, we have βj as in (11). Now we apply Theorem 2 for m0 = 2 and m1 = n/q e . If σmi (∅) = {j } then m /2

βj = δmii γimi

(49)

for some γi ∈ Q

(i = 0, 1),

where δmi is a fundamental discriminant dividing mi . Now the equations (48) and (49) are incompatible, since denoting by k(x) the square-free kernel of an integer x, we have m /2

n/2

k(δmii γimi ) = δmi = δn = k(δn γnn ). Therefore, σmi (∅) = {3 − j } (i = 0, 1) and we obtain m /2

β3−j = δmii γimi

(i = 0, 1).

We have δm0 = 1, hence β3−j ∈ Q[2,n/2q ] = Qn/q , which proves (11). Suppose now that δn has at least two distinct prime factors q1 and q2 and qiei n. Applying Theorem 2 for m0 = 2, mi = n/qiei (i = 1, 2) we obtain, as before, σmi (∅) = {3 − j } (i = 0, 1, 2). Then e

e

β3−j ∈ Q2 ∩

2 6

ei

Qn/2qi ,

i=1

Qn ,

which gives (10) with i = 3 − j . hence β3−j ∈ For ν = 2, let σn (∅) = {j }. If ε = 1 and δn = 1 or −4 we obtain (10) with i = j . If ε = −2n/2 and δn = 1 or −4 we consider m0 = 2, m1 = n/2 and obtain (12). If ε = −2n/2 and δn = 1, −4 we consider m0 = 4, m1 = n/2 and obtain (10) with i = 3 − j. If ε = 1 and δn has one odd prime factor q we consider m0 = 4, m1 = n/q e and obtain (13). If ε = 1 and δn has at least two odd prime factors q1 , q2 we consider m0 = 4, mi = n/qiei (i = 1, 2) and obtain (12) with j and 3 − j interchanged. For ν 3 let σn (∅) = {j } and n/2

βj = δn γnn . If δn = 1 or −4 we obtain the case (10) with i = j . If δn = ±8 we obtain the case (14). If δn has one odd prime factor q we consider m0 = 2ν , m1 = n/q e and obtain (15). If δn has at least two odd prime factors q1 and q2 we consider m0 = 2ν , mi = n/qiei (i = 1, 2) and obtain (10) with i = 3 − j or (14) with 3 − j in place of j .

1027


Sufficiency. If (10) holds then for each relevant divisor m of n we take σm = ci di , where ci , di are the cycles (∅ → {i}) and ({3 − i} → {1, 2}), respectively. If (11) holds, we take σm =

cj dj c3−j d3−j

if q | m, if q /| m.

If (12) holds, we take σm =

if 4 | m, if 4 /| m.

cj dj c3−j d3−j

If (13) holds, we take σm =

if q | m, or 4 /| m, if q /| m and 4 | m.

cj dj c3−j d3−j

If (14) holds, we take σm = cj dj . If (15) holds, we take σm =

cj dj c3−j d3−j

if q | m, if q /| m.

Deduction of Theorem 1 of [7] (necessity part) from Theorem 1 (above). Let n =

l j =0

e

pj j ,

where p0 = 2, pj are distinct odd primes and ej > 0 for j > 0. Applying Theorem 1 e above with m = pj j we infer that β=

(50)

k

ej

ej

aij pj /(ni ,pj )

αi

ej

pj

Γj

i=1

for some aij ∈ Z and Γj ∈ K(ζpej ) (for m = 1 the conclusion is trivial). By the theorem j

of Hasse [4] (see [8], Lemma 6) ej

(51)

p Γj j

ej

=

p εγj j

for some γj ∈ K, εj = 1 for j > 0

and

(52)

ε0 −τ ε0 ∈ 1, (−1)2 (ζ2τ

ε0 ∈ {1} if e0 1, ε0 ∈ {1, −1} if 1 < e0 < τ, 2ε0 −1 + ζ2−1 if e0 τ. τ + 2)

1028


We take integers u0 , . . . , ul satisfying the linear equation l n

uj = 1

e

j =0

pj j

and set γ =

l

u

γj j .

j =0

By (50) and (51) we have γn =

n

ej

pj

γj

nuj /pej j

−nu0 /2ε0

= βε0

j =0

k l

ej

−aij nuj /(ni ,pj )

αi

,

j =0 i=1

hence (53)

k

β

mi n/ni

αi

ε0

= εnu0 /2 γ n

i=1

K ∗.

for some mi ∈ Z, γ ∈ If e0 1, or e0 > τ , or ε0 = 1, or u0 is even, we obtain, by (51), condition (i) or (iv) of Theorem 1 of [7]. If 1 < e0 τ , ε = 1 and u0 is odd we apply Theorem 1 above with m = 2. We obtain a β= αi i γ 2 , 2|ni

which combined with (53) gives, by (52), l αii = −δ 2 2|ni

and β

k i=1

m n/n αi i i

=

−γ n n/2 n

γ1 − ζ2τ + ζ2−1 τ +2

if 1 < e0 < τ, if e0 = τ,

for some δ, γ1 ∈ K ∗ . These are just conditions (ii) and (iii) of Theorem 1 of [7]. The proof that conditions (i)–(iv) are sufficient is easy.

References [1] N. G. Chebotarev, Über einen Satz von Hilbert. Vestnik Ukr. Akad. Nauk, 1923, 3–7; Sobranie Sochineni˘ı I, Izd. Akad. Nauk SSSR, Moscow–Leningrad 1949–50, 14–17. [2] M. Filaseta, D. R. Richman, Sets which contain a quadratic residue modulo p for almost all p. Math. J. Okayama Univ. 31 (1989), 1–8. [3] M. Fried, Arithmetical properties of value sets of polynomials. Acta Arith. 15 (1969), 91–115.


1029

[4] H. Hasse, Zum Existenzsatz von Grunwald in der Klassenkörpertheorie. J. Reine Angew. Math. 188 (1950), 40–64. [5] −−, Vorlesungen über Klassenkörpertheorie. Physica-Verlag, Würzburg 1967. [6] W. H. Mills, Characters with preassigned values. Canad. J. Math. 15 (1963), 169–171. [7] A. Schinzel, On power residues and exponential congruences. Acta Arith. 27 (1975), 397–420; this collection: H4, 915–938. [8] −−, Abelian binomials, power residues and exponential congruences. Acta Arith. 32 (1977), 245–274; Addendum, ibid. 36 (1980), 101–104; this collection: H5, 939–970. [9] Th. Skolem, On the existence of a multiplicative basis for an arbitrary algebraic field. Norske Vid. Selsk. Forh. (Trondheim) 20 (1947), no. 2, 4–7.

Part I Primitive divisors


Commentary on I: Primitive divisors by C. L. Stewart

Let a and b be coprime integers with |a| > |b| > 0 and let n be a positive integer. A prime p is said to be a primitive divisor of a n − bn if p divides a n − bn but does not divide a m − bm for any positive integer m which is smaller than n. The study of primitive divisors had its origins in the work of Bang [1], Zsigmondy [18] and Birkhoff and Vandiver [3] from 1886, 1892 and 1904 respectively. It follows from their analysis that the primitive divisors of a n − bn are the prime factors of the n-th cyclotomic polynomial evaluated at a and b, Φn (a, b), with at most one exception. The exception, if it exists, is a prime factor of n. Gauss [7] and Dirichlet [6] factorized the polynomial Φn (x, 1) over a suitable quadratic number field. Aurifeuille and Le Lasseur (see [1]) deduced from it explicit non-trivial factorizations of the number Φn (x, y) for certain integers n, x and y. Factorizations of the type they considered are now known as Aurifeuillian factorizations. In a paper I1 written during a stay at Trinity College in Cambridge in 1961, Schinzel gave some new Aurifeuillian factorizations. In addition, he used Aurifeuillian factorizations to give conditions under which a n − bn has at least two primitive divisors. Stevenhagen [15] and Brent [4] have shown how to efficiently compute the factorizations given by Schinzel in I1. In [8], Granville and Pleasants show that Schinzel determined all possible such Aurifeuillian factorizations. One may extend the notion of a primitive divisor to sequences of Lucas numbers and sequences of Lehmer numbers. In 1913 Carmichael [5] proved that if un is the n-th term, for n > 12, of a Lucas sequence whose associated characteristic polynomial has real roots and coprime coefficients then un possesses a primitive divisor. Rotkiewicz [13], in 1962, generalized Schinzel’s argument of I1 to give conditions under which un has at least two primitive divisors. In 1930 Lehmer [10] introduced sequences which are more general than Lucas sequences but retain their striking divisibility properties and these sequences are now referred to as Lehmer sequences. Twenty-five years later Ward [17] established the analogue of Carmichael’s result for Lehmer sequences. In a sequence of three papers I2, I3 and I4 Schinzel used the Aurifeuillian factorizations from I1 to establish conditions under which Lucas or Lehmer numbers have at least k primitive prime factors with k equal to 2, 3, 4, 6 or 8. Let A and B be non-zero integers in an algebraic number field K and let n be a positive integer. A prime ideal of the ring of algebraic integers of K is said to be a primitive divisor

1034

I. Primitive divisors

of An − B n if it divides the ideal generated by An − B n but does not divide the ideal generated by Am − B m for any positive integer m with m < n. In I5 Schinzel proves that if A and B are non-zero coprime algebraic integers whose quotient is not a root of unity then An − B n has a primitive divisor provided that n exceeds N (d), a number which is effectively computable in terms of d where d is the degree of A/B over Q. In 1968 Postnikova and Schinzel [12] proved a weaker version of this result where N (d) was replaced by N (A, B), a number which is effectively computable in terms of A and B. The case d = 2 is of considerable interest since it gives information on non-degenerate Lucas and Lehmer sequences whose associated characteristic polynomial has coprime coefficients. In particular, if un is the n-th term of such a sequence and n exceeds N (2) then un has a primitive divisor. Schinzel [14] had earlier established that un has a primitive divisor if n exceeds a number which is effectively computable in terms of the coefficients of the associated characteristic polynomial of the sequence. Stewart [16] proved that one may take N(d) = max{2(2d − 1), e452 d 67 } and that there are only finitely many such Lehmer sequences whose n-th term, n > 6, n = 8, 10 or 12, does not possess a primitive divisor; for Lucas sequences the appropriate requirement is n > 4, n = 6. Further these sequences may be determined by solving certain Thue equations. Bilu, Hanrot and Voutier [2] used a theorem of Laurent, Mignotte and Nesterenko [9] concerning lower bounds for linear forms in the logarithms of two algebraic numbers, as elaborated by Mignotte [2], to help explicitly determine all such exceptional Lucas and Lehmer sequences. In particular, they proved that if n exceeds 30 and un is a Lucas or Lehmer number, from a sequence as above, then un has a primitive prime factor. Let A, B and d be as above and let k be a positive integer, ζk be a primitive k-th root of unity and K be an algebraic number field containing A, B and ζk . In I6 Schinzel proves that for each positive real number ε there exists a positive number c which depends on d and ε such that if n exceed c(1 + log k)1+ε then there is a prime ideal of the ring of j algebraic integers of K that divides An − ζk B n but does not divide Am − ζk B m for m < n and any integer j . The case when k = 1 is the main result of I5.

References [1] A. S. Bang, Taltheoretiske Undersøgelser. Tidsskrift for Mat. 4 (1886), 70–80, 130–137. [2] Yu. Bilu, G. Hanrot, P. M. Voutier, Existence of primitive divisors of Lucas and Lehmer numbers, with an appendix by M. Mignotte. J. Reine Angew. Math. 539 (2001), 75–122. [3] G. D. Birkhoff, H. S. Vandiver, On the integral divisors of a n − bn . Ann. of Math. (2) 5 (1904), 173–180. [4] R. P. Brent, On computing factors of cyclotomic polynomials. Math. Comp. 61 (1993), 131–149. [5] R. D. Carmichael, On the numerical factors of the arithmetic forms α n ± β n Ann. of Math. (2) 15 (1913), 30–70. [6] P. G. Lejeune Dirichlet, Vorlesungen über Zahlentheorie, 4th ed. Friedr. Vieweg & Sohn, Braunschweig 1894. [7] C. F. Gauss, Disquisitiones Arithmeticae. G. Fleischer, Leipzig 1801. [8] A. Granville, P. Pleasants, Aurifeuillian factorization. Math. Comp. 75 (2006), 497–508.

Primitive divisors

1035

[9] M. Laurent, M. Mignotte, Yu. Nesterenko, Formes linéaires en deux logarithmes et déterminants d’interpolation. J. Number Theory 55 (1995), 285–321. [10] D. H. Lehmer, An extended theory of Lucas’ functions. Ann. of Math. (2) 31 (1930), 419–448. [11] E. Lucas, Théorèmes d’arithmétique. Atti. R. Acad. Sci. Torino 13 (1877–78), 271–284. [12] L. P. Postnikova, A. Schinzel, Primitive divisors of the expression a n −bn in algebraic number fields. Mat. Sb. (N.S.) 75 (117) (1968), 171–177 (Russian); English transl.: Math. USSR-Sb. 4 (1968), 153–159. [13] A. Rotkiewicz, On Lucas numbers with two intrinsic divisors. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astr. Phys. 10 (1962), 229–232. [14] A. Schinzel, The intrinsic divisors of Lehmer numbers in the case of negative discriminant. Ark. Mat. 4 (1962), 413–416. [15] P. Stevenhagen, On Aurifeuillian factorizations. Nederl. Akad. Wetensch. Indag. Math. 49 (1987), 451–468. [16] C. L. Stewart, Primitive divisors of Lucas and Lehmer numbers. In: Transcendence Theory: Advances and Applications, Academic Press, London 1977, 79–92. [17] M. Ward, The intrinsic divisors of Lehmer numbers. Ann. of Math. (2) 62 (1955), 230–236. [18] K. Zsigmondy, Zur Theorie der Potenzreste. Monatsh. Math. 3 (1892), 265–284.

Originally published in Proceedings of the Cambridge Philosophical Society 58 (1962), 555–562


On primitive prime factors of a n − bn

1. Let a, b be relatively prime integers with |a| > |b| > 0. For any integer n > 0, let φn (a, b) denote the n-th cyclotomic polynomial, defined by φn (a, b) =

n

(a − ζnr b),

r=1 (r,n)=1

where ζn is a primitive n-th root of unity. A prime is called a primitive prime factor of a n − bn if it divides this number but does not divide a ν − bν for 0 < ν < n. Zsigmondy [14] proved, and Birkhoff and Vandiver [4] and Kanold [8] rediscovered, the following theorem (see [6], p. 195): the primitive prime factors of a n − bn coincide with the prime factors of φn (a, b), except for a possible prime q1 which may divide φn (a, b) (to the first power only) and also divide n, and may be a primitive prime factor of a σ − bσ , where σ = n/q1κ and (q1 , σ ) = 1. If there is such a prime q1 , then q1 = q(n), the greatest prime factor of n, since σ | (q1 −1), whence σ < q1 . There is at least one primitive prime factor of a n − bn except in the following cases: n = 1,

a − b = 1;

n = 3,

a = ±2,

b = ∓1;

n = 2,

a + b = ±2μ

n = 6,

a = ±2,

(μ 1); b = ±1.

If φn (a, b) is a prime, it is of course for n > 6 the only primitive prime factor of a n − bn . It is not obvious even that for every pair a, b there exists some n such that a n − bn has two primitive prime factors. In the present paper we give conditions which will ensure that a n −bn has at least two primitive prime factors, and in particular we prove that there are infinitely many n for which this happens. The last assertion is an immediate consequence of our main result, which follows. We use k(n) to denote the square-free kernel of n, that is, n divided by its greatest square factor. Theorem 2. Let

η=

1 if k(ab) ≡ 1 (mod 4), 2 if k(ab) ≡ 2 or 3 (mod 4).

Communicated by H. Davenport

I1. On primitive prime factors of a n − bn

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If n/ηk(ab) is an odd integer, then a n − bn has at least two primitive prime factors except in the following cases: n = 1; n = 2; n = 3;

a = ±(2α + 1)2 , b = ±(2α − 1)2 or 4a = ±(pα + 1)2 , 4b = ±(pα − 1)2 ; same but with ∓ for b instead of ±; |a| = ±3, b = ∓1 or a = ±4, b = ±1 or a = ±4, b = ∓3;

n = 4; n = 6; n = 12; n = 20;

|a| = 2, |b| = 1; a = ±3, b = ±1 or a = ±4, b = ∓1 or a = ±4, b = ±3; |a| = 2, |b| = 1 or |a| = 3, |b| = 2; |a| = 2, |b| = 1.

This theorem represents a continuation of the line of arithmetical investigations pursued by Aurifeuille and Le Lasseur [1], Lucas [11], Bickmore [3], Cunningham [5], Kra˘ıtchik ([9], [10], pp. 87–91), Rotkiewicz [13].

2. The proof of Theorem 2 is based on the following properties of the cyclotomic polynomials. We write φn (x) for φn (x, 1), and similarly for other polynomials later. Theorem 1. Let n > 1 be square-free and let m be a divisor of n such that n/m is odd. Then there exist polynomials Pn,m (x), Qn,m (x) with integral coefficients such that (1 ) 2 (x) − (−1|m)mxQ2 (x) φn (x) = Pn,m n,m

(1)

(m odd),

(2)

2 (−x) + (−1|m)mxQ2 (−x) (m odd), φ2n (x) = Pn,m n,m

(3)

2 (x) − mxQ2 (x) φ2n (x) = Pn,m n,m

(m even).

√ Further, these can be found from the following formulae (where c 0 if √ √ polynomials c 0 and c = i |c| if c < 0): 1/2 (4) Pn,m (x 2 ) − (−1|m)m xQn,m (x 2 ) = (x − ζns ) (x + ζnt ) = ψn,m (x) (m odd), s

t

1/2 xQn,m (−x 2 ) (5) Pn,m (−x ) − i (−1|m)m (x + iζns ) (x − iζnt ) = ψ2n,m (x) = 2

s

(m odd),

t

where the products are over(2 ) (6) (1 ) (2 )

0 < s < n,

0 < t < n,

(st, n) = 1,

(s|m) = 1,

(t|m) = −1;

(−1|m) is Jacobi’s symbol of quadratic character. If m = 1, the product over t is empty, and in (4) we get ψn,1 (x) = φn (x).

1038


and (7)

Pn,m (x 2 ) − m1/2 xQn,m (x 2 ) =

s (x − ζ4n ) = ψ2n,m (x) (m even), s

where the product is over (8)

(s, 4n) = 1,

0 < s < 4n,

(m|s) = 1.

The first part of this theorem when n = m was proved by Lucas [12], and was enunciated for all odd n, m by Cunningham ([5], p. 53), who also stated that the polynomials P , Q can be found by ‘conformal division’. A proof for the case n = 3m, m even (in our notation) was given recently by Beeger [2]. The second part of Theorem 1 seems to be new, as does the following lemma, on which the proof is based. Lemma 1. Let n > 1 be square-free and let m > 1 be an odd divisor of n. For ε = ±1, let 1/2 1 (9) A(ε) Zn,m (x) = (x − ζns ), n,m = 2 Yn,m (x) − ε{(−1|m)m} s

the product being over (10)

(s, n) = 1,

0 < s < n,

(s|m) = ε.

Then Yn,m , Zn,m have rational integral coefficients, and (11)

(−1) 2 2 1 φn (x) = A(1) n,m (x)An,m (x) = 4 {Yn,m − (−1|m)mZn,m }.

Proof. It follows from Dirichlet’s generalization ([7], supplement VII) of a well-known theorem of Gauss that the desired polynomials exist when m = n. The coefficients of (−1) 1/2 , and correA(1) m,m (x) and Am,m (x) are integers of the field generated by {(−1|m)m} sponding coefficients are algebraically conjugate. Put n = mk. In the product (9), but without the condition (s, n) = 1 in (10), put s = s + um, where 0 < s < m, We find that

0 u < k.

k (x − ζns ) = (x − ζns ζku ) = A(ε) m,m (x ). s

s

u

Hence (ε) k A(ε) n,m (x) = (Am,m (x ), φn (x)).

It follows that the coefficients of A(ε) n,m (x) are also integers of the field generated by {(−1|m)m}1/2 and that corresponding coefficients are algebraically conjugate. Since the integers of the field are expressible as

1/2 1 z , 2 y − {(−1|m)m} where y, z are rational integers, the polynomials Yn,m , Zn,m in (9) have rational integral coefficients.


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(11) follows immediately from (9), on dividing the values of r in φn (x) = (x − ζnr ) r

into two classes according as (r|m) = 1 or −1.

Proof of Theorem 1. Suppose first that n is odd and m = 1. We have (12) φn (x 2 ) = (x 2 − ζn2t ) = (x − ζnt )(x + ζnt ) = φn (x)φn (−x). (t,n)=1

(t,n)=1

Define polynomials Pn,1 , Qn,1 by φn (x) = Pn,1 (x 2 ) − xQn,1 (x 2 ). Then (4) holds for m = 1, since ψn,1 (x) = φn (x) as noted earlier. The identity (12) implies that 2 (x 2 ) − x 2 Qn,1 (x 2 ), φn (x 2 ) = Pn,1

and this gives (1) for m = 1 on replacing x 2 by x. Suppose, secondly, that n is odd and m > 1. By (12) and Lemma 1, (−1) (1) (−1) φn (x 2 ) = A(1) n,m (x)An,m (x)An,m (−x)An,m (−x).

(13) Put (14)

(−1) ψn,m (x) = A(1) n,m (x)An,m (−x) =

(x − ζns ) (x + ζnt ), s

t

with s, t as in (6). Express the polynomials Yn,m , Zn,m of Lemma 1 in the form Y (x) = T (x 2 ) + xU (x 2 ),

Z(x) = V (x 2 ) + xW (x 2 ).

Then we find that (15)

ψn,m (x) = Pn,m (x 2 ) − {(−1|m)m}1/2 xQn,m (x 2 ),

where Pn,m (x 2 ) = Qn,m (x 2 ) =

2 2 2 2 1 4 T − x U − (−1|m)m(V 1 2 (T W − U V ),

− x2W 2) ,

where T = T (x 2 ), etc. Since Y ≡ Z (mod 2) by (9), the polynomials Pn,m , Qn,m have integral coefficients. Now (15) gives (4), and (13), (14), (15) give (1) on replacing x 2 by x. Also (2) is a consequence of (1), in view of the identity (16)

φ2n (x) = φn (−x)

(n odd),

and (5) is a consequence of (4). We can now suppose that n, m are both even, say n = 2n1 , m = 2m1 , where n1 , m1 are necessarily odd. We suppose first that n1 > 1. By (12) and (16), φ2n (x) = φn (x 2 ) = φn1 (−x 2 ) = φn1 (ix)φn1 (−ix).

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Hence (17) φ2n (x 2 ) = φn1 (ζ82 x 2 )φn1 (ζ8−2 x 2 ) = φn1 (ζ8 x)φn1 (−ζ8 x)φn1 (ζ8−1 x)φn1 (−ζ8−1 x). It can be verified that the product on the right is the same as ψ2n,m (x)ψ2n,m (−x), where ψ2n,m (x) = ψ4n1 ,2m1 (x) is defined by the product in (7) with the conditions in (8). To do this, put s ≡ 8u + n1 v (mod 8n1 ), where v = 1, 3, 5, 7 and 0 < u < n1 ,

(u, n1 ) = 1.

The condition of quadratic character in (8) limits u to one of the two classes (mod m1 ) for each v. Considering cases according to the residues of m1 and n1 (mod 4), we find that ψ2n,m (x) = ψn1 ,m1 (αζ8 x)ψn1 ,m1 (βζ8−1 x), where α = ±1, β = ±1 and α = (−1|m1 )β. Using the definition of ψn1 ,m1 as a product in (4), we deduce from (17) that φ2n (x 2 ) = ψ2n,m (x)ψ2n,m (−x).

(18)

Using the polynomial expression for ψn1 ,m1 in (4) and putting Pn1 ,m1 (x) = K(x 2 ) + xL(x 2 ),

Qn1 ,m1 (x) = M(x 2 ) + xN (x 2 ),

we find that (19)

ψ2n,m (x) = Pn,m (x 2 ) − m1/2 xQn,m (x 2 ),

where Pn,m (x) = K 2 + x 2 L2 + m1 x(M 2 + x 2 N 2 ), Qn,m (x) = ±(KM + x 2 LN ) ± x(KN − ML), in which K stands for K(−x 2 ), etc., and the ± sign depends on the residue classes of n1 and m1 (mod 8). We now have (7), and (3) follows from (18) and (19) on replacing x 2 by x. This completes the proof if n1 > 1. If n1 = 1, then n = m = 2, and φ2n (x) = x 2 + 1, and we can take P2,2 (x) = x + 1, Q2,2 (x) = 1.

The preceding proof is analogous to Lucas’s proof of the case m = n. We note that in view of the identity (20)

∗

φn (x) = φn∗ (x n/n ),

where n∗ denotes the greatest square-free divisor of n, the assumption that n is square-free can be replaced, both in Lemma 1 and Theorem 1, by the weaker assumption that m is square-free.


1041

3. Proof of Theorem 2. Since the primitive prime factors of a n − bn coincide with those of |a|n − |b|n if ab > 0 or n ≡ 0 (mod 4), and those of |a|n/2 − |b|n/2 if ab < 0 and n ≡ 2 (mod 4), and those of |a|2n − |b|2n if ab < 0 and n ≡ 1 mod 2, it suffices to prove the theorem when a > b > 0. Put n = kl, where l is the product of those prime factors of n which do not divide ηk(ab), and write ν = ηk(ab)l. Since every prime factor of n is a prime factor of ν, we have (21)

φn (a, b) = φν (A, B),

where A = a n/ν ,

B = bn/ν .

The polynomials ψn,m (x), ψ2n,m (x) were defined in Theorem 1 when n > 1 is squarefree and n/m is an odd integer. We add the definition ψ1,1 (x) = x − 1. The hypotheses of Theorem 2 ensure that ψν,k(ab) is defined. Using the corresponding homogeneous forms, we put (22)

φn(ε) (a, b) = ψν,k(ab) (A1/2 , εB 1/2 )

for

ε = 1, −1.

Each of these is a rational integer; the quadratic irrationality on the left of (4), (5) or (7) disappears for the value of x in question, because of the definition of η. We have (23)

φn (a, b) = φn(1) (a, b)φn(−1) (a, b).

Since ψν,k(ab) (x)ψν,k(ab) (−x) = ±φν (x 2 ), the resultant R of the two polynomials on the left divides the discriminant of φν (x 2 ), and therefore also the discriminant of x 2ν − 1, which is (2ν)2ν . There exist polynomials χ (1) (x), χ (−1) (x) such that χ (1) (x)ψν,k(ab) (x) + χ (−1) (x)ψν,k(ab) (−x) = R identically in x. The coefficients of χ (1) , χ (−1) are expressible integrally in terms of the coefficients of ψν,k(ab) (x), and therefore involve only the quadratic irrational in (4), (5) or (7). On making the above relation homogeneous in x, y and putting x = A1/2 , y = B 1/2 , we find that the irrationality disappears, and from the resulting relation between integers we deduce that any common prime factor of φn(1) (a, b) and φn(−1) (a, b) must divide 2νB. By (23) and Zsigmondy’s theorem, quoted in §1, each prime factor of either φn(1) (a, b) or φn(−1) (a, b) is a primitive prime factor of a n − bn , except possibly for q(n), if this occurs to the first power only. Since this prime does not divide k(ab), it must equal q(l). If either φ (1) or φ (−1) is even, then 2 is a primitive prime factor of a n − bn , and this c can happen only if n = 1. Apart from this case, no prime factor of 2k(ab)lB can be a primitive prime factor of a n − bn , and consequently φ (1) , φ (−1) are relatively prime. In order to ensure the existence of two primitive prime factors of a n − bn , it is enough to have 1 if q(l) < q(n), (24) |φn(ε) (a, b)| > q(l) if q(l) = q(n),

1042


for ε = 1, −1, since then φ (1) , φ (−1) will have two distinct prime factors other than q(l). If n = 1 then k(ab) = 1 and a n − bn has two prime factors except when a = (2α + 1)2 , b = (2α − 1)2

4a = (pα + 1)2 , 4b = (pα − 1)2 ,

or

in accordance with the theorem. If n > 1, suppose first that l = 1 or 3. It follows from (4), (5) or (7) and (22) that c

(25)

|φn(ε) (a, b)| < (A1/2 + B 1/2 )φ(ν) (2A + 2B)φ(ν)/2 .

On the other hand, we have two lower bounds for φn (a, b). First, φn (a, b) = φν (A, B) > (A − B)φ(ν) . Secondly, ν A − ζνr B φν (A, B) = , φν (1, 1) 1 − ζνr r=1 (r,ν)=1

and since

A − ζνr B 2

= (A + B)2 + (A − B)2 cot 2 π r/ν 4

1 − ζνr

we obtain φν (A, B) >

1

φ(ν) 1 . 2A + 2B

It follows from (23) and (25) that

1/2 φ(ν) . |φn(ε) (a, b)| > max A1/2 − B 1/2 , 18 A + 18 B This implies that (24) holds when l = 1 or 3 except possibly if A1/2 − B 1/2 < 1 and A + B < 8, or if n = l = 3 and A1/2 − B 1/2 < 31/2 and A + B < 24, or if n > l = 3 and A1/2 − B 1/2 < 31/4 and A + B < 1921/2 . Direct examination of these cases leads to all the exceptions given in the theorem for 2 < n < 20. Suppose now that l 5 and put l = q(l)r. Put ν = ηk(ab)r = ν/q(l). It follows from (4), (5) or (7) and (22) that

q(l) q(l) (δ) φ (ε) = φν(δ) ,B /φν (A, B), A where δ = ±1 depends on ε and on the residue classes of k(ab) and of q(l) (mod 4). Using the inequalities

1/2 (x 1/2 − 1)φ(ν ) φν(δ) + 1)φ(ν ) (x) (x

(x > 1),


1043

we get for q(l) 3 |φ (ε) | >

c

Aq(l)/2 − B q(l)/2 A1/2 + B 1/2

φ(ν )

A + B + (AB)1/2 (A − B)Aq(l)/2−3/2 A + B + 2(AB)1/2

φ(ν ) > 2−1/2 (A − B)Aq(l)/2−3/2 .

φ(ν )

Since l is odd and square-free, we have q(l) 5, so 2−1/2 Aq(l)/2−3/2 > q(l) except when c either l = 5 or 15 and A 7 or l = 7, 21, 35 or 105 and A = 2 or 3. Direct examination of these cases leads to the last exception stated in the theorem. This completes the proof.

4. It follows from the identity φn (ch ) =

φnd (c) when

(h, n) = 1

d |h

and from (16) and Theorem 2 that if ab = ±ch , where c, h are integers and h 3 or k(c) is odd and h = 2, then for infinitely many n, a n − bn has three primitive prime factors. This suggests the following problems. Problem 1. For every pair a, b, does there exist n such that a n − bn has three primitive prime factors? Problem 2. Does there exist a pair a, b with ab = ±ch (h 2) such that a n − bn has three primitive prime factors for infinitely many n ?

5. In conclusion, we apply Theorem 2 to obtain lower bounds for q(a n − bn ), the greatest prime factor of a n − bn . We note first: Lemma 2. (i) If n > 2, and we exclude the case n = 3, a = ±2, b = ∓1, then q(a n − bn ) n + 1. (ii) If n > 2, n ≡ 0 (mod 4), and we exclude the cases n = 3, |a| = 2, |b| = 1 and n = 6, |a| = 2, |b| = 1, then q(a n − bn ) 2n + 1. Proof. Apart from the excluded cases, a n − bn has at least one primitive prime factor q, by Zsigmondy’s theorem. This is of the form nk + 1, and of the form 2nk + 1 if n is odd. Thus it remains only to prove (ii) when n ≡ 2 (mod 4). For this we observe that a n/2 − bn/2

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has a primitive prime factor q1 of the form nk + 1. Since q = q1 , one at least of q, q1 is 2n + 1.

Using Theorem 2 we shall prove Lemma 3. If n > 2 and k(ab) | n, and we exclude the cases n = 4, 6, 12, |a| = 2, |b| = 1, then q(a n − bn ) 2n + 1. Proof. By Lemma 2 we can suppose n ≡ 0 (mod 4). If k(ab) is odd, a n − bn has at least one primitive prime factor q by Zsigmondy’s theorem. This is of the form nk + 1, and so q ≡ 1 (mod 4k(ab)). Hence k(ab) is a quadratic residue (mod q), which implies that a (q−1)/2 − b(q−1)/2 is divisible by q. Since q is a primitive prime factor of a n − bn , it is impossible that q − 1 = n, hence q 2n + 1. The same argument applies if k(ab) is even and n/2k(ab) is even. If k(ab) is even and n/2k(ab) is odd, then apart from the exception of Theorem 2, which can be tested directly,

a n − bn has at least two primitive prime factors. One at least of these is 2n + 1. If k(ab) = ±2, we can combine Lemmas 2 and 3 to give Theorem 3. If k(ab) = ±2 and n > 2 and we exclude the cases n = 4, 6, 12, |a| = 2, |b| = 1, then q(a n − bn ) 2n + 1. The same result holds if k(ab) = ±1 or more generally ab = ±ch (h 2). It suggests Problem 3. Does there exist any pair a, b with ab = ±2c2 , ±ch (h 2) such that q(a n − bn ) > 2n for all sufficiently large n ? I conclude by expressing my thanks to Dr. B. J. Birch and Prof. H. Davenport for their kind assistance in the preparation of this paper, to Prof. T. Nagell and Mr. A. Rotkiewicz for their valuable suggestions, and to the Rockefeller Foundation whose fellowship I held when writing the paper.

References [1] A. Aurifeuille, H. Le Lasseur, See: E. Lucas, Théorèmes d’arithmétique. Atti R. Acad. Sc. Torino 13 (1877–8), 271–284. [2] N. G. W. H. Beeger, On a new quadratic form for certain cyclotomic polynomial. Nieuw Arch. Wiskunde (2) 23 (1951), 249–252. [3] C. E. Bickmore, On the numerical factors of a n − 1. Messenger of Math. (2) 25 (1895–6), 1–44; 26 (1896–7), 1–38. [4] G. D. Birkhoff, H. S. Vandiver, On the integral divisors of a n − bn . Ann. of Math. (2) 5 (1904), 173–180. [5] A. Cunningham, Factorisation of N = y y ∓ 1 and x xy ∓ y xy . Messenger of Math. (2) 45 (1915), 49–75.


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[6] L. E. Dickson, History of the Theory of Numbers, I. Washington 1919. [7] P. G. L. Dirichlet, Vorlesungen über Zahlentheorie, 4th ed. Friedr. Vieweg & Sohn, Braunschweig 1894. [8] H. J. Kanold, Sätze über Kreisteilungspolynome und ihre Anwendungen auf einige zahlentheoretische Probleme. J. Reine Angew. Math. 187 (1950), 169–182. [9] M. Kraïtchik, Décomposition de a n ± bn en facteurs dans le cas où nab est un carré parfait avec une table des décompositions numériques pour toutes les valeurs de a et b inférieures à 100. Paris, 1922. [10] −−, Recherches sur la Théorie des Nombres, I. Paris, 1924. [11] E. Lucas, Sur la série récurrente de Fermat. Bull. Bibl. Storia Sc. Mat. e Fis. 11 (1878), 783–789. [12] −−, Sur les formules de Cauchy et de Lejeune Dirichlet. Ass. Française pour l’Avanc. des Sci., Comptes Rendus 7 (1878), 164–173. [13] A. Rotkiewicz, The numbers of the form 5 (1961), 95–99 (Polish).

(4k + 1)4k+1 − 1 (4k + 3)4k+3 + 1 , . Prace Mat. 4k 4k + 4

[14] K. Zsigmondy, Zur Theorie der Potenzreste. Monatsh. Math. 3 (1892), 265–284.

Originally published in Acta Arithmetica VIII (1963), 213–223


On primitive prime factors of Lehmer numbers I

Lehmer numbers are called terms of the sequences n odd, (α n − β n )/(α − β), Pn (α, β) = n n 2 2 (α − β )/(α − β ), n even, where α and β are roots of the trinomial z2 − L1/2 z + M, and L and M are rational integers (cf. [4]). Without any essential loss of generality (cf. [9]) we can assume that (1)

L > 0,

M = 0,

K = L − 4M = 0.

Lehmer numbers constitute a generalization of the numbers a n − bn (a, b—rational integers). A prime p is called a primitive prime factor of a number a n − bn if p | a n − bn

but

p /| a k − bk

for

k < n.

A proper (not merely automatical) generalization of this notion for Lehmer numbers is the notion of a prime factor p such that p | Pn

but

p /| KLP3 · · · Pn−1

or, which is easily proved to be equivalent, p | Pn

but

p /| nP3 · · · Pn−1 .

D. H. Lehmer [4] calls such primes p primitive extrinsic prime factors of Pn . In a postscript to my paper [7] I stated erroneously that Lehmer calls them intrinsic divisors, the term which has been used in a different sense by M. Ward [9]. To simplify the terminology, I adopt in the present paper the following definition. Definition. A prime p is called a primitive prime factor of the number Pn if p | Pn but p /| KLP3 · · · Pn−1 . Assume that, besides the restrictions on L, M stated in (1), (2)

(L, M) = 1,

(i.e. β/α is not a root of unity).

L, M = 1, 1 , 2, 1 , 3, 1

I2. On primitive prime factors of Lehmer numbers I

1047

Then it follows from the results of papers [2], [7], [9] that for n = 1, 2, 3, 4, 6, Pn has a primitive prime factor except for K > 0

if n = 5, L, M = 1, −1 ,

n = 10, L, M = 5, 1 , n = 12, L, M = 1, −1 , 5, 1

for K < 0

if n n0 (L, M)

where n0 can be computed effectively. I proved in [6] a theorem about numbers a n − bn with two primitive prime factors. A. Rotkiewicz [5] generalized this theorem to so-called Lucas numbers (which correspond to Lehmer numbers for L1/2 being a rational integer) under the assumptions M > 0, K > 0. The main aim of the present paper is to generalize the above theorem to Lehmer numbers. To state the generalization in a possibly concise manner I introduce the following two sets M, N: M = L, M : (L, M) = 1; L, M = 12, −25 , 112, 25 or 1 |M| 15, 2M + 2|M| + 1 L

< min(64 + 2M − 2|M|, 2M + 2|M| + 4|M|1/2 + 1) ,

N = L, M : (L, M) = 1, L, M = 4, −1 , 8, 1 or

1 |M| 15, L = 2M + 2|M| + 1 .

As can easily be verified, set M consists of 184 and set N of 32 pairs L, M . For an integer n = 0, let k(n) denote the square-free kernel of n, that is, n divided by its greatest square factor. The following theorem holds.

Theorem 1. For L, M satisfying (1), (2), put κ = k M max(K, L) and 1 if κ ≡ 1 (mod 4), η= 2 if κ ≡ 2, 3 (mod 4). If n = 1, 2, 3, 4, 6 and n/ηκ is an odd integer, then Pn has at least two primitive prime factors except 1.

2.

for K > 0, if n = η|κ|, L, M ∈ M0 ⊂ M or n = 3η|κ|, L, M ∈ N0 ⊂ N or n = 5, L, M = 9, 1 or n = 10, L, M = 5, −1 or n = 20, L, M = 1, −2 , 9, 2 ; for K < 0, if n n1 (L, M).

Finite sets M0 , N0 and function n1 (L, M) can be effectively computed. Let us observe that the sequences Pn and P n corresponding to L, M and max(K, L), |M| , respectively, are connected by the relation Pn if M > 0 or n even, Pn = P 2n /P n if M < 0 and n odd.

1048


Therefore the primitive prime factors of Pn coincide with those of P n if M > 0 or n ≡ 0 (mod 4), with those of P n/2 if M < 0 and n ≡ 2 (mod 4) and with those of P 2n if M < 0 and n ≡ 1 (mod 2). The remarks that L, M ∈ M or N if and only if max(K, L), |M| ∈ M or N, respectively, sgn κ = sgn M, if κ is even, η’s corresponding to κ and −κ are equal; if κ is odd, the product of these η’s is 2,

show that it suffices to prove the theorem for M > 0, κ = k M max(K, L) = k(LM). Before proceeding further, we introduce some notation and recall some useful results from paper [6]. For any integer n > 0 let

1. 2. 3.

Qn (x, y) =

n

(x − ζnr y),

r=1 (r,n)=1

where ζn is a primitive n-th root of unity. Put Qn (x) = Qn (x, 1) and similarly for other polynomials later. Denote by q(n) the greatest prime factor of n. Further, for n satisfying the assumptions of Theorem 1, let l be the product of those prime factors of n which do not divide ηκ, and write ν = ηκl, A = α n/ν , B = β n/ν . To obtain conformity of notation with paper [6] one should make in the latter the following permutation of letters: Φ → Q, P → R, Q → S. Then by Theorem 1 of [6] and remark that ν > 2, Qν (x 2 ) = ψν,κ (x)ψν,κ (−x),

(3) where(1 ) (4)

(5)

ψν,κ (x) = Rκl,κ (x 2 ) − κ 1/2 xSκl,κ (x 2 ) (κ 1/2 > 0), ⎧ r ⎪ x − (r|κ)ζκl if κ ≡ 1 (mod 4), ⎪ ⎪ ⎪ (r,κl)=1 ⎪ ⎪ ⎨ x + i(r|κ)ζ r if κ ≡ 3 (mod 4), κl = (r,κl)=1 ⎪ ⎪ r ) ⎪ (x − ζ4κl if κ ≡ 2 (mod 4) ⎪ ⎪ ⎪(r,4κl)=1 ⎩ (κ|r)=1

c

and R = Rκl,κ , S = Sκl,κ are polynomials with rational integral coefficients. Let us put, similarly as in [6], for ε = ±1, 1/2 , εB 1/2 ), Q(ε) n (α, β) = ψν,κ (A

(6) where arg A1/2 = for ε = ±1 (7) (8) (1 )

1 2

arg A, arg B 1/2 =

1 2

arg B. Then, if α, β are real, α > β > 0, we have

1/2 ϕ(ν) |Q(ε) − B 1/2 , ( 18 A + 18 B)1/2 , n (α, β)| > max A )

ϕ(ν −1/2 (A − B)Aq(l)/2−3/2 (l 3, ν = ν/q(l)). |Q(ε) n (α, β)| > 2

(r|κ) is Jacobi’s symbol of quadratic character.


c

1049

These inequalities were proved in [6] under the assumption that α, β are rational integers; however, the proof does not change if α, β are arbitrary real numbers, α > β > 0. Now we shall prove three lemmas. Lemma 1. If n satisfies the assumptions of Theorem 1, M > 0, p | Qn (α, β) and p is not a primitive prime factor of Pn (α, β), then p2 /| Qn (α, β), and if n = 2r α (r prime), then p = q(n) = q(l). If n = 2r α (r prime), r | Qn (α, β) if and only if r | L. Proof. It follows from Theorems 3.3 and 3.4 of [4] that if the assumptions of the lemma are satisfied and n = 12, then p2 /| Qn (α, β) and p = q(n). On the other hand, as can easily be verified, Qn (α, β) =

ϕ(n)/2

ai Lϕ(n)/2−i M i

i=0

where a0 = 1 and aϕ(n)/2 = ±1, unless n = 2r α (r prime). For n = 2r α , aϕ(n)/2 = ±r, so that r | Qn (α, β) if and only if r | L. For n = 2r α we have, in view of (L, M) = 1, (p, LM) = 1 so (p, κ) = 1. Since all prime factors of n divide ηκl, the lemma is thus proved for all n = 12. If n = 12, then Qn (α, β) = L2 − 4LM + M 2 ; if p is an imprimitive prime factor of Pn (α, β), then L ≡ kM (mod p) for some k 4. Hence, if p | Qn (α, β), then in view of (L, M) = 1, p = 2 or 3. On the other hand, it follows from 12 = ηκl that κ is even, LM

is even and p = 2. Thus p = 3 = l and p2 /| Qn (α, β), which completes the proof. Lemma 2. If n satisfies the assumptions of Theorem 1, δ = k(L)−{ϕ(n)/4} and M > 0, (−1) then the numbers δQ(1) (α, β) are coprime rational integers (2 ). n (α, β) and δQn Proof. We show first that ψν,κ (x) (ν > 1) are reciprocal polynomials. For instance, let κ ≡ 3 (mod 4). We have by (5) −r r r ψν,κ (x −1 ) = x −1 + i(r|κ)ζκl = x −ϕ(ν) i(r|κ)ζκl x − i(r|κ)ζκl (r,κl)=1

=x

(r,κl)=1

−ϕ(ν) ϕ(ν)

i

(−1)

ϕ(ν)/2

(r,κl)=1

−r x + i(−r|κ)ζκl = x −ϕ(ν) ψν,κ (x).

(r,κl)=1

Since in view of (4) 1 ψν,κ (x 1/2 ) + ψν,κ (−x −1/2 ) , 2

1 Sκl,κ (x) = ψν,κ (x 1/2 ) − ψν,κ (−x −1/2 ) , 2(κx)1/2 it follows that the polynomials R, S are reciprocal. We now prove that these polynomials are of degrees 21 ϕ(ν) and 21 ϕ(ν) − 1, respectively. In fact Rκl,κ (x) =

(9) (2 )

Qν (x) = R 2 (x) − κxS 2 (x), [x] and {x} denote the integral and the fractional part of x, respectively.

1050


whence degree S < degree R = degree S < 21 ϕ(ν) − 1,

1 2

degree Qν = 21 ϕ(ν). On the other hand, supposing that

R(x) = x ϕ(ν)/2 + ax ϕ(ν)/2−1 + bx ϕ(ν)/2−2 + . . . we should find by comparing both sides of (9) that x ϕ(ν) − μ(ν)x ϕ(ν)−1 + . . . = x ϕ(ν) + 2ax ϕ(ν)−1 + . . . , whence μ(ν) = −2a = 0 and, in view of the definition of ν, κ ≡ 2 (mod 4). Since Qν (x) = Qν/2 (x 2 ), identity (9) gives again

x ϕ(ν) − μ 21 ν x ϕ(ν)−2 + . . . = x ϕ(ν) + 2bx ϕ(ν)−2 + . . . , μ( 21 ν) = −2b = 0, which is impossible, because 21 ν is square-free. It follows from the above that (x +y)−ϕ(ν)/2 R(x, y), (x +y)1−ϕ(ν)/2 S(x, y) are homogeneous symmetric functions of x, y of dimension 0; so they are rationally expressible in terms of (x + y)2 and xy, and thus (A + B)−ϕ(ν)/2 R(A, B), (A + B)−1−ϕ(ν)/2 S(A, B) are rationally expressible by (A+B)2 and AB. In their turn (A+B)2 , AB and (A+B)/(α+β) are rationally expressible by (α + β)2 and αβ. Therefore the numbers A + B 1 ϕ(ν) L { 1 ϕ(ν)} 1 1 4 2 (α + β)2[ 4 ϕ(ν)] (A + B)− 2 ϕ(ν) R(A, B), δR(A, B) = k(L) α+β A + B 1 ϕ(ν) 1 1 S(A, B) L { 41 ϕ(ν)} 2 = (α + β)2[ 4 ϕ(ν)] (A + B)−1− 2 ϕ(ν) S(A, B), δ c A+B k(L) α+β are rationally expressible by (α + β)2 = L and αβ = M and as such are rational. Since for ε = ±1 1/2 S(A, B) A + B AB 1/2 (α, β) = δR(A, B) ± κ(α + β)2 αβ δ δQ(ε) n α + β αβ A+B and the numbers AB 1/2 A+B , = ±(αβ)(n−ν)/2ν , α+β αβ

LM 1/2

1/2 κ(α + β)2 αβ =κ k(LM)

(ε)

are rational, the numbers δQn (α, β) are also rational. If ϕ(n) ≡ 0 (mod 4) or k(L) = 1 then δ = 1, and it is immediately evident from (4) and (6) that these numbers are algebraic integers, consequently they are then rational integers. Let ϕ(n) ≡ 0 (mod 4) and k(L) = 1. Since n = 1, 2, 4, we have n = rα

or

n = 2r α ,

r prime ≡ 3 (mod 4).

Since k(L) | κ | n, k(L) is odd, we get k(L) = κ = r, n = 2r α . We have to prove that 1/2 , it is clear from the numbers r −1/2 Q(ε) n (α, β) are algebraic integers. First, since κ = r formula (4) that their difference is integral. Now in view of formulae (3) and (6) (10)

(−1) Qn (α, β) = Q(1) (α, β); n (α, β)Qn

their product equals therefore r −1 Qn (α, β) and is integral by Lemma 1. Thus the numbers (ε) r −1/2 Q(ε) n (α, β) are themselves integral. So we have proved that the numbers δQn (α, β) (ε = ±1) are rational integers. It remains to prove that they are coprime.


1051

By identity (3) the resultant R of polynomials ψν,κ (x), ψν,κ (−x) divides the discriminant of Qν (x 2 ) and therefore also the discriminant of x 2ν − 1, which is (2ν)2ν . There exist polynomials χ (1) (x), χ (−1) (x) such that χ (1) (x)ψν,κ (x) + χ (−1) (x)ψν,κ (−x) = R identically in x. The coefficients of χ (1) , χ (−1) are expressible integrally in terms of the coefficients of ψν,κ (x) and therefore are algebraic integers. On making the above relation homogeneous in x, y and putting x = A1/2 , y = B 1/2 , we deduce that any common prime (−1) factor of δQ(1) (α, β) must divide 2νM. By Lemma 1 and (10) each n (α, β) and δQn prime factor of δQ(ε) n (α, β) (ε = ±1) is a primitive prime factor of Pn except possibly for q(n), which then occurs to the first power only. Since no prime factor of 2νM can be (1) (−1) a primitive prime factor of Pn , numbers δQn (α, β), δQn (α, β) are relatively prime. The proof of the lemma is thus complete.

Lemma 3. If χ (r) is an arbitrary character mod m, m > 1 and |x| = 1, then Π= |x − ζmr | < exp(2m1/2 log2 m). χ(r)=const=0

Proof (1 ). We can assume without loss of generality that arg ζm = 2π/m. Let e be the least positive exponent such that χ e+1 = χ . If e = 1 much stronger estimation for Π is known (cf. [1]), if e = ϕ(m) the lemma is satisfied trivially, and thus we can assume j ϕ(m) > e > 1. Let the product Π be taken over r such that χ (r) = ζe 0 . Order these r 1 integers r according to the magnitude of − arg x so that m 2π r r 1 1 ϕ(m) 1 k − arg x < . . . < − arg x k= . m 2π m 2π e Denote by Ni and Ni,j (1 i k, 0 j < e) the number of all non-negative integers r r 1 1 i j r < m such that − arg x − arg x and χ (r) = 0 or χ (r) = ζe , m 2π m 2π respectively. We have

m − ϕ(m) ri − 1 arg x − Ni < 2ν(m)−1 m1/2 (1 i k), (11)

c m 2π (12)

e−1 r

1 i

− arg x + Ni

< 1 Ni,j − m

m 2π

(1 i k).

j =0

c

On the other hand, from a well-known theorem of Schur [8] (for imprimitivite character see [3] and the Addendum (2 )), which we apply successively to characters (1 ) (2 )

The idea of this proof is due to P. Erd˝os. An earlier proof of the writer led to a weaker estimation for Π. p. 1055

1052


χ (r), χ 2 (r), . . . , χ e−1 (r), we get

e−1

−hj0 hj

1/2

ζe (13) N ζ log m i,j e < m

(1 h < e, 1 i k).

j =0

Adding inequalities (11), (12), (13), we find

eNi,j − ϕ(m) ri − 1 arg x < em1/2 log m 0

m 2π Since Ni,j0 = i, putting for brevity π

r

i

m

−

(1 i k).

1 i arg x − π = i we get for each i k 2π k

|i | πk −1 m1/2 log m. Now, if arg ζk = 2π/k, we find k−1

|x − ζmri | |1 − ζki |−1 =

i=1

=

k−1

i=1

1 ri

i

−1

sin arg x − π

sin π

2 m k

k−1

i

i

i

−1

|cos i | + |sin i | cot π

π + = π

sin

i sin k k k

k−1

i=1

i=1

[k/2] 1 k 2 −1 1/2 1/2 exp 2m log m 1 + (π k m log m) πi i i=1 i=1 k < exp 2m1/2 log m 1 + log . 2 [k/2]

Since, on the other hand,

k−1 i=1

Π 2

k−1

|1 − ζki | = k and k = ϕ(m)/e < m/2, we get

|x − ζmri | |1 − ζki |−1

i=1 c

This proves the lemma.

k−1

|1 − ζki |

i=1

m exp(2m1/2 log2 m). < m exp 2m1/2 log m 1 + log 4

Proof of Theorem 1. As we already know, we can assume that M > 0. Then, in view of formula (8) and Lemmas 1 and 2, in order to prove Theorem 1 for a given index n, it is enough to establish that ⎧ ⎪ if q(l) < q(n) and n = 2r α , r as below, ⎨1, (ε) 1/2 (14) |Qn (α, β)| > r , if n = 2r α , r = k(L) prime ≡ 3 (mod 4), ⎪ ⎩ q(l), if q(l) = q(n) and n = 2r α , r as above. The proof of this inequality is different if α, β are real (K > 0) and if they are complex (K < 0); consequently the proof is divided into two parts.


1053

1. K > 0. If n > ν = ηκl, thus n 3ν, we apply (7) and find 1/2 |Q(ε) − B 1/2 )ϕ(ν) (α 3/2 − β 3/2 )ϕ(ν) n (α, β)| > (A

ϕ(ηκ)ϕ(l)/2 = KL1/2 + M(L1/2 − 2M 1/2 ) > (KL1/2 )ϕ(ηκ)ϕ(l)/2 .

Now, as can easily be verified, (KL1/2 )ϕ(ηκ)/2 > 2 for all L, M, so that ϕ(l) |Q(ε) 2q(l)−1 q(l) n (α, β)| > 2

and inequality (14) holds. Thus we can assume that n = ν, A = α, B = β. We shall consider successively l = 1, l = 3 and l 5. If l = 1, we have to prove |Q(ε) n (α, β)| > 1

(15)

|Q(ε) n (α, β)|

>r

1/2

if

if

n = 2r, r as below,

n = 2r, r = k(L) prime ≡ 3 (mod 4).

(ε)

Now, if |Qn (α, β)| 1, we have by inequality (7) 1 > α 1/2 − β 1/2 = (L1/2 − 2M 1/2 )1/2 ,

1 > 18 α + 81 β = 18 L1/2 ,

so that L < 4M +4M 1/2 +1, L < 64. Since 4M < L, we get M 15 and L, M ∈ M. It remains to consider the case n = 2r, r prime ≡ 3 (mod 4), r 7 (since n = 6), k(L) = r, k(M) = 1. By (7) we have

ϕ(ν)/2 1/2 |Q(ε) − 2M 1/2 , 18 L1/2 ) . n (α, β)| > max(L Since ϕ(ν) = r − 1, it suffices to establish the inequality (16)

max(L1/2 − 2M 1/2 , 18 L1/2 ) > r 1/(r−1) .

Since r 7, r 1/(r−1) 71/6 < 21/2 , inequality (16) holds certainly if L > 128. By an easy enumeration of cases we verify that it holds for each pair L, M , with k(L) = r, c k(M) = 1, unless L, M ∈ M. Suppose now that l = 3. If q(n) > 3 it is again sufficient to prove (15). By (8) we have −1/2 |Q(ε) (α − β) 1 n (α, β)| > 2

unless 1 > 2−1/2 (α − β) = 2−1/2 K 1/2 , i.e. K = 1. Since, as we already know, |Q(ε) n (α, β)| > 1 unless L, M ∈ M, we find that, if q(n) > 3, inequality (14) holds unless L, M ∈ N. We have yet to consider the case q(n) = l = 3, i.e. n = 12, k(LM) = 2. We find directly (ε)

Q12 (α, β) = L − ε21/2 L1/2 M 1/2 − M and since M < 41 L, −1/2 3 )L. |Q(ε) n (α, β)| > ( 4 − 2

1054


Thus |Qn (α, β)| > 3 unless L 12(3 − 23/2 )−1 < 75. By an enumeration of cases we (ε) c find that |Qn (α, β)| > 3 unless L, M ∈ N. It remains to consider l 5. Here we notice first that for all L, M in question (ε)

2−1/2 K 1/2 α 5

or κ 2 or L, M = 9, 1 ,

2−1/2 K 1/2 α 51/2 or κ 5 or L, M = 9, 2 , 2−1/2 K 1/2 α 51/4 or L, M = 5, 1 , 9, 2 . It follows that, if L, M = 5, 1 , 9, 1 , 9, 2 , (2−1/2 K 1/2 α)ϕ(ηκ) > 5; hence also for all l 5 (2−1/2 K 1/2 α (q(l)−3)/2 )ϕ(ηκ) > q(l),

(17)

and inequality (14) follows by (8). If L, M = 5, 1 , 9, 1 , 9, 2 , we find directly (2−1/2 K 1/2 α 2 )ϕ(ηκ) > 7; hence (17) holds if q(l) 7. It remains to consider the cases L, M = 5, 1 , 9, 1 , 9, 2 , l = 5 or 15. Their direct examination leads to the exceptions stated in the theorem. The proof for K > 0 is complete. 2. K < 0. By the fundamental lemma of [7] (18)

|Qn (α, β)| > |α|ϕ(n)−2

ν(n) log2 n

for

n > N (α, β).

On the other hand, by (5) and (6), Q(ε) n (α, β) can easily be represented as the products of B ϕ(ν)/2 and 2 or 1 expressions of the form |x − ζmr |, where x = −A1/2 B −1/2 , ±iA1/2 B −1/2 , χ(r)=const=0

and χ (r) is a real character mod m = κ or 4κ, respectively. Since |A1/2 B −1/2 | = 1, m 2n, we get by Lemma 3

ϕ(n)/2 (19) |Q(ε) exp 4(2n)1/2 (log 2n)2 . n (α, β)| < |α| It follows from (10), (18) and (19) that for n > N(α, β)

ϕ(n)/2−2ν(n) log3 n |Q(ε) exp −4(2n)1/2 (log 2n)2 . n (α, β)| > |α| Since, however, if K < 0, |α| 21/2 and for n > 1040 log 2 1 ϕ(n) − 2ν(n) log3 n − 4(2n)1/2 (log 2n)2 > log n, 2 2

we find for n > max N (α, β), 1040 |Q(ε) n (α, β)| > n, which completes the proof.


1055

Let us remark that Theorem 1 implies the following Corollary. If k(LM) = 1, K > 0, n is odd > 3, then Pn has at least two primitive prime factors, except for n = 5, L, M = 9, 1 . It follows that all terms from the fifth onwards of the above sequences Pn are composite. c

Theorem 2. If k M max(K, L) = ±1, ±2, then q(Pn ) n + 1 for n n0 (L, M). The theorem follows at once from two lemmas.

c

Lemma 4. If Pn is an arbitrary Lehmer sequence and n runs through all numbers ≡ 0 (mod 4), then q(Pn ) n + 1 for n n0 (L, M). The proof is analogous to the proof of Lemma 2 of [6].

c

Lemma 5. If P n is an arbitrary Lehmer sequence and n runs through all numbers ≡ 0 (mod κ), κ = k M max(K, L) , then q(Pn ) n + 1 for n n0 (L, M). Proof. By Lemma 4 we can suppose n ≡ 0 (mod 4). If κ is odd, then Pn has at least one primitive prime factor q for n large enough, by the theorem quoted in the introduction. q is of the form nk + (KL|q) and so q ≡ (KL|q) (mod 4κ). Hence (LM|q) = 1, which in view of the formula (α/β)(q−(KL|q))/2 ≡ (LM|q) (mod q)

(20)

implies that P(q−(KL|q))/2 is divisible by q. Since q is a primitive prime factor of Pn , we cannot have q − (KL|q) = n, whence q 2n − 1. The same argument applies if κ is even and n/2κ is even. If the latter ratio is odd, then by Theorem 1 for n large enough Pn has at least two primitive prime factors. One at least

c of these is n + 1, which completes the proof.

Addendum* Since the exact analogue of Schur’s inequality for imprimitive characters is not explicitly proved in [3] nor apparently anywhere else we shall show Theorem A1. For every non-principal character χ mod m and all integers a, b

b

χ (n)

< m1/2 log m.

n=a

b

χ (n) . We shall need the following lemmas. Let S(χ) = max

a,b n=a

∗

Added in 2005

1056


Lemma A1. For a primitive character χ1 mod k > 1 S(χ1 )
2β and sin πβ > 3β for 0 < β < 21 and 0 < β < 16 , respectively, hence the right hand side is less than 7 7 1/2 dβ 2 1/2 1/6 dβ k + k 1/2 3 1/2k β 1/6 β k 2 1 2 1/2

= k log + k 1/2 log 3 = k 1/2 log k + k 1/2 log 3. 3 3 3 3 Lemma A2. For a primitive character χ1 mod k > 1 1 2 2 k log k log k + log log k + 1 + . S(χ1 ) < 1 + k 1/2 π 2 π k 1/2 log k − 1

Proof. See [3], p. 83.

Lemma A3. If χ is a character mod m induced by a primitive character χ1 mod k, then |μ(d)χ1 (d)|. S(χ ) S(χ1 ) d |m/k

Proof. See [3], p. 86. Lemma A4. If either m = 6k or (k, 6) = 1, then in the notation of Lemma A3 3 m 1/2 |μ(d)χ1 (d)| < . (A1) 2 k d |m/k

Proof. Assume first that m = 6k and let

l m γ pi i , where α 0, β 0, γi > 0 = 2 α 3β k i=1

and pi > 3 are distinct primes. We have |μ(d)χ1 (d)| |μ(d)| = 2l0 +l , (A2) d |m/k

where

⎧ ⎪ ⎨0 l0 = 1 ⎪ ⎩ 2

d |m/k

if α = β = 0, if α + β > 0, αβ = 0, if αβ > 0.

1057


Now 1
1. Thus the numbers Qn (α, β) (0 i < e) are relatively prime in pairs, and in order to prove the theorem it suffices, again by Lemma 1 of [3], to establish the inequality (18)

|Q(i) n (α, β)| > n (0 i < e).

To this end, notice that by Lemma 3 of [3] (19)

log |Q(i) n (α, β)|
N(α, β)

(20) log |Qn (α, β)| > ϕ(n) − 2ν(n) log3 n log |α|. It follows from (17), (19) and (20) that for n > N(α, β) ϕ(n) ν(n) 3 log |Q(i) (α, β)| > − 2 log n log |α| − 2e(e − 1)n1/2 log2 n. n e Since |α| 21/2 and for n > 1060 log 2 ϕ(n) − 2ν(n) log3 n − 2e(e − 1)n1/2 log2 n > log n e 2 inequality (18) certainly holds for

n > max 1060 , N (α, β)

(e 6)

and the theorem is proved.

References [1] H. Hasse, Vorlesungen über Zahlentheorie. Springer, Berlin 1950. [2] A. Schinzel, The intrinsic divisors of Lehmer numbers in the case of negative discriminant. Ark. Mat. 4 (1962), 413–416. [3] −−, On primitive prime factors of Lehmer numbers I. Acta Arith. 8 (1963), 213–223; this collection: I2, 1046–1058.

Originally published in Acta Arithmetica XV (1968), 49–70


On primitive prime factors of Lehmer numbers III

1. The main aim of this paper is to complete the results of [5], [7] and [8] concerning Lehmer numbers with a negative discriminant. About the case of a positive discriminant I have nothing new to say except that J. Brillhart and J. L. Selfridge have found explicitly the sets M0 and N0 occurring in Theorem 1 of [7]. The notation of [7] is retained. In particular ζn = exp(2πi/n), n odd, (α n − β n )/(α − β), Pn (α, β) = n n 2 2 (α − β )/(α − β ), n even, where α and β are roots of the trinomial z2 − L1/2 z + M and L and M are rational integers. ke (n) is the e-th powers-free kernel of n, n∗ is the product of all distinct prime factors of n, z is the complex conjugate of z. We assume L > 0 > K = L − 4M,

(1)

(L, M) = 1,

(2) set

L, M = 1, 1 , 2, 1 , 3, 1 ,

A = max 12, log(M min{k(−K), k(L))} ,

B = max{12, log M}

and prove Theorem 1. If n > 3 · 1014 A3 then Pn (α, β) has at least one primitive prime factor. Theorem 2. For L, M satisfying (1), (2) set 1 if k(LM) ≡ 1 mod 4, η= 2 if k(LM) ≡ 2 or 3 mod 4, 1 if k(KM) ≡ 1 mod 4, η1 = 2 if k(KM) ≡ 2 or 3 mod 4, Corrigendum, Acta Arith. 16 (1969), 101

I4. On primitive prime factors of Lehmer numbers III

1067

1 if k(KL) ≡ 1 mod 4, 4 if k(KL) ≡ 2 or 3 mod 4.

η2 =

If n > 3 · 1014 A3 and n ≡ ηk(LM) mod 2ηk(LM) or n ≡ η1 k(KM) mod 2η1 k(KM) or n ≡ 0 mod η2 k(KL), then Pn (α, β) has two primitive prime factors; if all three congruences hold then Pn (α, β) has four primitive prime factors.

√ Theorem 3. Let e = 3, 4 or 6 and ζe belong to the field Q KL . Set ⎧ ⎪ ⎨1 if K ≡ 0 mod 8, 1 if KL ≡ 0 mod 27, η3 = η4 = 2 if L ≡ 0 mod 8, ⎪ 3 if KL ≡ 0 mod 27; ⎩ 4 if KL ≡ 0 mod 8; ⎧ 1 if K ≡ 0 mod 27, M ≡ 1 mod 4 or L ≡ 0 mod 27, M ≡ 3 mod 4, ⎪ ⎪ ⎪ ⎨2 if K ≡ 0 mod 27, M ≡ 3 mod 4 or L ≡ 0 mod 27, M ≡ 1 mod 4, η6 = ⎪ 3 if K ≡ 6 mod 9, M ≡ 1 mod 4 or L ≡ 3 mod 9, M ≡ 3 mod 4, ⎪ ⎪ ⎩ 6 if K ≡ 6 mod 9, M ≡ 3 mod 4 or L ≡ 3 mod 9, M ≡ 1 mod 4. If n/ηe ke (M)∗ is an integer relatively prime to e, (2n, 8) > 3 · 1014 η3 B 3 for e = 3, L ≡ 0 mod 3, (n3 , 8) η + 1 e then Pn (α, β) has e + (e, 2) primitive prime factors. 4 n > 3 · 1014 ηe B 3 and n

Proofs of these theorems given in §§2, 3, 4 respectively require some facts already established in [6], [7], [8] and also an improved version of Lemma 1 of [8] stated below as Lemma 3. An application to the estimation of the greatest prime factor of a linear recurrence of the second order is given in §5. The result obtained completes Theorem 8 of [9]. Unfortunately, the proof of a related result of [7] concerning the greatest prime factor of certain special Lehmer numbers contains a gap, which I am unable to fill in (in c the present edition a weaker result has been proved).

2. Lemma 1. For n = 1, 2, 3, 4, 6 primitive prime factors of Pn (α, β) coincide with prime factors of Qn (α, β)/ n∗ , Qn (α, β) and are of the form nt ± 1. Proof. This follows from Theorems 3.2, 3.3 and 3.4 of [2]. Lemma 2. For n > 3 · 1014 A3 , (1) and (2) imply the inequality (3)

|Qn (α, β)| > n|α|11ϕ(n)/13 .

Proof. We have (4)

|Qn (α, β)| = |α|

ϕ(n)

μ(n/d)

β d

.

α − 1

d |n

1068


In order to estimate |(β/α)d − 1| we apply Theorem 2 of [9]. We set there α , α

8 1 √

9

1√ 1√ 1√ 2 Lk(K) + 2 Kk(K), 2 Lk(K) − 2 Kk(K) 9 81√ 1√ 1√ 1√ 2 Lk(L) + 2 Kk(L), 2 Lk(L) − 2 Kk(L)

=

if k(−K) k(L), if k(−K) > k(L);

β = β = 1,

√ α , α , β , β are integers of the field Q KL , and we obtain

β d − 1

−25 · 105 a13 (log n + 2)2 , (5) log 2 log

α where

c

a1 = max π, log max{|eD|1/4 , |α β |, |α β |, |α β |, |α β |} = max π, 21 log max |eD|1/2 , M min{k(−K), k(L)}

√ and D is the discriminant of the field Q KL . Clearly D 4k(−K)k(L) and an easy computation shows that 1 4

thus

log 4ek(−K)k(L) max π, 21 log M min{k(−K), k(L)} ,

a1 = max π, 21 log M min{k(−K), k(L)} .

Since by (1) log |α| =

1 2

log M we get from (4) and (5)

log |Qn (α, β)| − log n|α|11ϕ(n)/13

2 13 1 13

ϕ(n) log |α| − 3.2 · 106 · 2ν(n)−1 a13 (log n + 2)2 − 2ν(n)−1 log 2 − log n ϕ(n) log M − 3.3 · 106 · 2ν(n)−1 a13 (log n + 2)2 .

For n > 3 · 1014 A3 > 5 · 1017 we have in virtue of Theorem 15 of [4] n n > γ . (6) ϕ(n) > γ c e log log n + 5/(2 log log n) e log log n + 0.675 On the other hand, for every n

√ 2ν(n) < 39 6 n

(this can be proved elementarily). The functions fr (n) =

nr/6 (eγ log log n + 0.675)(log n + 2)2

(r = 1 or 5)

are increasing for n > e13 . If a1 = π we find 1 13 ϕ(n) log M 2ν(n)−1 a13 (log n + 2)2

>

log 2 log 2 f5 (n) f5 (5 · 1017 ) > 106.56 > 3.3 · 106 . 3 254π 254π 3


If a1 =

1069

log(M min{k(−K), k(L))} π we find n > 24 · 1014 a13 , 1 log(M −1 min{−K, L}) log M = − log(M min{−K, L}) c 2 2 log(M min{−K, L}) 2π − log 2 2π − log 2 log(M min{−K, L}) a1 , 4π 2π hence 1 2

1 13 ϕ(n) log M 2ν(n)−1 a13 (log n + 2)2

2π − log 2 2π − log 2 f5 (n) f5 (24 · 1014 a13 ) 507πa12 507π a12 2π − log 2 = (24 · 1014 )2/3 f1 (24 · 1014 a13 ) c 507π 2π − log 2 (24 · 1014 )2/3 f1 (24 · 1014 π 3 ) > 106.53 > 3.3 · 106 . 507π This completes the proof.

>

Proof of Theorem 1 follows at once from Lemmata 1 and 2.

3. Lemma 3. Let e, n be positive integers, n > 2, (e, 2n) = 1 or 2. Let χ be a character n(e, n) mod n(e, n) of order e with conductor f , where , e = 1. Set f ψn (χ ; x, y) =

n

r x − χ (r)ζn(e,n) y .

r=1 (r,n)=1

Then Qn (x e , y e ) =

(7)

ψ n (χ ; x, y) =

(8)

ψn (χ ; x, εy), εe =1 χ (−1)ϕ(n)/e ψn (χ ; y, x),

ψ(χ ; x, y) = R0 (x e , y e ) +

(9)

e−1

τ (χ i )x e−i y i Ri (x e , y e ),

i=1

where Ri are polynomials over Q(ζe ) and τ (χ i ) are normalized primitive Gaussian sums belonging to characters χ i . Proof. Formula (7) follows at once, since εe =1

ψn (χ; x, εy) =

n

n

e r re εy = ye . x − χ (r)ζn(e,n) x − ζn(e,n)

r=1 εe =1 (r,n)=1

r=1 (r,n)=1

To prove formula (8) we notice that the assumptions on e and f imply

3 2 n(e, n) = (e3 , n2 )n/(e2 , n), (e , n ), n/(e2 , n) = 1,

1070 c


hence χ = χ(e3 ,n2 ) χn/(e2 ,n) ,

(10)

where χ(e3 ,n2 ) and χn/(e2 ,n) are characters mod (e3 , n2 ) and mod n/(e2 , n), respectively, the former primitive; n 3 mod 8 if n = 4, e = 2, r≡ 1 mod (e3 , n2 ) otherwise; r=1 (r,n)=1 n

r≡

r=1 (r,n)=1

−1 mod n 1 mod n

if n has a primitive root, otherwise.

Besides n

−r ζn(e,n) = (−1)ϕ(n)/(e,n) ;

r=1 (r,n)=1

⎧ ⎪ 1 mod 2 ⎪ ⎪ ⎪ ⎪ ⎪p − 1 mod 2 ϕ(n) ⎨ e ≡ p − 1 ⎪ e ⎪ ⎪ mod 2 ⎪ ⎪ ⎪ ⎩ 2 0 mod 2

if n = 4, e = 2, if n = pμ , p odd, if n = 2pμ , p odd, e even, otherwise.

It follows hence n

−r χ (r)ζn(e,n)

r=1 (r,n)=1

⎧ ⎪ −χ (3) = χ (−1)ϕ(n)/e ⎪ ⎪ 2 ⎪ ⎪ χ (−1) = (−1)(p−1)/e = (−1)((p−1)/e) ⎪ ⎪ ⎪ ⎪ = χ (−1)ϕ(n)/e ⎨ (p−1)/2 = (−1)(p−1)/2 (−1)(p−1)/2 = χn/2 (−1)(−1) ⎪ ⎪ = 1 = χ (−1)ϕ(n)/e ⎪ ⎪ ⎪ ⎪ ⎪ χ (−1) = 1 = χ (−1)ϕ(n)/e ⎪ ⎪ ⎩ χ (1) = 1 = χ (−1)ϕ(n)/e

if n = 4, e = 2, if n = pμ , p odd, if n = 2pμ , p odd, e even, if n = 2pμ , p, e odd, otherwise

and we get ψ n (χ ; x, y) =

n

−r x − χ (r)ζn(e,n) y

r=1 (r,n)=1

= (−1)ϕ(n)

n r=1 (r,n)=1

−r χ (r)ζn(e,n)

n

r x y − χ (r)ζn(e,n)

r=1 (r,n)=1

= (−1)ϕ(n)/e ψn (χ ; y, x).

1071

I4. On primitive prime factors of Lehmer numbers III c

In the proof of (9) we shall denote by a1 , a2 , . . . , b1 , b2 , . . . , c1 , c2 , . . . , d1 , d2 , . . . numbers of the field Q(ζe ), by pi (ξ, η, . . . ) and si (ξ, η, . . . ) the i-th fundamental symmetric function and the sum of i-th powers of the indeterminates ξ, η, . . . , respectively. We have ψn (χ ; x, y) =

(11)

ϕ(n)

n−1 (−1)j x ϕ(n)−j y j pj χ (1)ζn(e,n) , . . . , χ(n − 1)ζn(e,n)

j =0

c

and by Newton’s formulae

pj =

(12)

aα1 α2 ...αk s1α1 s2α2 · · · skαk .

α1 +2α2 +...+kak =j

c

On the other hand, in the notation of [1], §20,

n−1 = (13) si χ (1)ζn(e,n) , . . . , χ(n − 1)ζn(e,n)

1 i . τ χ i |ζn(e,n) (n, e)

This is obvious if (n, e) = 1; if (n, e) = 2 we have by (10) χ (r + n) = χ(8,n2 ) (r + n)χn/(n,4) (r + n) = −χ(8,n2 ) (r)χn/(n,4) (r) = −χ (r);

(14) 2

n

ri χ i (r)ζ2n =

r=1 (r,n)=1

n

ri χ i (r)ζ2n +

r=1 (r,n)=1

n

(r+n)i

χ i (r + n)ζ2n

i = τ (χ i |ζ2n ).

r=1 (r,n)=1

Now, by the reduction theory for Gaussian sums, we have i τ (χ i |ζn(e,n) ) = bi τ (χ i ),

c

on the other hand, by the theory of Jacobi sums τ (χ i ) = ci τ (χ )i It follows by (13) c

(15)

with

ci = 0.

s1α1 s2α2 · · · skαk = dα1 α2 ...αk τ χ α1 +2α2 +...+kαk .

Formulae (11), (12) and (15) give (9) with Ri (x, y) =

j

(−1) aα1 α2 ...αk dα1 α2 ...αk x

ϕ(n)+i−j e

−

i+e−1 e

y

j −i e

.

0j ϕ(n) α1 +2α2 +...+kαk =j ≡i mod e

c

Corollary 1. Let n > 2, χ be a quadratic character mod n(n, 2) with conductor f , where n(n, 2)/f is odd, and let ψn have the meaning of Lemma 3. Then (16) (17)

Qn (x 2 , y 2 ) = ψn (χ ; x, y)ψn (χ ; x, −y), ) ψn (χ ; x, y) = R(x 2 , y 2 ) − χ (−1)f xyS(x 2 , y 2 ),

where R and S are polynomials with rational coefficients and (18)

R(x, y) = χ (−1)ϕ(n)/2 R(y, x),

S(x, y) = χ (−1)ϕ(n)/2+1 S(y, x).

1072


Besides, for n even, ε = ±1, 2n

ψn (χ ; x, εy) =

(19)

r (x − ζ2n y).

r=1 χ(r)=ε

Proof. Formulae (16), (17) √and (18) follow from (7), (9) and (8), respectively, on taking into account that τ (χ ) = χ (−1)f is irrational. Besides for n even, ε = ±1, r+(1−εχ(r))n/2

r = ζ2n εχ (r)ζ2n

and in virtue of (14) the sequence 1 − εχ (r) n (1 r < n, (r, n) = 1) 2 is a permutation of the sequence r (1 r < 2n, χ (r) = ε), which implies (19). r+

Lemma 4. Let χ be a quadratic character mod n with conductor f and Φn(ε) (χ ; x, y) = ωnε (χ )

n

(x − ζnr y),

r=1 χ(r)=ε

where ε = ±1 and ωn (χ ) =

⎧ n ⎪ ⎪ ζnr if f = 3, ⎨ r=1

⎪ χ(r)=1 ⎪ ⎩ 1

otherwise.

Then Qn (x, y) = Φn(1) (χ ; x, y)Φn(−1) (χ ; x, y), ) Φn(ε) (χ ; x, y) = T (x, y) − ε χ (−1)f U (x, y),

(20) (21)

where T , U are polynomials with rational coefficients and T (x, y) = x 2

ν−2

ν−2

, U (x, y) = y 2 T (x, y) = −T (y, x), U (x, y) = U (y, x)

if f = 4, n = 2ν , if f = 8, n = 2ν , χ(−1) = −1 or f = 4, n = 2μ+2 q ν

(22) T (x, y) = T (y, x),

c

or f = q, n = q ν , q prime ≡ 3 mod 4, U (x, y) = χ (−1)U (y, x) otherwise.

Besides we have (23) c

Φn(ε) (χ ; x 2 , y 2 )

=

(εχ(2))

Φn

(εχ(2))

(χ ; x, y) Φn

(ε) ωn (χ )−ε Φ2n (χ; x, y)

(χ ; x, −y) (n odd), (n even).

1073


Proof. Let n = 2μ m, where m/f is odd. If f is odd, there exists an integer s such that χ(s) = 1, 3 if f = 3, (s − 1, m) = σ = 1 otherwise. Hence m(n,2)

(s − 1)

r=

r=1 χ(r)=ε

m(n,2)

sr −

r=1 χ(r)=ε

m(n,2)

r ≡ 0 mod m(n, 2)

r=1 χ(r)=ε

and n

σ

r ≡ 0 mod n

if n is odd,

r ≡ 21 ϕ(m)m mod 2m

if n is even.

r=1 χ(r)=ε 2m

σ

r=1 χ(r)=ε

c

In the latter case σ

n

2m

2μ−1 −1

r=1 χ(r)=ε

k=0

r=σ

r=1 χ(r)=ε

(r + 2km) = σ 2

μ−1

2m

r + σ ϕ(m)m

r=1 χ(r)=ε

2μ−1 −1

k

k=0

≡ 2μ−2 ϕ(m)m + 2μ−2 ϕ(m)m(2μ−1 − 1) ≡ 22μ−3 ϕ(m)m ≡

ϕ(n)n mod n. 4

It follows that for f odd n

(24)

ζnrσ = (−1)(n−1)ϕ(n)/2 .

r=1 χ(r)=ε

√

In particular, ωn (χ )6 = 1 and ωn (χ ) belongs to Q ωn (χ ) = ωn (χ )2χ(2) , thus

χ (−1)f . Moreover, for n odd

Φn(ε) (χ; x 2 , y 2 ) = ωnε (χ )

n r=1 χ(r)=εχ(2)

(x 2 − ζn2r y 2 ) = ωn2εχ(2)

n r=1 χ(r)=εχ(2)

(x − ζnr y)

n

(x + ζnr y)

r=1 χ(r)=εχ(2)

= Φn(εχ(2)) (χ ; x, y)Φn(εχ(2)) (χ ; x, −y),

1074 c


which proves (23) for n odd. For n even we have n

Φn(ε) (χ ; x 2 , y 2 ) = ωn (χ )ε

(x 2 − ζnr y 2 ) = ωn (χ )ε

r=1 χ(r)=ε

= ωn (χ )ε

2n

n

r+n r (x − ζ2n y)(x − ζ2n y)

r=1 χ(r)=ε r (x − ζ2n y) = ωn (χ )−ε Φ2n (χ ; x, y). (ε)

r=1 χ(r)=ε

c

Since (20) is obvious, it remains to prove (21) and (22). For f odd χ is induced by (r|f ), thus (21) follows from Lemma 1 of [6] and the remark after formula (20) there. Further, by (24), Φ n (χ; x, y) = ωn (χ )−ε (−1)ϕ(n)/2 (ε)

n r=1 χ(r)=ε

= (−1)ϕ(n)/2

n

ζn−r

n

(y − ζnr x)

r=1 χ(r)=ε

ζn−rσ Φn(ε) (χ ; y, x) = (−1)nϕ(n)/2 Φn(ε) (χ ; y, x),

r=1 χ(r)=ε

which implies (22) since 21 nϕ(n) is odd only for n = q ν , q prime ≡ 3 mod 4. c For f even χ (m/2 + r) = −χ (r), hence by (23) and (19) μ μ if m = 4, x 2 − εζ4 y 2 (ε) (ε) 2μ 2μ Φn (χ; x, y) = Φm (χ ; x , y ) = μ μ 2 2 ψm/2 (χ ; x , εy ) if m > 4 and the lemma follows from Corollary 1 since for m > 4 ⎧ ν ⎪ ⎨−1 if f = 8, n = 2 , χ(−1) = −1 χ(−1)ϕ(m/2)/2 = or f = 4, n = 2μ+2 q ν , q prime ≡ 3 mod 4, ⎪ ⎩ 1 otherwise. c c

Remark. Lemma 4 can also be deduced from the results of [3]. One has only to rectify the formulae for λ3N and λ4N given on p. 192 there.

Lemma 5. If n ≡ ηk(LM) mod 2ηk(LM), χ is the character mod ηn induced by k(LM)|r , )

√ (ε = ±1) (25) Q(ε) n (α, β) = ψn χ ; α, ε β and δ = k(L){ϕ(n)/4} , then δ −1 Qn (α, β) and δ −1 Qn integers dividing Qn (α, β). (1)

(−1)

(α, β) are relatively prime rational

(ε)

Proof. The assertion is proved as Lemma 1 in [7]. One has only to verify that Qn (α, β) defined by formula (6) there coincide with Q(ε) n (α, β) defined here. Alternatively one can proceed as below in the proof of Lemma 6.


1075

Lemma 6. If n ≡ η1 k(KM) mod 2η1 k(KM), χ1 is the character mod η1 n induced by k(KM)|r , )

√ (26) Q(ε) (ε = ±1) n (α, β) = ψn χ1 ; α, ε β and δ1 = k(K){ϕ(n)/4} , (1)

(−1)

then δ1−1 Qn (α, β) and δ1−1 Qn Qn (α, β).

(α, β) are relatively prime rational integers dividing

Proof. Since χ1 (−1) = −1, it follows from (17) and (18) that the functions R(x, y)(x − y)ϕ(n)/2 and S(x, y)(x − y)ϕ(n)/2−1 are symmetric of even degree thus are expressible rationally by (x + y)2 and xy. Hence the numbers R(α, β)K ϕ(n)/4 and S(α, β)K ϕ(n)/4−1/2 are rational. Since : ) KM χ1 (−1)f (χ1 )αβK = η1 k(KM) k(KM)

√ √ is rational, it follows from (17) and (26) that the numbers K ϕ(n)/4 ψn χ1 ; α, ε β and also δ1 Q(ε) n (α, β) are rational. 2 They are also obviously algebraic integers, thus they are rational integers. δ12 Q(ε) n (α, β) are perfect squares and since they are divisible by a square-free number δ12 they are divisible by its square δ14 . Thus δ1−1 Q(ε) n (α, β) are rational integers (ε = ±1). Finally, they are relatively prime. Indeed, the resultant of ψn (χ1 ; x, y) and ψn (χ1 ; x, −y) by (16) divides the discriminant of Qn (x 2 , y 2 ) and a fortiori (2n)2n . Since by (2) (α, β) = 1, it −1 (−1) follows from (26) that any common prime factor of δ1−1 Q(1) (α, β) n (α, β) and δ1 Qn divides 2n. On the other hand, by Lemma 1 any prime factor of 2n divides Qn (α, β) at most in the first power. Since by (16) and (26) (−1) Qn (α, β) = Q(1) (α, β), n (α, β)Qn

we reach the desired conclusion.

Lemma 7. If n > 4, n ≡ 0 mod η2 k(KL), χ2 is the character mod n induced by k(KL)|r , ε = ±1, q denotes a prime ≡ 3 mod 4, ⎧ ε (ε) ν ⎪ ⎨ζ8 Φn (χ2 ; α, β) if n = 2 , k(KL) = −1, (−ε) (ε) (27) Qn (α, β) = Φn (χ2 ; α, β) if k(KL) ≡ 5 mod 8, n odd, ⎪ ⎩ (ε) Φn (χ2 ; α, β) otherwise; c ⎧√ ν 2 if n = 2 , k(KL) = −1, ⎪ ⎪ ⎪ ⎪√ ν ⎪ −2 if n = 2 , k(KL) = −2, ⎪ ⎪ ⎪ ⎨√−1 if n = 2μ q ν , k(KL) = −1, δ2 = √ ⎪ k(K) if n = q ν , k(KL) = −q, ⎪ ⎪ √ ⎪ ⎪ ⎪ k(L) if n = 2q ν , k(KL) = −q, ⎪ ⎪ ⎩ 1 otherwise, c c

1076

I. Primitive divisors (1)

(−1)

then δ2−1 Qn (α, β) and δ2−1 Qn Qn (α, β).

(α, β) are relatively prime rational integers dividing

(ε)

Proof. It is enough to prove that δ2 Qn (α, β) is rational for ε = ±1; the remainder can be proved like the corresponding part of Lemma 6. If n = 2ν (ν 3), k(KL) = −1 we have by (22) √ ε 2ν−2 ν−2 c − εζ4 β 2 ) δ2 Q(ε) n (α, β) = 2 ζ8 (α = α2

ν−2

+ β2

ν−2

ν−2 ν−2 √ α2 − β 2 + ε −KL , α2 − β 2

(ε)

thus δ2 Qn (α, β) can be expressed rationally in terms of (α + β)2 = L and αβ = M and is rational. If n = 2ν , k(KL) = −2 or n = 2μ q ν , k(KL) = −1 it follows from (22) √ that T (x, y)(x 2 −y 2 ) and U (x, y) are symmetric functions of even degree, hence T (α, β) KL and U (α, β) are rational. Since : ) η2 KL (28) χ2 (−1)f (χ2 )KL = k(KL) k(KL) √ √ (ε) and δ2 = k(KL), it follows from (21) and (27) that KL Φn (χ2 ; α, β) and δ2 Q(ε) n (α, β) are rational. If n = q ν , k(KL) = −q, it follows from (22) that T (x, y)(x − y) and

U (x, y)(x + y)−1

are symmetric functions of even degree, hence ) ) √ k(K) T (α, β) and k(K) U (α, β)/ KL are rational. In the remaining cases by (22) T (x, y)(x + y)ϕ(n)/2

and U (x, y)(x + y)ϕ(n)/2 (x 2 − y 2 )−1

are symmetric functions of even degree, thus

√ k(L){ϕ(n)/4} T (α, β) and k(L){ϕ(n)/4} U (α, β)/ KL

are rational. The desired conclusion follows from (21), (27) and (28).

Proof of Theorem 2. In order to prove the first part of the theorem it is enough to show in view of Lemmata 1, 5, 6 and 7 that for n > 3 · 1014 A3 (ε) (ε) (29) min |Q(ε) n (α, β)|, |Qn (α, β)|, |Qn (α, β)| > n (ε = ±1). c

Now by (25)–(27), (19), Lemma 3 of [7] and Lemma in the Addendum (page 1085) we have (α, β)|, |Q(−ε) (α, β)|, |Q(−ε) (α, β)| < |α|ϕ(n)/2 exp(4n1/2 log2 n). (30) max |Q(−ε) n n n

1077


√ √ Since |α| = M 2 we get by (3) and (6) for n > 3 · 1014 A3 (ε) (ε) log min |Q(ε) n (α, β)|, |Qn (α, β)|, |Qn (α, β)| − log n = log |Qn (α, β)| − log max |Q(−ε) (α, β)|, |Q(−ε) (α, β)|, |Q(−ε) (α, β)| − log n n n n ϕ(n) log |α| − 21 ϕ(n) log |α| − 4n1/2 log2 n 9 log 2 1/2 2 208 > n log n g(n) − , 52 9 log 2 >

11 13

where g(n) =

(30)

n1/2 . (eγ log log n + 0.675) log2 n

g(n) is an increasing function for n > e5 and g(3 · 1014 A3 ) > g(5 · 1017 ) > 5 · 104 >

(31)

208 , 9 log 2

thus (29) follows. c

To prove the second part of the theorem we show that if n satisfies all three congruences n ≡ ηk(LM) mod 2ηk(LM), n ≡ η1 k(KM) mod 2η1 k(KM) and n ≡ 0 mod η2 k(KL) then (32) Q2n (α, β) = Q(ε,θ) (α, β), n ε=±1 θ=±1

where (33) c

(α, β) = Q(ε,θ) n √ −1 δ0 = 1

c

δ0 Q2n (α, β) (−ε)

Qn

(−θ)

(α, β)Qn

(−εθ)

(α, β)Qn

, (α, β)

if n = 4q ν , q prime ≡ 3 mod 4, k(KL) = −1, otherwise;

(ε,θ)

Qn (α, β) are rational integers relatively prime in pairs except for n = q ν or 2q ν , when two of them have the greatest common factor q. It follows from Lemmata 5, 6 and 7 that for ε = ±1, θ = ±1 (δδ1 δ2 )−1 Q(−ε) (α, β)Q(−θ) (α, β)Q(−εθ) (α, β) n n n is rational. On the other hand δδ1 δ2 = (ε,θ)

This implies that Qn

(−1)n q δ0

if n = q ν or 2q ν , otherwise.

(α, β) is rational. Moreover, since χ2 = χ χ1 , we have by

1078


(25)–(27), (19) and (23) for n odd (θ)

(ε)

(α, β) = Q(ε,θ) n

δ0 Qn (α, β)Qn (α, β) (−εθ)

Qn (α, β) √ √ √ √ δ0 ψn (χ ; α, ε β)ψn (χ1 ; α, θ β) = (−εθ) √ √ √ √ (−εθ) Φn (χ χ1 ; α, β)Φn (χ χ1 , α, − β) n n ) 2 ) 2

√

√ εθ r α − ζn β α + ζnr β , = δ0 ωn (χ χ1 ) r=1 χ(r)=ε χ1 (r)=θ

c

r=1 χ(r)=−ε χ1 (r)=−θ

for n even (θ)

(ε)

(α, β) = Q(ε,θ) n =

δ0 Qn (α, β)Qn (α, β) (−εθ)

(α, β)

√ √ √ √ (ε) (θ) δ0 Φ2n (χ ; α, β)Φ2n (χ1 ; α, β) √ √ (−εθ) Φ2n (χ χ1 ; α, β)

(ε,θ)

Therefore Qn

Qn

= δ0

2n

√

) α − ζnr β .

r=1 χ(r)=ε χ1 (r)=θ

(α, β) is an algebraic integer and hence a rational integer. Since 2 2 Q(ε,θ) (α, β)Q(ε,−θ) (α, β) = Q(ε) n n n (α, β) δ0 ,

(34)

2 2 Q(ε,θ) (α, β)Q(−ε,θ) (α, β) = Q(θ) n n n (α, β) δ0 ,

Q(ε,θ) (α, β)Q(−ε,−θ) (α, β) = Q(εθ) (α, β)2 δ02 , n n n (ε ,θ1 )

the greatest common factor of Qn 1 c divides at least two of the numbers

2 (−1) Q(1) (α, β)2 , n (α, β) , Qn

(ε ,θ2 ))

(α, β) and Qn 2

(α, β) for ε1 , θ1 = ε2 , θ2

2 (−1) (α, β)2 , Q(1) n (α, β) , Qn

(1) Qn (α, β)2 , Q(−1) (α, β)2 , n

equal to |δ 2 |, |δ12 |, |δ22 |, respectively. However these numbers are {1, q, q}, {q, 1, q} or {1, 1, 1} according to whether n = q ν , 2q ν or otherwise. It follows that Q(ε,θ) (α, β) are n relatively prime in pairs except for n = q ν or 2q ν , when (35) shows that two of them have the greatest common factor q. Now, by (3), (6), (30)–(32) and (34) we have for n > 3 · 1014 A3

3 22 ϕ(n) log |α| − ϕ(n) log |α| − 12n1/2 log2 n log Q(ε,θ) (α, β) − log n2 > n 13 2 5 ϕ(n) log 2 − 12n1/2 log2 n 52 5 log 2 1/2 2 624 > n log n g(n) − > 0. 52 5 log 2 In virtue of (33) and Lemma 1 the theorem follows.


1079

√ Corollary 2. If e = 1, 2, 3, 4 or 6, ζe belongs to the field Q KL and n > 3 · 1014 A3 e nt ± 1, relatively prime to then α n − ζei β n has a rational prime factor of the form (i, e) αe − β e . Proof. For e = 1 or 2 the corollary follows at once from the divisibility Qne (α, β)|α n −ζe β n , Lemma 1 and Lemma 2. For e > 2 since i/(i,e)

ζei = ζe/(i,e) , it is enough to consider the case i = ±1. Then n i n Q(i) ne (α, β) | α − ζe β ,

Q(i) ne (α, β) | Qne (α, β)

and the corollary follows from Lemma 1 and (29).

4.

c

In this and in the next section we call an integer a + bζe of the field Q(ζe ) normalized if e = 3 or 6, a ≡ −1 mod 3, b ≡ 0 mod 3 or e = 4, a ≡ 1 mod 4, b ≡ 0 mod 2, semi-normalized if either a + bζe or −(a + bζe ) is normalized. Two normalized integers of Q(ζe ) which divide each other are equal. Lemma 8. Let e = 3, 4 or 6 and ω be a semi-normalized integer of Q(ζe ) such that (ω, ω) = 1. Then there exists a character χ of order e, even for e = 6, such that 4ke (ωω)∗ if e = 6, ωω ≡ 3 mod 4, f (χ ) = ke (ωω)∗ otherwise, τ (χ i ) = cie ωe−i ωi , Proof. Let ω = ±ω0e

e−1

where ci ∈ Q(ζe ).

ωkk , where each ωk is a product of distinct normalized irrational

k=1 ωk ’s

primes of Q(ζe ) and are relatively prime in pairs. In virtue of Lemma 2 of [8] there exists for each ωk a character χk of order e such that f (χki ) = ωk ωk ,

i τ (χki ) = χk (−1)ei/(2,e) ωe−i k ωk

Consider the character χ0 =

e−1 k=1

(0 < i < e).

χkk . In virtue of well known theorems we have

f (χ0 ) =

e−1 k=1

ωk ωk = ke (ωω)∗ ,

1080


τ (χ0i ) =

e−1 k=1

e−e{ki/e} e{ki/e} ωk

χk (−1)eki/(2,e) ωk

k=1 ki≡0 mod e

= χ0 (−1)ei/(2,e) × =

e−1

τ (χkki )e = e−1

e−1

ωke−ki ωkki k

k=1 e−1

−ke+e+e[ki/e] −e[ki/e] ωk

ωk

k=1 ki≡0 mod e

ω−ke+ki ωkki k

k=1 ki≡0 mod e χ0 (−1)ei/(2,e) cie ωe−i ωi ,

where ci = ±ω0−1

e−1

−k−[−ki/e] −[ki/e] ωk .

ωk

k=1

For e = 3 or 4 we have χ0

(−1)e/(2,e)

χ0 (−1) =

5

= 1. For e = 6

χk (−1)k = (−1)(f (χ1 χ3 χ5 )−1)/6 ,

k=1

thus χ0 (−1) = −1 only if f (χ1 χ3 χ5 ) = k(ωω) ≡ ωω ≡ 3 mod 4. Set now χ4 χ0 if e = 6, ωω ≡ 3 mod 4, χ= χ0 otherwise, where χ4 is the primitive character mod 4. For e = 6, ωω ≡ 3 mod 4 we have χ (−1) = 1,

f (χ ) = 4ke (ωω)∗ ,

τ (χ i )e = τ (χ4i )e τ (χ0i )e = (−1)i χ0 (−1)i cie ωe−i ωi = cie ωe−i ωi . Thus the character χ satisfies the conditions of the lemma.

Lemma 9. Let e, ω and χ have the meaning of Lemma 8 and ε run through e-th roots of unity. If m(m, e)/f (χ ) is an integer relatively prime to e, χ (−1)ϕ(m)/2e is any square root ϕ(m)/e , c of χ (−1) (35)

ϕ(m)/2e Q(ε) ψm (χ ; ω1/e , εω1/e ), m (ω, ω) = χ (−1)

χ is considered as a character mod m(m, e) and m > 3 · 1014 max3 {12, log ωω}, then (ε) Q(ε) m (ω, ω) are rational integers, relatively prime in pairs and |Qm (ω, ω)| > m. Proof. We have χ (−1)ϕ(m)/2 = 1 and by Lemma 8

e τ (χ i )(ω1/e )e−i (εω1/e )i = cie (ωω)e , thus χ (−1)ϕ(m)/2e ∈ Q(ζe ) and

τ (χ i )(ω1/e )e−i (εω1/e )i ∈ Q(ζe ).

1081


It follows hence that χ (−1)ϕ(m)/2e R0 (ω, ω) ∈ Q(ζe ) and χ (−1)ϕ(m)/2e τ (χ i )(ω1/e )e−i (εω1/e )i Ri (ω, ω) ∈ Q(ζe )

(0 < i < e),

thus by (9) and (36) Q(ε) m (ω, ω) ∈ Q(ζe ). (ε)

On the other hand, Qm (ω, ω) is real because by (8) χ(−1)ϕ(m)/2e ψm (χ ; ω1/e , εω1/e ) = χ (−1)−ϕ(m)/2e ψm (χ ; ω1/e , ε−1 ω1/e ) = χ (−1)ϕ(m)/2e ψm (χ ; ε−1 ω1/e , ω1/e ) = χ (−1)ϕ(m)/2e ψm (χ ; ω1/e , εω1/e ). (ε)

Since Qm (ω, ω) is obviously an algebraic integer it is a rational integer. To prove that (θ) Q(ε) m (ω, ω) and Qm (ω, ω) are relatively prime for ε = θ we notice that by (7) the resultant of ψm (χ; x, εy) and ψm (χ ; x, θy) divides the discriminant of Qm (x e , y e ) and a fortiori (em)em . Since (ω, ω) = 1 it follows by (36) that any common prime factor of Q(ε) m (ω, ω) and Q(θ) m (ω, ω) divides em. On the other hand, by Lemma 1, any prime factor of 6m divides Qm (ω, ω) at most in first power. Since (36) Qm (ω, ω) = Q(ε) m (ω, ω) ε

we reach the desired conclusion. Now, if (m, e) = 1 −ϕ(m)/2e

χ (−1)

Q(ε) m (ω, ω)

θ e =1

c

c

=ω

ϕ(m)/e

m ω1/e r − ζm . εθ ω1/e

r=1 χ(r)=θ

Therefore, by Lemma 3 of [7] ϕ(m)/e |Q(ε) exp(2em1/2 log2 m). m (ω, ω)| |ω|

c

c

If (m, e) = 2 the same conclusion follows from the lemma in the Addendum.

On the other hand, since k (ω + ω)2 = 1 we have by Lemma 3 for m > 3 · 1014 max3 {12, log ωω} > 5 · 1017 |Qm (ω, ω)| > m|ω|11ϕ(m)/13 .

1082


It follows by (6), (31), (32) and (37) that for m in question log |Q(ε) m (ω, ω)| − log m 11 e−1 > ϕ(m) log |ω| − ϕ(m) log |ω| − 2e(e − 1)m1/2 log2 m 13 e 1 ϕ(m) log |ω| − 60m1/2 log2 m 78 log 2 1/2 2 9360 m log m g(m) − > 0. 156 log 2

This completes the proof. Proof of Theorem 3. We set for e = 3 or 6 ⎧ ⎪ α, n if K ≡ 0 mod 27, ⎪ ⎪ ⎪ ; < ⎪ ⎪ (2n, 8) ⎪ ⎪ if L ≡ 0 mod 27, ⎪ ⎨ ζ4 α, n (n3 , 8) ω, m = = s n > ⎪ ζ3 α, if K ≡ 6 mod 9, ⎪ ⎪ ⎪ 3 ⎪ ; < ⎪ ⎪ n (2n, 8) ⎪ s ⎪ ⎩ ζ12 α, · 3 if L ≡ 6 mod 9; 3 (n , 8) for e = 4

c

⎧ ⎪ ⎨α, n ω, m = ζ4 α, n/2 ⎪ ⎩ ζ4 α 2 , n/4

if K ≡ 0 mod 8, if L ≡ 0 mod 8, if KL ≡ 0 mod 8.

It can be verified that for a suitably chosen s = ±1, ω is a semi-normalized integer of Q(ζe ) and m > 3 · 1014 B 3 . Moreover if KL ≡ 0 mod e3 /(8, e3 ), Qm (ω, ω) (37) ±Qn (α, β) = s −s Qm (ω, ω)Qm (ζe ω, ζe ω) otherwise. c

Since ωω = M and n/ηe ke (M)∗ , e = 1, ω and m satisfy the assumptions of Lemma 9. Therefore by (37) Qm (ω, ω) has e pairwise relatively prime factors > m and by Lemma 1 Qm (ω, ω) has e distinct prime factor ≡ ±1 mod m. These primes clearly do not divide n and again by Lemma 1 they are primitive prime factors of Pn (α, β). If KL ≡ 0 mod e3 /(8, e3 ) we have e = e + (e, 2)[(ηe + 1)/4] and the theorem is proved. Otherwise the resultant of Qm (x, y) and Qm (ζes x, ζe−s y) divides the discriminant of their product Qn (x, y) and a fortiori nn . The same applies to the greatest common divisor of Qm (ω, ω) and Qm (ζes ω, ζe−s ω). Therefore, the primitive prime factors mentioned beforehand do not divide Qm (ζes ω, ζe−s ω). By Lemma 2 we have for m > 3 · 1014 B 3

Qm (ζ s ω, ζ −s ω) > m, e e thus for e = 3 we get from Lemma 1 and (38)

η + 1 e 4

4 = e + (e, 2)


1083

primitive prime factors of Pn (α, β). Finally if e = 4 or 6 and KL ≡ 0 mod e3 /8, Pm (ζes ω, ζe−s ω) has by Theorem 2 two primitive prime factors. These factors by Lemma 1 divide Qm (ζes ω, ζe−s ω), thus we get from (38) η + 1 e e + 2 = e + (e, 2) 4

primitive prime factors of Pn (α, β).

5. Theorem 4. Let un be a recurrence of the second order given by the formula un = Ωωn + Ω ωn , where ω and ω satisfy z2 − P z + Q = 0, P , Q, u0 , u1 are rational integers, (38)

Δ = P 2 − 4Q < 0,

P 2 = Q, 2Q, 3Q

and ω/ω , Ω/Ω√ are multiplicatively dependent. If e is the number of roots of unity contained in Q( Δ), u and v are the least in absolute value integers satisfying (ω/ω )eu/2 = (−Ω/Ω )ev/2 , v > 0, n > 0 and nv + u > 3 · 1014 max3 12, log 2Q2 (P 2 , Q)−1 , then (39)

q(un ) nv + u − 1 (q denotes the greatest prime factor). Proof. Let r and s be integers such that ru − sv = σ = (u, v). It follows from (40) that (ω/ω )u/σ (−Ω/Ω )−v/σ is a root of unity, hence by the definition of e ω eu/σ Ω −ev/σ − =1 ω Ω and by the choice of u and v, σ 2. It follows further from (40) that ω σ e/2 Ω −erv/2 ω esv/2 = − ω Ω ω whence ev/σ Ω r ω s ωe = . (40) e ω Ω ω √ The number (Ω/Ω )r (ω /ω)s is a quotient of two conjugates in Q( Δ) and is different from ±1 since by (39) ω/ω is not a root of unity. Therefore, it can be represented in the form √ √ (L1/2 + K 1/2 )/2 , where L, K are rational integers, L > 0, K < 0, Q( KL) = Q( Δ) 1/2 1/2 (L − K )/2

1084


and (4L, L − K) = 4. Set L − K = 4M,

(L1/2 + K 1/2 )/2 = α, (L1/2 − K 1/2 )/2 = β. √ √ α e and β e are relatively prime integers of Q( Δ) semi-normalized if Q( Δ) = Q(ζe ). Also ωe (P 2 , Q)−e/2 and ωe (P 2 , Q)−e/2 are such integers and since by (41) α ev/σ ωe (P 2 , Q)−e/2 = , ωe (P 2 , Q)−e/2 β ev/σ we get ωe (P 2 , Q)−e/2 = ±α ev/σ ,

ωe (P 2 , Q)−e/2 = ±β ev/σ ,

−μ

ω = ζ2e (P 2 , Q)1/2 α v/σ , ω = ζ2e (P 2 , Q)1/2 β v/σ .

Since (Ω 2 Δ, Ω 2 Δ) = (2u1 − P u0 )2 , u21 − P u1 u0 + Qu20 = Δ1 is a rational integer and by (40) 2 Ω Δ/Δ1 ev/2 α euv/σ = , Ω 2 Δ/Δ1 β (41)

μ

it follows as before that (42)

8

9 −ν 1/2 u/σ ν Δ1/2 β u/σ if u 0, Δ α , ζ ζ 2e 2e 1 1 Ω(ω − ω ), Ω (ω − ω) = 8 −ν 1/2 9 ν Δ1/2 α |u|/σ if u < 0. ζ2e Δ1 β |u|/σ , ζ2e 1

Thus we obtain −(n−1)μ−ν

un = ζ2e

nμ+ν

1/2

Δ1 (P 2 , Q)(n−1)/2 (αβ)(|u|−u)/2σ

α (nv+u)/σ − ζe β (nv+u)/σ . μ v/σ v/σ α − ζe β

Since ω/ω is not a root of unity, by (41) α/β also is not such a root, hence L, M = 1, 1 , 2, 1 , 3, 1 . Further, it follows from (42) that M = αβ ωω (P 2 , Q)−1 = Q(P 2 , Q)−1 . Since min{−K, L} 2M, we get A max{12, log 2Q2 (P 2 , Q)−2 } nμ+ν

and in virtue of Corollary 2 α (nv+u)/σ − ζe β (nv+u)/σ has a rational prime factor p of e nv + u the form · t ± 1 relatively prime to α e − β e . (nμ + ν, e) σ

Since (nv + u)/σ, v/σ = 1, the highest common factor of α (nv+u)/σ − ζenμ+ν β (nv+u)/σ

and

α v/σ − ζeμ β v/σ μ

divides α e − β e . Thus p is relatively prime to α v/σ − ζe β v/σ , we have p | un and q(un ) p nv + u − 1 except possibly if c

(43)

σ = 2,

nμ + ν ≡ 0 mod e.


In that case we have by the choice of u, v ω eu/4 ω

1085

Ω ev/4 = − , Ω

hence by (42), (43) νv/2 ≡ μu/2 mod 2 and by (38) (nv + u)/σ is odd. The prime p being of the form (nv + u)t/2 ± 1 must be at least nv + u − 1, which completes the proof.

Addendum* We shall prove the following lemma used in the proof of Theorem 2 instead of Lemma 3 of [7], if (e, n) = 2. Lemma. Let χ be a character mod 2n of exponent e ≡ n ≡ 0 mod 2 and let |x| = 1. Then

n

(x − χ (r)ζ r ) exp(2en1/2 log2 n). 2n

r=1

Proof. The left hand side does not exceed 2ϕ(n) , hence we may assume that 2ϕ(n) > exp(2en1/2 log2 n), which gives (A1) n1/2 log2 n1/2

2n1/2 log2 n ϕ(n) > , e log 2 16e > > 23e, n1/2 > 23e(log 23e)2 , log 2

n > 529e2 (log 23e)4 ,

(A2)

log n > 12.

For a non-negative integer d < e let rd1 < rd2 < . . . < rdkd be all integers r such that 1 r n and χ (r) = ζed . Let Ni and Nij be the number of r rdi such that χ (r) = 0 and χ (r) = ζei , respectively. We have Ni +

(A3) Ni = rdi −

δ |2n

∗

Added in 2005

μ(δ)

r di

δ

e−1

Nij − rdi = 0,

j =0

= rdi

r μ(δ) di 1− + , μ(δ) δ δ δ |2n

δ |2n

1086 hence (A4)


r

di 1 = 2ν(2n)−1 < (2n)1/2 . 2n − ϕ(2n) − Ni max

ε=±1 2n δ |2n μ(δ)=ε

On the other hand, by Theorem A1 from Addendum to paper I2 (p. 1055), which we apply successively to characters χ , χ 2 , . . . , χ e−1 , we have

e−1

−hd hj

(A5) Nij ζe < (2n)1/2 log 2n (1 h < e, 1 i kd ).

ζe j =0

Adding the inequalities (A4) and (A5) to the equality (A3) we obtain

ϕ(2n)

rdi < e(2n)1/2 log 2n.

eNid − 2n Since Nid = i, it follows that

r

e(2n)1/2 log 2n i

di

(A6) − .

< 2n ϕ(2n)/e ϕ(2n) j

Defining N0 and N0j as the number of r n such that χ (r) = 0 and χ (r) = ζe , respectively, and arguing similarly we obtain

n kd

e(2n)1/2 log 2n

− ,

<

2n ϕ(2n)/e ϕ(2n) hence ϕ(n) ϕ(n) + (2n)1/2 log 2n > kd > − (2n)1/2 log 2n. e e

(A7) Now, put

l=

(A8)

? ϕ(n) e

@ − (2n)1/2 log 2n .

It follows from (A1) and (A2) that ϕ(n) ϕ(n) e − >l> , e 2 2e hence we may choose an integer m prime to e/2 such that (A9)

ϕ(n) e ϕ(n) m> − > l > m − (2n)1/2 log 2n. e e 2 It follows from (A1), (A2) and (A10) that (A10)

m > π(m − l).

(A11) Let x0 =

m π

arg x + 21 ]. For i l we have rdi 1 i − x0 − arg x = + ρdi , 2n 2π 2m

1087


where, by (A5), (A7)–(A9) and (A1) (A12)

|ρdi |

1 e(2n)1/2 log 2n i(ϕ(n)/e − m) + + ϕ(2n) 2ϕ(n)m/e 4m 1/2 e(2n) log 2n e/2 e e(2n)1/2 log 3n + + < . 2ϕ(n) 2ϕ(n)/e 2ϕ(n) 2ϕ(n)

Now, we have kd n e−1

e−1

rdi

x − χ (r)ζ r =

x − ζ d ζ rdi 2 2kd −l x − ζed ζ2n , e 2n 2n r=1

where the product and

d=0 i=1

d

and later the sum

d

i ≡ εm + x0 mod 2m

(A13)

d

d=0

are taken over all integers i such that 1 i l d=ε

if

e 2

(ε = 0 or 1).

Hence, by (A7) and (A8), e−1 n

x − χ (r)ζ r 22e(2n)1/2 log 2n+1

x − ζ d ζ rdi . e 2n 2n

(A14)

r=1

d=0

d

On the other hand, P = = =

e−1 d

x − ζ d ζ rdi 1 − ζ d ζ i−x0 −1 e 2n

e 2m

d=0 e−1

d −1 π d rdi 1 π

− arg x sin + (i − x0 )

sin π + π d e 2n 2 e 2m

d=0 e−1 d=0 e−1 d=0

d −1 π d π π

(i − x0 ) + πρdi sin + (i − x0 )

sin π + d e 2m e 2m

d

d π

|cos πρdi | + |sin πρdi | cot π + (i − x0 ) . e 2m

A A However, if A πx A = 0 we have |cot x| and for i occurring in

1 AxA A π A π

d

we have Ad A 1 A A (i − x0 )A = 0, A + e 2m

hence by (A12) (A15)

P

e−1 d=0

d

|ρdi | 1 + Ad A A + i−x0 A e 2m

exp

e−1 e(2n)1/2 log 3n . Ad A d A + i−x0 A d=0

e

2m

1088


Now, if i 1 − x0 d2 i 2 − x0 d1 + ≡ + mod 1, e 2m e 2m

0 dν < e, 1 iν l,

where

then e e (i1 − x0 ) ≡ (i2 − x0 ) mod m, 2 2 and, since (m, 2e ) = 1, i1 ≡ i2 mod m. In view of 1 iν l < m, this gives i1 = i2 , hence d1 ≡ d2 mod e; d1 = d2 . Therefore, (A16)

e−1 em−1 e(2n)1/2 log 3n/2ϕ(n) e(2n)1/2 log 3n/2ϕ(n) Ad A d A + i−x0 A

j/em

=

j =1

2m

e

d=0

e(2n)1/2 log 3n

1+

ϕ(n)

em/2−1 j =1

em j

e(2n)1/2 log 3n · ϕ(n) log ϕ(n) ϕ(n)

e(2n)1/2 log2 n + log 23 · log n .

On the other hand, (A17)

e−1 d

d=0

e−1

−1

1 − ζ d ζ i−x0 =

1 − ζ d ζ i−x0 ·

1 − ζ d ζ i−x0

, e 2m

d=0

d

e 2m

e 2m

d

where the products d and d are taken over all integers i satisfying (A13) such that 1 i m and l < i m, respectively. Denoting by ix the only positive integer i m such that i ≡ x0 mod m we have further (A18)

e−1 d=0

d

m e−1

1 − ζ d ζ i−x0 =

1 − ζ d ζ i−x0

e 2m

e−1 d=0

e 2m

d=0 i −x d≡ x m 0 · 2e mod e

m m−1

e−1 f j

1 − ζme(i−x0 )/2 · |1 − ζe | = e |1 − ζm | = em, f =1

i=1 i=ix

(A19)

1 − ζ d ζ ix −x0

e 2m

i=1 d=0 i=ix

=

e−1

j =1

m e−1

1 − ζ d ζ i−x0 −1 =

1 − ζ d ζ i−x0 −1 e 2m e 2m

d

i=l+1 d=0 i=ix

=

m

1 − ζme(i−x0 )/2 −1 . i=l+1 i=ix


1089

When i runs through integers from l + 1 to m except ix , 2e (i − x0 ) gives distinct non-zero residues mod m. Hence, by (A11), m m−l m−l

πj −1 m

1 − ζme(i−x0 )/2 −1 min 1, 2 sin m πj j =1

i=l+1 i=ix

j =1

m +m−l π(m − l) n 1/2 1/2 n exp (2n)1/2 log 2n · log √ log2 n . + (2n)1/2 log 2n < exp 2 2π 2 log 2n exp (m − l) log

It follows now from (A2) and (A14)–(A19) that

n

r

x − χ(r)ζ2n

r=1 1/2 log 2n+1

22e(2n)

1/2 2 exp e(2n)1/2 log2 n + log 23 log n · em · exp n2 log n

exp 2 log 2 · e(2n)1/2 log 2n + log 2

1/2 2 log n + e(2n)1/2 log2 n + log 23 · log n + log n2 + n2

< exp 2en1/2 log2 n .

References [1] H. Hasse, Vorlesungen über Zahlentheorie. Springer, Berlin 1964. [2] D. H. Lehmer, An extended theory of Lucas’ functions. Ann. of Math. (2) 31 (1930), 419–448. [3] T. Nagell, Contributions à la théorie des corps et des polynômes cyclotomiques. Ark. Mat. 5 (1964), 153–192. [4] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers. Illinois J. Math. 6 (1962), 64–94. [5] A. Schinzel, The intrinsic divisors of Lehmer numbers in the case of negative discriminant. Ark. Mat. 4 (1962), 413–416. [6] −−, On primitive prime factors of a n − bn . Proc. Cambridge Philos. Soc. 58 (1962), 555–562; this collection: I1, 1036–1045. [7] −−, On primitive prime factors of Lehmer numbers I. Acta Arith. 8 (1963), 213–223; this collection: I2, 1046–1058. [8] −−, On primitive prime factors of Lehmer numbers II. Acta Arith. 8 (1963), 251–257; this collection: I3, 1059–1065. [9] −−, On two theorems of Gelfond and some of their applications. Acta Arith. 13 (1967), 177–236; Corrigendum 16 (1969), 101; Addendum 56 (1996), 181.

Originally published in Journal für die reine und angewandte Mathematik 268/269 (1974), 27–33


Primitive divisors of the expression An − B n in algebraic number fields

To Professor Helmut Hasse on his 75th birthday

Let A, B be non-zero integers of an algebraic number field K of degree l. A prime ideal P of K is called a primitive divisor of An − B n if P | An − B n but P /| Am − B m for A m < n. It has been proved in [3] that if (A, B) = 1 and is not a root of unity then the B primitive divisors exist for all n > n0 (A, B) and the question has been raised whether the same is true for n > n0 (K). A certain step in this direction was made by E. H. Grossman who proved in 1972 the following theorem (unpublished): Let E(A, B) be the set of positive integers n such that An −B n does not have a primitive divisor. Then Card{n ∈ E(A, B) : n x} logm x for x > x0 (m, l), where logm x denotes the m-fold iterated logarithm. The aim of this paper is to give an affirmative answer to the question and in fact to prove the following stronger A is not a root of unity then An − B n has a primitive B A divisor for all n > n0 (d), where d is the degree of and n0 (d) is effectively computable. B

Theorem 1. If (A, B) = 1 and

The theorem is best possible up√ to the order of the function n0 (d); an absolute constant cannot be expected since for A = d 2, B = 1, Ad − B d = 1 has no primitive divisor. The proof is based on four lemmata, the critical one being an easy consequence of the recent deep theorem of Baker [1], which we quote below with some changes in the notation: Let α1 , . . . , αk be non-zero algebraic numbers with degrees at most d and let the heights of α1 , . . . , αk−1 and αk be at most H and H ( 2) respectively. For some effectively computable number C > 0 depending only on k, d and H the inequalities 0 < |m1 log α1 + . . . + mk log αk | < C − log H log M have no solution in rational integers m1 , . . . , mk with absolute values at most M( 2) (the logarithms have their principal values).

I5. Primitive divisors of An − B n

A

1091

α A = , where α, β ∈ K0 ; α, β are integers and (α, β) = d. B B β Let S be a set of all isomorphic injections of K0 in the complex field and set α w = log max |α σ |, |β σ | − log N d β

Let Q

= K0 ,

σ ∈S

where N denotes the absolute norm in K0 . Clearly w

α β

is independent of the choice of

α, β in K0 . α is not a root of unity, then β α log |α n − β n | = n log |β| + O d + w log n, β

Lemma 1. If |α| = |β| but

where the constant in the symbol O depends only on d and is effectively computable. α , thus log α1 = π i, log α2 = ϑi, β where −π < ϑ π. It follows that for a suitable constant C depending only upon d the inequality Proof. We set in Baker’s theorem k = 2; α1 = −1, α2 =

0 < |πm + ϑn| < C − log H log M where H is the height of α, has no solution in rational integers m, n with absolute values α at most M. However πm + ϑn = 0 since is not a root of unity. On the other hand, if β |m| > n then |πm + ϑn| π. Hence we can take M = n and we obtain Aϑ A A A π A nA C − log H log M π where x is the distance of x from the nearest integer. Since A nϑ A nϑ A A 2 |enϑi − 1| = 2 sin 2A A 2 π we get log |α n − β n | = n log |β| + log |enϑi − 1| = n log |β| + O(log H log n). σ (β x − α σ ) are rational integers The coefficients of the irreducible polynomial N d−1 (1)

σ ∈S

and their absolute values do not exceed (|β σ | + |α σ |) 2d ew(α/β) . Nd−1 σ ∈S

1092


It follows that

α , log H = O d + w β

which together with (1) implies the lemma. Lemma 2. If |α| = |β| then

α log |α n − β n | = n log max{|α|, |β|} + O d 2 + dw , β

where the constant in the symbol O is absolute and effectively computable. Proof. Suppose without loss of generality that |α| < |β|. We have for n 2

α 2

α n

(2) 2|β|n |α n − β n | = |β|n

− 1

|β|n 1 −

β β

α 2

and it remains to estimate 1 − . Let T be the set of all isomorphic injections of K0 K 0 , β where “bar” denotes the complex conjugation, and let x τ0 = x, x τ1 = x. We have

α 2

(αα)τ − (ββ)τ −1 1 − = NK0 K¯ 0 /Q (αα − ββ) |β|−2 β Nd2|T |/|S| 21−|T |

τ ∈T τ =τ0

−1 max |α τ α τ τ1 |, |β τ β τ τ1 |

τ ∈T

2|T |/|S| −|T |

2

> Nd

−1 max |α τ |, |β τ | max{|α τ τ1 |, |β τ τ1 |}−1 .

τ ∈T

τ ∈T

When τ runs over T , and run |T |/|S| d − 1 times over α σ (σ ∈ S). Hence

α 2

1 − 2−|T | e−2w(α/β)|T |/|S| 2−d(d−1) e−2(d−1)w(α/β) β ατ

α τ τ1

and by (2) the lemma follows.

divides Lemma 3. Let ξ be a number of the field K0 and let p be a prime ideal of K 0 which e the rational prime p to the power e (e = ordp p > 0). If ordp (ξ − 1) > then p−1 ordp (ξ n − 1) = ordp (ξ − 1) + ordp n.

Proof. See [3], Lemma 1.

Lemma 4. Let Φn (x, y) be the n-th cyclotomic polynomial in homogeneous form. If P is a prime ideal of K, n > 2(2d − 1), P | Φn (A, B), and P is not a primitive divisor of An − B n , then ordP Φn (A, B) ordP n.


1093

A , p be the prime ideal of K0 divisible by P to the power e0 and let λi be B the least exponent λ for which

Proof. Let γ =

pi | γ λ − 1. It is obvious that pi | γ λ − 1 is equivalent to λi | λ. We have (γ m − 1)μ(n/m) . Φn (A, B) = B ϕ(n) Φn (γ , 1) = B ϕ(n) m|n

On the other hand, since P | Φn (A, B) and (A, B) = 1, P /| B. Hence n μ ordp (γ m − 1). (3) ordP Φn (A, B) = e0 ordp Φn (γ , 1) = e0 m m|n

e = k. If m is not a multiple of λ1 then p−1 ordp (γ m − 1) = 0. If m is a multiple of λi , but not a multiple of λi+1 then

We use the notation of Lemma 3 and set

ordp (γ m − 1) = i

(i k).

Further, if m is a multiple of λk+1 then by Lemma 3 ordp (γ m − 1) = ordp (γ λk+1 − 1) + ordp

m . λk+1

Hence (4)

ordp Φn (γ , 1) =

m n n μ μ + ordp (γ λk+1 − 1) − k m m i=1 m|n λi |m

m|n λk+1 |m

+

m|n λk+1 |m

n

μ

m

ordp

m . λk+1

We note that λk+1 < n. In fact, if k = 0 then the conditions that P | An − B n and P is not a primitive divisor imply that there is a number m < n such that P | Am − B m and then p | γ m − 1, i.e., λ1 < n. If k > 0 we have by Euler’s theorem for the field K0 p j p ϕ(pk ) k k γ γ ϕ(p ) ≡ 1 mod pk , γ ϕ(p )p = −1 , j j =0

k ordp γ ϕ(p )p − 1 min{k + e, pk} > k, thus λk+1 ϕ(pk )p. On the other hand, k > 0 implies p e + 1 d + 1 and since N p pd/e we have

λk+1 pϕ(pk ) p N (pk ) − 1 p pd/(p−1) − 1 2(2d − 1)

1094


(the last inequality requires some elementary, but tedious calculations). It follows that λ1 λ2 . . . λk+1 < n. Hence

n

μ

m|n λi |m

m

ordp Φn (γ , 1) =

= 0 (i = 1, 2, . . . , k + 1) and by (4) n

μ

m

m|n λk+1 |m

n

=

μ

m|n λk+1 |m ordp (n/m)=0

=


=

ordp


m

m λk+1

ordp

m λk+1

+

n

μ


m

ordp

m λk+1

n n mp m μ ordp ordp +μ mp λk+1 m λk+1 ⎧ ⎨e n ordp p = μ ⎩0 mp

n is a power of p, λk+1 otherwise. if

It follows by (3) that ordP Φn (A, B) e0 e ordp

n λk+1

ordP n.

Remark. Lemma 4 is an improvement of Lemma 2 of [3] where 2l (2l − 1) occurs instead of 2(2d − 1). The possibility of replacing 2l (2l − 1) by 2l(2l − 1) was first observed by E. Grossman (unpublished).

Proof of Theorem 1. In order to apply Lemma 4 we estimate NK/Q Φn (A, B). Clearly Φn (A, B) = B ϕ(n) Φn

and since

B β

A

α B ϕ(n) Φn (α, β) , 1 = B ϕ(n) Φn ,1 = B β β

= d−1 we have

Φn (A, B) = d−ϕ(n) Φn (α, β) ,


1095

d log |NK/Q Φn (A, B)| = log |N Φn (α, β)| − ϕ(n) log N d l n

= log (α σ )m − (β σ )m − ϕ(n) log N d μ m σ ∈S m|n α n (5) σ σ = μ m log max{|α |, |β |} + O d + w log m m β σ ∈S m|n

− ϕ(n) log N d α α = ϕ(n)w +O d +w 2ν(n) log n, β β

c

where the constant in O depends only on d and is effectively computable. Now, by the α theorem of Blanksby and Montgomery [2] if is an integer β σ α 1 1

α

w max σ , 1 log 1 + = log > . β β 52d log 6d 52d log 6d + 1 σ ∈S

α If is not an integer, then (β) = d and β α w log Nβ − log N d log 2. β Thus in both cases (6)

w

α β

>

1 . 52d log 6d + 1

We have also (cf. [3]) (7)

ϕ(n) 2ν(n)

B

n . 30

It follows from (5), (6) and (7) that for n > n0 (d) |NK/Q Φn (A, B)| > nl and this by Lemma 4 implies the theorem. We note for further use that for n > n1 (d) (8)

11

A

l

|NK/Q Φn (A, B)| > nl e 13 ϕ(n)w( B )( d ) .

A Corollary 1. If (A, B) = D and is not a root of unity then An − B n has a primitive B

A divisor for all n > max n0 (d), ϕ(d) , where d is the degree of and d is the maximal B A ideal of Q divisible by D. B Proof. The ideal D is principal, equal say to (Δ) in a certain field K1 . Set A1 = AΔ−1 , B1 = BΔ−1 and apply Theorem 1. It follows that An1 − B1n has a primitive divisor in K1 for all n > n0 (d). On the other hand, ideals (A1 ) and (B1 ) are defined already in the field

1096 A


A

in its reduced form. Set d = d1 d2 B B where each prime factor of d1 divides A1 B1 and (d2 , A1 B1 ) = 1. By Euler’s theorem

A ϕ(d2 )

ϕ(d ) ϕ(d ) ϕ(d) ϕ(d) − 1 and a fortiori d2 | A1 2 − B1 2 | A1 − B1 . It for the field K0 , d2

B follows that for n > ϕ(d) a primitive divisor of An1 − B1n is prime to d2 and since it is obviously prime to d1 , it is a primitive divisor of An − B n in K1 . The corresponding prime

ideal of K is a primitive divisor of An − B n in K. Q

as the numerator and the denominator of

Corollary 2. Let K, M be rational integers, (9)

L > 0 > K = L − 4M,

(L, M) = 1,

L, M = 1, 1 , 2, 1 , 3, 1 ,

let α, β be the roots of the trinomial z2 − L1/2 z + M and set ⎧ n (α − β n ) ⎪ ⎪ if n is odd, ⎨ (α − β) Pn (α, β) = (α n − β n ) ⎪ ⎪ ⎩ 2 if n is even. (α − β 2 ) There exists an effectively computable absolute constant c0 such that Pn (α, β) for n > c0 has a primitive prime factor (i.e. a prime factor not dividing KLP1 · · · Pn−1 ). α α is not a root of unity. Since is of degree 2 it is enough β β to take c1 = n0 (2) and to observe that a primitive divisor of Pn (α, β) in Q(α, β) divides a rational prime which has the asserted property.

Proof. By (9), (α, β) = 1 and

Corollary 2 is an improvement of Theorem 1 of [6], where the corresponding property of proved for n > n0 (L, M) (given explicitly). Since for n > 2, Φn (α, β) ∈ Q Pn (α, β) was α and w = 2 log |α| the inequality (8) takes the form β |Φn (α, β)| > n|α|11ϕ(n)/13

for

n > c1

which replaces Lemma 2 of [6]. This allows to improve Theorems 2 and 3 of that paper. Let ke (n) be the e-th powers-free kernel of n, k2 (n) = k(n), n∗ be the product of all distinct prime factors of n. We have Theorem 2. For L, M satisfying (9) set 1 if k(LM) ≡ 1 mod 4, η= 2 if k(LM) ≡ 2 or 3 mod 4, 1 if k(KM) ≡ 1 mod 4, η1 = 2 if k(KM) ≡ 2 or 3 mod 4, 1 if k(KL) ≡ 1 mod 4, η2 = 4 if k(KL) ≡ 2 or 3 mod 4.


1097

There exists an effectively computable constant c2 with the following property. If n > c2 and n ≡ ηk(LM) mod 2ηk(LM) or n ≡ η1 k(KM) mod 2η1 k(KM) or n ≡ 0 mod η2 k(KL) then Pn (α, β) has two primitive prime factors; if all three congruences hold then Pn (α, β) has four primitive prime factors. Theorem 3. Let e = 3, 4 or 6 and exp η3 =

1 if KL ≡ 0 mod 27, 3 if KL ≡ 0 mod 27;

√ 2πi belong to the field Q( KL). Set e ⎧ ⎪ ⎨1 if K ≡ 0 mod 8, η4 = 2 if L ≡ 0 mod 8, η6 = ηη3 . ⎪ ⎩ 4 if KL ≡ 0 mod 8;

n There exists an effectively computable constant c3 with the following property. If ηe ke (M)∗ η + 1 e is an integer prime to e and n > c3 then Pn (α, β) has e + (e, 2) primitive prime 4 factors. An inspection of the proofs given in [6] shows that it is enough to take c2 = max(c1 , 5 · 1017 ),

c3 = max(c1 , 3 · 1018 ).

I take this opportunity to mention papers of L. Rédei [4] and H. Sachs [5] dealing with the problem considered in [3] and not quoted in that paper. They have been brought to my attention by Dr. K. Szymiczek. Both papers contain results from which Lemma 4 above could easily be deduced, but the corresponding proofs are longer than the proof of that lemma.

References [1] A. Baker, A sharpening of the bounds for linear forms in logarithms. Acta Arith. 21 (1972), 117–129. [2] P. E. Blanksby, H. L. Montgomery, Algebraic integers near the unit circle. Acta Arith. 18 (1971), 355–369. [3] L. P. Postnikova, A. Schinzel, Primitive divisors of the expression a n − bn in algebraic number fields. Mat. Sb. (N.S.) 75 (1968), 171–177 (Russian); English transl.: Math. USSR-Sb. 4 (1968), 153–159. [4] L. Rédei, Über die algebraisch-zahlentheoretische Verallgemeinerung eines elementarzahlentheoretischen Satzes von Zsigmondy. Acta Sci. Math. Szeged 19 (1958), 98–126. [5] H. Sachs, Untersuchungen über das Problem der eigentlichen Teiler. Wiss. Z. Martin-LutherUniv. Halle-Wittenberg. Math.-Nat. Reihe 6 (1956/57), 223–259. [6] A. Schinzel, On primitive prime factors of Lehmer numbers III. Acta Arith. 15 (1968), 49–70; Corrigendum, ibid. 16 (1969), 101; this collection: I4, 1066–1089.

Originally published in Mathematics of Computation 61 (1993), 441–444


An extension of the theorem on primitive divisors in algebraic number fields

In memory of D. H. Lehmer Abstract. The theorem about primitive divisors in algebraic number fields is generalized in the following manner. Let A, B be algebraic integers, (A, B) = 1, AB = 0, A/B not a root of unity, and ζk a primitive root of unity of order k. For all sufficiently large n, the number An − ζk B n has a j prime ideal factor that does not divide Am − ζk B m for arbitrary m < n and j < k.

The analogue of Zsigmondy’s theorem in algebraic number fields [3] asserts the following. If A, B are algebraic integers, (A, B) = 1, AB = 0, and A/B of degree d is not a root of unity, there exists a constant n0 (d) such that for n > n0 (d), An − B n has a prime ideal factor that does not divide Am − B m for m < n. This theorem will be extended as follows: Theorem. Let K be an algebraic number field, A, B integers of K, (A, B) = 1, AB = 0, A/B of degree d not a root of unity, and ζk a primitive k-th root of unity in K. For every ε > 0 there exists a constant c(d, ε) such that if n > c(d, ε)(1 + log k)1+ε , there exists a j prime ideal of K that divides An − ζk B n , but does not divide Am − ζk B m for m < n and arbitrary j . The above theorem implies the finiteness of the number of solutions of generalized cyclotomic equations considered by Browkin ([1], p. 236). The proof will follow closely the proof given in [3]. Let Q(A/B) = K0 , A/B = α/β, where α, β ∈ K0 , α, β are integers, and (α, β) = d. Let S and S0 be the set of all isomorphic injections of K0 (ζk ) and K0 , respectively, in the complex field, and set w(α/β) = log max{|α σ |, |β σ |} − log N d, σ ∈S0

where N denotes the absolute norm in K0 . Here, w(α/β) is the logarithm of the Mahler measure of α/β and so it is independent of the choice of α, β in K0 . Lemma 1. If |α| = |β|, but α/β is not a root of unity, then log |α n − ζk β n | = n log |β| + O(d + w(α/β)) log kn, where the constant in the O-symbol depends only on d and is effectively computable.

I6. An extension of the theorem on primitive divisors

1099

Lemma 2. If |α| = |β|, then log |α n − ζk β n | = n log max{|α|, |β|} + O(d 2 + dw(α/β)), where the constant in the O-symbol is absolute and effectively computable. The next lemma is just quoted from [3], where it occurs as Lemma 4. Lemma 3. Let φn (x, y) be the n-th cyclotomic polynomial in homogeneous form. If P is a prime ideal of K, n > 2(2d − 1), P | φn (A, B), and P is not a primitive divisor of An − B n , then ordP φn (A, B) ordP n. Finally, we prove Lemma 4. Let

ψn (x, y; ζk ) =

j

(x − ζkn y).

(j,n)=1 j ≡1 mod k

We have

ψn (x, y; ζk ) =

(1)

(x n/m − ζkm y n/m )μ(m) ,

m|n (m,k)=1

where mm ≡ 1 mod k and deg ψn = ϕ(n)

(k, n) . ϕ((k, n))

Proof. The right hand side of (1) can be written as

n/m−1

i m (x − ζn/m ζkn/m y)μ(m) .

m|n i=0 (m,k)=1 j

A factor x − ζkn y occurs in this product with the exponent

E=

m|n (m,k)=1

Now, n/m−1

1=

i=0 m(ki+m)≡j mod kn

n/m−1

μ(m)

1.

i=0 m(ki+m)≡j mod kn

⎧ ⎪ ⎪ ⎪ ⎨

n/m−1

1

if m | j,

i=0 ⎪ ⎪ki+m≡j/m mod kn/m

⎪ ⎩ 0

otherwise,

and if m | j , n/m−1 i=0 ki+m≡j/m mod kn/m

1=

1 0

if j ≡ 1 mod k, otherwise.

1100


Hence, E= and finally

⎧ ⎨ ⎩0

E=

μ(m)

if j ≡ 1 mod k,

m|n, m|j

1 0

otherwise,

if (n, j ) = 1, j ≡ 1 mod k, otherwise,

which proves the first part of the lemma. In order to prove the second part, we notice that there are exactly ϕ(n) integers j kn such that (n, j ) = 1, j ≡ 1 mod k.

(k, n) positive ϕ((k, n))

Lemma 5. For every ε > 0 there exists c(d, ε) such that, if n > c(d, ε)(1 + log k)1+ε , then we have

NK/Q ψn (A, B; ζk ) > (nk)[K:Q] .

(2) Proof. By Lemma 4,

ψn (A, B; ζk ) =

c

B ϕ(n)(k,n)/ϕ((k,n)) β

ψ(α, β; ζk ),

and since (B/β) = d−1 , we have

ψn (A, B; ζk ) = d−ϕ(n)(k,n)/ϕ((k,n)) ψn (α, β; ζk ), 1 log |NK/Q ψn (A, B; ζk )| [K : K0 (ζk )] (k, n) = log |NK0 (ζk )/Q ψn (α, β; ζk )| − [K0 (ζk ) : K0 ]ϕ(n) log N d ϕ((k, n))

σ n/m

μ(m) log (α ) − ζ m (β σ )n/m

= k

σ ∈S

m|n (m,k)=1

(k, n) − [K0 (ζk ) : K0 ]ϕ(n) log N d ϕ((k, n)) α n = log max{|α σ |, |β σ |} + O d + w log kn μ(m) m β σ ∈S

m|n (m,k)=1

(k, n) − [K0 (ζk ) : K0 ]ϕ(n) log N d ϕ((k, n)) α α (k, n) = [K0 (ζk ) : K0 ] ϕ(n) w +O d +w 2ν(n) log kn , ϕ((k, n)) β β

I6. An extension of the theorem on primitive divisors

1101

where the constant in O depends only on d and is effectively computable. Now, by Dobrowolski’s theorem [2], if α/β is an integer, then α α σ log log ed 3 log log ed 3

w c2 = log max σ , 1 log 1 + c1 , β β log d log d σ ∈S0

where c1 and c2 are absolute constants. If α/β is not an integer, then (β) = d and α w log Nβ − log N d log 2. β Thus, in both cases, α log log ed 3 w , c2 β log d provided c2 log 2. Since for every ε > 0 ϕ(n) > c3 (ε)n1−ε , 2ν(n) it follows that for n > c(d, ε)(1 + log k)1+ε

log NK/Q ψn (A, B; ζk ) > [K : Q] log nk,

which proves the lemma. c

Proof of the Theorem. By Lemma 5, for n > c(d, ε)(1 + log k)1+ε we have (2), and thus ψn (A, B; ζk ) has a prime ideal factor P in K such that ordP ψn (A, B; ζk ) > ordP kn. But P | ψn (A, B; ζk ) | φkn (A, B), hence by Lemma 3 we have that P is a primitive prime j divisor of Akn − B kn and thus does not divide Am − ζk B m for m < n and arbitrary j . On the other hand, P | ψn (A, B; ζk ) | An − ζk B n ,

thus P has the desired property.

References [1] J. Browkin, K-theory, cyclotomic equations, and Clausen’s function. In: Structural Properties of Polylogarithms, Math. Surveys Monogr. 37, Amer. Math. Soc., Providence 1991, Chapter 11, 233–273. [2] E. Dobrowolski, On a question of Lehmer and the number of irreducible factors of a polynomial. Acta Arith. 34 (1979), 391–401. [3] A. Schinzel, Primitive divisors of the expression An − B n in algebraic number fields. J. Reine Angew. Math. 268/269 (1974), 27–33; this collection: I5, 1090–1097.

Part J Prime numbers


Commentary on J: Prime numbers by Jerzy Kaczorowski

Papers J1, J2, J3 and J5 concern one of the most challenging open problems in the prime number theory. Hypothesis H is the common generalization of a conjecture of Dickson dated 1904 (the case of linear polynomials) and a conjecture of Buniakowski dated 1857 (the case of a single polynomial). Hypothesis H is sometimes formulated as a conjecture about prime values taken by polynomials in many variables (compare [43], Chapter 6). Lemma 4 in J5 shows that polynomials in one variable already represent the general case. The only instance where Hypothesis H is proved is still the classical Dirichlet theorem on primes in arithmetic progressions, which corresponds to the case of a single linear polynomial. Close approximations to it can be achieved by sieve methods. As early as 1937, G. Ricci [65] proved that if polynomials f1 (x), . . . , fs (x) satisfy conditions of the Hypothesis H, there exists a positive integer k such that all numbers f1 (m), . . . , fs (m) are Pk -almost primes (i.e. products of at most k primes) for infinitely many positive integers m. Value of k depends on degrees of the polynomials involved and can be made explicit. For instance H.-E. Richert [66] proved that an irreducible polynomial of degree d 1 with integer coefficients, positive leading term and without fixed divisor attains infinitely many values which are Pd+1 -almost primes (see also A. A. Bukhshtab [5]). The reader is referred to the classical treatise by H. Halberstam and H.-E. Richert [32] for a variety of similar results. For special polynomials dependence on d can be improved. For instance H. Iwaniec [44] proved that there are infinitely many positive integers m such that m2 +1 is a P2 -almost prime. Adopting terminology from G. H. Hardy and J. E. Littlewood [33], conjecture formulated in J3 is conjugated to Hypothesis H. There is no single instance where it is verified unconditionally except the “trivial” case k = 0, deg(g) = 1, when it reduces to the Dirichlet prime number theorem. The closest approximations concern binary Goldbach problem. H. L. Montgomery and R. C. Vaughan [55] proved that the number of even integers not exceeding x which are not sums of two primes is O(x θ ) for certain θ < 1 (θ = 0.914 is admissible (see [50]), J. Pintz [62] announced θ = 2/3). Generalized Riemann Hypothesis implies θ = 1/2 + ε for every ε > 0 as shown by G. Hardy and J. E. Littlewood [34]. The reader is referred to [62] for a very detailed survey on research done in connection with the Goldbach problem. Another result which should be mentioned here as a close approximation to a special case of the Hypothesis H, is the famous theorem of J. R. Chen [6] saying that every

1106

J. Prime numbers

sufficiently large even integer is a sum of a prime and a product of at most two primes (i.e. a P2 -number). Sierpiński’s conjecture that for every k > 1 there exists a positive integer m for which the equation ϕ(x) = m has exactly k solutions was proved for even k by K. Ford and S. Konyagin [19] and by K. Ford [18] in the full generality. (Theorem C14 from J2 shows that Sierpiński’s conjecture follows from Hypothesis H.) A trivial consequence of Hypothesis H is that there are arbitrary long arithmetic progressions formed by primes. This was proved unconditionally by B. Green and T. Tao [30]. Hypothesis H implies that for every positive integer k, there exist infinitely many pairs of primes (p, p ), such that p = p + 2k. In particular, denoting by pn the n-th prime, we have pn+1 − pn lim inf = 0. n→∞ log n This was proved recently by D. A. Goldston, J. Pintz and C.Y.Yıldırım [26]. The method of the proof seems to give more. It is observed in [27] that a suitably extended version of the Bombieri–Vinogradov prime number theorem would imply that the difference pn+1 − pn is bounded for infinitely many n’s. Artin’s conjecture on primitive roots, another consequence of Hypothesis H, is still unproved at present (2005). Nevertheless, C. Hooley [41] proved that it follows from the Extended Riemann hypothesis (for the Dedekind zeta functions). For a weaker sufficient condition see [25]. Riemann Hypothesis for function fields is true, as proved by A. Weil [74], consequently the function field analog of the Artin conjecture is true as well. Using an idea of Gupta and Ram Murty [31], D. R. Heath-Brown [36] proved that for nonzero integers q, r and s which are multiplicatively independent and such that none of q, r, s, −3qr, −3qs, −3rs, qrs is a square, the Artin conjecture holds for at least one of them. See also [56]. For Artin’s conjecture in algebraic number fields see W. Narkiewicz [58]. It is proved in J1 that Hypothesis H implies that every positive rational number can be written in the form p+1 (1) , q +1 where p and q are primes (Theorem C2.1 ). C. Badea [1] proved that for every positive rational r there exists a number K = K(r) such that a+1 , b+1 for infinitely many a and b that are PK -almost primes. Moreover, every sufficiently large integer n can be written in the form (1) with q prime and p a P3 -number, p n357/200 . It is also known that the set of positive integers of the form (1) with p and q prime has a positive upper density, see P. D. T. A. Elliott [15], [14]. P. T. Bateman and R. A. Horn’s heuristic arguments for the asymptotic formula in Hypothesis H are based on probabilistic arguments ([4]). Expected main term of the asymptotic formula agrees with what follows from the circle method disregarding estimates of the minor arcs, see the great classic by G. H. Hardy and J. E. Littlewood Partitio Numerorum, III [33]. Upper estimates of the proper size can be obtained by sieve methods, r=

Prime numbers

1107

see mentioned above book by H. Halberstam and H.-E. Richert. Sieve techniques provide lower estimates of the proper size for almost prime values of polynomials. Limitations to the equi-distribution of prime values of polynomials was discovered by J. Friedlander and A. Granville [21] by applying a method of H. Maier [52]. Suppose Hypothesis H is true for a polynomial f of degree d > 1. Then the sequence of the prime values of f is sparse in the sense that the number of f (n) x, f (n) prime, is x θ with θ < 1 (θ = 1/d works). Primes of the form x 4 + y 2 provide an example of a sparse sequence of primes with θ = 3/4, as proved by H. Iwaniec and J. Friedlander [22] and [23]. Similarly, primes of the form x 3 + 2y 3 provide an example of a sparse set of primes with θ = 2/3, see D. R. Heath-Brown [37]. For an arbitrary binary cubic forms see D. R. Heath-Brown, B. Z. Moroz [38]. One can also judge how far existing methods are apart from what is needed for a proof of the Hypothesis H by considering Pyatetski-Shapiro prime number theorem. Let πc (x) denote the number of primes p x which are of the form [nc ]. Pyatetski-Shapiro [64] proved that πc (x) ∼

x 1/c log x

as x → ∞ for 1 c < 12/11. Observe that this provides a sparse sequence of primes with exponent θ = 1/c, so one would expect that a method capable for treating the simplest non-linear case of the Hypothesis H, that means the case of the polynomial x 2 +1, should give the range 1 c 2 in Pyatetski-Shapiro’s prime number theorem. Initial result from [64] was subject of a sequence of improvements by G. A. Kolesnik [46], [47], D. R. Heath-Brown [35] and H. Liu and J. Rivat [51], who was able to get 1 c < 15/13. See also [11], [10], [49] and [67]. About the G. H. Hardy and J. E. Littlewood conjecture implicitly formulated in [33] (compare C12.2 in J1) that π(x + y) π(x) + π(y) for x, y 2: P. Dusart [13] proved that it holds for 2 x y 75 x log x log2 x. In general, Hardy–Littlewood’s conjecture is not compatible with a special form of the Hypothesis H (k-tuples conjecture): inequality π(x + y) aπ(x/a) + π(y) is not valid for 1 a 2, see [39] and [7]. For other results see [72], [73], [28], [60], [61], [24] and [54]. The classical theorem of G. Rabinowitsch links prime values √ of the polynomial x 2 + x + A and divisibility theory in the quadratic number field Q( 1 − 4A). For prime values of quadratic polynomials see [29]. Let q(n) denote the greatest prime divisor of an integer n. The problem of estimating q(f (x)) goes back to C. L. Siegel [69], who proved that q(f (x)) → ∞ as x → ∞ for every irreducible f ∈ Z[x] of degree d > 1. In Section 5 of J4 the problem of estimating q(f (x)), where f ∈ Z[x] is a fixed quadratic or cubic polynomial is addressed. As a consequence of results on solvability of certain Diophantine equations, it is proved that if deg(f ) = 2 and f is not a square of a linear polynomial or deg(f ) = 3 and f is a binomial, then (2)

lim inf x→∞

q(f (x)) c(f ), log log x

1108

J. Prime numbers

where c(f ) > 0 is an effective constant which depends on f (see also [45]). The same holds for arbitrary irreducible polynomial f ∈ Z[x] of degree d > 2 as shown by S. V. Kotov [48]. Theorems 13, 14, 15 from Section 5 of J4 concern Ω-estimates for 1/q(f (x)). It is proved for instance (Theorem 15) that for any polynomial f ∈ Z[x] of degree > 1 one has lim inf x→∞

log q(f (x)) c1 (f ), log |f (x)|

with an explicit constant c1 (f ). In cases when f is a binomial, the estimate can be improved (Theorems 13 and 14 in J4). For smooth values of polynomials with integer coefficients see N. M. Timofeev [71], A. Hildebrand [40], A. Balog, I. Ruzsa [2], C. Dartyge [8], G. Martin [53] and C. Dartyge, G. Martin, G. Tenenbaum [9]. Distribution of smooth shifted primes is considered in É. Fouvry, G. Tenenbaum [20], C. Pomerance, I. E. Shparlinski [63] and W. D. Bauka, A. Harcharras and I. E. Shparlinski [3]. There is a great difference between estimates for q(f (x)) holding for all sufficiently large x and Ω-estimates for this quantity, the latter are much sharper. For example, as a consequence of Richert’s theorem [66] for every irreducible polynomial f ∈ Z[x] of degree d 1 with positive leading term and no fixed prime divisors, we have q(f (x)) = Ω(x d/(d+1) ). If d = 2 the exponent can be improved from 2/3 to 1, as easily follows from Iwaniec theorem from [44]. Hypothesis H gives exponent d for every f satisfying our conditions. Theorems of type (2) are probably very hard to improve. The famous abc-conjecture by J. Oesterlé and D. Masser (see [59]) predicts that for every ε > 0 there is a constant c(ε) > 0 such that for every non-zero relatively prime integers a, b, c satisfying a + b = c one has 1+ε p . max(|a|, |b|, |c|) c(ε) p|abc

Let f (x) = ad x d + ad−1 x d−1 + . . . + a0 , ad > 0, be a polynomial from Z[x] of degree d 2 which is not of the form b d a x− da for certain a, b ∈ Z, a > 0. We have (3)

d d add−1 f (x) = (dad x + ad−1 )d + h(x),

where h ∈ Z[x] is a non-zero polynomial of degree d − 2. Taking sufficiently large integer x, writing

D(x) = g.c.d. d d add−1 f (x), (dad x + ad−1 )d , |h(x)|

1109

Prime numbers

and applying abc-conjecture we obtain from (3) f (x) (da x + a )d |h(x)| xd d d−1 max , , D(x) D(x) D(x) D(x) 1+ε

p , exp (1 + ε)ϑ q(f (x))

(da x+a )h(x) d d−1

p D 2 (x)

where ϑ(ξ ) =

log p

pξ

is the familiar theta function from the theory of prime numbers. By the Prime Number Theorem we have ϑ(ξ ) ∼ ξ as ξ → ∞. Therefore the last expression is d−1 1+ε x e(1+ε)q(f (x)) D 2 (x) and consequently lim inf x→∞

q(f (x)) 1. log x

Hence, even making so strong assumption as the abc-conjecture, one reduces just one log in (2). Sharper estimates can be obtained for the greatest prime divisor of the product f (n). nx

Let us denote this quantity by P (x, f ). It was first considered in case of f0 (x) = x 2 + 1 by P. Chebyshev, who proved that P (x, f0 )/x → ∞ as x → ∞. This was improved and generalized by many authors, compare [57], [16], [42], [12], [17], [70]. The sharpest result for general polynomial was achieved by G. Tenenbaum [70]:

P (x, f ) > x exp (log x)α for every α < 2 −log 4 = 0.61370 . . . . For f (x) = f0 (x) = x 2 +1 the best result belongs to J. -M. Deshouillers and H. Iwaniec: P (x, f0 ) > x 1.202 . See also [68] and [42].

References [1] C. Badea, Note on a conjecture of P. D. T. A. Elliott. Arch. Math. (Brno) 23 (1987), 89–94. [2] A. Balog, I. Z. Ruzsa, On an additive property of stable sets. In: Sieve Methods, Exponential Sums, and their Applications in Number Theory (Cardiff, 1995), London Math. Soc. Lecture Note Ser. 237, Cambridge Univ. Press, Cambridge 1997, 55–63.

1110

J. Prime numbers

[3] W. D. Banks, A. Harcharras, I. E. Shparlinski, Smooth values of shifted primes in arithmetic progressions. Michigan Math. J. 52 (2004), 603–618. [4] P. T. Bateman, R. A. Horn, A heuristic asymptotic formula concerning the distribution of prime numbers. Math. Comp. 16 (1962), 363–367. [5] A. A. Bukhshtab, Combinatorial strengthening of the sieve of Eratosthenes method. Uspekhi Mat. Nauk 22 (1967), no. 3 (135), 199–226 (Russian); English transl.: Russian Math. Surveys 22 (1967), 205–233. [6] J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes. Sci. Sinica 16 (1973), 157–176. [7] D. A. Clark, N. C. Jarvis, Dense admissible sequences. Math. Comp. 70 (2001), 1713–1718. [8] C. Dartyge, Entiers de la forme n2 + 1 sans grand facteur premier. Acta Math. Hungar. 72 (1996), 1–34. [9] C. Dartyge, G. Martin, G. Tenenbaum, Polynomial values free of large prime factors. Period. Math. Hungar. 43 (2001), 111–119. [10] J.-M. Deshouillers, Répartition de nombre premiers de la forme [nc ]. In: Journées Arithmétiques (Grenoble 1973), Bull. Soc. Math. France Mém. 37, Paris, 1974, 49–52. [11] −−, Nombres premiers de la forme [nc ]. C. R. Acad. Sci. Paris Sér. A-B 282 (1976), A131–A133. [12] J.-M. Deshouillers, H. Iwaniec, On the greatest prime factor of n2 + 1. Ann. Inst. Fourier (Grenoble) 32:4 (1982), 1–11. [13] P. Dusart, Sur la conjecture π(x + y) π(x) + π(y). Acta Arith. 102 (2002), 295–308. [14] P. D. T. A. Elliott, A conjecture of Kátai. Acta Arith. 26 (1974/5), 11–20. [15] −−, Arithmetic Functions and Integer Products. Springer, New York 1985. [16] P. Erd˝os, On the greatest prime factor of xk=1 f (k). J. London Math. Soc. 27 (1952), 379–384. [17] P. Erd˝os, A. Schinzel, On the greatest prime factor of xk=1 f (k). Acta Arith. 55 (1990), 191–200. [18] K. Ford, The number of solutions of φ(x) = m. Ann. of Math. (2) 150 (1999), 283–311. [19] K. Ford, S. Konyagin, On two conjectures of Sierpiński concerning the arithmetic functions σ and φ. In: Number Theory in Progress, Vol. 2 (Zakopane–Ko´scielisko, 1997), de Gruyter, Berlin 1999, 795–803. [20] É. Fouvry, G. Tenenbaum, Entiers sans grand facteur premier en progressions arithmetiques. Proc. London Math. Soc. (3) 63 (1991), 449–494. [21] J. Friedlander, A. Granville, Limitations to the equi-distribution of primes IV. Proc. Roy. Soc. London Ser. A 435 (1991), no. 1893, 197–204. [22] J. Friedlander, H. Iwaniec, The polynomial X2 + Y 4 captures its primes. Ann. of Math. (2) 148 (1998), 945–1040. [23] −−, −−, Asymptotic sieve for primes. Ann. of Math. (2) 148 (1998), 1041–1065. [24] R. Garunkštis, On some inequalities concerning π(x). Experiment. Math. 11 (2002), 297–301. [25] L. J. Goldstein, On the generalized density hypothesis I. In: Analytic Number Theory (Philadelphia, 1980), Lecture Notes in Math. 899, Springer, Berlin 1981, 107–128. [26] D. A. Goldston, J. Pintz, C. Y. Yıldırım, Small gaps between primes II. Preprint 2005. [27] D. A. Goldston, Y. Motohashi, J. Pintz, C. Y. Yıldırım, Small gaps between primes exist. Proc. Japan Acad. Ser. A Math. Sci. 82 (2006), 61–65.

Prime numbers

1111

[28] D. M. Gordon, G. Rodemich, Dense admissible sets. In:Algorithmic Number Theory (Portland, 1998), Lecture Notes in Comput. Sci. 1423, Springer, Berlin 1998, 216–225. [29] A. Granville, R. A. Mollin, Rabinowitsch revisited. Acta Arith. 96 (2000), 139–153. [30] B. Green, T. Tao, The primes contain arbitrarily long arithmetic progressions. Ann. of Math. (2), to appear. [31] R. Gupta, M. Ram Murty, A remark on Artin’s conjecture. Invent. Math. 78 (1984), 127–130. [32] H. Halberstam, H.-E. Richert, Sieve Methods. Academic Press, London–New York 1974. [33] G. H. Hardy, J. E. Littlewood, Some problems of ‘Partitio numerorum’ III. On the expression of a number as a sum of primes. Acta Math. 44 (1923), 1–70. [34] −−, −−, Some problems in ‘Partitio numerorum’ V. A further contribution to the study of Goldbach’s problem. Proc. London Math. Soc. (2) 22 (1924), 46–56. [35] D. R. Heath-Brown, The Pjateckiˇı–Šapiro prime number theorem. J. Number Theory 16 (1983), 242–266. [36] −−, Artin’s conjecture for primitive roots. Quart. J. Math. Oxford Ser. (2) 37 (1986), 27–38. [37] −−, Primes represented by x 3 + 2y 3 . Acta Math. 186 (2001), 1–84. [38] D. R. Heath-Brown, B. Z. Moroz, Primes represented by binary cubic forms. Proc. London Math. Soc. (3) 84 (2002), 257–288. [39] D. Hensley, I. Richards, Primes in intervals. Acta Arith. 25 (1973/74), 375–391. [40] A. Hildebrand, On integer sets containing strings of consecutive integers. Mathematika 36 (1989), 60–70. [41] C. Hooley, On Artin’s conjecture. J. Reine Angew. Math. 225 (1967), 209–220. [42] −−, On the greatest prime factor of a quadratic polynomial. Acta Math. 117 (1967), 281–299. [43] M. N. Huxley, The Distribution of Prime Numbers. Large Sieves and Zero-density Theorems. Clarendon Press, Oxford 1972. [44] H. Iwaniec, Almost primes represented by quadratic polynomials. Invent. Math. 47 (1978), 171–188. [45] M. Keates, On the greatest prime factor of a polynomial. Proc. Edinburgh Math. Soc. (2) 16 (1969), 301–303. [46] G.A. Kolesnik, The distribution of primes in sequences of the form [nc ]. Mat. Zametki 2 (1967), 117–128 (Russian). [47] −−, Primes of the form [nc ]. Pacific J. Math. 118 (1985), 437–447. [48] S. V. Kotov, The greatest prime factor of a polynomial. Mat. Zametki 13 (1973), 515–522 (Russian); English transl.: Math. Notes 13 (1973), 313–317. [49] D. Leitmann, D. Wolke, Primzahlen der Gestalt [f (n)]. Math. Z. 145 (1975), 81–92. [50] H. Z. Li, The exceptional set of Goldbach numbers II. Acta Arith. 92 (2000), 71–88. [51] H. Q. Liu, J. Rivat, On the Pjateckiˇı-Šapiro prime number theorem. Bull. London Math. Soc. 24 (1992), 143–147. [52] H. Maier, Primes in short intervals. Michigan Math. J. 32 (1985), 221–225. [53] G. Martin, An asymptotic formula for the number of smooth values of a polynomial. J. Number Theory 93 (2002), 108–182. [54] G. Mincu, A few inequalities involving π(x). An. Univ. Bucure¸sti Mat. 52 (2003), 55–64.

1112

J. Prime numbers

[55] H. L. Montgomery, R. C. Vaughan, The exceptional set in Goldbach’s problem. Acta Arith. 27 (1975), 353–370. [56] M. Ram Murty, Artin’s conjecture and elliptic analogues. In: Sieve Methods, Exponential Sums, and their Applications in Number Theory (Cardiff, 1995), London Math. Soc. Lecture Note Ser. 237, Cambridge Univ. Press, Cambridge 1997, 325–344. [57] T. Nagell, Généralisation d’un théorème de Tchébycheff . J. Math. Pure Appl. (8) 4 (1921), 343–356. [58] W. Narkiewicz, A note on Artin’s conjecture in algebraic number fields. J. Reine Angew. Math. 381 (1987), 110–115. [59] J. Oesterlé, Nouvelles approches du “théorème” de Fermat. Séminaire Bourbaki, Vol. 1987/88. Astérisque 161–162 (1988), Exp. 694, 165–186. [60] L. Panaitopol, Inequalities concerning the function π(x): applications. Acta Arith. 94 (2000), 373–381. [61] −−, Checking the Hardy–Littlewood conjecture in special cases. Rev. Roumaine Math. Pures Appl. 46 (2001), 465–470. [62] J. Pintz, Recent results on the Goldbach conjecture. Preprint 2005. [63] C. Pomerance, I. E. Shparlinski, Smooth orders and cryptographic applications. In: Algorithmic Number Theory (Sydney, 2002), Lecture Notes in Comput. Sci. 2369, Springer, Berlin 2002, 338–348. [64] I. I. Pyatetskii-Shapiro, On the distribution of prime numbers of the form [f (n)]. Mat. Sbornik (N.S.) 33 (75) (1953), 559–566 (Russian). [65] G. Ricci, Su la congettura di Goldbach e la costante di Schnirelmann. Ann. Scuola Norm. Sup. Pisa (2) 6 (1937), 71–116. [66] H.-E. Richert, Selberg’s sieve with weights. Mathematika 16 (1969), 1–22. [67] G. J. Rieger, Über ein additives Problem mit Primzahlen. Arch. Math. (Basel) 21 (1970), 54–58. [68] T. N. Shorey, R. Tijdeman, On the greatest prime factor of polynomials at integer points. Compositio Math. 33 (1976), 187–195. [69] C. L. Siegel, The integer solutions of equation y 2 = ax n + bx n−1 + . . . + k. J. London Math. Soc. 1 (1926), 66–68. [70] G. Tenenbaum, Sur une question d’Erd˝os et Schinzel II. Inventiones Math. 99 (1990), 215–224. [71] N. M. Timofeev, Polynomials with small prime divisors. Tashkent. Gos. Univ. Nauchn. Trudy 548 Voprosy Mat. (1977), 87–91 (Russian). [72] V. St. ¸ Udrescu, Some remarks concerning the conjecture π(x + y) π(x) + π(y). Rev. Roumaine Math. Pures Appl. 20 (1975), 1201–1209 (insert). [73] T. Vehka, I. Richards, Explicit construction of an admissible set for the conjecture that sometimes π(x + y) > π(x) + π(y). Notices Amer. Math. Soc. 26 (1979), A-453. [74] A. Weil, On the Riemann hypothesis in functionfields. Proc. Nat. Acad. Sci. U.S.A. 27 (1941), 345–347.


Originally published in Acta Arithmetica IV (1958), 185–208

Sur certaines hypothèses concernant les nombres premiers with W. Sierpiński (Warszawa)

La répartition des nombres premiers parmi les nombres naturels n’est pas encore suffisamment étudiée : c’est pourquoi depuis les temps les plus anciens on a énoncé diverses hypothèses concernant les nombres premiers. Plusieurs de ces hypothèses se sont montrées fausses ; quelques unes d’elles ne sont pas encore mises en défaut, et il y en a qui sont vérifiées pour tous les nombres ne dépassant pas un nombre très grand. Une de plus anciennes hypothèses sur les nombres premiers, ayant au moins 25 siècles, était celle des Chinois : un nombre naturel n > 1 est premier si et seulement si le nombre 2n − 2 est divisible par n (1 ). La nécessité de cette condition a été démontrée il y a quelques centaines d’années. En 1681 Leibniz a essayé de démontrer qu’elle est suffisante, mais sa démonstration était basée sur un raisonnement faux, et en 1819 on a trouvé que l’hypothèse des Chinois était fausse, puisque le nombre 2341 − 2 (qui a 103 chiffres) est divisible par 341, bien que le nombre 341 = 11 · 31 ne soit pas premier. Ensuite on a démontré (de nos temps) qu’il existe une infinité de nombres composés n pour lesquels le nombre 2n − 2 est divisible par n, impairs aussi bien que pairs. (Le plus petit de ces nombres pairs est le nombre n = 161038 = 2 · 73 · 1103 trouvé en 1950 par D. H. Lehmer). n P. Fermat supposait premiers tous les nombres Fn = 22 + 1, où n = 0, 1, 2, . . . . Cela est vrai pour n = 0, 1, 2, 3 et 4, mais, comme l’a trouvé L. Euler en 1772, le nombre F5 (qui a 10 chiffres) est composé, car il est divisible par 641. Maintenant nous connaissons 29 nombres Fn composés, pour n = 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 18, 23, 36, 38, 39, 55, 63, 73, 117, 125, 144, 150, 207, 226, 228, 268, 284, 316, 452. On peut donc énoncer l’hypothèse qu’il existe une infinité de nombres Fn composés. On a même énoncé l’hypothèse plus forte : les nombres Fn premiers sont en nombre fini. Ce sont peut-être seulement ceux que connaissait Fermat, à savoir les nombres Fn pour n 4. Le plus petit nombre Fn dont nous ne sachions pas s’il est premier ou non est F13 . Le plus grand nombre Fn composé connu est F452 dont le plus petit diviseur premier est le nombre 27 · 2455 + 1 (voir [14]). Erratum Acta Arith. 5 (1959), 259. (1 ) Ancient Chinese mathematicians never made this conjecture, see [12a], p. 54, footnote d.

1114

J. Prime numbers

Le fait que le nombre F16 est composé met en défaut l’hypothèse que tous les nombres de la suite infinie 2

2

22

2 + 1, 22 + 1, 22 + 1, 22 + 1, 22 + 1, 22

22

2

+ 1, . . .

sont premiers, puisque F16 est le cinquième terme de cette suite. Quant aux nombres de Mersenne Mn = 2n − 1 on a énoncé l’hypothèse que si le nombre Mn est premier, le nombre MMn est aussi premier. Or, d’après un calcul qui a été fait en 1953 par D. J. Wheeler, le nombre MM13 = 28191 − 1 (qui a 2466 chiffres) est composé, bien que le nombre M13 soit premier. On a encore énoncé l’hypothèse que les nombres qn (n = 0, 1, 2, . . . ), où q0 = 2 et qk+1 = 2qk − 1 pour k = 0, 1, 2, . . . , sont tous premiers. Cela est vrai pour 0 n 4. Or, le nombre q5 a plus de 1037 chiffres et nous ne savons pas s’il est premier ou non. En 1742 Ch. Goldbach a énoncé l’hypothèse que tout nombre pair > 4 est la somme de deux nombres premiers impairs. On peut énoncer l’hypothèse G un peu plus forte : tout nombre pair > 6 est la somme de deux nombres premiers distincts. On peut démontrer que l’hypothèse G équivaut à l’hypothèse que tout nombre naturel > 17 est la somme de trois nombres premiers distincts. Or, de l’hypothèse de Goldbach A. Schinzel a déduit que tout nombre impair > 17 est la somme de trois nombres premiers distincts. En 1937 J. Vinogradoff a démontré que tout nombre impair suffisamment grand est la somme de trois nombres premiers impairs. Quant à l’hypothèse G, S. Gołaszewski et B. Leszczyński l’ont vérifiée pour tous les nombres pairs 50000. On a aussi énoncé l’hypothèse que le nombre des décompositions d’un nombre pair 2n en une somme de deux nombres premiers tend vers l’infini avec n (cf. [10], Conjecture A). Il est probable que les nombres pairs > 188 ont plus de 10 décompositions et que les nombres pairs > 4574 donnent plus de 100 décompositions. Nous déduirons de l’hypothèse G quelques conséquences. P1 . Tout nombre impair est de la forme n − ϕ(n) où n est un nombre naturel. Démonstration de l’implication G → P1 . On a 1 = 2−ϕ(2), 3 = 9−ϕ(9), 5 = 25 − ϕ(25). Si m est un nombre impair > 5 on a m + 1 > 6 et de G résulte l’existence des nombres premiers distincts p et q tels que m + 1 = p + q et on a pq − ϕ(pq) = pq − (p − 1)(q − 1) = p + q − 1 = m, donc m = n − ϕ(n) pour n = pq. L’implication G → P1 se trouve ainsi démontrée.

P2 . Tout nombre impair m > 7 est de la forme σ (n) − n, où n est un nombre impair > m. Démonstration de l’implication G → P2 . Si m est un nombre impair > 7, il résulte de G qu’il existe des nombres premiers distincts p et q < p tels que m − 1 = p + q, et on a σ (pq) − pq = (p + 1)(q + 1) − pq = p + q + 1 = m. Comme m est impair > 7, les nombres p et q sont impairs, q 3, donc pq 3p = 2p + p > p + q + 1 = m et en posant n = pq on obtient un nombre impair n > m tel que m = σ (n) − n. L’implication G → P2 est ainsi démontrée.

P. Erd˝os a posé la question s’il existe une infinité de nombres naturels qui ne sont pas termes de la suite σ (n) − n. (Tels sont par exemple les nombres 2 et 5). Une question

J1. Sur certaines hypothèses concernant les nombres premiers

1115

analogue peut être posée pour la suite n − ϕ(n). (Les quatre nombres naturels les plus petits qui ne sont pas termes de cette suite sont 10, 26, 34 et 50). P2.1 . Il existe des suites aussi longues que l’on veut (1)

n, f (n), ff (n), fff (n), . . . ,

où f (n) = σ (n) − n,

dont le dernier terme est 1. Démonstration de l’implication P2 → P2.1 . D’après P2 pour tout nombre impair m > 7 il existe un nombre impair n = g(m) > m, tel que f (n) = m. Pour tout n impair > 7 la suite infinie de nombres naturels n, g(n), gg(n), . . . est donc croissante. k étant un nombre naturel, posons n = g k (11). Nous obtenons ainsi la suite c

n = g k (11), f (n) = g k−1 (11), . . . , f k (n) = 11, f k+1 (n) = 1 (puisque f (11) = σ (11)−11 = 1) qui a k +2 termes dont le dernier est = 1. L’implication P2 → P2.1 se trouve ainsi démontrée.

P2.2 . Il existe un infinité de nombres naturels n tels que la suite infinie (1) est périodique. Démonstration de l’implication P2 → P2.2 . Soit g(m) la fonction définie dans la démonstration de l’implication P2 → P2.1 et posons pour k naturels n = g k (25). Nous obtiendrons la suite n = g k (25), f (n) = g k−1 (25), . . . , f k (n) = 25, f k+i (n) = 6 pour i = 1, 2, . . . (puisque f (25) = 6 et f (6) = 6). La suite infinie (1) a donc ici k nombres impairs suivis d’une infinité de nombres 6. Il est à remarquer que L. E. Dickson a énoncé l’hypothèse que pour tout nombre naturel n > 1 la suite (1) ou bien se termine par le nombre 1 ou bien elle est périodique (Dickson [5]; cf. Catalan [3]). On voit sans peine que l’on peut exprimer cette hypothèse en disant que la suite (1) est toujours bornée. On ne sait pas s’il existe une infinité de nombres naturels n pour lesquels la suite (1) est périodique et la période est pure (comme par exemple pour n = 220, où la période est formée de deux termes ou pour n = 12496, où la période est formée de 5 termes). En 1950 G. Giuga a énoncé l’hypothèse que pour qu’un nombre naturel p > 1 soit premier, il faut et il suffit que le nombre 1p−1 + 2p−1 + . . . + (p − 1)p−1 + 1 soit divisible par p. (On démontre sans peine que cette condition est nécessaire). Il affirme que cette hypothèse est vraie pour tous les nombres < 101000 . Hypothèse de A. Schinzel. A. Schinzel a énoncé hypothèse H0 suivante :

c

H0 . s étant un nombre naturel et f1 (x), f2 (x), . . . , fs (x) des polynômes irréductibles en x à coefficients entiers, où le coefficient de la plus haute puissance de x est positif, et satisfaisant à la condition

1116 c

J. Prime numbers

Il n’existe aucun entier > 1 qui divise le produit f1 (x)f2 (x) · · · fs (x) quel que soit l’entier x, alors il existe au moins un nombre naturel x pour lequel les nombres f1 (x), f2 (x), . . . , . . . , fs (x) sont tous premiers. C.

On démontre sans peine que l’hypothèse H0 équivaut à l’hypothèse H suivante : H. s étant un nombre naturel et f1 (x), f2 (x), . . . , fs (x) des polynômes en x satisfaisant aux conditions de l’hypothèse H0 , il existe une infinité de hboxnombres naturels x pour lesquels les nombres f1 (x), f2 (x), . . . , fs (x) sont premiers. En effet, supposons que l’hypothèse H0 soit vraie et soient f1 (x), f2 (x), . . . , fs (x) des polynômes satisfaisant aux conditions de l’hypothèse H0 . On démontre sans peine que, quel que soit le nombre naturel k, les polynômes f1 (x + k), f2 (x + k), …, fs (x + k) satisfont aussi aux conditions de l’hypothèse H0 . D’après H0 il existe donc un nombre naturel x tel que les nombres f1 (x + k), f2 (x + k), …, fs (x + k) sont tous premiers et, comme on le prouve aisément, pour k suffisamment grand tous ces nombres premiers sont aussi grands que l’on veut. On a donc H0 → H et comme, d’autre part, on a évidemment H → H0 , l’equivalence H0 ≡ H se trouve démontrée. Quant à l’hypothèse H il est à remarquer que du théorème 1 du travail de G. Ricci [13] on déduit sans peine que si les polynômes f1 (x), f2 (x), . . . , fs (x) satisfont aux conditions de l’hypothèse H0 , il existe une constante C dépendant de f1 , f2 , . . . , fs telle que pour une infinité de nombres naturels x chacun des nombres f1 (x), f2 (x), . . . , fs (x) a au plus C diviseurs premiers. Nous déduirons maintenant de l’hypothèse H plusieurs conséquences. C1 . Si s est un nombre naturel, a1 < a2 . . . < as des entiers et si les binômes fi (x) = x+ai (i = 1, 2, . . . , s) satisfont à la condition C, il existe une infinité de nombres naturels x pour lesquels f1 (x), f2 (x), . . . , fs (x) sont des nombres premiers consécutifs. Démonstration de l’implication H → C1 . Nos binômes étant irréductibles et satisfaisant à la condition C, il résulte de H qu’il existe une infinité de nombres naturels x pour lesquels les nombres fi (x) (i = 1, 2, . . . , s) sont premiers. Soit h as − 2a1 + 2 un tel nombre naturel et posons c

(2)

b=

(h + as )! (h + a1 )! (h + a2 ) · · · (h + as )

et gi (x) = bx + h + ai

pour

i = 1, 2, . . . , s.

On a 2(h + ai ) = h + h + 2ai h + h + 2a1 h + as + 2 > h + as et, le nombre h + ai = fi (h) étant premier, les facteurs de (h + as )! autres que h + ai , étant < 2(h + ai ), ne sont pas divisibles par h + ai et il en résulte que (b, h + ai ) = 1. Supposons maintenant qu’il existe un nombre premier p tel que p | g1 (x)g2 (x) · · · gs (x) pour

x = 0, 1, 2, . . . , p − 1.


1117

On a donc p | g1 (0)g2 (0) · · · gs (0) = (h + a1 )(h + a2 ) · · · (h + as ) et tous ces facteurs étant premiers, il existe un nombre naturel k s tel que p = h + ak et d’après (2) et h+as < 2(h+ak ) = 2p on en conclut que p ne divise pas b. Il existe donc pour tout nombre naturel i s un seul nombre x de la suite 0, 1, 2, . . . , p −1, tel que p | bx +h+ai = gi (x) et il résulte tout de suite de p | g1 (x)g2 (x) · · · gs (x) pour x = 0, 1, 2, . . . , p −1 que p s, donc h + ak s, et comme, d’autre part, h + ak h + a1 as − a1 + 2 s + 1 (puisque les entries a1 , a2 , . . . , as vont en croissant) on aboutit à une contradiction. Les binômes irréductibles gi (x) (i = 1, 2, . . . , s) satisfont donc à la condition C et, d’aprés H, il existe une infinité de nombres naturels x tels que les nombres gi (x) (i = 1, 2, . . . , s) sont premiers. Si pour un tel x ces nombres premiers n’étaient pas consécutifs, il existerait un entier j tel que a1 j as et j = a1 , a2 , . . . , as tel que le nombre q = bx + h + j > h + j serait premier. Or, comme a1 j as c et j = a1 , a2 , . . . , as , on a, d’aprés (2), h + j | b, donc h + j | q > h + j , ce qui est impossible, puisque h + j > h + a1 qui est premier. L’implication H → C1 se trouve ainsi démontrée.

c

C1.1 . Tout nombre pair peut être représenté d’une infinité de manières comme la différence de deux nombres premiers consécutifs. Démonstration de l’implication C1 → C1.1 . Soit f1 (x) = x, f2 (x) = x + 2n (où n est un nombre naturel donné). Comme

f1 (1)f2 (1), f1 (2)f2 (2) = 2n + 1, 2(2 + 2n) = 1, il résulte de C1 qu’il existe une infinité de nombres naturels x tels que x et x + 2n sont deux nombres premiers consécutifs, oit x = pk , x + 2n = pk+1 (où pi désigne le i-ème nombre premier), d’oú 2n = pk+1 − pk . Cela prouve que C1 → C1.1 (cf. Hardy and Littlewood [10], Conjecture B).

C1.2 . m étant un nombre naturel donné, il existe 2m nombres premiers consécutifs formant m couples de nombres jumeaux. Démonstration de l’implication C1 → C1.2 . Soit f2i−1 (x) = x + (2m)! (i − 1), f2i (x) = x + (2m)! (i − 1) + 2

pour

i = 1, 2, . . . , m

et P (x) = f1 (x)f2 (x) · · · f2m (x). Soit p un nombre premier tel que p | P (x) pour x = 0, 1, . . . , p − 1. Comme P (x) est un polynôme en x de degré 2m où le coefficient de x 2m est = 1, d’après le théorème de Lagrange la congruence P (x) ≡ 0 (mod p) a au plus 2m racines. Or, comme P (x) ≡ 0 (mod p) pour x = 0, 1, . . . , p − 1, on en conclut que p 2m. Mais P (1) est évidemment un nombre impair et comme p | P (1), on trouve p > 2. D’autre part, d’après p 2m on a p | (2m)! i pour i entier et comme p | P (2), on trouve p | 23m , ce qui est impossible. Les binômes fj (x) (j = 1, 2, . . . , 2m) satisfont donc à la condition C et il résulte de C1

1118

J. Prime numbers

qu’il existe une infinité de nombres naturels x tels que fj (x) (j = 1, 2, . . . , 2m) sont des nombres premiers consécutifs, fj (x) = pk+j −1 pour j = 1, 2, . . . , 2m. On a donc pk+2i−1 − pk+2i−2 = 2 pour i = 1, 2, . . . , n et l’implication C1 → C1.2 se trouve démontrée.

On peut démontrer pareillement qu’il existe pour tout m naturel 4m + 1 nombres premiers consécutifs dont les 2m premiers et de même les 2m derniers donnent m couples de nombres jumeaux. V. Thébault a démontré [18] que si n > 1 termes d’une progression arithmétique de raison r sont des nombres premiers > n, alors r est divisible par tout nombre premier n. Or, nous démontrerons que C1 entraîne la conséquence suivante : C1.4 . Si r est un nombre naturel divisible par tout nombre premier n, où n est un nombre naturel donnée > 1, il existe une infinité de systèmes de n nombres premiers consécutifs formant une progression arithmétique de raison r. Démonstration de l’implication C1 → C1.4 . Soit fi (x) = x + ir pour i = 0, 1, 2, . . . , . . . , n − 1. S’il existait un nombre premier p tel que p | f0 (x)f1 (x) · · · fn−1 (x) pour x = 0, 1, 2, . . . , p − 1, il résulterait du théorème de Lagrange que p n, donc p | r. D’autre part on a p | f0 (1)f1 (1) · · · fn−1 (1) = 1(1 + r)(1 + 2r) · · · (1 + (n − 1)r) et vu que p | r on trouve p | 1, ce qui est impossible. La condition C est donc satisfaite et il résulte de C1 qu’il existe une infinité de nombres naturels x tels que les nombres fi (x) (i = 1, 2, . . . , n) sont des nombres premiers consécutifs. Nous avons ainsi démontré que C1 → C1.4 .

En particulier, pour n = 3, il résulte de C1.4 qu’il existe pour tout nombre naturel h une infinité de nombres naturels k tels que pk+1 − pk = pk+2 − pk+1 = 6h. Il en résulte qu’il existe une infinité de progressions arithmétiques formées de trois nombres premiers consécutifs. Or, d’après L. E. Dickson ([6], p. 425) Moritz Cantor a énoncé l’hypothèse ([2]) que trois nombres premiers consécutifs dont aucun n’est le nombre 3 ne peuvent pas former de progression arithmétique. En 1955 A. Schinzel a remarqué que cette hypothèse est en défaut puisque 47, 53 et 59 sont trois nombres premiers consécutifs formant une progression arithmétique de raison 6. Parmi les nombres < 1000 on trouve plusieurs telles progressions dont les premiers termes sont respectivement 151, 167, 367, 557, 587, 601, 647, 727, 941, 971. Les nombres 199, 211 et 223 et pareillement les nombres 1499, 1511 et 1523 forment des progressions arithmétiques de raison 12 composées de nombres premiers consécutifs et les nombres 251, 257, 263, 269

et

1741, 1747, 1753, 1759

forment des progressions arithmétiques de raison 6 composées chacune de quatre nombres premiers consécutifs. D’après C1.4 (pour n = 4) il existe une infinité de telles progressions. Nous déduirons maintenant de l’hypothèse H la conséquence suivante :


1119

C2 . a, b, c étant des nombres naturels tels que (a, b) = (a, c) = (b, c) = 1 et 2 | abc, l’équation ap − bq = c a une infinité de solutions en nombres premiers p et q. (Cette hypothèse a été énoncée par Hardy et Littlewood [10], p. 45, Conjecture D). Démonstration de l’implication H → C2 . a, b, c étant des nombres naturels tels que (a, b) = (a, c) = (b, c) = 1 et 2 | abc, il existe, on le sait, des nombres naturels r et s tels que ar − bs = c. Soit f1 (x) = bx + r, f2 (x) = ax + s, on a donc f1 (x)f2 (x) = abx 2 + (ar + bs)x + rs. S’il existait un nombre premier p tel que p | f1 (x)f2 (x) pour tout entier x, on aurait (pour x = 0) p | rs, donc (pour x = ±1) p | ab ± (ar + bs), d’où p | 2ab et p | 2(ar + bs). Si l’on avait p = 2, on aurait, d’après p | rs, 2 | r ou bien 2 | s. Si 2 | r, on ne peut avoir 2 | s, puisqu’alors il viendrait 2 | ar ± bs, donc 2 | ab et 2 | c, contrairement à (ab, c) = 1. Donc, si 2 | r, s est impair et de p | ab + (ar + bs) il résulte que 2 | (a + 1)b, donc ou bien a est impair ou bien b est pair. Si b était pair, alors, d’après ar − bs = c, c serait pair, contrairement à (b, c) = 1. Donc b est impair et a impair et aussi c = ar − bs impair, contrairement à 2 | abc. Donc r ne peut pas être pair ; s est donc pair et comme plus haut on démontre que cela implique une contradiction. On a donc p = 2, par conséquent p | ab et p | ar + bs et, comme p | rs, d’après p | ar 2 + brs on trouve p | ar 2 , d’où p | ar et, comme en vertu de p | ars + bs 2 on a p | bs 2 , d’où p | bs, il vient p | ar − bs = c, ce qui est impossible, puisque (ab, c) = 1. Les binômes f1 (x) et f2 (x) satisfont donc à la condition C et il existe une infinité de nombres naturels x tels que p = f1 (x) et q = f2 (x) sont des nombres premiers, donc bx + r = p et ax + s = q, ce qui donne ap − bq = ar − bs = c. L’implication H → C2 se trouve ainsi démontrée.

Voici maintenant une conséquence de C2 : C2.1 . Tout nombre rationnel positif peut être représenté d’une infinité de manières sous la forme (p + 1)/(q + 1) ainsi que sous la forme (p − 1)/(q − 1), où p et q sont des nombres premiers. Démonstration de l’implication C2 → C2.1 . Soit r un nombre rationnel > 1 ; on peut le représenter sous la forme r = b/a, où a et b sont des nombres naturels, b > a ; (a, b) = 1 et il en résulte que (a, b − a) = 1 et on a évidemment 2 | ab(b − a). D’après C2 il existe donc une infinité de systèmes de deux nombres premiers p et q tels que ap − bq = b − a, d’où b/a = (p + 1)/(q + 1). Si r était rationnel, 0 < r < 1, on aurait r = a/b où b > a et on trouverait a/b = (q +1)/(p +1). Pour la forme (p −1)/(q −1) la démonstration serait analogue, en partant de l’équation ap − bq = a − b pour a > b. Pour r = 1 la proposition C2.1 est évidente. En particulier, pour r = 2 il résulte de C2.1 qu’il existe une infinité de nombres premiers p pour lesquels le nombre 2p + 1, respectivement le nombre 2p − 1 est premier.

Si p et 2p + 1 sont premiers, on a ϕ(2p + 1) = 2p, donc de C2.1 résulte la proposition suivante : C2.1.1 . La suite 21 ϕ(n) (n = 1, 2, . . . ) contient une infinité de nombres premiers.

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Soit k un nombre naturel pair. D’après C2.1 il existe une infinité de nombres premiers p > k tels que 2p−1 est un nombre premier. k étant pair, on a k = 2l. Or, pour

tout l naturel on a ϕ(4l) = 2ϕ(2l), donc ϕ(4lp) = 2ϕ(l)ϕ(p) = 2(p − 1)ϕ(l) et ϕ 2l(2p − 1) = c ϕ(2l)ϕ(2p−1) = (2p−2)ϕ(2l) donc ϕ(4lp−2l) = ϕ(4lp) et l’équation ϕ(x +k) = ϕ(x) est remplie pour x = 4lp − 2l, k = 2l. On a ainsi la proposition suivante : C2.1.2 . L’équation ϕ(x + k) = ϕ(x), où k est un nombre naturel pair, a une infinité de solutions. Pour k impairs l’étude de cette équation est beaucoup plus compliquée : voir A. Schinzel [16]. Il résulte tout de suite de C2.1 qu’il existe pour tout nombre rationnel r > 0 une infinité de couples de nombres naturels x et y tels que σ (x)/σ (y) = r (on peut prendre pour x et y des nombres premiers). Une propriété analogue de la fonction ϕ peut aisément être démontrée sans faire appel à l’hypothèse H. En effet, si r = l/m, où l et m sont des nombres naturels et (l, m) = 1 et si k est un nombre naturel quelconque tel que (k, lm) = 1, on a ϕ(l 2 mk)/ϕ(lm2 k) = l/m = r. Or, il résulte tout de suite C2.1 que, pour tout nombre rationnel r > 0, l’équation ϕ(x)/ϕ(y) = r a une infinité de solutions en nombres premiers x et y. P. Erd˝os a démontré d’une façon élémentaire l’existence des suites infinies mk et nk (k = 1, 2, . . . ) de nombres naturels tels que mk /nk → +∞ et ϕ(mk ) = ϕ(nk ) pour k = 1, 2, . . . . Sa méthode n’est pas applicable à la fonction σ . Or, C2.1 entraîne le corollaire suivant : C2.1.3 . Quel que soit le nombre naturel k, il existe des nombres naturels m et n tels que σ (m) = σ (n) et m/n > k. Démonstration de l’implication C2.1 → C2.1.3 . Comme on sait, il existe pour tout nombre naturel k un nombre naturel l tel que σ (l)/ l > 2k (ce qui résulte par exemple de l’inégalité σ (n!) 1 1 1 + + ... + pour n = 1, 2, . . . n! 1 2 n et de la divergence de la série harmonique). Or, d’après C2.1 (pour r = σ (l)) il existe des nombres premiers p > l et q > l tels que σ (p) p+1 = = σ (l). σ (q) q +1 Posons m = p, n = lq. On aura donc σ (n) = σ (lq) = σ (l)σ (q) = σ (p) = σ (m) donc σ (m) = σ (n), et m p p σ (q) σ (l) p = = · · > · 2k > k. n lq σ (p) q l p+1

C3 . Si a, b et c sont des entiers, a > 0, (a, b, c) = 1 et les nombres a + b et c ne sont pas simultanément pairs, et b2 − 4ac n’est pas un carré, il existe une infinité de nombres premiers de la forme ax 2 + bx + c. (Cf. Hardy et Littlewood [10], p. 48, Conjecture F).


1121

Démonstration de l’implication H → C3 . Comme b2 − 4ac n’est pas un carré, le trinôme ax 2 + bx + c est irréductible. Il remplit aussi la condition C, puisque

f (0), f (1), f (2) = (c, a + b + c, 4a + 2b + c) = (c, a + b, 2a) = (c, a + b, a) = (c, b, a) = 1. c

L’implication H → C3 se trouve ainsi démontrée.

C3.1 . Si k est un entier et −k n’est pas un carré, il existe une infinité de nombres premiers de la forme x 2 + k. (Pour k = 1 cf. Hardy et Littlewood [10], p. 48, Conjecture E). Pour déduire C3.1 de C3 il suffit de poser, dans C3 , a = 1, b = 0, c = k.

c

C3.1.1 . Tout nombre naturel pair√est d’une infinité de manières somme de deux nombres premiers conjugués du corps Q( −1). Démonstration de l’implication C3.1 → C3.1.1 . Pour k naturel donné il existe, d’après C3.1 , une infinité de nombres premiers > 2 de la forme p = x 2 + k 2 ; ces nombres sont, on le voit sans peine, de la forme 4t + 1, et on a p √ = (k + xi)(k − xi) où k + xi et k − xi sont des nombres premiers conjugués du corps Q( −1), et 2k = (k + xi) + (k − xi). Quant aux nombres impairs, on peut démontrer que √ tout nombre naturel impair < 29 est la somme de deux nombres premiers du corps Q( −1), mais il existe une infinité de nombres impairs qui ne sont pas de telles sommes, par exemple tous les nombres 170k +29 et tous les nombres 130k + 33 où k = 0, 1, 2, . . . . Il est à remarquer que sans avoir recours à l’hypothèse H nous ne savons pas démontrer non seulement qu’il existe une infinité de nombres premiers de la forme x 2 + 1, où x est un nombre naturel, mais aussi qu’il existe une infinité de nombres premiers de la forme x 2 + y 2 + 1, où x et y sont des nombres naturels. Cependant on sait démontrer qu’il existe une infinité de nombres premiers de la forme x 2 + y 2 + z2 + 1, où x, y, z sont des nombres naturels : tels sont, par exemple, tous les nombres premiers de la forme 8k + 7.

C4 . L’équation ax 2 + bx + c = dy, où a, b, c, d sont des entiers, a > 0 et d > 0, a une infinité de solutions en nombres premiers x et y si et seulement si Δ = b2 − 4ac n’est pas un carré (d’un nombre entier) et si elle a au moins une solution en nombres entiers x0 , y0 tels que (x0 y0 , 6ad) = 1. Démonstration de l’implication H → C4 . Nous prouverons sans avoir recours à l’hypothèse H que la condition est nécessaire. Si l’équation ax 2 + bx + c = dy a une infinité de solutions en nombres premiers, il existe des nombres premiers x0 et y0 plus grands que 6ad et tels que ax02 + bx0 + c = dy0 et alors on a (x0 y0 , 6ad) = 1. Si Δ était un carré, soit b2 − 4ac = k 2 , où k est un entier 0, on aurait, comme on le vérifie aisément 4ady0 = (2ax0 + b + k)(2ax0 + b − k). Or, on déduit sans peine de cette égalité que pour x0 premiers suffisamment grands le nombre y0 ne peut pas être premier. La condition de C4 est donc nécessaire. Supposons maintenant que le nombre Δ ne soit pas un carré et que x0 et y0 soient des entiers tels que ax02 + bx0 + c = dy0 et

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(x0 y0 , 6ad) = 1. Posons f1 (x) = dx + x0 ,

f2 (x) = adx 2 + (2ax0 + b)x + y0 .

Les polynômes f1 et f2 sont irréductibles, puisque (2ax0 + b)2 − 4ady0 = (2ax0 + b)2 − 4a(ax02 + bx0 + c) = b2 − 4ac = Δ et, d’après l’hypothèse, Δ n’est pas un carré (d’un nombre rationnel). S’il existait un nombre premier p tel que p | f1 (x)f2 (x) pour x entiers, alors, en vertu du théorème de Lagrange, on aurait ou bien p 3 ou bien p | ad 2 , donc toujours p | 6ad 2 et p | f1 (0)f2 (0) = x0 y0 et, comme (x0 y0 , 6ad) = 1, d’où (x0 y0 , 6ad 2 ) = 1, on aurait p | 1, ce qui est impossible. Les polynômes f1 (x) et f2 (x) satisfont donc aux conditions de l’hypothèse H, par conséquent pour une infinité de nombres naturels x les nombres f1 (x) = p et f2 (x) = q sont premiers et on vérifie sans peine que ap 2 + bp + c = dq.

L’implication H → C4 est ainsi démontrée. C4.1 . Tout nombre rationnel r > 1 peut être représenté d’une infinité de manières sous la forme (p2 − 1)/(q − 1), où p et q sont des nombres premiers. Démonstration de l’implication C4 → C4.1 . Soit r un nombre rationnel > 1, donc r = d/a, où a et d sont des nombres naturels, d > a. Posons, dans C4 , b = 0, c = d − a. On aura donc b2 − 4ac = −4a(d − a) < 0, ce qui n’est pas un carré. Or, les nombres x0 = y0 = 1 sont tels que (x0 y0 , 6ad) = 1 et ax02 + (d − a) = dy0 . En vertu de C4 il existe donc une infinité de nombres premiers p et q tels que ap2 + (d − a) = dq, d’où

(p2 − 1)/(q − 1) = d/a = r, ce qui prouve que C4 → C4.1 . C4.1.1 . Il existe une infinité de triangles orthogonaux de côtés naturels dont deux sont des nombres premiers. Démonstration de l’implication C4.1 → C4.1.1 . Pour r = 2 il résulte de C4.1 que l’équation p2 = 2q − 1 a une infinité de solutions en nombres premiers. Or, cette équation équivaut évidemment à l’équation p 2 + (q − 1)2 = q 2 . On a donc C4.1 → C4.1.1 . Voici quelques triangles satisfaisant aux conditions de C4.1.1 : (3, 4, 5), c

(5, 12, 13),

(11, 60, 61),

(29,420,421),

(19, 180, 181),

(61,1860,1861).

Dans Scripta Mathematica 22 (1956), p. 158, Curiosum 435 (G. An interesting observation) on trouve l’observation qu’il existe un grand nombre de cas où pour p premier l’addition de l’unité au nombre triangulaire d’ordre p, respectivement la soustraction du nombre 2 donne un nombre premier, par exemple t3 + 1 = 7, t7 + 1 = 29, t5 − 2 = 13. Nous déduirons de l’hypothèse H les conséquences C4.2 et C4.3 suivantes : C4.2 . Il existe une infinité de nombres premiers p tels que 21 p(p + 1) + 1 est un nombre premier.


1123

Démonstration de l’implication C4 → C4.2 . Posons, dans C4 , a = b = 1, c = d = 2. Le nombre b2 − 4ac = −7 n’est pas un carré. L’équation x 2 + x + 2 = 2y admet la solution x0 = −1, y0 = 1, qui remplit la condition (x0 y0 , 6ad) = 1, et la proposition C4.2 résulte immédiatement de C4 .

C4.3 . Il existe une infinité de nombres premiers p tels que le nombre 21 p(p + 1) − 2 est premier. Démonstration de l’implication C4 → C4.3 . Posons, dans C4 , a = b = 1, c = −4, d = 2. Le nombre b2 −4ac = 17 n’est pas un carré. L’équation x 2 +x −4 = 2y admet la solution x0 = 1, y0 = −1, telle que (x0 y0 , 6ad) = 1, donc C4 entraîne immédiatement C4.3 .

C4.4 . La suite σ (n) (n = 1, 2, . . . ) contient une infinité de nombres premiers. Démonstration de l’implication C4 → C4.4 . Posons, dans C4 , a = b = c = d = 1. Le nombre b2 − 4ac = −3 n’est pas un carré. L’équation x 2 + x + 1 = y admet la solution x0 = −1, y0 = 1, où (x0 y0 , 6ad) = 1, et, comme pour p premiers on a σ (p2 ) = p2 +p+1, C4 entraîne la proposition C4.4 .

C5 . Tout nombre naturel peut être représenté d’une infinité de manières sous la forme σ (x) − σ (y) (où x et y sont des nombres naturels). Démonstration de l’implication H → C5 . Si n est pair, il existe, d’après C1.1 , une infinité de nombres premiers p et q tels que p−q = n, d’où σ (p)−σ (q) = (p+1)−(q +1) = n. Or, si n est impair, posons, dans C4 , a = b = d = 1, c = n. Le nombre b2 −4ac = 1−4n < 0 n’est pas un carré. Si 3 | n, alors, n étant impair, on a (n+2, 6) = 1 et pour x0 = 1, y0 = n+2 on a x02 +x0 +n = y0 et (x0 y0 , 6ad) = (n+2, 6) = 1. Si l’on n’a pas 3 | n, alors (n, 6) = 1 et pour x0 = −1, y0 = n on trouve x02 + x0 + n = y0 et (x0 y0 , 6ad) = (−n, 6) = 1. D’après C4 il existe donc une infinité de nombres premiers p et q tels que p 2 + p + n = q, d’où σ (q) − σ (p2 ) = q + 1 − (p 2 + p + 1) = q − p 2 − p = n. On a donc H → C5 .

Il est à remarquer que pour la fonction ϕ la proposition analogue à C5 est fausse, car on peut démontrer d’une façon élémentaire qu’aucun des nombres 2 · 7n − 1 (n = 1, 2, . . . ) n’est de la forme ϕ(x)−ϕ(y), mais, comme pour p et q premiers on a ϕ(p)−ϕ(q) = p−q, on déduit de C1.1 que tout nombre pair est de la forme ϕ(x) − ϕ(y). C6 . n étant un nombre impair > 1, k un entier donné quelconque qui n’est pas une puissance d’un entier à l’exposant d > 1 et d | n, il existe une infinité de nombres premiers de la forme x n + k, où x est un nombre naturel (pour n = 3 cf. Hardy et Littlewood [10], p. 50, Conjecture K). Si, en outre k est pair, il existe une infinité de nombres premiers p tels que pn + k est un nombre premier.

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J. Prime numbers

Démonstration de l’implication H → C6 . n étant un nombre impair et k n’étant pas n une puissance d’un entier d > 1 et d | n, le polynôme f1 (x) = x + k est

à l’exposant irréductible. Or, on a f1 (0), f1 (1) = (k, k + 1) = 1 et on déduit de H la première partie de C6 . Si k est pair, alors, en posant f2 (x) = x on a f1 (−1)f2 (−1), f1 (1)f2 (1) = (k − 1, k + 1) = 1, la condition C est encore remplie et H entraîne la deuxième partie de C6 .

Il est à remarquer que sans l’aide de l’hypothèse H nous ne savons démontrer même pas l’existence d’une infinité de nombres premiers de la forme x 3 + y 3 + z3 , où x, y et z sont des entiers. On sait cependant démontrer (sans l’aide de l’hypothèse H) l’existence d’une infinité de nombres premiers de la forme x 3 + y 3 + z3 + t 3 où x, y, z, t sont des entiers : tels sont, par exemple, tous les nombres de la forme 9k ± 1. C7 . Il existe une infinité de nombres naturels n tels que chacun des nombres n, n + 1, n + 2 est le produit de deux nombres premiers distincts. Démonstration de l’implication H → C7 . Soit f1 (x) = 10x + 1, f2 (x) = 15x + 2, f3 (x) = 6x + 1. On a ici a = f1 (0)f2 (0)f3 (0) = 2 et b = f1 (1)f2 (1)f3 (1) = 11 · 17 · 7, donc (a, b) = 1 et il résulte de H qu’il existe une infinité de nombres naturels x tels que les nombres p = 10x + 1, q = 15x + 2, r = 6x + 1 sont premiers. Pour n = 3p on trouve n + 1 = 3p + 1 = 2(15x + 2) = 2q, n + 2 = 2q + 1 = 30x + 5 = 5(6x + 1) = 5r et p 11 > 3, q 17 > 2, r 7 > 5, d’où il résulte que chacun des nombres n, n + 1, n + 2 est le produit de deux nombres premiers distincts. De C7 résulte tout de suite l’existence d’une infinité de nombres naturels n tels que les nombres n, n + 1 et n + 2 ont le même nombre de diviseurs.

Or, il n’existe pas quatre nombres naturels consécutifs dont chacun serait le produit de nombres premiers distincts, un de ces nombres étant toujours divisible par 4. C8 . Il existe pour tout nombre naturel s un nombre naturel ms tel que chacune des équations ϕ(x) = ms et σ (x) = ms a plus de s solutions. (Ce problème a été posé par P. Erd˝os). Démonstration de l’implication H → C8 . Posons fi (x) = 2i x + 1 et gi (x) = 2i x − 1 (i = 0, 1, . . . , 2s + 1). Comme f0 (0)f1 (0) · · · f2s+1 (0)g0 (0)g1 (0) · · · g2s+1 (0) = 1, les polynômes fi et gi (i = 0, 1, . . . , 2s + 1) satisfont à la condition C et d’après H, il existe un nombre naturel x tel que les nombres fi (x) et gi (x) pour i = 0, 1, . . . , 2s + 1 sont premiers. Posons ai = fi (x)f2s−i+1 (x),

bi = gi (x)g2s−i+1 (x),

(i = 0, 1, . . . , 2s + 1)

fi (x) et f2s−i+1 (x), respectivement gi (x) et g2s−i+1 (x) (pour i = 0, 1, . . . , 2s + 1) étant des nombres premiers distincts, on a

fi (x), f2s−i+1 (x) = 1 et gi (x), g2s−i+1 (x) = 1 pour i = 0, 1, . . . , 2s + 1, donc, pour i = 0, 1, . . . , 2s + 1 :

ϕ(ai ) = ϕ fi (x) ϕ f2s−i+1 (x) = 2i x22s−i+1 x = 22s+1 x 2 ,

σ (bi ) = σ gi (x) σ g2s−i+1 (x) = 2i x22s−i+1 x = 22s+1 x 2 .


1125

Les nombres ai (i = 0, 1, . . . , s) et de même les nombres bi (i = 0, 1, . . . , s) étant distincts, l’implication H → C8 se trouve démontrée.

Il est à remarquer qu’une proposition analogue pour la fonction ϕ a été démontrée sans avoir recours à l’hypothèse H par P. Erd˝os ([7], p. 213) et que, selon son avis, une modification de sa démonstration permettrait de démontrer une proposition analogue pour la fonction σ . Or, une démonstration tout à fait élémentaire pour la fonction ϕ a été donnée par A. Schinzel [15]. Sans avoir recours à l’hypothèse H nous ne savons par démontrer que l’équation ϕ(x) = σ (y) a une infinité de solutions en nombres naturels x et y. C9 . Il existe une infinité de nombres premiers p pour lesquels le nombre 2p − 1 est composé. Démonstration de l’implication H → C9 . Soit f1 (x) = 4x − 1, f2 (x) = 8x − 1. Il résulte de H qu’il existe une infinité de nombres naturels x pour lesquels les nombres p = 4x − 1 et q = 8x − 1 sont premiers. Mais alors on a q − 1 = 2p et, comme on sait, q | 2p − 1 et, si x > 1, on a 2p − 1 > q et le nombre 2p − 1 est composé. Il résulte donc l’hypothèse H qu’il existe une infinité de nombres de Mersenne Mp = 2p − 1 composés dont les indices p sont des nombres premiers.

Un nombre naturel composé n est dit absolument pseudo-premier si pour tout entier a on a n | a n − a. C10 . Il existe une infinité de nombres absolument pseudo-premiers. Démonstration de l’implication H → C10 . Soit f1 (x) = 6x + 1, f2 (x) = 12x + 1, f3 (x) = 18x + 1. Comme f1 (0)f2 (0)f3 (0) = 1, il résulte de H qu’il existe une infinité de c nombres naturels x tels que chacun des nombres p = 6x + 1, q = 12x + 1, r = 18x + 1 est premier et alors on le sait, le nombre pqr est absolument pseudo-premier (il est donc aussi un nombre de Carmichael) (voir [4], p. 271).

C11 (Hypothèse de E. Artin). Tout nombre entier g = −1 qui n’est pas un carré est racine primitive pour une infinité de nombres premiers. Démonstration de l’implication H → C11 . Soit g = a 2 b, où a est un nombre naturel, b un entier qui n’est divisible par aucun carré > 1. Comme g n’est pas un carré, on a b = 1. Soit b1 le plus grand diviseur impair de b. Nous prouverons d’abord qu’il existe des binômes f1 (x) et f2 (x) satisfaisant à la condition C et tels que 1◦ 2◦

quel que soit le nombre naturel x, b est un non-résidu quadratique pour f1 (x); f1 (x) − 1 = 2f2 (x) si b = 3 et f1 (x) − 1 = 4f2 (x) si b = 3.

Si b < 0, soit f1 (x) = −4bx − 1, f2 (x) = −2bx − 1. La condition 2◦ est évidemment remplie et, comme f1 (0)f2 (0) = 1, les binômes f1 (x) et f2 (x) satisfont à la condition C.

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Si b est pair, on a f1 (x) ≡ −1 (mod 8) et le symbole de Jacobi 2 =1 f1 (x) et

b f1 (x)

f1 (x) b1 =− = −(−1)(b1 −1)/2 f1 (x) b1 −1 = −(−1)(b1 −1)/2 = −1, b1

=

−b1 f1 (x)

ce qui prouve que b est un non-résidu quadratique pour f1 (x), c’est-à-dire que la condition 1◦ est remplie. Si b est impair, on a b = −b1 et on parvient au même résultat. Si b > 0 et b est pair, on a b = 2b1 . Soit f1 (x) = 4bx + 2b − 1, f2 (x) = 2bx + b − 1, 2 P (x) = f1 (x)f2 (x).

On a P (1) + P (−1) − 2P (0) = 16b , P (0) = (2b − 1)(b − 1) et, b étant pair, on a P (1) + P (−1) − 2P (0), P (0) = 1 et on en conclut que la condition C est remplie. La condition 2◦ est évidemment aussi remplie. Comme b = 2b1 = 2(2k + 1), on trouve 2 = −1 f1 (x) ≡ 3 (mod 8), d’où f1 (x) et

b f1 (x)

b1 b1 f1 (x) =− = −(−1)(b1 −1)/2 f1 (x) f1 (x) b1 −1 = −1, = −(−1)(b1 −1)/2 b1 =

2 f1 (x)

ce qui prouve que b est un non-résidu quadratique pour f1 (x) et la condition 1◦ est remplie. Soit maintenant b un nombre impair > 3, donc b = q1 q2 · · · qk , où qi sont des nombres premiers (i = 1, 2, . . . , k), q1 < q2 < . . . < qk et qk > 3. Le nombre premier qk a donc au moins deux non-résidus quadratiques et l’un d’eux est n0 ≡ −1 (mod qk ). Le système des deux congruences n ≡ −1 (mod 4q1 q2 · · · qk−1 ) et n ≡ −n0 (mod qk ) a évidemment une solution n = n1 . Soit f1 (x) = 4bx + n1 ,

f2 (x) = 2bx + 21 (n1 − 1),

P (x) = f1 (x)f2 (x).

On trouve sans peine P (0) =

1 2

n1 (n1 − 1),

P (1) + P (−1) − 2P (0) = 16b2 .

Or, comme n1 ≡ −1 (mod 4q1 q2 · · · qk−1 ), d’où 21 (n1 −1) ≡ −1 (mod 2q1 q2 · · · qk−1 ), et n1 ≡ 0 (mod qk ) (puisque n1 ≡ −n0 (mod qk ) et n0 est un non-résidu quadratique pour qk ) et 21 (n1 − 1) ≡ 0 (mod qk ) (puisque n1 − 1 ≡ −n0 − 1 ≡ 0 (mod qk )), on a (4b, n1 ) = 1

et 2b, 21 (n1 − 1) = 1, d’où 16b2 , 21 n1 (n1 − 1) = 1, donc P (0), P (1) + P (−1) − 2P (0) = 1, d’où il résulte que les binômes f1 (x) et f2 (x) satisfont à la condition C. Or, la condition 2◦ est évidemment remplie. Or, on a f1 (x) ≡ −1 (mod 4q1 q2 · · · qk−1 ) et


1127

f1 (x) ≡ n1 (mod qk ), d’où b −f1 (x) −n1 f1 (x) −n1 = (−1)(b−1)/2 = = f1 (x) b b q1 q2 · · · qk−1 qk 1 n0 = = −1. q1 q2 · · · qk−1 qk Le nombre b est donc un non-résidu quadratique pour f (x) et la condition 1◦ est remplie. Dans le cas b = 3 soit f1 (x) = 12x + 5, f2 (x) = 3x + 1. Ici on vérifie sans peine que les conditions C, 1◦ et 2◦ sont remplies. Il résulte de l’hypothèse H qu’il existe une infinité de nombres naturels x tels que les nombres f1 (x) et f2 (x) sont tous les deux premiers. Soit x un de ces nombres, tel que f1 (x) > g 4 . Si g appartenait modulo f1 (x) à un exposant < f1 (x) − 1, on aurait, d’après 2◦ , f1 (x) | g (f1 (x)−1)/2 − 1 ou bien f1 (x) | g 4 − 1. Or, vu le théorème d’Euler relatif au symbole de Legendre, l’égalité g = a 2 b et la condition 1◦ , on a b g (f1 (x)−1)/2 g ≡ ≡ −1 (mod f1 (x)), ≡ f1 (x) f1 (x) ce qui est incompatible avec f1 (x) | g (f1 (x)−1)/2 − 1 (puisque f1 (x) est impair). On a donc f1 (x) | g 4 −1, ce qui est impossible vu que f1 (x) > g 4 > 1. g est donc une racine primitive pour le module f1 (x). L’hypothèse de Artin est donc une conséquence de l’hypothèse H.

Nous étudierons maintenant la fonction (x) = lim [π(y + x) − π(y)]. y→∞

(Cf. Hardy et Littlewood [10], p. 52–68). On a (1) = (2) = 1, mais nous ne connaissons pas des valeurs (x) pour aucun nombre naturel x > 2. Il sera utile d’introduire la fonction auxiliaire (x) = max ϕ(x!, y + x) − ϕ(x!, y) 0y<x!

où ϕ(m, n) désigne le nombre de nombres naturels ne dépassant pas n et premiers avec m. De la définition de la fonction (x) résultent les lemmes suivantes : Lemme 1.

min z, ϕ(z!, y + x) − ϕ(z!, y) y,z=1,2,... max min y, π(y + x) − π(y) (x).

(x) =

max

y=1,2,...

Lemme 2. (x + 1) (x). Lemme 3. (x) + (y) (x + y). Lemme 4. (x) ϕ(x). Nous démontrerons maintenant :

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Théorème 1. (1) = (2) = 1, (3) = (4) = (5) = (6) = 2, (7) = (8) = 3, (9) = . . . = (12) = 4, (13) = . . . = (16) = 5, (17) = . . . = (20) = 6, (21) = . . . = (26) = 7, (27) = . . . = (30) = 8, (31) = (32) = 9, (33) = . . . = (36) = 10. Démonstration. D’après de lemme 4 on trouve (2) 1, (6) 2, (12) 4, (30) 8. En vertu du lemme 3 on a (8) (6) + (2) 3, (32) (30) + (2) 9, (36) (30) + (6) 10. Enfin il est facile de démontrer que parmi 16 nombres naturels consécutifs quelconques il y a au plus 5 nombres qui ne sont divisibles par aucun des nombres 2, 3 et 5, parmi 20 nombres naturels consécutifs quelconques il y a au plus 6 tels nombres et parmi 26 nombres naturels consécutifs quelconques il y a au plus 7 nombres qui ne sont divisibles par aucun des nombres 2, 3, 5 et 7. Donc (16) 5, (20) 6, (26) 7. D’autre part on a π(1+1)−π(1) = 1, π(3+2)−π(2) = 2, π(7+4)−π(4) = 3, π(9+4)−π(4) = 4, π(13+6)−π(6) = 5, π(17+6)−π(6) = 6, π(21+10)−π(10) = 7, π(27 + 10) − π(10) = 8, π(31 + 10) − π(10) = 9, π(33 + 10) − π(10) = 10. Donc, en vertu du lemme 1 on a (1) 1, (3) 2, (7) 3, (9) 4, (13) 5, (17) 6, (21) 7, (27) 8, (31) 9, (33) 10. La fonction (x) étant monotone (lemme 2), notre théorème résulte sans peine des inégalités obtenues.

Appelons k-jumeaux (en allemand k-linge) k nombres premiers k < q1 < q2 < . . . . . . < qk tels que (qk − q1 ) = k − 1. Ainsi deux nombres premiers q1 > 2 et q2 seront 2-jumeaux si (q2 − q1 ) = 1, c’est-à-dire q2 − q1 = 2. Les nombres premiers q1 , q2 , q3 tel que 3 < q1 < q2 < q3 et (q3 − q1 ) = 2 seront appelés 3-jumeaux etc. Les données numériques concernant les nombres k-jumeaux ont été données pour k = 2, qk 106 par G. H. Hardy et J. E. Littlewood ([10], p. 44), pour k = 3, qk 106 par G. H. Hardy et J. E. Littlewood ([10], p. 63), pour k = 4, qk 106 par G. H. Hardy et J. E. Littlewood ([10], p. 63), pour k = 4, 106 < qk 2 · 106 par Ch. Sexton [17], pour k = 4, 2 · 106 < qk 3 · 106 par W. A. Golubew [8], pour k = 4, 3 · 106 < qk 5 · 106 par W. A. Golubew [9], pour k = 5, qk 2 · 106 par W. A. Golubew [8], pour k = 5, 2 · 106 < qk 5 · 106 par W. A. Golubew [9], pour k = 6, qk 14 · 106 par W. A. Golubew [9]. Le problème si pour tout k naturel il existe une infinité de nombres k-jumeaux équivaut, comme on le démontre sans peine, au problème si l’on a pour tout x, (x) = (x) : l’hypothèse H résout donc ce problème positivement (voir plus loin C12 ). Théorème 2. (57) = (58) = (59) = (60) = 15. Démonstration. L. Aubry a démontré (voir L. E. Dickson [6], p. 355) que parmi 30 nombres impairs consécutifs il y a au plus 15 nombres qui ne sont divisibles par aucun des nombres 3, 5 et 7. Il en résulte que (60) 15. D’autre part on a π(57 + 16) − π(16) = 15, donc (57) 15. Vu le lemme 2 on a donc (57) = . . . = (60) = 15.

Théorème 3. On a (95) = . . . = (100) = 23.


1129

Démonstration. A. Schinzel a démontré (dans un article qui paraître ailleurs (2 )) que parmi 100 nombres naturels consécutifs quelconques il y a au plus 23 nombres qui ne sont divisibles par aucun nombre premier 17, d’où résulte tout de suite que (100) 23. c D’autre part on a ϕ(23!, 4083966+95)−ϕ(23!, 4083966) = 23, donc d’après le lemme 1 : (95) 23. La fonction (x) étant monotone on en obtient le théorème 3.

En vertu du lemme 1, le théorème 3 donne (100) 23, ce qui est incompatible avec l’inégalité (97) 24 qui a été déduite à la p. 67 du travail cité de Hardy et Littlewood [10] de leur hypothèse X. Or, cette déduction était fausse, car ces auteurs affirment qu’aucun des nombres premiers 17 et 113 ne donne le reste 8 mod 17, ce qui n’est pas vrai, puisque 59 ≡ 8 (mod 17). Théorème 4. On a (x) π(x) pour 1 < x 132. Démonstration. Vu le théorème 1 nous avons (2) = 1 = π(2), (6) = 2 = π(3), (8) = 3 < π(7), (12) = 4 = π(9), (16) = 5 < π(13), (20) = 6 < π(17), (26) = 7 < π(21), (30) = 8 < π(27), (32) = 9 < π(31), (36) = 10 < π(33), et, les fonctions (x) et π(x) étant monotones, cela prouve le théorème 4 pour 1 < x 36. Or, en vertu du lemme 3 on a (38) (30) + (8)

= 8 + 3 = 11 < π(37),

(42) (30) + (12) = 8 + 4 = 12 = π(39), (46) (30) + (16) = 8 + 5 = 13 < π(43), (50) (30) + (20) = 8 + 6 = 14 < π(47). D’après le théorème 2 on a (60) = 15 = π(51). En vertu du lemme 3 on trouve (62) (60) + (2)

= 15 + 1 = 16 < π(61),

(66) (60) + (6)

= 15 + 2 = 17 < π(63), (68) (60) + (8) = 15 + 3 = 18 < π(67), (72) (60) + (12) = 15 + 4 = 19 = π(69), (76) (60) + (16) = 15 + 5 = 20 < π(73), (80) (60) + (20) = 15 + 6 = 21 = π(77), (86) (60) + (26) = 15 + 7 = 22 = π(81). En vertu du théorème 3 on a (100) 23 = π(87). En vertu du lemme 3 on trouve (102) (100) + (2) (106) (100) + (6)

= 23 + 1 = 24 < π(101), = 23 + 2 = 25 < π(103),

(108) (100) + (8) = 23 + 3 = 26 < π(107), (112) (100) + (12) = 23 + 4 = 27 < π(109), (116) (100) + (16) = 23 + 5 = 28 < π(113), (2 ) Voir J2, p. 1139.

1130

J. Prime numbers

(120) (100) + (20) = 23 + 6 = 29 < π(117), (126) (100) + (26) = 23 + 7 = 30 < π(121), (130) (100) + (30) = 23 + 8 = 31 < π(127), (132) (100) + (32) = 23 + 9 = 32 = π(131). Les fonctions (x) et π(x) étant monotones, nous en concluons que le théorème 4 est vrai pour 1 < x 132.

Corollaire 1. (x) π(x) pour 1 < x 132. La démonstration résulte du lemme 1 et du théorème 4. Hardy et Littlewood ont énoncé ([10], p. 54) l’hypothèse que (x) π(x) quel que soit le nombre x > 1. Corollaire 2. Si x > 1, y > 1 et si l’un au moins des nombres x et y est 132, on a π(x + y) π(x) + π(y). Démonstration. Sans nuire à la généralité nous pouvons supposer que x < y, 1 < x 132. En vertu du théorème 4 on a donc (x) π(x). Or, en vertu du lemme 1 on a, pour tout nombre y,

min y, π(x + y) − π(y) π(x). Comme y x π(x + y) − π(y), on a π(x + y) − π(y) π(x), c’est-à-dire π(x + y) π(x) + π(y).

Il est à remarquer que E. Landau [12] a démontré que pour x suffisament grands on a π(2x) < 2π(x). Nous appliquerons maintenant l’hypothèse H à l’étude de la fonction (x). C12 . (x) = (x) pour x naturels. Démonstration de H → C12 . D’après le lemme 1 il suffit de prouver que (x) (x). Dans ce but supposons que pour x naturel donné s = (x). D’après la définition de (x) il existe un entier y tel que 0 y < x! et que s = ϕ(x!, y + x) − ϕ(x!, y). Évidemment on a s x et il existe s entiers croissants a1 , a2 , . . . , as , où 0 a1 < as x tels que (y + ai , x!) = 1 pour i = 1, 2, . . . , s. Soit fi (ξ ) = ξ + ai pour i = 1, 2, . . . , s, P (ξ ) =

s

fi (ξ ).

i=1

Si p est un nombre premier tel que p | P (ξ ) pour ξ entiers, on a, d’après le théorème de Lagrange, p s x, donc p | x! et, d’après (y + ai , x!) = 1, (y + ai , p) = 1 pour s

i = 1, 2, . . . , s, et comme P (y) = (y + ai ), cela donne P (y), p = 1, contrairement à p | P (y).

i=1


1131

La condition C est donc remplie et d’après H il existe une infinité de nombres naturels ξ tels que le nombres ξ + ai (i = 1, 2, . . . , s) sont tous premiers. Comme 0 a1 < as x, il en résulte que π(ξ + x) − π(ξ ) s = (x) pour une infinité de nombres naturels ξ et, vu la définition de la fonction (x) cela donne (x) (x). L’implication H → C12 se trouve ainsi démontrée.

C12.1 . (1) = (2) = 1, (3) = . . . = (6) = 2, (7) = (8) = 3, (9) = . . . = (12) = 4, (13) = . . . = (16) = 5, (17) = . . . = (20) = 6, (21) = . . . = (26) = 7, (27) = . . . = (30) = 8, (31) = (32) = 9, (33) = . . . = (36) = 10, (57) = . . . = (60) = 15, (95) = . . . = (100) = 23. C12.1 est une conséquence immédiate de C12 et des théorèmes 1, 2 et 3. C12.2 . L’hypothèse de Hardy et Littlewood suivant laquelle (x) π(x) pour x naturels > 1 équivaut à l’inégalité (∗)

π(x + y) π(x) + π(y) pour x > 1, y > 1.

Démonstration de C12 → C12.2 . L’inégalité (∗) entraîne tout de suite l’inégalité (x) π(x) (sans avoir recours à l’hypothèse H). Supposons maintenant que (x) π(x) pour x naturels > 1 et soient x et y deux nombres naturels > 1. Sans diminuer la généralité du raisonnement nous pouvons supposer que 1 < x y. Comme (x) π(x), on a, d’après C12 , (x) π(x), donc d’après le

lemme 1, pour tout y, min y, π(x + y) − π(x) π(x). Or, y x π(x + y) − y, donc π(x + y) − π(y) π(x), c’est-à-dire π(x + y) π(x) + π(y).

Il est intéressant qu’on ne puisse démontrer par le calcul ni la fausseté de l’hypothèse H ni celle de l’hypothèse de Hardy–Littlewood sur la fonction (x). (Quant à cette dernière, si l’inégalité (x) 2 avait lieu pour un x quelconque, on aurait lim (pk+1 − pk ) < ∞). k→∞

Il est cependant possible qu’on puisse trouver des nombres x et y plus grands que 1 pour lesquels π(x + y) > π(x) + π(y), ce qui prouverait que l’hypothèse H et l’hypothèse de Hardy–Littlewood sur la fonction (x) ne peuvent pas être simultanément vraies. Hypothèse H1 de W. Sierpinski. ´ Si pour un nombre naturel n > 1 les nombres 1, 2, 3, . . . , n2 sont rangés successivement en n lignes, n nombres dans chaque ligne, alors chaque ligne contient au moins un nombre premier. La proposition que la deuxième ligne contient au moins un nombre premier équivaut évidemment au théorème de Tchebycheff que pour n naturels > 1 il existe entre n et 2n au moins un nombre premier. La proposition que pour n 9 chacune des 9 premières lignes contient au moins un nombre premier peut sans peine être déduite du théorème de R. Breusch [1] d’après lequel pour x 48 il y a entre x et 98 x au moins un nombre premier. Ensuite il est facile de déduire du théorème d’Hadamard–de la Vallée Poussin sur les nombres premiers que pour tout k et n n0 (k) chacune des k premières lignes contient au moins un nombre premier.

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J. Prime numbers

On a ici lim n0 (k) = +∞ et le problème se pose si le plus grand nombre n pour lequel k→∞

il n’existe aucun nombre premier entre (k − 1)n et kn tend vers +∞ avec k. Par la méthode de Brun on pourrait démontrer (voir G. Ricci [13]), que chacune des lignes de notre carré contient un nombre dont le nombre des diviseurs premiers est limité par une constante universelle. Consequence. Entre deux carrés consécutifs il existe au moins deux nombres premiers distincts. En effet, pour démontrer cette implication, il suffit de remarquer que si n est un nombre naturel > 1 les nombres naturels consécutifs (n−1)2 , (n−1)2 +1, . . . , n2 forment les deux dernières lignes dans notre carré composé des nombres 1, 2, . . . , n2 . En observant que dans tout intervalle fermé dont les extrémités sont les cubes de nombres naturels consécutifs, il y a au moins deux carrés distincts, on en déduit tout de suite qu’entre deux cubes de nombres naturels consécutifs il y a au moins deux nombres premiers. Cette proposition n’est pas encore démontrée sans avoir recours à l’hypothèse H1 , mais on a démontré que pour n naturels suffisamment grands il existe entre n3 et (n + 1)3 au moins un nombre premier. (On ne sait pourtant pas si cela est vrai pour tout n naturel). Remarquons que l’hypothèse H1 pour les nombre n premiers résulte tout de suite de l’hypothèse suivante énoncée en 1932 par R. Haussner : entre deux multiples consécutifs 2 il existe au moins un nombre d’un nombre premier pi qui sont tous les deux inférieurs à pi+1 premier ([11], p. 192). Pour n = 7, par exemple, il résulte de l’hypothèse de R. Haussner que non seulement chacune des 7 lignes de notre carré des nombres 1, 2, . . . , 49, mais aussi les 10 lignes suivantes (dont la première contient sept nombres 50, 51, . . . , 56 et la dernière les nombres 113, 114, . . . , 119) contient chacune au moins un nombre premier. Il est intéressant de remarquer ici que la ligne suivante la 18-ème, formée des nombres 120, 121, . . . , 126, ne contient aucun nombre premier. Hypothèse H2 deA. Schinzel. Si pour un nombre naturel n > 1 les nombres 1, 2, 3, . . . , n2 sont rangés en n lignes, n nombres dans chaque ligne, alors, si (k, n) = 1, la k-ième collonne contient au moins un nombre premier.

Nous ne savons pas quel sera le sort de nos hypothèses, cependant nous pensons que même si elles seront mises en défaut, cela ne sera pas sans profit pour la théorie des nombres.

Travaux cités [1] R. Breusch, Zur Verallgemeinerung des Bertrandschen Postulates, daß zwischen x und 2x stets Primzahlen liegen. Math. Z. 34 (1932), 505–526. [2] M. Cantor, Ueber arithmetische Progressionen von Primzahlen. Z. Math. Phys. 6 (1861), 340–343.


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[3] E. Catalan, Propositions et questions diverses. Bull. Soc. Math. France 16 (1888), 128–129. [4] J. Chernick, On Fermat’s simple theorem. Bull. Amer. Math. Soc. 45 (1939), 269–274. [5] L. E. Dickson, Theorems and tables on the sum of the divisors of a number. Quart. J. Math. 44 (1913), 264–288. [6] L. E. Dickson, History of the Theory of Numbers. Chelsea, New York 1952. [7] P. Erd˝os, On the normal number of prime factors of p − 1 and some related problems concerning Euler’s φ-function. Quart. J. Math. Oxford Ser. 6 (1935), 205–213. [8] W.A. Golubew, Abzählung von „Vierlingen” von 2000000 bis 3000000 und von „Fünflingen” von 0 bis 2000000. Anz. Österreich. Akad. Wiss. Math.-Naturwiss. Kl., 1956, 153–157. [9] −−, Abzählung von „Vierlingen” und „Fünflingen” bis zu 5000000 und von „Sechslingen” von 0 bis 14000000. Anz. Österreich. Akad. Wiss. Math.-Naturwiss. Kl., 1957, 82–87. [10] G. H. Hardy, J. E. Littlewood, Some problems of ‘Partitio numerorum’ III. On the expression of a number as a sum of primes. Acta Math. 44 (1923), 1–70. [11] R. Haussner, Über die Verteilung von Lücken- und Primzahlen. J. Reine Angew. Math. 168 (1932), 192. [12] E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen, T. 1. Teubner, Leipzig 1909. [12a] J. Needham, Science and Civilization in China, vol. 3. Mathematics and the Sciences of the Heavens and the Earth. Cambridge Univ. Press, New York 1959. [13] G. Ricci, Su la congettura di Goldbach e la costante di Schnirelmann. Ann. R. Scuola Norm. Sup. Pisa (2) 6 (1937), 71–116. [14] R. M. Robinson, Factors of Fermat numbers. Math. Tables Aids Comput. 11 (1957), 21–22. [15] A. Schinzel, Sur un problème concernant la fonction ϕ(n). Czechoslovak Math. J. 6 (1956), 164–165; this collection: G3, 875–876. [16] −−, Sur l’équation ϕ(x + k) = ϕ(x). Acta Arith. 4 (1958), 181–184. [17] Ch. Sexton, Abzählung von „Vierlingen” von 1000000 bis 2000000. Anz. Österreich. Akad. Wiss. Math.-Naturwiss. Kl., 1955, 236–239. [18] V. Thébault, Sur les nombres premiers impairs. C. R. Acad. Sci. Paris 218 (1944), 223–224.

Originally published in Acta Arithmetica VII (1961), 1–8


Remarks on the paper “Sur certaines hypothèses concernant les nombres premiers”

In the paper [14] mentioned in the title some historical inaccuracies are committed which ought to be corrected, besides some new results strictly connected with the above paper arise, which seem to the writer worthy of mention. This is the aim of the present paper. To begin with, as kindly pointed out by Professor P. T. Bateman, Hypothesis H coincides for the case of linear polynomials fi with a conjecture of L. E. Dickson announced in [7]. Therefore, it is easy to see that C1 , C2 , C7 –C12 are consequences of Dickson’s conjecture. On the other hand, as Dickson quoted in [8], Vol. I, p. 333, V. Bouniakowsky conjectured ([1]) that if d is the greatest fixed divisor of a given irreducible polynomial f (x) (with integral coefficients, the highest coefficient > 0) then the polynomial f (x)/d represents infinitely many primes. This conjecture of Bouniakowsky implies Hypothesis H for the case s = 1 and therefore C3 and the first part of C6 . Now we shall deduce Bouniakowsky’s conjecture from Hypothesis H. For further use we shall deduce the following stronger proposition. C13 . Let F1 (x), F2 (x), . . . , Fs (x), G1 (x), G2 (x), . . . , Gt (x) be irreducible integervalued polynomials of positive degree with the highest coefficient > 0. If there does not exist any integer > 1 dividing the product F1 (x)F2 (x) · · · Fs (x) for every x and if Gj (x) ≡ Fi (x) for all i s, j t, then there exist infinitely many positive integers x such that the numbers F1 (x), F2 (x), . . . , Fs (x) are primes and the numbers G1 (x), G2 (x), . . . , Gt (x) are composite. Proof of the implication H → C13 . Let Fi = Φi /di , Gj = Γj /ej , where Φi , Γi are polynomials with integral coefficients, di , ej are positive integers. Let further d = d1 d2 · · · ds , e = e1 e2 · · · et , F = F1 F2 · · · Fs , Φ = Φ1 Φ2 · · · Φs , d = p1α1 p2α2 · · · pkαk . Since the polynomial F has no fixed divisor > 1, there exist integers xi such that F (xi ) ≡ 0 (mod pi ). We can assume that the polynomials Fi , Gj (i s, j t) are algebraically coprime, because otherwise either Gj = Fi or Gj (x) would be composite for all sufficiently large x. We have then (F, Gj ) = 1 (j t) and there exist polynomials aj (x), bj (x) with integral

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J2. Remarks on J1

coefficients and an integer cj = 0 such that aj (x)F (x) − bj (x)Gj (x) = cj .

(1) c

Let c = c1 c2 · · · ct . Since every non-constant polynomial possesses infinitely many prime divisors, there exist primes qj /| cde such that qj | G(yj ) for some integer yj . Let q = q1 q2 · · · qt . In virtue of the Chinese Remainder Theorem, there exist integers z satisfying the following system of congruences (2)

c

z ≡ xi (mod piαi +1 ),

i s,

z ≡ yj (mod qj ),

j t,

let z0 be any of them. Let us consider polynomials fi (x) = Fi (dqx + z0 ) =

Φi (dqx + z0 ) . di

Since di | dq and Φi (z0 )/di = Fi (z0 ) is an integer, polynomials fi have integral coefficients and the highest coefficient > 0. Besides, they are irreducible. We shall show that f (x) = f1 (x) · · · fs (x) has no fixed divisor > 1. Suppose that prime p is such a divisor. We have by (2), since piαi +1 /| d, f (0) = F (z0 ) ≡ F (xi ) ≡ 0 (mod pi ),

i s,

and since qj /| e, Gj (z0 ) ≡ Gj (yj ) ≡ 0 (mod qj ). It follows hence by (1), because qj /| c, that f (0) = F (z0 ) ≡ 0 (mod qj ). Therefore, we must have (p, dq) = 1. On the other hand, by the assumption about F , there exists an integer zp such that F (zp ) ≡ 0 (mod p). Let x0 be a root of the congruence dqx + z0 ≡ zp (mod p). Since (d, p) = 1, we have Φ(zp ) Φ(dqx0 + z0 ) ≡ = F (zp ) ≡ 0 (mod p). d d Our supposition about p is therefore false and the polynomials fi satisfy the conditions of Hypothesis H. Thus, there exist infinitely many integers x such that numbers c fi (x) = Fi (dqx + z0 ) are primes. Meanwhile for every x f (x0 ) = F (dqx0 + z0 ) =

Gj (dqx + z0 ) =

Γj (dqx + z0 ) Γj (z0 ) ≡ = Gj (z0 ) ≡ 0 (mod qj ) ej ej

then for sufficiently large x numbers Gj (dqx + z0 ) are composite.

1136

J. Prime numbers

Now we shall deduce C14 . For every k > 1 there exist infinitely many numbers mk such that the equation ϕ(y) = mk has exactly k solutions. Proof of the implication H → C14 . Consider first k even, k = 2l, and put in H fi (x) = 2x 2i−1 + 1

(i = 1, 2, . . . , 2l),

f2l+1 (x) = x.

The polynomials fi (x) are irreducible, their highest coefficient is > 0 and since f1 (−1)f2 (−1) · · · f2l+1 (−1) = −1, they satisfy the conditions of Hypothesis H. Therefore, there exist infinitely many integers x such that all fi (x) are primes. Consider for such x > 5 the equation ϕ(y) = mk = 4x 4l .

(3)

Since x is odd, mk ≡ 0 (mod 8), y may have only one of the following forms: pα , 2pα , α 4p , pα q β , 2pα q β , where p and q are primes > 2. If α > 1 we should have p(p−1) | 4x 4l ,

whence as x is prime > 5, p = x and x − 1 | 4x 4l , which is impossible. Therefore, there is α = 1 and similarly, β = 1. y = p or 2p is impossible since then p = ϕ(y) + 1 = 4x 4l + 1 ≡ 0 (mod 5). y = 4p is also impossible, because then p = 21 ϕ(y) + 1 = 2x 4l + 1 ≡ 0 (mod 3). In the case y = pq or 2pq, we get (p − 1)(q − 1) = 4x 4l , whence p = 2x n + 1,

q = 2x 4l−n + 1.

Since for n even 2x n + 1 ≡ 0 (mod 3), there remains the only possibility (4)

y = (2x 2i−1 + 1)(2x 4l−2i+1 + 1) = fi (x)f2l−i+1 (x)

or y = 2fi (x)f2l−i+1 (x)

(i = 1, 2, . . . , l).

Since the numbers fi (x) are primes, the 2l values y given by formulae (4) satisfy (3), which completes the proof for even k. Consider now odd k, k = 2l + 3 (l = 0, 1, . . . ) and put in C13 Fi (x) = 2x 6i−3 + 1,

Fl+i (x) = 6x 6i−1 + 1

F2l+1 (x) = x, Gj (x) = 2x 6j −5 + 1, G2l+1 (x) = 2x

6l+1

F2l+2 (x) = 6x

Gl+j (x) = 2x 6j −1 + 1,

(i = 1, 2, . . . , l),

6l+2

+ 1,

(j = 1, 2, . . . , l),

G2l+2 (x) = 12x 6l+2 + 1.

1137

J2. Remarks on J1

The polynomials Fi are irreducible and satisfy other conditions of C13 , because in view of F (−1) = −5l · 7,

F (1) = 3l · 7l+1 ,

F (2) ≡ 0 (mod 7),

F (x) has no fixed divisor > 1. Since Gj = Fi (i, j 2l + 2), there exist by C13 infinitely many integers x such that numbers Fi (x) are primes and numbers Gj (x) are composite (i, j 2l + 2). Observe that the numbers 2x n + 1 are composite for all positive n 6l + 2, n = 6i − 3, because for n even 2x n + 1 ≡ 0 (mod 3). Consider for x of the above kind the equation ϕ(y) = mk = 12x 6l+2 .

(5) c

By much the same arguments as in case of (3), we infer that y may have only one of the following forms: p, 2p, 4p, pq, 2pq, where p, q are primes > 2 (the possibility y = 9q or 18q fails, because we should have then q = 16 ϕ(y) + 1 = 2x 6l+2 + 1). It cannot be y = p or 2p, because then p = ϕ(y) + 1 = 12x 6l+2 + 1, which is composite. The case y = 4p gives (6)

p = 21 ϕ(y) + 1 = 6x 6l+2 + 1 = F2l+2 (x).

In the case y = pq or 2pq, we get (p − 1)(q − 1) = 12x 6l+2 , whence p − 1 = 2x n , q − 1 = 6x 6l+2−n (0 n 6l + 2) or p, q change places. The numbers 2x n + 1 being composite (0 < n 6l + 2, n = 6i − 3), the only two possibilities remain 1◦

y = 3(6x 6l+2 + 1) = 3F2l+2 (x) or y = 6F2l+2 (x);

2◦

y = (2x 6i−3 + 1)(6x 6(l−i)+5 + 1) = Fi (x)F2l−i+1 (x) or y = 2Fi (x)F2l−i+1 (x)

(i = 1, 2, . . . , l).

The numbers Fi (x) being primes, the 2l + 2 values y given above satisfy (5), which together with (6) gives exactly 2l + 3 solutions of (5).

C15 . For every k 1, there exist infinitely many numbers nk such that the equation σ (y) = nk has exactly k solutions. Proof of the implication H → C15 . Put in H, c

fi (x) = 2(2x + 1)2i − 1 (i = 1, 2, . . . , 2k),

f2k+1 (x) = x,

f2k+2 (x) = 2x + 1.

The polynomials fi (x) are irreducible, their highest coefficient is > 0 and since f1 (−1)f2 (−1) · · · f2k+2 (−1) = 1, they satisfy the conditions of Hypothesis H. Therefore, there exist infinitely many integers x such that all fi (x) are primes and since x − 1)/(2x + 1)4k+2 tends to infinity, infinitely many of them satisfy the inequality c (2 x 2 − 1 > 4(2x + 1)4k+2 .

1138

J. Prime numbers

Consider for any such x the equation σ (y) = nk = 4(2x + 1)4k+2 . Suppose that pα | y, p α+1 /| y, where p is prime, α > 1. It follows from the above equation pα+1 − 1

that

4(2x + 1)4k+2 . p−1 In virtue of the theorem of Zsigmondy (cf. [8], Vol. I, p. 195), (p α+1 − 1)/(p − 1) has at least one prime factor of the form (α + 1)l + 1. Since α + 1 > 2 and the numbers x and 2x + 1 are primes, we clearly must have (α + 1)l + 1 = 2x + 1,

α + 1 = x,

hence 2x − 1

px − 1 4(2x + 1)4k+2 , p−1

which contradicts the assumption about x. The obtained contradiction proves that y is square-free, and since nk ≡ 0 (mod 3), nk ≡ 0 (mod 8), y may have only one of the forms p, pq where p and q are primes, 2 < p < q. y = p is impossible since then p = σ (y) − 1 = 4(2x + 1)4k+2 − 1 ≡ 0 (mod 3). In the case y = pq we get (p + 1)(q + 1) = 4(2x + 1)4k+2 , whence c

p = 2(2x + 1)n − 1,

q = 2(2x + 1)4k+2−n − 1,

0 < n < 2k + 1.

Since x ≡ −1 (mod 3), 2(2x + 1)n − 1 ≡ 0 (mod 3) for all odd n, there remains the only possibility

y = 2(2x + 1)2i − 1 2(2x + 1)4k+2−2i − 1 = fi (x)f2k+1−i (x) (i = 1, 2, . . . , k). Since the numbers fi (x) are primes, the k values of y given above satisfy the equation

σ (y) = nk , which completes the proof. P. Erd˝os proved without any conjecture that if there exists one mk such that the equation ϕ(y) = mk has exactly k solutions, then there exist infinitely many such mk ([9], Theorem 4), and the analogous theorem for the equation σ (y) = nk (l.c., p. 12). For k = 1 the well known conjecture of Carmichael states that such a number mk does not exist and for k > 1 W. Sierpiński conjectured that mk and nk exist (cf. [9], p. 12). We have just deduced this conjecture from Hypothesis H; by more complicated arguments we could also deduce that for every pair k, l , where k = 1, l 0, there exist infinitely many numbers m such that the equation ϕ(y) = m has exactly k solutions and the equation σ (y) = m has exactly l solutions.

J2. Remarks on J1

1139

On page 191 (1 ) paper [14] contains two historical mistakes. The theorem about the difference of arithmetical progression formed by primes, ascribed to V. Thébault, was proved earlier by M. Cantor ([2]). On the other hand, the disproving of the M. Cantor conjecture about progressions formed by consecutive primes, ascribed to the writer, was made much earlier by F. H. Loud (cf. [4]). Part of the paper [14] concerning functions , was covered to some extent by the results of H. Smith’s paper [16]. It is easy to notice that the function Δn considered by Smith is connected with function by the condition (Δn) = n − 1 < (1 + Δn) and “k-tuples” considered by him just correspond “nombres k-jumeaux” of [14]. Theorem 1 of [14] follows from the table given for Δn by Smith, his results further imply the following equalities

(7)

(37) = . . . = (42) = 11,

(43) = . . . = (48) = 12,

(49) = (50) = 13,

(51) = . . . = (56) = 14,

(57) = . . . = (60) = 15,

(61) = . . . = (66) = 16,

(67) = . . . = (70) = 17,

(71) = . . . = (76) = 18,

(77) = . . . = (80) = 19,

(81) = . . . = (84) = 20,

(85) = . . . = (90) = 21, (95) = . . . = (100) = 23,

(91) = . . . = (94) = 22, (101) = . . . = (110) = 24,

(111) = . . . = (114) = 25,

(115) = (116) = 26,

thus, in particular Theorems 2 and 3 of [14]. It dispenses the writer of the duty of publishing mentioned in [14] the laborious proof that (100) 23. From formulae (7) it immediately follows that (x) π(x) for 36 < x 116. Paper [14] contained a proof that (x) π(x) for 1 < x 132. Owing to Smith’s results, one can prove the stronger Theorem. (x) π(x) for 1 < x 146. Proof. It is sufficient to prove the above inequality for 132 < x 146. Profiting by Lemma 3 of [14] we get (140) (114) + (26) = 25 + 7 = 32 = π(133), (146) (114) + (32) = 25 + 9 = 34 = π(141), which in view of monotonicity of functions and π gives the desired result. Analogously, as in [14], we obtain Corollary. If x > 1, y > 1 and if at least one of the numbers x and y is 146, then π(x + y) π(x) + π(y).

(8) (1 )

Page 1118 in this collection.

1140

J. Prime numbers

As to inequality (8), it was verified by E. Łukasiak for 1 < x, y < 1223 = p201 . H. Smith gave also in [16], numerical data concerning k-tuples for 7 k 15, qk 137 · 106 . One may remark that there was omitted the 15-tuple formed by primes 17, . . . , 73. As to Hypothesis H1 of [14], we shall give the following remarks. L. Skula noticed (written communication) that if H1 is true, then also the intervals [n2 + 1, n2 + n] and [n2 + n + 1, n2 + 2n] contain primes. of the conjectures that for On the other hand Hypothesis H1 is a simple consequence √ all x 117 there is a prime between x and x + x or that for all x 8 there is a prime between x and x + log2 x (cf. H. Cramér [6]). Since these conjectures hold for ([17]) and D. H. Lehmer ([11]) x 20.3 · 106 , as can be verified owing to A. E. Western √ tables, Hypothesis H1 holds for all n 4500 < 103 20.3. As to Hypothesis H2 , it was verified by A. Gorzelewski for n 100. Finally, it seems interesting to review 17 conjectures concerning primes, written out by R. D. Carmichael from Dickson’s book [8]: 13 from Volume I ([3], p. 401) and 4 from Volume II ([5], p. 76). One of these conjectures ([3], 14) is already proved ([13], [15]), 3 are consequences of Hypothesis H ([3], 6, 8, 11), 2 are consequences of Hypothesis H1 ([3], 12, 13), 4 are various modifications of Goldbach conjecture ([3], 9; [5], 1, 2, 3), 7 are false. Among these latter: 2 are mentioned in [14], Schaffler’s and Cantor’s conjectures ([3], 7, 10), 3 concerning Mersenne primes Mn ([3], 1, 2, 3) are wrong respectively for n = 13, 263, 607, one concerning primitive roots ([3], 15) was recently disproved by A. Makowski ([12]) and one ([5], 4) we shall disprove now. It states that every prime 18n ± 1, or else its triple, is expressible in the form x 3 − 3xy 2 ± y 3 . If it is true, then for all z, the form x 3 − 3xy 2 ± y 3 represents at least 1 c π ( z) numbers z (π (x) is the number of primes 18n ± 1 x). But this is incompatible 3 with Siegel’s theorem (cf. [10], p. 139).

References [1] V. Bouniakowsky, Nouveaux théorèmes relatifs à la distinction des nombres premiers et à la décomposition des entiers en facteurs. Mém. Acad. Sci. St. Pétersbourg (6), Sci. Math. Phys. 6 (1857), 305–329. [2] M. Cantor, Ueber arithmetische Progressionen von Primzahlen. Z. Math. Phys. 6 (1861), 340–343. [3] R. D. Carmichael, Review of volume I “History of the Theory of Numbers”. Amer. Math. Monthly 26 (1919), 396–403. [4] −−, Note on prime numbers. Amer. Math. Monthly 27 (1920), 71. [5] R. D. Carmichael, Review of volume II “History of the Theory of Numbers”. Amer. Math. Monthly 28 (1921), 72–78. [6] H. Cramér, On the order of magnitude of the difference between consecutive prime numbers. Acta Arith. 2 (1936), 23–46. [7] L. E. Dickson, A new extension of Dirichlet’s theorem on prime numbers. Messenger of Math. (2) 33 (1904), 155–161. [8] −−, History of the Theory of Numbers. Chelsea, New York 1952.

J2. Remarks on J1

1141

[9] P. Erd˝os, Some remarks on Euler’s ϕ-function. Acta Arith. 4 (1958), 10–19. [10] P. Erd˝os, K. Mahler, On the number of integers which can be represented by a binary form. J. London Math. Soc. 13 (1938), 134–139. [11] D. H. Lehmer, Tables concerning the distribution of primes up to 37 millions. Mimeographed, 1957. Deposited in the UMT file. [12] A. Makowski, On a conjecture of Murphy. An. Soc. Paran. Mat. (2) 3 (1960), 13. [13] S. S. Pillai, On some empirical theorem of Scherk. J. Indian Math. Soc. 17 (1927–28), 164–171. [14] A. Schinzel, W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers. Acta Arith. 4 (1958), 185–208; Erratum, ibid. 5 (1959), 259; this collection: J1, 1113–1133. [15] W. Sierpiński, Sur une propriété des nombres premiers. Bull. Soc. Roy. Sci. Liège 21 (1952), 537–539. [16] H. F. Smith, On a generalization of the prime pair problem. Math. Tables Aids Comput. 11 (1957), 249–254. [17] A. E. Western, Note on the magnitude of the difference between successive primes. J. London Math. Soc. 9 (1934), 276–278.

Originally published in Mathematics of Computation 17 (1963), 445–447


A remark on a paper of Bateman and Horn

Let f1 , f2 , . . . , fk be distinct irreducible polynomials with integral coefficients and the highest coefficient positive, such that f (x) = f1 (x)f2 (x) · · · fk (x) has no fixed divisor > 1. Denote by P (N ) the number of positive integers x N such that all numbers f1 (x), f2 (x), . . . , fk (x) are primes. P. T. Bateman and R. A. Horn [1] recently gave the heuristic asymptotic formula for P (N): ω(p) 1 −k N −1 (h 1 − h · · · h ) , 1 − (1) P (N ) ∼ 1 2 k p p logk N p where hi is the degree of fi and ω(p) is the number of solutions of the congruence f (x) ≡ 0 (mod p). Formula (1) contains as particular cases six conjectures from a well known paper of Hardy and Littlewood [3] called by the latter Conjectures B, D, E, F, K, P, as well as their conditional theorem X1. This is evident except for Conjecture D, which concerns the number of solutions of the equation (2)

ap − bp = k

(a > 0, b > 0, (a, b) = 1)

in primes p, p with p n. In order to apply formula (1) here one should put f1 (x) = n − u0 u0 + bx, f2 (x) = v0 + ax, N = , where u0 , v0 are fixed integers such that b au0 − bv0 = k. Conjectures denoted by Hardy and Littlewood by J, M, N are of distinctly different character; besides the first has been proved by S. Chowla [2] and Yu. V. Linnik [4]. Conjecture A (a strong form of Goldbach’s Conjecture) is a particular case of C, Conjectures H and I are particular cases of G. It remains therefore to consider Conjectures C, G, L, which are, according to Hardy and Littlewood, conjugate to Conjectures D, F, K, respectively. We quote them below for the convenience of a reader, with slight changes in the notation (e.g. p, p denote primes). Conjecture C. If a, b are fixed positive integers and (a, b) = 1 and P (k) is the number of representations of k in the form k = ap + bp

J3. A remark on a paper of Bateman and Horn

then

k P (k) = o (log k)2

1143

unless (k, a) = 1, (k, b) = 1, and one and only one of k, a, b is even. But if these conditions are satisfied then p − 1 2C2 k P (k) ∼ , ab (log k)2 p−2 where C2 =

∞ p=3

1 1− (p − 1)2

and the first product extends over all odd primes p which divide k, a or b.

c

Conjecture G. Suppose that a and b are integers, and a > 0, and let P (n) be the number of representations of n in the form am2 + bm + p. Then if n, a, b have a common factor, or if n and a + b are both even, or if b2 + 4an is a square then √ n . P (n) = o log n In all other cases

√ " 1 b2 + 4an p n P (n) ∼ √ 1− , p−1 p−1 p a log n p3 p/| a

where p is a common odd prime divisor of a and b, and " is 1 if a + b is odd and 2 if a + b is even. Conjecture L. Every large number n is either a cube or the sum of a prime and a (positive) cube. The number P (n) of representations is given asymptotically by 1 n1/3 1− (n)p , P (n) ∼ log n p p−1 where p ≡ 1 (mod 3), p /| n, and (n)p is equal to 1 or to − 21 according as n is or is not a cubic residue of p. A comparison of formula (1) with the above formulas of paper [3] suggests forcibly the following conjecture. Let polynomials f1 , f2 , . . . , fk (k 0), f = f1 f2 · · · fk satisfy the same conditions as above. Let g be a polynomial with integral coefficients and the highest coefficient positive. Let n be a positive integer such that n−g(x) is irreducible and f (x)(n−g(x)) has no fixed divisor > 1. Denote by N (n) = N the number of positive integers x such that n−g(x) > 0 and by P (n) the number of x’s such that all numbers f1 (x), f2 (x), . . . , fk (x) and n − g(x)

1144

J. Prime numbers

are primes. Then for large n we have (3)

P (n) ∼

N logk+1 N

(h0 h1 · · · hk )−1

1−

p

ω(p) 1 −k−1 , 1− p p

where h0 is the degree of g and ω(p) is the number of solutions of the congruence f (x)(n − g(x)) ≡ 0 (mod p). Conjectures C, G, L and therefore also A, H, I are particular cases of formula (3). To see this, as far as C is concerned, one should put k − al , b where l is an integer such that al ≡ k (mod b), −b < l 0. Conjecture A has been extensively verified ([3], p. 37). I have had no possibility to verify by computation the agreement of formula (3) with reality in other cases. For such comparisons one should CN replace N(log N )−k−1 by 2 (log u)−k−1 du, as is pointed out in [3]. I conclude with expressing my thanks to the referee for his valuable suggestions. f1 (x) = bx + l,

g(x) = ax,

n=

References [1] P. T. Bateman, R. A. Horn, A heuristic asymptotic formula concerning the distribution of prime numbers. Math. Comp. 16 (1962), 363–367. [2] S. Chowla, The representation of a number as a sum of four squares and a prime. Acta Arith. 1 (1935), 115–122. [3] G. H. Hardy, J. E. Littlewood, Some problems of ‘Partitio numerorum’ III. On the expression of a number as a sum of primes. Acta Math. 44 (1923), 1–70. [4] Yu. V. Linnik, An asymptotic formula in an additive problem of Hardy–Littlewood. Izv. Akad. Nauk SSSR Ser. Mat. 24 (1960), 629–706 (Russian).

Originally published in Acta Arithmetica XIII (1967), 228–236


On two theorems of Gelfond and some of their applications

5. The greatest prime factor of a quadratic or cubic polynomial One of the consequences of Theorem 10 merits to be stated as a separate theorem (q(n) denotes the greatest prime factor of n). Theorem 11. If ν = 2 or 3, A and E are non-zero integers then ⎧4 if ν = 2 and AE is not a perfect square ⎪ ⎪ ⎪7 q(Ax ν − E) ⎨ or ν = 3 and A2 E is a perfect cube, 2 lim ⎪ if ν = 2 and AE is a perfect square, ⎪ x=∞ log log x ⎪ ⎩ 73 2 14 if ν = 3 and A E is not a perfect cube.

Proof. Since Ax ν − E = A1−ν (Ax)ν − Aν−1 E we apply Theorem 10 case (iv)(1 ) with εP1 = Aν−1 E and obtain the assertion except in the case A2 E being a perfect cube. In this case we set A2 E = F 3 and since

q(y 3 − A2 E) q(y 2 + F y + F 2 ) = q (2y + F )2 + 3F 2 we apply Theorem 10 case (iv) with εP1 = −3F 2 .

Corollary 7. If f (x) is any quadratic polynomial without a double root, then

4 q f (x) if f is irreducible, 27 lim x=∞ log log x 7 if f is reducible. Proof is obtained by reducing f (x) to the canonical form.

Theorem 11 can be improved if ν = 2, E | 4 or ν = 3, E | 3. The latter case was done by Nagell [18], cf. [19]. We prove (1 )

See Acta Arith. 13 (1967), p. 221.

1146

J. Prime numbers

Theorem 12. If A = 0 is an integer and E | 4, then q(Ax 2 − E) 4 if AE is not a perfect square, lim 2 if AE is a perfect square. x=∞ log log x Proof. It is sufficient to prove the theorem for A > 0 square-free and (A, E) = 1. Let Ax 2 − E = d > AE 2 and let d0 be the square-free kernel of d. Clearly p. (164) do p|d

The primes p dividing d have the property that AE is mod p a quadratic residue. If AE is not a perfect square the density of primes with that property is 1/2, hence by the prime number theorem p exp δ1 q(d) + o(q(d)) (165) p|d

where

δ1 =

1 1 2

if AE is a perfect square, otherwise.

On the other hand, d = d0 d12 ,

(Ax)2 − Ad0 d12 = AE.

Since (Ax)2 − AE > (AE)2 , Ad0 is not a perfect square. Moreover if E = ±4 we may assume Ad0 d1 odd. Let U1 , V1 be the least positive solution of the equation U 2 − Ad0 V 2 = AE

(166) and consider the recurrence

un = Ωωn + Ω ω , n

(167) where

) )

v

v ω = |AE|−1 U1 + V1 Ad0 , ω = |AE|−1 U1 − V1 Ad0 , ) )

Ω = U1 + V1 Ad0 /2, Ω = −U1 + V1 Ad0 /2

and v = 1 if AE = 1 or 4 or E = −d0 or −4d0 , v = 2 otherwise. It follows from Theorems 11 and 13 of√[19] that if E | 2, ω is the least greater than 1 totally positive unit of the√ ring generated by Ad0 and if E = 4, ω is the least such unit of the field R generated by Ad0 . Hence ω does not exceed the sixth power of the fundamental unit of R. Applying (157)(2 ) with D = Ad0 or 4Ad0 we get from (164) and (165)

) log ω = O d0 log d0 exp 21 δ1 q(d) + o(q(d)) . (2 )

See ibid., p. 225.

1147

J4. Two theorems of Gelfond

It follows further from the quoted theorems of [19] that all the positive integers V satisfying (166) for a suitable integer U , are contained in {un }. Thus in particular |d1 | = un . Since

ω/ω

=

(−Ω/Ω )v ,

it follows from Theorem 8 (3 ) that

q(d) q(d1 ) nv

or

24 nv.

Now, by (167) log un = n log ω + O(1) and we get

log d = log d0 + 2 log |d1 | δ1 q(d) + o q(d) + q(d) exp 21 δ1 q(d) + o q(d)

= exp 21 δ1 q(d) + o q(d) . Solving this inequality with respect to q(d) we obtain the theorem.

The theorems which follow go in the direction opposite to that of Theorems 11 and 12. Theorem 13. If ν, A, E are non-zero integers, ν 2, then ⎧ 2ν ⎪ ⎪ e−γ ⎪ ⎪ ϕ(2ν) ⎪ ⎪ ⎨ −γ log q(Ax ν − E) log log log x 2e lim ⎪ log |Ax ν − E| e−γ ⎪ x=∞ ⎪ ⎪ ν ⎪ ⎪ ⎩e−γ ϕ(ν) where γ is Euler’s constant and ϕ is Euler’s function.

if AE < −1, if AE = −1, if AE = 1, if AE > 1,

Proof. We assume without loss of generality A > 0, set for positive integers n: ⎧ −1 ν−1 2n A (A E) if AE < −1, ⎪ ⎪ ⎪ ⎨22n−1 if AE = −1, xn = n ⎪ 2 if AE = 1, ⎪ ⎪ ⎩ −1 ν−1 n A (A E) if AE > 1 and find log log log xn = log log n + o(1). On the other hand,

⎧ ν−1 2νn−1 − 1 if AE < −1, (A E) ⎪ ⎪ ⎪ ⎨(−2ν )2n−1 − 1 if AE = −1, Axnν − E = E × νn ⎪ if AE = 1, 2 −1 ⎪ ⎪ ⎩ ν−1 νn−1 (A E) − 1 if AE > 1.

(3 )

See ibid., p. 217.

1148

J. Prime numbers

Denoting by Xδ the δ-th cyclotomic polynomial and by d(δ) the number of divisors of δ we have for any positive integers g > 1 and m gm − 1 = Xδ (g) δ |m

and by [3], p. 178 q(g m − 1) max |Xδ (g)| max g ϕ(δ)+d(δ) g ϕ(m)+d(m) . δ |m

δ |m

It follows that

ϕ(kn − 1) + d(kn − 1) log log n log q(Axnν − E) log log log xn lim , lim log |Axnν − E| kn n=∞ n=∞

where k = 2ν if AE < −1, k = 2 if AE = −1, k = 1 if AE = 1 and k = ν if AE > 1. Now, a standard argument (cf. [14], §59) shows that ϕ(kn − 1) log log n k = e−γ . kn ϕ(k) n=∞ lim

Since d(kn − 1) log log n =0 n=∞ kn lim

the theorem follows. If ν = 2, E | 4, Theorem 13 can be improved to the following Theorem 14. If A, E, r, s are integers, Ar = 0, E | 4, then

log q A(rx + s)2 − E log log log x lim < ∞. log |A(rx + s)2 − E| x=∞

Proof. We assume without loss of generality that A > 0, r > 0, s > |E| and set √ √ √ √ s A + As 2 − E s A − As 2 − E α= , β= . √ √ |E| |E| ) Then A(As 2 − E) generates a real quadratic field and α 2 is a unit of this field. Let l be the least positive exponent such that α 2l ≡ 1 mod r(α + β). We set for positive integers n xn =

√ s |E| √ α 2ln+1 + β 2ln+1 − . r 2r A

√ |E| α 2ln+1 + β 2ln+1 s can be expressed rationally √ (α+β) = and the quotient r α+β 2r A in terms of (α + β)2 = 4As 2 /E and αβ = ±1, thus xn is rational. Moreover by the choice

We have

1149


of l α 2ln+1 + β 2ln+1 ≡ 1 mod r, α+β thus xn is an integer. Since α > |β|, we have log log log xn = log log n + o(1),

log A(rxn + s)2 − E = 2ln log α + O(1). On the other hand, A(rxn + s)2 − E =

2 |E| 2ln+1 − β 2ln+1 = (As 2 − E) Xδ2 (α, β), α 4 δ |2ln+1 δ>1

where Xδ (α, β) = β ϕ(δ) Xδ

α . β

Since Xδ (α, β) can be for δ > 2 expressed rationally in terms of (α + β)2 and αβ, all factors on the right hand side are rational integers and we get

q A(rxn + s)2 − E max q(As 2 − E), max |Xδ (α, β)| δ |2ln+1 δ>1

max q(As 2 − E), α ϕ(2ln+1)+d(2ln+1) . It follows like in the proof of Theorem 13:

log q A(rxn + s)2 − E log log log xn

lim log A(rxn + s)2 − E n=∞

ϕ(2ln + 1) + d(2ln + 1) log log n 2l = e−γ < ∞.

lim 2ln ϕ(2l) n=∞

Theorems 13 and 14 do not say anything about q f (x) for a general quadratic polynomial f (x). A much weaker but more general result is the following Theorem 15. If f (x) is any polynomial of degree ν > 1 with integer coefficients, then ⎧ 1

⎪ ⎨ 2 P (4) for ν = 2, log q f (x) 21 P (6) for ν = 3, lim ⎪ ⎩ x=∞ log |f (x)| P (ν) for ν > 3, where P (ν) =

∞

1−

i=1

1 , ui

u1 = ν − 1, ui+1 = u2i − 2.

In the proof of this theorem we denote by S the set of all polynomials with integer coefficients and the leading coefficient positive.

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J. Prime numbers

Lemma 10. If F (x) ∈ S is a polynomial of degree d there exists a polynomial H (x) ∈ S of degree d − 1 such that F H (x) has a factor G(x) ∈ S of degree d 2 − 2d. Proof. Let F (x) = a0 x d + . . . + ad . We set for any integer k a−1 a1 1 − − k = a0 1 − x − xHk (x) , Gk (x) = x d F x (d − 1)a0 (d − 1)a0 a1 − k = 0, Hk (x) is of where Hk (x) is a polynomial, Hk (0) = dk and if F − (d − 1)a0 degree d − 1 with the leading coefficient a1 −1 −k . −a0 F − (d − 1)a0 Clearly

Gk (x) 1 a1 + − −k (168) F Hk (x) − k ≡ F − a0 x x (d − 1)a0 a1 1 − ≡F − k ≡ 0 mod Gk (x). x (d − 1)a0 We choose k such that

(−1)d F −

a1 −k >0 (d − 1)a0

and set

H (x) = Hk (−1)d−1 (d − 1)2 a02 x − k.

It is easy to verify that H (x) ∈ S. On the other hand, in view of (168), F H (x) is divisible

by Gk (−1)d−1 (d −1)2 a02 x . The complementary factor of F H (x) is of degree d 2 −2d and its suitable multiple belonging to S can be taken as G(x).

Lemma 11. If f (x) satisfies the assumptions of Theorem 15, then for any

positive integer n there exists a polynomial hn (x) ∈ S of degree u1 u2 · · · un such that f hn (x) has a factor gn (x) ∈ S of degree un+1 + 1. Proof by induction with respect to n. For n = 1 the assertion follows from Lemma 10 on setting there F = ±f . Assume that f hn (x) has a factor gn (x) ∈ S of degree un+1 + 1. Applying

Lemma 10 with F = gn (x) we find a polynomial H (x) ∈ S of degree un+1 such that gn H (x) has a factor gn+1 (x) ∈ S of degree (un+1 + 1)2 − 2(un+1 + 1) = u2n+1 − 1 = un+2 + 1.

Clearly gn+1 (x)

is also a factor of F hn (H (x)) and we complete the proof by taking hn+1 (x) = hn H (x) .


1151

Proof of Theorem 15. It follows easily by induction that un+1 + 1 = ν

n

(ui − 1)

(n = 1, 2, . . . ).

i=1

un+1 + 1 tends to P (ν) decreasing monotonically. Since P (ν) P (4) = νu1 u2 · · · un 0.55 . . . > 21 for ν > 3, we have Hence

un+1 + 1 > νu1 · · · un − un+1 − 1. By Gauss’s Lemma we can assume that in Lemma 11 both polynomials gn (x) and f hn (x) /gn (x) have integer coefficients. It follows that for ν > 3

log q f (x) log q f (hn (x))

lim lim x=∞ log |f (x)| x=∞ log |f hn (x) |

log max |gn (x)|, |f hn (x) /gn (x)|

lim log |f hn (x) | x=∞ max{un+1 + 1, νu1 · · · un − un+1 − 1} un+1 + 1 = = . νu1 u2 · · · un νu1 u2 · · · un Since the last inequality holds for every n, we get

log q f (x) P (ν) lim x=∞ log |f (x)|

(ν > 3).

It remains to consider ν = 2 and ν = 3. If ν = 2 we have

f x + f (x) + f x + f (x) = f (x) 1 + f (x) + 21 f (x)f (x) f1 (x), where f1 (x) is a quartic polynomial with integer coefficients. It follows by the already proved part of the theorem

log q f1 (x) P (4) lim x=∞ log |f1 (x)| and

log max |f (x)|, 1 + f (x) + 21 f (x)f (x) , q f1 (x) log q f (x)

lim lim log f x + f (x) + f x + f (x)

x=∞ log |f (x)| x=∞ max 41 , 41 , 21 P (4) = 21 P (4).

If ν = 3 there exists by Lemma 10 a polynomial H (x) ∈ S such that

f H (x) = G1 (x)G2 (x), again where G1 , G2 are cubic polynomials with integer coefficients. Applying

Lemma 10 with F (x) = ±G1 (x) we find a polynomial H1 (x) ∈ S such that G1 H1 (x) = G3 (x)G4 (x), where G3 , G4 are cubic polynomials with integer coefficients. It follows by

1152

J. Prime numbers

the already proved part of the theorem

log q G2 (H1 (x))

P (6) x=∞ log |G2 H1 (x) |

and since f H H1 (x) = G2 H1 (x) G3 (x)G4 (x)

log q f (x) log max q G2 (H1 (x)) , |G3 (x)|, |G4 (x)|

lim lim log f H (H1 (x))

x=∞ log |f (x)| x=∞ max 21 P (6), 41 , 41 = lim

1 2

P (6).

This completes the proof. The above proof of Theorem 15 suggests the following

Problem. Does there exist for any polynomial f (x) ∈ S and any ε > 0 a polynomial h(x) ∈ S of degree d such that the degree of each irreducible factor of f h(x) is less than εd ? I do not know the answer to this problem even for f (x) = 4x 2 + 4x + 9, ε = 21 .

Addendum* ∞

1 , u1 3, ui+1 = ui i=1 u2i − 2. I have overlooked that already in 1929 A. Ostrowski [A] gave the value of this ( u21 − 4 (l.c., formula (7.10)). Hence Theorem 15 takes the form: product as u1 + 1 In the formulation of Theorem 15 occurs the product

1−

Theorem 15 . If f (x) is any polynomial of degree ν > 1 with integer coefficients then ⎧ √ 1

⎪ for ν = 2, ⎨ 4 √5 log q f (x) 41 21 lim for ν = 3, ⎪ log x ⎩) x=∞ 2 (ν − 1) − 4 for ν > 3.

References [3] G. D. Birkhoff, H. S. Vandiver, On the integral divisors of a n −bn . Ann. of Math. (2), 5 (1904), 173–180. [14] E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen. Reprint, Chelsea, New York 1953. *

Acta Arith. 56 (1990), 181


1153

[18] T. Nagell, Über den größten Primteiler gewisser Polynome dritten Grades. Math. Ann. 114 (1937), 284–292. [19] −−, Contributions to the theory of a category of Diophantine equations of the second degree with two unknowns. Nova Acta Soc. Sci. Upsal. (4) 16 (1955), no. 2. [A] A. Ostrowski, Ueber einige Verallgemeinerungen des Eulerschen Produktes. Verh. Naturforsch. Ges. Basel 40 (1929), 153–214; Collected Mathematical Papers, vol. 3, Birkhäuser, Basel 1984, 352–413.

Originally published in Acta Arithmetica XXXVIII (1980), 285–322


On the relation between two conjectures on polynomials

1. The aim of this paper is to establish a relation between the conjecture H on simultaneous representation of primes by several irreducible polynomials (see [12] and [5]) and a conjecture on Diophantine equations with parameters that we shall denote by C. Both conjectures involve the notion of the fixed divisor of a polynomial, i.e. the greatest common divisor of all values the polynomial takes for integral values of the arguments. The conjectures run as follows. H. Let f1 (x), . . . , fk (x) be irreducible polynomials with integral coefficients and the k leading coefficients positive such that fj (x) has the fixed divisor 1. Then there exist j =1

infinitely many positive integers x such that all numbers fj (x) are primes. C. Let F (x, y) ∈ Z[x, y] be a form such that (1)

F (x, y) = F1 (ax + by, cx + dy) for any F1 ∈ Z[x, y] and any a, b, c, d ∈ Z

a b

= ±1. implies

c d

If f (t1 , . . . , tr ) ∈ Z[t1 , . . . , tr ] has the fixed divisor equal to its content and the equation (2)

F (x, y) = f (t1 , . . . , tr )

is soluble in integers x, y for all integral vectors [t1 , . . . , tr ] then there exist polynomials X, Y ∈ Z[t1 , . . . , tr ] such that identically

(3) F X(t1 , . . . , tr ), Y (t1 , . . . , tr ) = f (t1 , . . . , tr ). A conjecture similar to C has been proposed by Chowla [3]. He has made no assumption (1) but required F and f to be irreducible and have the fixed divisor 1. The following example shows that this is not enough: F (x, y) = x 2 + 3y 2 ,

f (t1 , t2 ) = t12 + t1 t2 + t22 .

In this example the set of values of F (x, y) and of f (t1 , t2 ) is the same, but F and f are not equivalent by unimodular transformation, which answers in the negative a question of

J5. Relation between two conjectures on polynomials

1155

Chowla (ibid., p. 73) repeated in [9]. The condition imposed in C on the fixed divisor of f is essential, as the following example shows F (x, y) = 2x 2 y 3 ,

f (t) = t 3 (t + 1)4 .

Here the solutions of the equations (2) are given by x = 2(t + 1)2 , y = 21 t

if

t ≡ 0 mod 2,

x=

if

t ≡ 1 mod 2,

1 4 (t

+ 1) , y = 2t 2

but there are no integer-valued polynomials X(t), Y (t) satisfying (3). Another example with F primitive is given at the end of Section 2. One special case of C corresponding to F = x 2 + y 2 has been proved in [3] and [4]. Chowla has also indicated how his conjecture for F (x, y) quadratic should follow from the special case k = 1 of H. We shall extend these results in the following two theorems. Theorem 1. C holds if F (x, y) = x k y l (k 1, l 1) or if F is quadratic and equivalent (properly or improperly) to every form in its genus. For such and for no other quadratic F C extends to all polynomials f ∈ Z[t1 , . . . , tr ]. Theorem 2. H implies C if F is a quadratic form or a reducible cubic form. We shall see (Corollary to Lemma 3) that C implies the following, less precise but more general assertion. D. Let F (x, y) ∈ Z[x, y] be any form and f ∈ Z[t1 , . . . , tr ] any polynomial. If the equation (2) is soluble in integers x, y for all integral vectors [t1 , . . . , tr ] then there exist polynomials X, Y ∈ Q[t1 , . . . , tr ] satisfying (3). D has been proved for F = x n and any r in [7] and [11] also for any irreducible quadratic F and r = 1 in [4], r > 1 in [14]; for reducible quadratic F it follows easily. We shall show Theorem 3. H implies D if F factorizes into two relatively prime factors in an imaginary quadratic field. In virtue of Theorem 3 H implies D for F = x n + y n . By a modification of the proof of that theorem in this special case we shall show yet Theorem 4. H implies C if F (x, y) = x n + y n (n 2). For n = 2 and for no other n in question C extends to all polynomials f ∈ Z[t1 , . . . , tr ]. At the cost of considerable technical complications indicated briefly later one can extend Theorem 2 to all forms F splitting completely over a cyclic field except those with all zeros conjugate and real. The quantitative version of H formulated by Bateman and Horn [1] (see also [5]) implies C in the exceptional case at least for r = 1. Similarly Theorem 3 can be extended to all forms F that factorize into two distinct complex conjugate factors over an imaginary cyclic field.

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J. Prime numbers

2. In the sequel we shall use the vector notation and write t instead of [t1 , . . . , tr ], t instead of [t2 , . . . , tr ], t for max |ti |. We shall denote the content of a polynomial f by 1ir

C(f ), its total degree by |f | and call a form F satisfying (1) primary. The letters N, Z, Q denote the set of positive integers, the ring of integers and the rational field, respectively. For a fixed field K N denotes the norm from K to Q or from K(t) to Q(t). The content of a polynomial over K is an ideal of K but if K = Q it is often identified with the positive c generator of this ideal. All considered forms are defined over Z unless stated to the contrary. Lemma 1. Let P ∈ Z[t], p be a prime dividing neither the leading coefficient nor the discriminant of P . If t0 ∈ Z, P (t0 ) ≡ 0 mod p then either P (t0 ) ≡ 0 mod p2 or P (t0 + p) ≡ 0 mod p2 . Proof. Denoting the leading coefficient of P by a, the discriminant of P by D and its derivative by P we have P (t)U (t) + P (t)V (t) = aD, where U, V ∈ Z[t]. Setting t = t0 we infer from P (t0 ) ≡ 0 mod p, aD ≡ 0 mod p that P (t0 ) ≡ 0 mod p. Now from the expansion P (t0 + p) = P (t0 ) + P (t0 )p +

P (t0 ) 2 p + ... 2

we get P (t0 + p) − P (t0 ) ≡ 0 mod p2 , whence the assertion.

Lemma 2. If a quadratic form F is primary then F = AG(x, y),

where A ∈ Z, G(x, y) ∈ Z[x, y],

A is square-free, the discriminant Δ of G is either 1 or fundamental and

Δ p

= −1 for

every prime factor p of A. Proof. If G is reducible, G = (ax + by)(a x + b y) we have F (x, y) = (Aax + Aby)(a x + b y) and by (1)

a A

a

b

= ±1, b

A = ±1

and

a Δ =

a

2 b

= 1. b

2 2 If G is irreducible, √ let G = ax + bxy + cy , and let ω1 , ω2 be a basis of the ideal b+ Δ a = a, . Then we have for suitable integers a1 , a2 , b1 , b2 2

a = a1 ω1 + a2 ω2 , √ b+ Δ = b1 ω1 + b2 ω2 . 2


1157

√ Let K = Q( Δ) and let us set F1 (x, y) = Aa −1 N (xω1 + yω2 ). Since Na = |a| and (ω1 , ω2 ) ≡ 0 mod a we have F1 (x, y) ∈ Z[x, y]. On the other hand

√ b+ Δ y = (a1 x + b1 y)ω1 + (a2 x + b2 y)ω2 , ax + 2

hence F (x, y) = F1 (a1 x + b1 y, a2 x + b2 y) and by (1)

a1

a2

b1

= ±1. b2

√ b+ Δ is itself a basis for a and by a well known result It follows that a, 2

√

a b + Δ

1 2√ , |a| = √ abs

|d|

a b − Δ

2

c

where d is the discriminant of K. It follows that Δ = d is a fundamental discriminant. If Δ A is not square-free or for some p | A we have = 0 or 1 then for a suitable prime p ideal p: N p | A. Let pa have an integral basis [Ω1 , Ω2 ] and let us set F1 (x, y) = Aa −1 N p−2 N (xΩ1 + yΩ2 ). Since N(Ω1 , Ω2 ) = |a|N p we have F1 (x, y) ∈ Z[x, y]. On the other hand ωi N p = ci Ω1 + di Ω2

(i = 1, 2)

for suitable ci , di ∈ Z, hence (ω1 x + ω2 y)Np = (c1 x + c2 y)Ω1 + (d1 x + d2 y)Ω2 and we get F (x, y) = F1 (c1 x + c2 y, d1 x + d2 y).

1158

J. Prime numbers

Now by (1)

c1

d1

c2

= ±1, d2

hence [ω1 N p, ω2 N p] is a basis for ap and aN p = ap, a contradiction.

Remark. Similarly one can show that if a primary form F (x, y) is irreducible and F (ϑ, 1) = 0 then [1, ϑ] can be extended to a basis of the ideal (1, ϑ). Proof of Theorem 1. Consider first F (x, y) = x k y l and let (4)

f (t) = c

n

fν (t)eν

ν=1

be the canonical factorization of f into primitive irreducible polynomials with integral coefficients. In view of the condition on the fixed divisor of f for every prime factor p of c there exists a vector t p ∈ Zr such that n

fν (t p )eν ≡ 0 mod p.

ν=1

It follows from (2) with t = t p that ordp c = kα + lβ, where α = ordp x, β = ordp y, and we get (5)

c = ±ξ k ηl ,

ξ, η ∈ Z.

On the other hand we can assume that f (t) depends upon t1 . Let a0 (t ), D(t ) be the n leading coefficient and the discriminant respectively of fν (t) with respect to t1 . We ν=1

have a0 D = 0 and there exists a vector t 0 ∈ Zr−1 such that a0 (t 0 )D(t 0 ) = 0.

For every ν n there exists a prime p and an integer t0 such that (6)

fν (t0 , t 0 ) ≡ 0 mod p,

ca0 (t 0 )D(t 0 ) ≡ 0 mod p.

Put (7)

P (t) =

n

fν (t, t 0 ).

ν=1

Since a0 (t 0 ) = 0, the discriminant of P (t) equals D(t 0 ). Hence by (6) and Lemma 1 there exists a t1 ∈ Z such that P (t1 ) ≡ 0 mod p,

P (t1 ) ≡ 0 mod p2 .


1159

We infer from (4), (5) and (6) that (8) fν (t1 , t 0 ) ≡ 0 mod p,

fν (t1 , t 0 ) ≡ 0 mod p2 ,

fμ (t1 , t 0 ) ≡ 0 mod p (μ = ν).

It follows from (2) with t = [t1 , t 0 ], (6) and (8) that eν = kαν + lβν ,

(9)

where αν = ordp x, βν = ordp y. Take now X0 (t) = ξ

n

fν (t)αν ,

Y0 (t) = η

ν=1

n

fν (t)βν .

ν=1

It follows from (5) and (9) that X0 (t)k Y0 (t)l = ±f (t). If the sign on the right hand side is positive we take X = X0 , Y = Y0 . If the sign is negative and either k or l is odd, we take X = ±X0 , Y = ±Y0 . If the sign is negative and k, l are both even we get a contradiction. Indeed, by (5) c < 0, by (9) eν ≡ 0 mod 2, hence by (4) f (t) 0. Taking t ∈ Zr such that f (t) = 0 we get from (2) x k y l < 0, which is impossible. Consider now the case of F quadratic. By Lemma 2 F is of the form AG(x, y), where Δ = −1 for every A is square-free, G(x, y) is a primitive form with discriminant Δ, p prime factor p of A and either Δ = 1 or Δ is fundamental. In the first case F (x, y) is equivalent to xy and for the latter form one can take X(t) = f (t), Y (t) = 1. In the second case, if G(ϑ, 1) = 0, K = Q(ϑ) and a is the ideal (1, ϑ), we have N (x − ϑy) . Na Changing, if necessary, the sign of A we can assume that G(x, y) =

A N (x − ϑy). Na Na f (t) for all t ∈ Zr implies, by Theorem 1 The solubility of the equation N (ω) = A of [14], the existence of a polynomial ω(t) ∈ K[t] such that F (x, y) =

(10)

Na f (t). N ω(t) = A

(11) Let b = C(ω) and let ba

−1

=

j i=1

ba−1

pai i

j i=1

i pb i

k

qici

i=1

be the factorization of in prime ideals of K. Here pi are distinct pairwise nonconjugate prime ideals of degree 1 in K, pi is conjugate to pi and qi are prime ideals of degree 2 in K. Since AN (ba−1 ) ∈ Z and A has only prime ideal factors of degree 2 in K,

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J. Prime numbers

we get ai + bi 0 2ci + 1 0

(1 i j ), (1 i k),

hence (12)

max{0, ai } + min{0, bi } 0,

max{0, bi } + min{0, ai } 0

ci 0

(1 i j ),

(1 i k).

Let us consider the ideal c=

j

min(0,bi )−min(0,ai ) min(0,ai )−min(0,bi ) pi .

pi

i=1

Since F is equivalent to every form in its genus the same is true about G, thus there is only one narrow class in the genus of a, or there are two such classes represented by a and a . In any case the principal genus consists only of the principal class and the class of a2 . Since 2 pi ∼ p−1 i , c belongs to the principal genus and we get c ∼ 1 or c ∼ a . In the former case let c = (γ1 ) with γ1 totally positive and consider the polynomial ω1 (t) = γ1 ω(t). We have C(ω1 ) = (γ1 )C(ω) = cb = a

j

max{0,ai }+min{0,bi }

pi

i=1

j

max{0,bi }+min{0,ai }

pi

i=1

k

qici

i=1

and by (12) C(ω1 ) ≡ 0 mod a. It follows that all the coefficients of ω1 are in a and since, by Lemma 2, [1, ϑ] is a basis of a, we get ω1 (t) = X1 (t) − ϑY1 (t), where X1 , Y1 ∈ Z[t]. It follows now from (10) and (11) that

A A N ω1 (t) = N γ1 N ω(t) = N c · f (t) = f (t). F X1 (t), Y1 (t) = Na Na In the case c ∼ a2 let ca−1 a = (γ2 ) with γ2 totally positive and consider the polynomial ω2 (t) = γ2 ω(t). We have C(ω2 ) = (γ2 )C(ω) = ca−1 a b = a

j i=1

and by (12) C(ω2 ) ≡

0 mod a .

max{0,ai }+min{0,bi }

pi

j i=1

max{0,bi }+min{0,ai }

pi

k i=1

qici ,


1161

Since [1, ϑ ] is a basis of a , we infer that ω2 (t) = X2 (t) − ϑ Y2 (t), where X2 , Y2 ∈ Z[t]. Since N γ2 = 1, it follows as before that

F X2 (t), Y2 (t) = f (t). It remains to prove that if there is a form inequivalent to F in the genus of F , then C does not extend to all polynomials f ∈ Z[t]. For this purpose let us observe that there exists then in K a class C of ideals such that C 2 is neither the principal class nor the class of a2 . Choose in C −1 a prime ideal p of degree 1 with N p = p. There exists a prime ideal q such that p2 aq is principal, equal to, say (α). Consider the polynomials ω(t) = α

(13)

tp − t , p

f (t) =

A N ω(t). Na

We have C(f ) = hence f (t) ∈ Z[t]. Also, since ω(t) = x − ϑy and

|A| |N α| = |A|N q ∈ Z, Na p 2

tp − t ∈ Z for all t ∈ Z we have for all t ∈ Z: ω(t) ∈ a; p f (t) = F (x, y)

for suitable x, y ∈ Z. On the other hand, suppose that

(14) f (t) = F X(t), Y (t) , X, Y ∈ Z[t] and let x, y be the leading coefficients of X, Y . Then comparing the leading coefficients on both sides of (14) we get by (13) A A Nα = F (x, y) = N (x − ϑy), 2 Na p Na

Nq = N

(x − ϑy) . a

Since q is a prime ideal, x − ϑy ∈ a, it follows that (x − ϑy) =q a

or

q .

Hence aq ∼ 1 or aq−1 ∼ 1. By the choice of q this gives p2 ∼ 1 or p2 a2 ∼ 1 contrary to the choice of p.

Remark. The above proof seems to suggest that if F satisfies (1) and for all t ∈ Zr the equation (2) is soluble in integers x, y, then there exist integer-valued polynomials X(t), Y (t) satisfying (3) identically. The following example shows that this is not the case. √ √ 1 + −23 , Let F (x, y) = x 2 + xy + 6y 2 , K = Q( −23), ω = 2

f (t) = N ( 21 ω4 − ω)t 2 + ω − 8 .

1162

J. Prime numbers

The discriminant of F is −23 hence F is primary. Further, f (t) ∈ Z[t] since (2, ω) ( 21 ω4 − ω, ω − 8) = with ω conjugate to ω. (2, ω ) Moreover the equation F (x, y) = f (t) is soluble in integers x, y for all t ∈ Z. Indeed if t ≡ 0 mod 2 we can take

x + yω = 21 ω4 − ω t 2 + ω − 8 and if t ≡ 1 mod 2 we can take x + yω =

−3 − −3 +

√ √

−23 −23

( 21 ω4 − ω)t 2 + ω − 8 .

The number on the right hand side is an integer in K since for t ≡ 1 mod 2 ( 21 ω4 − ω)t 2 + ω − 8 ≡ 21 ω4 − 8 mod 4(ω4 − 2ω) and we have in K the factorization into prime ideals √

(2) = pp , (ω) = pq, (−3 + −23)/2 = p3 . On the other hand, the polynomial ( 21 ω4 − ω)t 2 + ω − 8 is irreducible over K since 8−ω 62 N 1 is not a square in Q. Therefore, if integer-valued polynomials X(t), = 4 381 2ω − ω Y (t) satisfied

F X(t), Y (t) = f (t) identically, we should have either X(t) + Y (t)ω = γ ( 21 ω4 − ω)t 2 + γ (ω − 8) or X(t) + Y (t)ω = γ ( 21 ω4 − ω)t 2 + γ (ω − 8) for some γ ∈ K with N γ = 1. Taking t = 0 and 1 we should get γ ( 21 ω4 − ω, ω − 8) p p integral, hence (γ ) integral and (γ ) = . However the ideal on the right hand side is p p not principal.

3. c

Lemma 3. Every form F (x, y) with at least two distinct zeros (in P1 (C)) can be repre a b

= 0. sented as F1 (ax + by, cx + dy), where F1 is primary, a, b, c, d ∈ Z and

c d

a b

= 0. Let F ∗ be the product Proof. Suppose that F (x, y) = G(ax + by, cx + dy),

c d

∗ for G. It c of all projectively distinct primitive irreducible factors of F and similarly G


1163

follows that F ∗ = ±C −1 G∗ (ax + by, cx + dy),

where C = C G∗ (ax + by, cx + dy) | C(F ). Hence

|F ∗ |(|F ∗ |−1)

∗ 2−2|F ∗ | ∗ a b

disc G ·

disc F = C c d

a b

is bounded. Take now a and since disc F ∗ = 0, |F ∗ | > 1 the absolute value of

c d

a b

is maximal. G must representation of F (x, y) as G(ax + by, cx + dy), where abs

c d

c be primary, otherwise representing it as G1 (a1 x + b1 y, c1 x + d1 y) we should obtain a representation of F as G1 (αx + βy, γ x + δy) with

a b

α β

a b

a1 b1

,

> abs

abs

= abs

· abs

c c1 d1

c d

γ δ

c d

a1 b1

= 0. In the latter case, however, G and hence

contrary to the choice of G, unless

c1 d1

also F should have only one zero, contrary to the assumption.

c

Corollary. C implies D. Proof. Let F (x, y) ∈ Z[x, y] be any form, f (t) ∈ Z[t] any polynomial and suppose that for all t ∈ Zr there exist x, y ∈ Z satisfying F (x, y) = f (t). If F (x, y) = const or f (t) = const, D is trivial. If F (x, y) has only one zero, we take without loss of generality F (x, y) = a(bx + cy)n , where b = 0. Applying Theorem 3 of [13] to the equation n n c au = f (t) we infer the existence of a polynomial U (t) ∈ Q[t] such that aU (t) = f (t). It suffices to take X(t) = b−1 U (t), Y (t) = 0. If F (x, y) has at least two distinct

zeros

then, by Lemma 3, F (x, y) = F1 (ax + by,

a b

= 0. On the other hand there exists a vector cx + dy), where F1 is primary and

c d

t 0 ∈ Zr such that f (t 0 ) = e = 0. Consider now the equation F1 (x, y) = f (et + t 0 ). The polynomial on the right hand side has both the content and the fixed divisor equal to |e|, hence by C there exist polynomials X1 , Y1 ∈ Z[t] such that F1 X1 (t), Y1 (t) = f (et +t 0 ). Determining X(t), Y (t) from the equations t − t t − t 0 0 , cX(t) + dY (t) = Y1 aX(t) + bY (t) = X1 e e we get

X(t), Y (t) ∈ Q[t], F X(t), Y (t) = f (t), thus D holds.

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J. Prime numbers

Lemma 4. H implies the following. Let fν ∈ Z[t] (1 ν n) be distinct irreducible polynomials such that their leading n forms hν (t) all assume a positive value for a t ∈ Nr and that fν (t) has the fixed ν=1

divisor 1. Then for any B there exists a t ∈ Nr such that fν (t) are distinct primes > B. Proof. The condition that fν are irreducible and distinct implies that they are prime to each other. Indeed, otherwise two of them would differ by a constant factor c = 1. The n numerator and the denominator of c would divide fν (t) for all t hence c = −1. But ν=1 this contradicts the condition on hν . Let us choose an a ∈ Nr such that (15)

hν (a) > 0

(1 ν n)

and let n

a = |h1 | + |h2 | + . . . + |hr | ! hν (a). ν=1

Since f (t) =

n

fν (t)

ν=1

has the fixed divisor 1 we infer from the Chinese Remainder Theorem the existence of a τ ∈ Zr such that

(16) f (τ ), a = 1. Consider the polynomials fν (ax + at + τ ) (1 ν n). They are irreducible as polynomials in x, t and prime to each other. Consequently the resultant Rμ,ν (t) of fμ (ax + at + τ ) and fν (ax + at + τ ) is non-zero for all μ < ν n. By Hilbert’s irreducibility theorem there exists a t 0 ∈ Zr such that fν (ax + at 0 + τ ) (1 ν n) are all irreducible as polynomials in x and n

(17)

Rμ,ν (t 0 ) = 0.

μ |B| +

n

|Rμ,ν (t 0 )|.

μ B. They are distinct since the common value of fμ (ax + at 0 + τ ) and fν (ax + at 0 + τ ) would have to divide Rμ,ν (t 0 ) which is impossible by (17) and (18).

Lemma 5. Let K be the rational field or a quadratic field, Δ be the discriminant of K and let ϕν ∈ K[t] (1 ν n) be polynomials irreducible over K and prime to each other. If (19)

the fixed divisor of

n

N ϕν (t) equals

ν=1

n

N C(ϕν )

ν=1

then for every M ∈ N, there exists a μ ∈ N prime to M with no prime ideal factor of degree 1 in K and τ ∈ Zr with the following property. Let ψν (t) = ϕν (μt + τ ) (1 ν n). n N ψ (t ) ν 1 H implies the existence of a N C(ψ ν=1

ν) ψν (t 2 ) are prime in K, distinct and do t 2 ∈ Nr such that t 2 ≡ t 1 mod m, all the ideals C(ψν ) not divide A. Moreover, either μ = 1, τ = 0 have the above property (this happens for K = Q) or there is a sequence of pairs μi , τ i with the above property such that (μi , μh ) = 1 for i = h, and the number of distinct μi x is greater than cx 1/n / log x for a certain c > 0 and all x > x0 .

For any A ∈ N, t 1 ∈ Zr and m ∈ N prime to Δ

Proof. We begin with a remark concerning the fixed divisor that we shall use twice. If P ∈ Z[t] has the fixed divisor d then any fixed prime divisor p of P (mt + a) divides dm. Indeed, if p /| d then there exists a u ∈ Zr such that P (u) ≡ 0 mod p and if p /| m there exists a v ∈ Zr such that mv + a ≡ u mod p, hence P (mv + a) ≡ 0 mod p. Now we proceed to the proof of the lemma. Let ϕν (t) = aν fν (t) N ϕν (t) = aν fν (t)

(ν k), (k < ν n),

where fν ∈ Z[t] are irreducible over Q and (aν ) = C(ϕν )

(ν k),

|aν | = N C(ϕν )

(k < ν n).

(If K = Q we take k = 0.) Let hν be the leading form of fν . We can choose the signs

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J. Prime numbers

of aν so that for a suitable t ∈ Nr : hν (t) > 0 for all ν n. We have n k n N ϕν (t) =± fν2 (t) fν (t) N C(ϕν )

(20)

ν=1

ν=1

ν=k+1

and (19) implies on an application of the Chinese Remainder Theorem that for a suitable τ 0 ∈ Zr n Δ, fν (τ 0 ) = 1.

(21)

ν=1

Let fν (τ 0 ) ≡ ν mod Δ, ν > 0 (ν k). Without loss of generality we may assume that Δ Δ (22) = 1 (1 ν j ), = −1 (j < ν k). c ν ν Since ϕν are prime to each other (23)

(fλ , fν ) = 1

unless λ = ν or λ > k, ν > k and ϕλ /ϕν ∈ K,

where ϕν is conjugate to ϕν over Q(t). n fν are prime to each other. Let t = [t, t ], a0 (t ) In particular, f1 , . . . , fj and ν=j +1

be the leading coefficient of

n

fν (t), D(t ) the discriminant of

ν=1

resultant of

j ν=1

(24)

fν (t),

n ν=j +1

j

fν (t) and R(t ) the

ν=1

fν (t) with respect to t. It follows that a0 DR = 0.

Since fν (t) are irreducible over K for ν j we infer by Hilbert’s irreducibility theorem that there exists a τ ∈ Zr−1 such that fν (t, τ ) are irreducible over K for ν j and (25)

a0 (τ )D(τ )R(τ ) = 0.

Let fν (ϑν , τ ) = 0 and K ν = Q(ϑν ) (ν j ). We have K ⊂ K ν and by Bauer’s theorem there exist for each ν j infinitely many primes with a prime ideal factor of degree 1 in K ν , but not in K. Choose for each ν j a different prime pν with the above property and such that (26)

pν /| Ma0 (τ )D(τ )R(τ ).

Since pν does not split in K we have Δ (27) = −1 pν

(ν j ).

On the other hand, since pν has a prime ideal factor of degree 1 in K ν , by Dedekind’s theorem, there exists an integer u such that fν (u, τ ) ≡ 0 mod pν .

1167


j fi (t, τ ) equals D(τ ) ≡ 0 mod pν . Since By (25) and (26) the discriminant of j i=1 a0 (τ ) ≡ 0 mod pν and fi (u, τ ) ≡ 0 mod pν we infer from Lemma 1 that either i=1

j

fi (u, τ ) ≡ 0 mod pν2

i=1

or j

fi (u + pν , τ ) ≡ 0 mod pν2 .

i=1

Therefore, there exists an integer τν such that fν (τν , τ ) ≡ 0 mod pν ,

(28)

j

(29)

fi (τν , τ ) ≡ 0 mod pν2 .

i=1

Moreover, since by (25) and (26) the resultant of

fi (t, τ ) and

i=1

to R(τ ) ≡ 0 mod pν , we have (30)

j

n

n i=j +1

fi (t, τ ) is equal

fi (τν , τ ) ≡ 0 mod pν .

i=j +1

Let us choose τ ≡ τν mod pν2 (1 ν j ) and set (31)

μ=

j

pν ,

τ = [τ, τ ].

ν=1

By (28)–(30) we have fν (τ ) ≡ 0 mod pν ,

(32)

n

(33)

fi (τ ) ≡ 0 mod pν2 .

i=1

We shall show that n

fi (μt + τ ) = P (μt + τ )

i=1

has the fixed divisor d equal to p1 p2 · · · pj . Indeed by (19) and (20) the fixed divisor of P (t) equals 1, hence d consists of prime factors of μ. However by (33) d ≡ 0 mod pν2

(ν j ).

On the other hand by (31) and (32) fν (μt + τ ) ≡ fν (τ ) ≡ 0 mod pν .

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J. Prime numbers

Thus d = p1 p2 · · · pj , the polynomials gν (t) = pν−1 fν (μt + τ ) gν (t) = fν (μt + τ )

(34)

have integral coefficients,

n

(ν j ), (j < ν n)

gν (t) has the fixed divisor 1 and a fortiori the content 1.

ν=1

Moreover by (23) gλ = gν

(35)

unless λ = ν or λ > k, ν > k and ϕλ /ϕν ∈ K.

It follows that ψν (t) = aν pν gν (t)

(ν j ),

ψν (t) = aν gν (t)

(j < ν k),

N ψν (t) = aν gν (t)

(k < ν n),

(36) (37) where besides (38)

C(ψν ) = (aν pν ) (ν j ),

C(ψν ) = (aν )

N C(ψν ) = |aν |

(39)

(j < ν k),

(k < ν n).

It follows that n k n N ψν (t) =± gν2 (t) gν (t). N C(ψν )

ν=1

ν=1

If now for a t 1 ∈ Zr we have

ν=k+1

n N ψν (t 1 ) m, Δ =1 N C(ψν ) ν=1

there exists a t 0 ∈

Zr

satisfying t 0 ≡ t 1 mod m,

(40)

μt 0 + τ ≡ τ 0 mod Δ.

Since m,

n

n gν (t 0 ) = m, gν (t 1 ) = 1

ν=1

ν=1

and by (34) and (21) Δ,

n ν=1

n n gν (t 0 ) = Δ, gν (0) = Δ, fν (τ 0 ) = 1, ν=1

it follows that n ν=1

gν (Δmt + t 0 )

ν=1


1169

has the fixed divisor 1. The polynomials gν (Δmt + t 0 ) are irreducible and their leading forms all take a positive value for a suitable t ∈ Nr in virtue of the corresponding property of fν (t). By Lemma 4 H implies the existence of an x ∈ Nr such that gν (Δmx + t 0 ) are primes greater than |A| and gλ (Δmx + t 0 ) = gν (Δmx + t 0 )

(41)

unless

gλ = gν .

Taking t 2 = Δmx + t 0 we get from (40) t 2 ≡ t 1 mod m,

(42)

μt 2 + τ ≡ τ 0 mod Δ.

Thus by (34) pν gν (t 2 ) = fν (μt 2 + τ ) ≡ fν (τ 0 ) ≡ ν mod Δ gν (t 2 ) = fν (μt 2 + τ ) ≡ fν (τ 0 ) ≡ ν mod Δ and we infer from (22) and (27) that Δ = −1 gν (t 2 )

(ν j ), (j < ν k)

(ν k).

Hence, for ν k, gν (t 2 ) are prime

in K not dividing A and in virtue of (36) and (38) the ψν (t 2 ) same applies to the ideals aν = . The remaining ideals aν (k < ν n) are prime c C(ψν ) and do not divide A in virtue of (37) and (39). Assuming λ = ν,

aλ = aν ,

we get by (35) and (41) for a suitable γ ∈ K λ > k,

ν > k,

ϕλ = γ ϕν , ψλ = γ ψν , C(ψλ ) = (γ )C(ψν ),

ψν (t 2 ) ψν (t 2 ) = , C(ψν ) C(ψν )

thus the ideal aν is ambiguous. By Dedekind’s theorem aν | Δ, hence by (37) and (39) gν (t 2 ) | Δ. However by (34) and (42) gν (t 2 ) = fν (μt 2 + τ ) ≡ fν (τ 0 ) mod Δ and we get a contradiction with (21). The contradiction shows that the ideals aν are distinct and the proof of the first part of the lemma is complete. To prove the second part we note that if j = 0 then (31) gives μ = 1. The value of τ is then irrelevant and can be taken 0. Therefore assume that j > 0 and that we have already defined μ1 , τ 1 , . . . , μi−1 , τ i−1 (i 1), each μi with j prime factors. Then we replace in the above proof M by Mμ1 · · · μi−1 and define μi , τ i by (31). It is clear that the sequence thus obtained satisfies (μi , μh ) = 1 for i = h. Denote by P (K ν ) the set of primes with a prime ideal factor of degree 1 in K ν . By Bauer’s theorem P (K ν ) \ P (K)

1170

J. Prime numbers

has a positive density, say, δν . Computing μi from (31) we take pν to be the least element of P (K ν ) \ P (K) different from ω + j (i − 1) + ν − 1 given primes, where ω is the number of prime factors of Ma0 (τ )D(τ )R(τ ). Hence for i > i0 we have pν 2δν−1 j i log j i and μi =

j

pν (c−1 j i log j i)j ,

c=

ν=1

1 2

j

δν1/j .

ν=1

Since the number of solutions of the inequality (c−1 j i log j i)j x in positive integers i is for x large enough at least is at least

cx 1/j , the number of distinct μi x log x − 1

cx 1/n cx 1/j − i0 > log x − 1 log x

(x > x0 )

which completes the proof. Remark. The lemma extends to all cyclic fields. c c

Lemma 6. Let K be any field, K an algebraic closure of K, f ∈ K[t] a non-zero polynomial. If a form F ∈ K[x, y] has at least three distinct zeros in P1 (K) then there exist no more than |F |3 3|f | pairs X(t), Y (t) such that X, Y ∈ K[t], X, Y linearly independent over K and

(43) F X(t), Y (t) = f (t). Proof. Without loss of generality we may assume that K is algebraically closed. By a linear transformation we can transform F to the form

F (x, y) = x k y l G(x, y), k 1, l 1, G(x, y), xy = 1. Let us assign two solutions X1 , Y1 and X2 , Y2 of (43) to the same class if X2 = ξ X1 , Y2 = ηY1 for some ξ, η ∈ K \ {0}. The number of classes does not exceed the number of pairs of monic polynomials x, y ∈ K[t] such that xy | f (t), which is clearly bounded by 3|f | . The number of polynomials in one class can be estimated as follows. If F (ξ X1 , ηY1 ) = F (X1 , Y1 ) then

X X 1 1 ,η = F ,1 F ξ Y1 Y1


1171

and since X1 /Y1 takes in K infinitely many values we have identically F (ξ u, η) = F (u, 1). Hence ξ k ηl G(ξ u, η) = G(u, 1) and the comparison of the leading coefficients and of the constant terms on both sides gives ξ k ηl ξ |G| = 1,

ξ k ηl η|G| = 1.

It follows that ξ |G| = η|G| ,

ξ |G|(k+l+|G|) = 1,

ξ |G| |F | = 1.

Thus there are |G| |F | possibilities for ξ and for each ξ at most |G| possibilities for η, which gives at most |F | |G|2 |F |3 possibilities for ξ, η . The lemma follows.

Lemma 7. If F (x, y) ∈ Z[x, y] is a non-singular cubic form, then for every integer 3 a = 0 the number of solutions of the

equation F (x, y) = az in integers x, y, z such that b (x, y, z) = 1 and 1 z Z is O (log Z) , where b is a constant depending on F and a. Proof. It is enough to estimate the number of solutions with |x| |y|. Assume that F (x, y) = az3 ,

|x| |y|.

If F (1, 0) = 0 we have |F (x, y)| |y| hence h = max |x|, |y|, |z| Z 3 , where the constant in the symbol depends on a, later also on F . If F (1, 0) = 0 let

(44)

(45)

1zZ

F (x, y) = a0

3

and

(x − ξl y),

l=1

where ξ1 is the real zero of F nearest to x/y. Since F (x, y) = 0 we have by Thue’s theorem |x − ξ1 y| |y|−3/2 . On the other hand |x − ξ2 y| |x − ξ3 y| y 2 . Hence by (44) and (45) |a|z3 = |F (x, y)| y 1/2

and

h Z6.

Since F (x, y) = az3 represents in projective coordinates a curve of genus

1, in virtue of a theorem of Néron (see [8], p. 82), the number of solutions of (44) is O (log Z 6 )g/2+1 where g is the rank of the curve.

Remark. The lemma extends to all forms F with at least three distinct zeros. If the genus of the curve F (x, y) = az|F | is greater than 1 one needs a theorem of Mumford [10]. Lemma 8. Let K be any field, U a finite subset of K and P ∈ K[t], P = 0. The equation P (t) = 0 has no more than |P | |U |r−1 solutions t ∈ U r , where |U | is the number of elements of U .

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J. Prime numbers

Proof (by induction on r). For r = 1 the assertion is obvious. Assume that it holds for polynomials in r − 1 variables and let P (t) =

p

Pi (t )t1

p−i

.

i=0

The solutions of P (t) = 0 are of two kinds: satisfying P0 (t ) = 0 and P0 (t ) = 0. Since t1 can take at most |U | values, by the inductive assumption the number of solutions of the first kind does not exceed |P0 | |U |r−1 . Similarly since t can take at most |U |r−1 values the number of solutions of the second kind does not exceed p|U |r−1 . However |P0 | + p |P | and the proof is complete.

Remark. A different proof can be obtained by an adaptation of the proof given by Schmidt for the special case K = U (see [17], p. 147, Lemma 3A). Lemma 9. If f (t), g(t) ∈ Q[t], g(t) | f (t)n and the fixed divisor of f (t) equals C(f ) then the fixed divisor of g(t) equals C(g). Proof. Let the fixed divisor of g be C(g)d, d ∈ N and let f (t)n = g(t)h(t). Clearly for all t ∈ Zr f (t)n is divisible by C(g)dC(h) = dC(f n ) = dC(f )n . On the other hand the fixed divisor of f (t)n is C(f )n . Hence d = 1.

Proof of Theorem 2. Consider first the case where F is a quadratic form. Then by Lemma 2 F (x, y) = A(ax 2 + bxy + cy 2 ),

where

A, a, b, c ∈ Z

and either Δ = b2 − 4ac = 1 or Δ is a fundamental discriminant. Since the fixed divisor of f (t) equals C(f )√we have A | C(f ) and we can assume without loss of generality that A = 1. Let K = Q( Δ), f (t) = l

(46)

n

ϕν (t)eν

ν=1

c

be a factorization of f (t) over K into irreducible factors such that ϕν are distinct and have the coefficient of the first term in the inverse lexicographical order equal to 1. Since the fixed divisor of f equals C(f ) the condition (19) is satisfied in virtue of Lemma 9. Let μ, τ be parameters whose existence for {ϕν } and M = a is asserted in Lemma 5 and let ψν = ϕν (μt + τ )

(1 ν n).

It follows that f (μt + τ ) = l

(47)

n

ψν (t)eν

ν=1

and (48)

B = |l|

n ν=1

C(ψν )eν = C f (μt + τ ) ∈ N,

1173


where an ideal in Q is identified with its positive generator. If Δ = 1, F is equivalent to xy and Theorem 1 applies. Assume that Δ = 1, thus K is a quadratic field.

Taking m = 1 in ψ (t ) ν 2 Lemma 5 we infer that H implies the existence of a t 2 ∈ Zr such that are distinct C(ψν ) prime ideals of K not dividing B. By the assumption there exist x0 , y0 ∈ Z such that ax02 + bx0 y0 + cy02 = f (μt 2 + τ ).

(49)

Hence, after a transformation √ b+ Δ y0 ax0 + b + √Δ 2 = |f (μt 2 + τ )|, where a = a, . N a 2 It follows from (47) and (48) that for an integral ideal b and some αν 0 α √ n n ψν (t 2 ) ν ψν (t 2 ) b + Δ −1 , b, y0 a = b = 1. (50) ax0 + 2 C(ψν )αν C(ψν ) ν=1

ν=1

ϕνeν

ϕνeν

ϕν

On the other hand

f (t) implies

f (t), where is conjugate to ϕν with respect to Q(t). If ϕν ∈ / Q[t] we have ϕν = ϕν and since ϕν has the coefficient of the leading term equal to 1, by (46) ϕν = ϕλ ,

eν = eλ ,

ψν = ψλ

for a λ = ν.

Thus without loss of generality we may assume that for a certain k ≡ n mod 2 (51) ϕν = ϕν ,

eν = eν ,

ψν = ψν ,

ν = ν

where

(1 ν k),

ν = ν − (−1)n−ν Hence by (48) |ax02

+ bx0 y0 + cy02 |

n ψν (t 2 ) αν +αν

= Nb

ν=1

C(ψν )

(k < ν n).

n ψν (t 2 ) =1 N b, C(ψν )

,

ν=1

and a comparison with (49) gives (52)

αν + αν = eν

(1 ν n).

Let us define now X(t), Y (t) by the equation √ √ n b + Δ ϕν (t) αν b+ Δ . Y (t) = ax0 + y0 (53) ϑ(t) = aX(t) + 2 2 ψν (t 2 ) ν=1

The polynomials X(t), Y (t) have integral coefficients since by (50) √ n

b + Δ C(ψν ) αν

y0 = ab, C ϑ(νt + τ ) = ax0 + 2 ψν (t 2 ) c ν=1 μ|ϑ| C(ϑ) ≡ 0 mod a and (μ, a) = 1 implies C(ϑ) ≡ 0 mod a.

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J. Prime numbers

On the other hand, by (53), (49), (51), (52), (46) and (47)

F X(t), Y (t) = aX(t)2 + bX(t)Y (t) + cY (t)2 n ϕν (t)ϕν (t) αν = (ax02 + bx0 y0 + cy02 ) ψν (t 2 )ψν (t 2 ) ν=1

n ϕν (t) αν +αν = f (μt 2 + τ ) ψν (t 2 )

= f (μt 2 + τ )

ν=1 n ν=1

ϕν (t) eν = f (t). ψν (t 2 )

Assume now that F is a reducible cubic form. If F is singular we have F = (ax + by)2 · (cx + dy), hence by the condition (1)

a b

c d = ±1, F is equivalent to x 2 y and Theorem 1 applies. If F is non-singular we have (54)

F (x, y) = (a0 x + b0 y)F1 (x, y),

where F1 is a non-singular primitive quadratic form. By Lemma 3 we have (55)

F1 (x, y) = G(a1 x + b1 y, a2 x + b2 y),

where G is primary and primitive. Let us put G(x, y) = ex 2 + gxy + hy 2 . By Lemma 2, the discriminant Δ = g 2 − 4eh equals 1 or is fundamental. The condition that F is primary implies that

a0 a1 a1 a2 a2 a0

= 1.

, , (56) d=

b0 b1 b1 b2 b2 b0

Otherwise, by a classical result on integral matrices (see [2], p. 52) the linear forms ai x+bi y (0 i 2) would be √ expressible integrally in terms of two linear forms with determinant d > 1. Let K = Q( Δ) and let the factorization of f (t) over K be given by (46). Since the fixed divisor of f (t) equals C(f ) the condition (19) is satisfied in virtue of Lemma 9. By Lemma 6 the equation

(57) F X(t), Y (t) = f (t) has only finitely many solutions in polynomials X(t), Y (t) ∈ Q[t] that are linearly independent. Let M be a positive integer such that MX, MY ∈ Z[t] for all of them. We apply Lemma 5 to the sequence {ϕν } with this M. Let μ, τ be any parameters with the property asserted in that lemma and let ψν (t) = ϕν (μt + τ ). We have again the formulae (47) and (48).

1175


We shall deduce from H the existence of polynomials x(t), y(t) ∈ Z[t] such that F x(t), y(t) = f (μt + τ ). This suffices to prove the theorem. Indeed the polynomials t − τ t − τ X(t) = x , Y (t) = y μ μ satisfy (57) and on one hand μ|x| X, μ|y| Y ∈ Z[t], on the other hand, if X, Y are linearly independent, we have by the choice of M MX, MY ∈ Z[t]. Since (μ, M) = 1 we get X, Y ∈ Z[t]. If X, Y are linearly dependent, then F X(t), Y (t) = f (t) = C0 (f )f0 (t)3 , C(f0 ) = 1 and X(t) = ξ ζ −1 f0 (t),

Y (t) = ηζ −1 f0 (t), F (ξ, η) = C(f )ζ , 3

ξ, η, ζ ∈ Z, |f0 |

ζ |μ

(ξ, η, ζ ) = 1;

.

If the above holds for all pairs μi , τ i of the sequence mentioned in the last assertion of Lemma 5, then, using the obvious notation, we infer from (μi , μh ) = 1 that either |f | |ζi | = |ζh | for i = h or there exists an i with |ζi | = 1. In the former case since |ζi | μi 0 Z 1/|f0 |n the number of distinct |ζi | Z is Ω , which contradicts Lemma 7. Therefore, log Z the latter case holds and Xi , Yi ∈ Z[t]. In order to deduce the existence of x(t), y(t) we shall consider successively the cases Δ = 1, Δ < 0, Δ > 1. If Δ = 1, by (47), (48) and Lemma 5, H implies the existence for every t 1 ∈ Zr and |ψν (t 2 )| every m prime to f (μt 1 + τ ) of a t 2 ≡ t 1 mod m such that are distinct primes C(ψν ) not dividing B (1 ν n). On the other hand, since a unimodular transformation of G does not affect the condition (56), we can assume G(x, y) = xy. By the assumption of C there exist integers x, y such that F (x, y) = f (μt 2 + τ ) and it follows from (47), (48), (54) and (55) that for suitable integers ci and nonnegative integers αiν (0 i 2, 1 ν n) ai x + bi y = ci

(58)

n ψν (t 2 ) αiν ν=1

C(ψν )

,

c0 c1 c2 = B sgn l, α0ν + α1ν + α2ν = eν . The set S of systems {ci }, {αiν } satisfying (59) is finite. It follows from (58) that (60) Ds (t 2 ) = 0, (59)

s∈S

1176

J. Prime numbers

where for s = {ci }, {αiν } : Ds (t) = det[ai , bi , Ψis (t)]0i2 ,

Ψis (t) = ci

n ψν (t) αiν ν=1

C(ψν )

.

Since Ψis (t 2 ) ≡ Ψis (t 1 ) mod m, Ds (t 2 ) ≡ Ds (t 1 ) mod m and (60) gives Ds (t 1 ) ≡ 0 mod m. s∈S

The latter congruence holds for all m prime to f (μt 1 + τ ), hence f (μt 1 + τ ) Ds (t 1 ) = 0 s∈S

and since t 1 is an arbitrary integral vector f (μt + τ )

Ds (t) = 0

s∈S

identically. However f (μt + τ ) = 0, thus there exists an s ∈ S such that Ds (t) = 0. By (56) the rank of the matrix [ai , bi ]0i2 is two, thus the system of equations ai x + bi y = Ψis (t)

(0 i 2)

is soluble in polynomials x, y ∈ Q[t]. Moreover, by Cramer’s formulae

ai bi ai bi

x,

aj bj aj bj y ∈ Z[t]

(0 i j 2)

and again by (56) x, y ∈ Z[t]. On the other hand, by (59), (48) and (47) F (x, y) =

2 i=0

(ai x + bi y) =

2 i=0

ci

n ψν (t) αiν ν=1

C(ψν )

= B sgn l

n ψν (t) eν ν=1

=l

n

C(ψν )

ψν (t)eν = f (μt + τ ).

ν=1

Let us consider now the case Δ = 1. Then, by Lemma 5 and (47), (48), H implies the r existence for every

t 1 ∈ Z and every m prime to Δf (μt 1 + τ ) of a t 2 ≡ t 1 mod m such ψν (t 2 ) that the ideals (ν n) are prime in K, distinct and do not divide B. By the C(ψν ) assumption of C there exist integers x, y such that F (x, y) = f (μt 2 + τ ) and it follows from (47), (48), (51) and (55) that for suitable integral ideals a, b and


1177

nonnegative integers αν , βν (1 ν n)

(61)

n ψν (t 2 ) αν (a0 x + b0 y) = a , C(ψν ) ν=1 √ n ψν (t 2 ) βν g+ Δ −1 , (a2 x + b2 y) g = b e(a1 x + b1 y) + 2 C(ψν ) ν=1

g + √Δ where g = e, , 2 αν + βν + βν = eν

aN b = (B),

(62)

(1 ν n),

ν is defined in (51). We get a0 x + b 0 y = α e(a1 x + b1 y) +

ψν (t 2 )αν ,

ν=1

√

(63)

n

g+ Δ ψν (t 2 )βν , (a2 x + b2 y) = β 2 n

ν=1

where (α) = a

(64)

n

C(ψν )−αν ,

(β) = gb

ν=1

n

C(ψν )−βν

ν=1

and by (47) and (62) αNβ = le.

(65)

Since a is integral, Ψ0 (t; α, αν ) = α

n

ψν (t)αν has integral coefficients. On the other

ν=1

hand, by (51) and (62) αν = αν (k < ν n) and by (65) α ∈ Q, hence Ψ0 (t; α, αν ) ∈ Z[t]. Similarly, since b is integral, β

n

ψν (t)βν ∈ g[t] and we get

ν=1

(66)

β

n ν=1

ψν (t)

βν

√ g+ Δ = eΨ1 (t; β, βν ) + Ψ2 (t; β, βν ), 2

where Ψi (t; β, βν ) ∈ Z[t] (i = 1, 2). The equations (63) take the form (67)

a0 x + b0 y = Ψ0 (t 2 ; α, αν ), ai x + bi y = Ψi (t 2 ; β, βν ) (i = 1, 2).

1178

J. Prime numbers

For a system s = [α, β, {αν }, {βν }] we put Ψ0s (t) = Ψ0 (t; α, αν ),

Ψis (t) = Ψi (t; β, βν )

(i = 1, 2)

and denote by S the set of all such systems satisfying (62) and (64). If Δ < 0 the set S is finite. It follows from (67) that (68) Ds (t 2 ) = 0, s∈S

where

Ds (t) = det ai , bi , Ψis (t) 0i2 .

Since Ψis (t 2 ) ≡ Ψis (t 1 ) mod m we infer from (68), as in the case Δ = 1 from (60), that for a suitable s ∈ S the system of equations ai x + bi y = Ψis (t)

(0 i 2)

is soluble in polynomials x, y ∈ Z[t]. By (54), (55), (66), (47), (65) and (51) we get √ g+ Δ F (x, y) = (a0 x + b0 y)N e(a1 x + b1 y) + (a2 x + b2 y) e−1 2 √ g+ Δ = Ψ0s (t)N eΨ1s (t) + Ψ2s (t) e−1 2 n n =α ψν (t)αν N β ψν (t)βν e−1 ν=1

= αNβe−1

n

ν=1

ψν (t)αν +βν +βν = l

ν=1

n

ψν (t)eν = f (μt + τ ).

ν=1

If Δ > 0 the set S is infinite. We can however divide it into finitely many classes assigning two systems [α, β, {αν }, {βν }] and [α, γ , {αν }, {βν }] to the same class if ±γ /β is a totally positive unit of K. Then every class contains exactly one system satisfying 1 β < ε,

(69)

where ε > 1 is the fundamental totally positive unit. Denoting the set of all systems satisfying (62), (64) and (68) by S0 we infer from (67) the existence of a σ ∈ Z such that Dσ s (t 2 ) = 0, s∈S0

where for s = [α, β, {αν }, {βν }]

a0

Dσ s (t) =

a1

a2

b0 b1 b2

Ψ0 (t; α, αν )

Ψ1 (t; εσ β, βν )

. Ψ2 (t; εσ β, βν )

Since Dσ s (t 2 ) ≡ Dσ s (t 1 ) mod m for all s we conclude that Dσ s (t 1 ) ≡ 0 mod m (70) s∈S0

where σ depends on m.


1179

We have an identity √ √ g+ Δ g+ Δ u eΨ1s (t) + Ψ2s (t) = eΦ1s (t, u) + Φ2s (t, u), 2 2

(71) where

c

g g 1 Δ − g 2 u − u−1 −1 Φ1s (t, u) = 1+ √ u 1− √ +u Ψ1s (t) + Ψ2s (t), √ 2 4e Δ Δ Δ u − u−1 1 g g Φ2s (t, u) = e √ u 1+ √ Ψ1s (t) + + u−1 1 − √ Ψ2s (t). 2 Δ Δ Δ

Since ε is conjugate to ε −1 Φis (t, εσ ) ∈ Q[t] (i = 1, 2) and by (71) Ψi (t; εσ β, βν ) = Φis (t, εσ )

(i = 1, 2).

The congruence (70) takes the form Es (t 1 , εσ ) ≡ 0 mod m, (72) s∈S0

where

a0

Es (t, u) =

a1

a2

b0 Ψ0s (t)

b1 Φ1s (t, u)

. (73) b2 Φ2s (t, u)

However uEs (t, u) ∈ Q[t, u] and hence u|S0 | Es (t 1 , u) ∈ Q[u]. s∈S0

c

Since the congruence (72) is soluble for all m prime to Δf (μt 1 + τ ), it follows from Theorem 6 of [15] that the equation Es (t 1 , εσ ) = 0 f (μt 1 + τ ) s∈S0 r is soluble in integers σ . Thus for every f (μt 1 + τ ) = 0 or f (μt 1 + τ ) = 0 t 1 ∈ Z either and there exist a σ ∈ Z and an s = α, β, {αν }, {βν } ∈ S0 such that Es (t 1 , εσ ) = 0. In the latter case it follows from (71) and (73) that

b0 Ψ0s (t 1 ) a0

√ √

n g+ Δ g+ Δ

εσ β ψν (t 1 )βν

a2 eb1 + b2

ea1 +

2√ 2√ ν=1

n

βν

ea1 + g − Δ a2 eb1 + g − Δ b2 ε −σ β ψ (t ) ν 1

2 2 ν=1 √ = −e Δ Es (t 1 , εσ ) = 0

1180

J. Prime numbers

and ε σ β

n

ψν (t 1 )βν satisfies the quadratic equation

ν=1

Lz2 − KΨ0s (t 1 )z − L Nβ

n

ψν (t 1 )βν +βν = 0,

ν=1

where

(74)

β ,

L

are conjugate to β, L, respectively,

a0 b0

√ √

L=

g− Δ g − Δ

, a2 eb1 + b2

ea1 + 2 2

√ √

ea + g + Δ a eb + g + Δ b

1 2 1 2

2√ 2√

K =

ea + g − Δ a eb + g − Δ b

1 2 1 2

2 2

Since e[a0 , b0 ] = 0 we have L = 0 by (56), and (75)

εσ β

n

ψν (t 1 )βν t 1 |f |

ν=1

where denotes the maximum modulus of the conjugates and the constant in the symbol depends on F, f, μ, τ , s. On the other hand, by (47), (51), (62), (65) and (69) β

n

ψν (t 1 )βν t 1 |f |/2 .

ν=1

Since f (μt 1 + τ ) = 0 whence by (64)

n

βν

N β ψν (t 1 ) N gb 1

ν=1

we get β −1

n

ψν (t 1 )−βν < t 1 |f |/2 .

ν=1

This together with (75) implies ε|σ | = ε σ t 1 3|f |/2 ,

|σ |

3 log t 1 |f | + , 2 log ε

where is a constant depending on F, f, μ, τ but independent of s (S0 is finite). Let us choose now a positive integer T so large that log T (76) 2T + 1 > |f |(|S0 | + 1) 3|f | + 2 + 1 . log ε


1181

If t 1 runs through all integral vectors satisfying t 1 T , σ runs through integers satisfying |σ |

log T 3 |f | + . 2 log ε

The number of vectors in question is (2T + 1)r , the number of integers does not exceed log T 3|f | + 2 + 1, hence there is an integer σ0 that corresponds to at least log ε −1 log T + 2 + 1 (2T + 1)r 3|f | log ε different vectors t 1 satisfying t 1 T . By (76) we get more than |f |(|S0 | + 1) × (2T + 1)r−1 such vectors satisfying the equation Es (t 1 , εσ0 ) = 0. f (μt 1 + τ ) s∈S0

Since by (62), (71) and (73) the degree of Es (t, εσ0 ) does not exceed |f |, the degree of the polynomial on the left hand side does not exceed |f |(|S0 | + 1) and Lemma 8 shows that Es (t, εσ0 ) = 0 f (μt + τ ) s∈S0

identically. Therefore, there exists an s ∈ S0 such that Es (t, εσ0 ) = 0 and by (56) the system of equations a0 x + b0 y = Ψ0s (t), ai x + bi y = Φis (t) (i = 1, 2) is soluble in polynomials x, y ∈ Z[t]. By (54), (55), (71), (66), (47), (62), (65) and (51) we get for these polynomials √ g+ Δ (a2 x + b2 y) e−1 F (x, y) = (a0 x + b0 y)N e(a1 x + b1 y) + 2 √ g+ Δ (a2 x + b2 y) e−1 = Ψ0s (t)N ε −σ0 e(a1 x + b1 y) + ε −σ0 2 √ g+ Δ = Ψ0s (t)N eΨis (t) + Ψ2s (t) e−1 = f (μt + τ ) 2 and the proof is complete.

Remark. For the proof of a more general result mentioned in the introduction one needs more general versions of Lemmata 2, 5 and 7 and Theorem 7 of [16] instead of Theorem 6 of [15]. In the difficult case of an irreducible form F with all zeros real Theorem 7 of [16] does not suffice, but Skolem’s conjecture on exponential congruences would do (see [18]). One could avoid this step in the proof provided it were known that the number of vectors t satisfying t T and the conditions of Lemma 4 grows faster than T r−1 (log T )|F | . For r = 1 much more has been conjectured by Bateman and Horn [1].

1182

J. Prime numbers

4. The next lemma is a refinement of Lemma 1 of [13]. Lemma 10. Let P ∈ Q[t, u] be a polynomial such that for no ϕ ∈ Q(t)

P t, ϕ(t) = 0 identically. Then there exists a t 1 ∈ Zr such that for any M ∈ N there exists an m ∈ N prime to M such that for all t ∈ Zr , t ≡ t 1 mod m and all u ∈ Q P (t, u) = 0. Proof. Following the proof of Lemma 1 in [13] we take m = q1 · · · qk , where in the notation of that paper the primes qi are chosen not to divide M.

Lemma 11. Let G, H ∈ Q[x, y] be relatively prime forms, p, gi , hi ∈ Q[t] (i I ) arbitrary polynomials, p = 0. If for every t 1 ∈ Zr and for every integer m prime to p(t) there are an i I , a t 2 ∈ Zr , t 2 ≡ t 1 mod m and x, y ∈ Q satisfying (77)

H (x, y) = hi (t 2 )

G(x, y) = gi (t 2 ),

then there exist a j I and polynomials X, Y ∈ Q[t] such that G(X, Y ) = gj ,

H (X, Y ) = hj .

Proof. If G(x, y) − gi (t), H (x, y) − hi (t) had a common factor d(x, y, t) = const then the leading forms of d with respect to x, y would divide G(x, y) and H (x, y). Thus for each i I

G(x, y) − gi (t), H (x, y) − hi (t) = 1. Let Ri (t, x), Si (t, y) be the resultants of G(x, y) − gi (t) and H (x, y) − hi (t) with respect to y and x respectively. It follows from the construction of resultants that the leading coefficients of Ri in x and of Si in y are equal to the resultants of G(1, z), H (1, z) and of G(z, 1), H (z, 1) respectively. Hence these leading coefficients are independent of t. Let (78)

(79)

ri

x − Ri (t) ,

Ri (t, x) = Ri0 (t, x)

=1 si

Si (t, y) = Si0 (t, y)

y − Siσ (t) ,

σ =1

c

where Ri0 and Si0 have no factor linear in x or y respectively. If for some triple (i, , σ ) with i I , 1 ri , 1 σ si G(Ri , Siσ ) = gi the lemma follows.

and

H (Ri , Siσ ) = hi ,


1183

Therefore, suppose that for each triple (i, , σ ) in question G(Ri , Siσ ) = gi Then c

(80)

H (Ri , Siσ ) = hi .

or

2 2 Tiσ = G(Ri , Siσ ) − gi + H (Ri , Siσ ) − hi = 0

and we set in Lemma 10 (81)

P (t, u) = p(t)

I

Ri0 (t, u)Si0 (t, u)

si ri

Tiσ (t).

=1 σ =1

i=1

By that lemma with M = 1 there exist an m ∈ N and a t 1 ∈ Zr such that if t ≡ t 1 mod m and u ∈ Q we have P (t, u) = 0.

(82)

In particular, taking t = t 1 we get p(t 1 ) = 0. Applying Lemma 10 again with M

= p(t1 ) we infer the existence of an integer m with the above property satisfying m, p(t 1 ) = 1. However now by the assumption there exist an i I , a t 2 ≡ t 1 mod m and x, y ∈ Q such that (77) holds. By the fundamental property of resultants we have Ri (t, x) = 0 = Si (t, y) and in view of (78), (79), (81) and (82) there exist , σ such that 1 ri , 1 σ si , x = Ri (t 2 ),

y = Siσ (t 2 ).

It follows from (77) and (80) that Tiσ (t 2 ) = 0,

contrary to (81) and (82).

Remark. Lemma 11 extends to any system of forms G1 , . . . , Gk ∈ Q[x1 , . . . , xk ] without a common non-trivial zero. Proof of Theorem 3. If f = 0 the theorem is trivially true. If f = 0 let f (t 0 ) = e = 0. We set f0 (t) = f (et + t 0 ) and find as in the proof of Corollary to Lemma 3 that the fixed divisor of f0 (t) equals C(f0 ). (If the fixed divisor of f equals C(f ) we can take directly e = 1, t 0 = 0.) Let K be the least field over which F factorizes into two coprime factors c G, H and let (83)

f0 (t) = l

n

ϕν (t)eν

ν=1

be a factorization of f over K into irreducible factors such that ϕν are distinct and have the coefficient of the first term in the inverse lexicographical order equal to 1. Since the fixed divisor of f0 (t) equals C(f0 ) the polynomials ϕν satisfy (19) in virtue of Lemma 9. Let μ, τ be parameters whose existence for {ϕν } and μ = 1 is asserted in Lemma 5 and let ψν = ϕν (μt + τ )

(1 ν n).

1184

J. Prime numbers

It follows that f0 (μt + τ ) = l

(84)

n

ψν (t)eν

ν=1

and (85)

B = |l|

n

C(ψν )eν = C f0 (μt + τ ) ∈ N,

ν=1

where an ideal in Q is identified with its positive generator. Consider first the case where K = Q and let (86)

f0 (μt + τ ) = gi (t)hi (t)

(1 i I )

be all possible factorizations of the left hand side into two factors with integral coefficients.

H implies that if m, f0 (μt 1 + τ ) = 1 there exist an i I , a t 2 ≡ t 1 mod m and x, y ∈ Z such that G(x, y) = gi (t 2 ), H (x, y) = hi (t 2 ).

Indeed, by (84) and (85), the condition m, f0 (μt 1 + τ ) = 1 implies

(87)

m,

n ψν (t 1 ) =1 C(ψν )

ν=1

|ψν (t 2 )| C(ψν ) (ν n) are distinct primes not dividing B. By the assumption of D there exist x, y ∈ Z such that and, by Lemma 5, H implies the existence of a t 2 ∈ Zr , t 2 ≡ t 1 mod m such that

G(x, y)H (x, y) = F (x, y) = f0 (μt 2 + τ ) and it follows from (84) and (85) that for some a, b, αν , βν ∈ Z, αν 0, βν 0, we have n ψν (t 2 ) αν

G(x, y) = a

ν=1

C(ψν )

ab = B sgn l,

,

H (x, y) = b

n ψν (t 2 ) βν ν=1

αν + βν = eν

C(ψν )

,

(1 ν n).

Taking gi (t) = a c

n ψν (t) αν ν=1

C(ψν )

,

hi (t) = b

n ψν (t) βν ν=1

C(ψν )

we get (86) and (87). Now we apply Lemma 11 with p(t) = f0 (μt + τ ) and we get the existence of X0 , Y0 ∈ Q[t] satisfying G(X0 , Y0 ) = gj ,

H (X0 , Y0 ) = hj

1185


for some j I . Setting (88)

X(t) = X0

t − eτ − t 0 , eμ

Y (t) = Y0

t − eτ − t 0 , eμ

we get by (86) t − eτ − t t − eτ − t t − t

0 0 0 F X(t), Y (t) = gj hj = f0 = f (t). eμ eμ e Consider now the case where K is an imaginary quadratic field with discriminant Δ. Then v (89) F (x, y) = N Φ(x, y), w where v, w ∈ Z, (v, w) = 1, Φ ∈ K[x, y] has integral coefficients and

(90) Φ(x, y), Φ (x, y) = 1, where Φ is conjugate to Φ over Q(x, y). Let w (91) f0 (μt + τ ) = ηi ηi (t) (i I ) v be all the factorizations of the left hand side into two conjugate polynomials with integral coefficients in K. Since K has finitely many units the number of such factorizations is finite. H implies that if m, Δf0 (μt 1 + τ ) = 1 there exist an i I , a t 2 ≡ t 1 mod m and x, y ∈ Z such that Φ(x, y) = ηi (t 2 ).

(92) Indeed, by (84) and (85), we have (93)

e n w ψν (t) ν . f0 (μt + τ ) = B v v C(ψν )eν

w

ν=1

Since, by Lemma 5, n

n N ψ (t) n ν has the fixed divisor 1, ψν (t)eν has the fixed divisor ν=1 N C(ψν ) ν=1

C(ψν )eν . On the other hand, for every t ∈ Zr

ν=1

w f0 (μt + τ ) = N Φ(x, y) ∈ Z, v hence (94)

w A ∈ Z. v

By (84) and (85) the condition m, Δf0 (μt 1 + τ ) = 1 implies n N ψν (t 1 ) m, Δ =1 N C(ψν ) ν=1

ψν (t 2 ) (ν n) and, by Lemma 5, H implies the existence of a t 2 ≡ t 1 mod m such that C(ψν )

1186

J. Prime numbers

are distinct prime ideals not dividing wB. By the assumption of D there exist x0 , y0 ∈ Z such that w w (95) N Φ(x0 , y0 ) = F (x0 , y0 ) = f (μt 2 + τ ) v v and it follows from (93) and (94) that for an integral ideal b and some integers αν 0 α n n

ψν (t 2 ) ν ψν (t 2 ) Φ(x0 , y0 ) = b , b, = 1. C(ψν )αν C(ψν ) ν=1

ν=1

On the other hand, in full analogy with (51), we can assume that for a certain k ≡ n mod 2 (96)

ψν = ψν ,

eν = eν ,

Hence NΦ(x0 , y0 ) = Nb

ν = ν (ν k),

n ψν (t 2 ) αν +αν ν=1

C(ψν )

ν = ν − (−1)n−ν (ν > k). n ψν (t 2 ) N b, =1 C(ψν )

,

ν=1

and a comparison with (93) gives αν + αν = eν

(97)

(1 ν n).

Now let us put η(t) = Φ(x0 , y0 )

(98)

n ψν (t) αν . ψν (t 2 )

ν=1

The polynomial η(t) has integral coefficients in K since n

C(ψν )αν

α = b. C(η) = Φ(x0 , y0 ) ψν (t 2 ) ν ν=1

Moreover, by (95), (96), (97) and (84) n ψν (t)ψν (t) αν η(t)η (t) = N Φ(x0 , y0 ) ψν (t 2 )ψν (t 2 ) ν=1 n n ψν (t) αν +αν ψν (t) eν w w = f0 (μt 2 + τ ) = f0 (μt 2 + τ ) v ψν (t 2 ) v ψν (t 2 ) ν=1 ν=1 w = f0 (μt + τ ). v Hence η(t) = ηi (t) for an i I and (92) follows immediately from (98). Now we apply Lemma 11 with p(t) = Δf0 (μt + τ ),

√ G(x, y) = Φ(x, y) + Φ (x, y), H (x, y) = Φ(x, y) − Φ (x, y) / Δ

and we get the existence of X0 , Y0 ∈ Q[t] satisfying (99)

Φ(X0 , Y0 ) = ηj ,

Φ (X0 , Y0 ) = ηj


1187

for a j I . Using again the transformation (88) we get by (89) and (90)

w t − eτ − t 0 t − eτ − t 0 w t − t 0 F X(t), Y (t) = ηj ηj = f0 = f (t).

v eμ eμ v e Lemma 12. Let k ∈ N be odd, ai (t) ∈ Z[t] (0 i k), a0 (t) = 1, x(t) ∈ Q[t]. If k−1 k (100) ai (t)x(t)k−1−i = 0 i+1 i=0

then x(t) ∈ Z[t]. Proof. Suppose that C(x) ∈ / Z. Then for some prime p ordp C(x) = −c −1. The function ordp C(P ) is a valuation of the ring Q[t] (see [6], p. 171). In virtue of the properties of valuations (100) implies

k ordp kC(x)k−1 min ordp C(ai )C(x)k−1−i , 0 1 we take further K = Q, μ = 1, τ = 0, α α α j α x 2 i −y 2 G(x, y) = x 2 + y 2 , H (x, y) = i+j =k−1

and we get from (86) and (88) that for some polynomials g, h ∈ Z[t] and X, Y ∈ Q[t] g(t)h(t) = f (t), (102)

G(X, Y ) = g,

H (X, Y ) = h.

1188

J. Prime numbers

However k−1 k α G(X, Y )i (−X 2 )k−1−i H (X, Y ) = i+1 i=0

hence taking in Lemma 12 ai (t) = g(t)i

(0 i < k − 1),

ak−1 (t) = −h(t),

x(t) = −X(t)2

α

we get from (102) that α

−X(t)2 ∈ Z[t]. Thus X(t) ∈ Z[t] and by symmetry Y (t) ∈ Z[t]. Moreover X(t)n + Y (t)n = G(X, Y )H (X, Y ) = f (t). If k = 1 we take in the proof of Theorem 3 K = Q(ζ4 ), Φ(x, y) = x 2

(103)

c

α−1

+ ζ4 y 2

α−1

,

v/w = 1,

where ζq is a primitive q-th root of unity. By Lemma 5 μ factorizes over K into prime ideals of degree 2. By (92) and (99) for some polynomials η ∈ Z[ζ4 , t] and X0 , Y0 ∈ Q[t] (104) (105)

η(t)η (t) = f (μt + τ ), η conjugate to η over Q(t),

Φ X0 (t), Y0 (t) = η(t).

Let us set (106)

ϑ(t) = η

t − τ , μ

X(t) = X0

t − τ , μ

Y (t) = Y0

t − τ . μ

We have μ|η| ϑ(t) ∈ Z[ζ4 , t] hence, if p is a prime ideal of K in the denominator of C(ϑ), p | μ and p = p . However by (104) ϑ(t)ϑ (t) = f (t), hence ordp C(ϑ) =

1 2

N C(ϑ) = C(f ) ∈ Z

ordp C(f ) 0 and ϑ(t) ∈ Z[ζ4 , t].

Now (103), (105) and (106) imply X(t)2

α−1

, Y (t)2

α−1

∈ Z[t];

X(t), Y (t) ∈ Z[t]

and we get by (104) α

α

X(t)2 + Y (t)2 = f (t). The proof of the first part of the theorem is complete. In order to prove the second part it is enough to consider the case n > 2 (for n = 2 the assertion is contained in Theorem 1).


1189

Let p be a prime satisfying (107)

p ≡ 1 mod 2α+1 ,

p ≡ 1 mod 2n if

n = 2α

and let us choose an integer c such that cn + 1 ≡ 0 mod pn ,

c = −1

if

α = 0.

Consider now the polynomial f (t) = u(t)n + v(t)n ,

(108) where u(t) =

t (t − 1) · · · (t − p + 1) , p

v(t) = cu(t) + pn−1 .

It is easily seen that f (t) ∈ Z[t] and pn if α > 0, (109) |f | = p(n − 1) if α = 0. c Moreover, since polynomials u(t), v(t) are integer-valued the equation x n + y n = f (t) is soluble in x, y ∈ Z for all t ∈ Z. On the other hand, suppose that (110)

X(t)n + Y (t)n = f (t),

X, Y ∈ Z[t].

Since X(t)n + Y (t)n =

n−1

2i+1 X(t) − ζ2n Y (t)

i=0

we have

|f |

n max{|X|, |Y |} if α > 0, (n − 1) max{|X|, |Y |} if α = 0.

Hence by (109) (111)

max{|X|, |Y |} p.

Taking i = 0, 1, . . . , p − 1 we get u(i) = 0 hence (112)

X(i)n + Y (i)n = pn(n−1) .

If n = 2α , α > 1 or n = 3 by special cases of Fermat’s last theorem (111) implies (113)

X(i)Y (i) = 0

(0 i < p).

If n > 3, by Zsigmondy’s theorem either X(i)Y (i) = 0 or X(i) = ±Y (i) or X(i)n + Y (i)n has the so-called primitive prime factor ≡ 1 mod 2n. The last two possibilities are incompatible with (107) and (112) hence (113) holds for all n > 2. By (112) if X(i) = 0, Y (i) = p n−1 for α = 0, Y (i) = ±p n−1 for α > 0. In view of symmetry between X and Y we may assume that there is a set S ⊂ {0, 1, . . . , p − 1} with the following

1190

J. Prime numbers

properties |S|

p+1 , 2(n, 2)

X(i) = 0,

(If n is even we can replace Y by −Y .) Let P (t) =

Y (i) = pn−1

for i ∈ S.

(t − i). i∈S

It follows that (114)

|P |

p+1 , 2(n, 2)

X(t) ≡ 0 mod P (t),

Y (t) ≡ pn−1 mod P (t)

and we get from (108) and (110) Y (t)n ≡ v(t)n mod P (t)n . Since Y (t) ≡ v(t) mod P (t) and (v, P ) = 1 we obtain Y (t) ≡ v(t) mod P (t)n . However by (111) max{|Y |, |v|} p < n|P | hence Y (t) = v(t) ∈ / Z[t].

References [1] P. T. Bateman, R. A. Horn, A heuristic asymptotic formula concerning the distribution of prime numbers. Math. Comp. 16 (1962), 363–367. [2] A. Châtelet, Leçons sur la théorie des nombres. Paris 1913. [3] S. Chowla, Some problems of elementary number theory. J. Reine Angew. Math. 222 (1966), 71–74. [4] H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107–116; this collection: A6, 27–35. [5] H. Halberstam, H.-E. Richert, Sieve Methods. Academic Press, London–New York 1974. [6] H. Hasse, Zahlentheorie. Akademie-Verlag, Berlin 1963. [7] T. Kojima, Note on number-theoretical properties of algebraic functions. Tôhoku Math. J. 8 (1915), 24–37. [8] S. Lang, Diophantine Geometry. Interscience, New York–London 1962. [9] W. J. LeVeque, A brief survey of Diophantine equations. In: Studies in Number Theory, Math. Assoc. Amer., Buffalo 1969, 4–24. [10] D. Mumford, A remark on Mordell’s conjecture. Amer. J. Math. 87 (1965), 1007–1016. [11] P. Ribenboim, Polynomials whose values are powers. J. Reine Angew. Math. 268/269 (1974), 34–40.


1191

[12] A. Schinzel, W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers. Acta Arith. 4 (1958), 185–208; Erratum ibid. 5 (1959), 259; this collection: J1, 1113–1133. [13] A. Schinzel, On Hilbert’s Irreducibility Theorem. Ann. Polon. Math. 16 (1965), 333–340; this collection: F1, 839–845. [14] −−, On a theorem of Bauer and some of its applications II. Acta Arith. 22 (1973), 221–231; this collection: C5, 210–220. [15] −−, Abelian binomials, power residues and exponential congruences. Acta Arith. 32 (1977), 245–274; this collection: H5, 939–970. [16] −−, Addendum and corrigendum to [15]. Ibid. 36 (1980), 101–104 (in this collection included into H5). [17] W. M. Schmidt, Equations over Finite Fields. An Elementary Approach. Lecture Notes in Math. 536, Springer, Berlin 1976. [18] Th. Skolem, Anwendung exponentieller Kongruenzen zum Beweis der Unlösbarkeit gewisser diophantischer Gleichungen. Vid. Akad. Avh. Oslo I 1937 nr. 12.

Part K Analytic number theory


Commentary on K: Analytic number theory by Jerzy Kaczorowski

1. Papers K1 and K4 concern the following closely related topics: values of Dirichlet L-functions at 1, class numbers of quadratic number fields, existence of the Siegel zero and character sums. Problems related to them occupy central position in number theory and attracted attention of many leading mathematicians. Study of the ideal class group goes back to C. F. Gauss who used the language of binary quadratic forms (like in K1). Let d be a fundamental discriminant. Famous Dirichlet class √ number formula relates h(d), the class number of Q( d), to L(1, χd ), where d χd (n) = n is the Kronecker symbol. For instance if d < −4 we have √ h(d) = π −1 dL(1, χd ). We refer to W. Narkiewicz [19] for the basic theory. The fact that the size of L(1, χ) is related to the exceptional zero was first observed by H. Hecke, who proved that if such a zero does not exist then (log |d|)−1 , see e.g. [17]. In particular for d < 0 √ L(1, χd ) −1 we have then h(d) |d|(log |d|) . The main theorem of K1 makes this relation very explicit, see also D. Goldfeld [10] and A. Granville, H. M. Stark [12]. E. Landau [17] introduced the idea of twisting L-functions by Dirichlet characters which proved to be very useful later on. Using it he proved that if L(β, χ ) = L(β , χ ) = 0 for certain real β and β and two primitive real characters χ (mod |d|) and χ (mod |d |), then min(β, β ) 1 − c(log(|dd |))−1 , c being a positive constant. Hence if real zeros exist they are very rare. This is a simple instance of a general repulsion principle saying roughly that if an L-function has a real zero close to 1 then for many other L-functions with comparable conductors such zeros cannot exist. This was mastered in papers by M. Deuring [7] and H. Heilbronn [16] and is known under the name the “Deuring–Heilbronn phenomenon”. Yu. V. Linnik [18] exploited this in a clever way in his famous work on the least prime in an arithmetic progression. C. L. Siegel [24] proved that L(1, χ) |d|−ε for every primitive character χ (mod |d|) and every positive ε. This implies that h(d) |d|1/2−ε

1196

K. Analytic number theory

as d → −∞. In particular, for every positive integer h0 there are finitely many quadratic imaginary fields with the class number equal to h0 . Unfortunately Siegel’s theorem is ineffective and consequently it cannot be used for determining fields with a given class number. In particular, it does not help in finding all fundamental discriminants d < 0 with h(d) = 1. This was done by K. Heegner [15] and H. M. Stark [25], who used arithmetic of elliptic curves, and independently by A. Baker [2], who used his theory of linear form in logarithms of algebraic numbers. See also [26]. Basing on ideas by J. Friedlander [9] and D. Goldfeld [11], B. Gross and D. Zagier [13] gave effective lower estimate for h(d). Their method uses elliptic curves whose L-functions have central zero of the proper order. Choosing an elliptic curve of rank 3 and conductor 5077, J. Oesterlé [20] proved that for a negative fundamental discriminant 1 2 h(d) 1− √ log |d|. 55 p p|d

Assuming the Modified Generalized Riemann Hypothesis (zeros are all on the critical line or on the real axis) lower estimates for L(1, χd ) and h(d) can be much improved, see P. Sarnak, A. Zaharescu [23]. For instance assuming MGRH for all L-functions of elliptic curves one has L(1, χd ) |d|−2/5−ε , where χd (37) = −1, with an effective implied constant. Siegel’s theorem can be also improved making other assumptions on the distribution of zeros of L-functions. B. Conrey and H. Iwaniec [6] proved that ) h(d) |d|(log |d|)−A for some constant A > 0 if the gap between consecutive zeros on the critical line is smaller than the average for sufficiently many pairs of zeros. Connections between class numbers of quadratic number fields and character sums are well known. For instance if d < −4 is a fundamental discriminant, then 1 1 kχd (k) = χd (k), h(d) = − |d| 2 − χd (2) 0 2|α| > −β + d > 0. (1 )

β is not to be confused with Siegel’s zero.

K1. On Siegel’s zero

1205

We can assume without loss of generality that α > 0. Now, for any form f ∈ C, there exists a properly unimodular transformation p r T = q s taking (α, β, γ ) into f . The first column of this transformation can be made to consist of positive rational integers by Theorem 79 of [5]. If f satisfies (10), we infer from αp2 + βpq + γ q 2 = a

(12) that

√

p + β − d

2α

√

β+ d

q = a αp + 2

−1

√ −1 1√ 1 q

d ·2 dq = q −1 4 2

and by Lemma 16, p. 175 from [5], p/q is a convergent of the continued fraction expansion for √ −β + d . ω= 2α From this point onwards, we shall use the notation of Perron’s monograph [7]. Since by (11) ω−1 > 1

c

and

0 > (ω )−1 > −1,

ω−1 is a reduced quadratic surd and it has a pure periodic expansion into a continued fraction. Hence ω = [0, b1 , b2 , . . . , bk ] where the bar denotes the primitive period. The corresponding complete quotients form again a periodic sequence √ Pν + d , ω0 = ω, ων = Qν where for all ν 1, ων is reduced, (13)

ων = ων+k ,

and k is the least number with the said property. Lemma 3. Let [0, b1 , b2 , . . . , bk ] be the continued fraction for ω defined above. Then √ [k,2] 1 d 2 min √ , bν + 1 |a| 2 d ν=2 (a,b,c)∈C √ dbν 2

where the sum on the left is taken over all (a, b, c) in the class C satisfying (10). Proof. If Aj /Bj is the j -th convergent of ω, we have by formula (18), §20 of [7] (Aν−1 Q0 − Bν−1 P0 )2 − d(Bν−1 )2 = (−1)ν Q0 Qν

1206


which gives on simplification 2 αA2ν−1 + βAν−1 Bν−1 + γ Bν−1 = (−1)ν Qν /2.

(14)

Similarly, eliminating Qν from formulae (16) and (17) in §20 of [7], we get c

(15)

2αAν−1 Aν−2 + β(Aν−1 Bν−2 + Bν−1 Aν−2 ) + 2αBν−1 Bν−2 = (−1)ν−1 Pν .

Let p = Aν−1 , q = Bν−1 (ν 1). By (12) a = (−1)ν Qν /2. Hence, by formula (1) of §6 of [7]

Aν−1

Bν−1 and since

it follows that

Aν−2

= (−1)ν Bν−2

Aν−1

Bν−1

Aν−1 T = Bν−1

Aν−2 Bν−2

r

=1 s

1 0

t , (−1)ν

t ∈ Z.

Thus we find using (14) and (15) Aν−1 Aν−2 1 t f = (α, β, γ ) Bν−1 Bν−2 0 (−1)ν Qν Qν−1 1 = (−1)ν , (−1)ν−1 Pν , (−1)ν 0 2 2 In order to make f satisfy (10) we must choose Pν 1 t = (−1)ν + . Qν 2 Thus f is uniquely determined by ων and, in view of (13), we have (16)

(a,b,c)∈C

1 |a|

[k,2]

2(Qν )−1 .

ν=1√ Qν < 21 d

Since ων is reduced, we have further for ν in question √ √ √ √ 2 d d Pν + d d > > 2. Qν Qν Qν Hence for bν = [ων ],

t . (−1)ν

1207


we get the inequalities

√

d > bν 2,

bν + 1 >

√ d/Qν ,

and by (16), Lemma 3 follows. Now, let ε0 be the least totally positive unit ε0 > 1 of the ring Z[σ ] where ⎧ √ 1 ⎪ ⎪ d if d ≡ 0 (mod 4), ⎨ 2 √ σ = ⎪ ⎪ ⎩ 1 + d if d ≡ 1 (mod 4). 2 By Theorem 7 of Chapter IV of [6] √ u+v d , ε0 = 2 where for l = [k, 2], v = (ql−1 , pl−1 − ql−2 , pl−2 ),

u = pl−1 + ql−2

and pj , qj are the numerator and denominator, respectively, of the j -th convergent for ω−1 . Moreover, since ω−1 satisfies the equation −γ ω−2 − βω−1 − α = 0

(−γ > 0),

we find from formula (1) of §2 of Chapter IV of [6] that ql−2 − pl−1 = −βv,

−pl−2 = −αv.

Hence

√ √ pl−1 + ql−2 β+ d pl−2 d + = ql−2 + pl−2 . ε0 = 2 2α 2α Since pj = Bj +1 , qj = Aj +1 , we get √ √ √ d β+ d Al−1 β+ d (17) ε0 = Bl−1 Bl−1 ω + = Bl−1 . + Bl−1 2α 2α α Now, −1 = bl + ω1−1 = bl + ω, ωl = bl + ωl+1

ωl = bl + ω

and since ωl is reduced 0 > bl + ω > −1 √ √ d β+ d < . bl = [−ω ] = 2aα α Thus (17) gives ε0 > bl Bl−1 >

l ν=1

bν ,

1208


and by (16)

(18)

(a,b,c)∈C

1 2 2 (xi + 1) = √ M √ max |a| d d

where maximum is taken over all non-decreasing sequences of at most l real numbers satisfying √ 2 xi 21 d − 1 = D, xi ε0 . c Let (x1 , x2 , . . . , xm ) be a point in which the maximum is taken with the least number m. We assert that the sequence contains at most one term x with 2 < x < D. Indeed, if we c had 2 < xi xi+1 < D, we could replace the numbers xi , xi+1 by xi , min(xi /2, D/xi+1 )

xi+1 min

x

i

,

D xi+1

2 and the sum (xi + 1) would increase. Also, if we had x1 = x2 = x3 = 2, we could replace them by x1 = 8, and the sum (xi + 1) would remain the same while m would decrease. Let ε0 log(ε0 /4) = D e θ, where e = . c 4 log D Using d > 676, we get ⎧ √ 1 ⎪ e d + max(4θ + 1, 2θ + 4) ⎪ ⎨2 √ 1 M = 2 e d + 2θ + 4 ⎪ ⎪ ⎩1 √ 2e d + θ + 7

if 4θ < D, if 2θ < D 4θ, if D 2θ.

Now, e=

log ε0 log 4θ − . log D log D

Since for 1 x y, y(log x/ log y) x − 1, and for d > 676, D/ log D 12/ log 12 > 4.8, we obtain if 4θ < D, M− c

1 √ log ε0 log 4θ log 4θ = max(4θ + 1, 2θ + 4) − D − d 2 log D log D log D < max(4θ + 1, 2θ + 4) − max(4θ − 1, 6) 2,

if 2θ < D 4θ , M−

1 √ log ε0 log 2θ log 2 log 4θ d = 2θ + 4 − D −D − 2 log D log D log D log D < 2θ + 4 − 2θ + 1 − 3 − 1 = 1,

1209


if D 2θ, M−

log θ log 4 log 4θ 1 √ log ε0 =θ +7−D −D − d 2 log D log D log D log D < θ + 7 − θ + 1 − 6 − 1 = 1.

This together with (18) gives the theorem.

4. Proof of Corollary. We can assume 1 − β < (log |d|)−2 . It follows then by Theorem 1 that, for every η > 0, there exists c(η) such that if d > c(η) 6 L(1, χ) η (19) 1 − β 2 . 1− π 2 1/a Let h0 be the number of classes of forms in question. For d < −4, we have π h0 L(1, χ) = √ , |d| and by Theorem 2 1 h0 . |a| Hence by (19) 1−β

1 η 6 6 h0 π > − η 1 − √ √ . π 2 h0 |d| 2 π |d|

For d > 0, we have L(1, χ) =

h0 log ε0 . √ d

Now, for any class C of forms (a,b,c)∈C

1 = |a| 1 4

√(a,b,c)∈C dab>−a

1 + a

(−a,b,−c)∈C √ d−ab>a

1 . |a|

If (a, b, c) runs through C, (−a, b, −c) runs through another class which we denote by −C (it may happen that −C = C). If C1 = C2 , then −C1 = −C2 . Hence

C 1 4

√ (a,b,c)∈C d|a||b|>−|a|

1 1 =2 |a| a

1210


and by Theorem 2

1 4 h0 log ε0 h0 log ε0 1 + < , 1 + O √ √ √

a 2 log 21 d − 1 log d d d √ where the constant in the O-symbol is effective. (Note that ε0 > 21 d.) This together with (19) gives the corollary.

References [1] H. Davenport, Multiplicative Number Theory. Markham, Chicago 1967. [2] D. Goldfeld, An asymptotic formula relating the Siegel zero and the class number of quadratic fields. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 2 (1975), 611–615. [3] W. Haneke, Über die reellen Nullstellen der Dirichletschen L-Reihen. Acta Arith. 22 (1973), 391–421. [4] A. E. Ingham, The Distribution of Prime Numbers. Cambridge Univ. Press, Cambridge 1932. [5] B. W. Jones, The Arithmetic Theory of Quadratic Forms. The Math. Assoc. of America, Buffalo 1950. [6] S. Lang, Introduction to Diophantine Approximations. Addison–Wesley, Reading 1966. [7] O. Perron, Die Lehre von den Kettenbrüchen, Band I. Teubner, Stuttgart 1954. [8] J. Pintz, Elementary methods in the theory of L-functions II. On the greatest real zero of a real L-function. Acta Arith. 31 (1976), 273–289. [9] K. Prachar, Primzahlverteilung. Springer, Berlin–Göttingen–Heidelberg 1957. [10] E. Titchmarsh, The Zeta Function of Riemann. Cambridge 1930.

Originally published in Proceedings of the Indian Academy of Sciences Mathematical Sciences 97 (1987), 297–303


Multiplicative properties of the partition function with E. Wirsing (Ulm)

Abstract. A lower bound for the multiplicatively independent values of p(n) for N n < N + R is given. The proof depends on the Hardy–Ramanujan formula and is of an elementary nature.

1. Introduction P. Erd˝os andA. Ivić [1] in their study of the number of non-isomorphicAbelian groups of a given order needed a lower estimate for the number m(N ) of multiplicatively independent values of the partition function p(n) in 1 n N . The first named author has proved (see [1], Lemma 2) that, denoting the number of prime divisors of n by ω(n), one has N ω p(n) → ∞, n=1

whence m(N ) → ∞, as N → ∞. In the present paper we give an explicit estimate for m(N ), at the same time eliminating from the proof the appeal to “linear forms in logarithms”. What we actually treat is the number m(N, N + R) of multiplicatively independent values of p(n) in N n < N + R, where R is relatively small compared to N .

2. Theorem. There is an N0 such that m(N, N + R) R

log N − log R 3 2

log N + R log 2

for N N0 and all R ∈ N. The same lower bound applies to ω

Nn 23 log N (log N − 1 − log R). Thus in the proof and in applications like Corollary 2 below there is no need to consider larger R. Corollary 1. m(N, N + R) Corollary 2. m(N, N + R)

2 3

− o(1) R if R = o(log N ) as N → ∞.

1 − o(1) log N if log 2

R → ∞ as N → ∞. log N

In particular we may state that N log N p(n) m(N ) (1 − ε) ω log 2

if

N N0 (ε).

n=1

Concerning the paper of Erd˝os and Ivić we have Corollary 3. Let a(n) be the number of non-isomorphic Abelian groups of order n and C(x) the number of distinct values of a(n) for n x. Then for every ε > 0 and x x1 (ε) log C(x)

(log log x)2 . log 16 + ε

Similarly for the number D(x) of distinct values a(n) x with any n one has, if x x2 (ε), log D(x)

(log log x)2 . log 4 + ε

Proof of Corollary 3. Notation as√in [1]. Our Corollary 2 allows to pick p(ki ) multiplicatively independent, k1 , . . . , kt log x, √ with t (log 4 + ε)∼ log log x. The construction gives C(x) r t where 2rt log(rt) log x. Here r = (log x)1/2−ε is admissible if x x1 , which gives our proposition. For the estimate of D(x) take k1 , . . . , kt log x with t (log 2 + ε) ∼ log log x. Now D(x) r t upon the condition that r log p(ki ) log x, which because of log p(k) i √ √ k log x is again satisfied by r ∼ (log x)1/2−ε .

3. The proof uses the Hardy–Ramanujan formula for p(n) (see (1) below) to construct a large number of distinct linear combinations of the numbers log p(n) with bounded integral coefficients. If, on the other hand, the dimension of this Z-module were too small it would contain too few elements with bounded height. This mechanism is rather unspecific. Thus √ the theorem as it stands applies to any function q(n) = p(n) + O[exp(c n)] where

K2. Multiplicative properties of the partition function

1213

√ c < π 2/3, and similar results can be proved whenever an arithmetic function q(n) allows an expansion log q(n) = a0 nα0 + a1 nα1 + . . . + b log n + remainder, where a0 > 0, α0 > 0, α0 ∈ / N, α0 > α1 > . . ., αi − αi+1 1, and where the remainder term is small enough. The factor 2/3 in Corollary 2 would become (α0 + 1)−1 . Lemma 1. Write Δ for the forward difference and put Δr := |Δr log p(N )|. Then, as N → ∞, we have

1/2 1/2−r

N Δr = c2 r!

(1 + O(N −1/3 )) r

with some constant c2 , uniformly in 0 r N 1/6 . Proof. The Hardy–Ramanujan formula (see [2]) gives (1)

p(n) = c1 f (n −

where

f (x) :=

The actual values

√ 1 24 ) + O[exp(c3 n)],

√ 1 x) , exp(c √ 2 x

c2 > c3 > 0.

) 1 c2 c1 = √ , c2 = π 2/3 , c3 = 2 π 8 are mostly irrelevant for our purpose. Keeping the abbreviation x = n − 1/24 we find √ √ c1 1 p(n) = c2 − √ exp(c2 x) + O[exp(c3 x)] 2x x √ √ c1 1 = c2 − √ exp(c2 x) 1 + O[exp(−c4 x)] 2x x

with some c4 > 0. Therefore

√ log p(n) = g(x) + O[exp(−c4 n)],

where

√ c 1 c2 1 g(x) = c2 x + log + log 1 − √ − log x 2 c2 x

—apart from the log x term—is a power series that converges for all large x, g(x) =

∞

ai x −i/2 − log x.

i=−1

Actually a−1 = c2 ,

|ai | c2−i

for

i 1.

1214


The case r = 0 of the lemma is now obvious. For r 1 the generalized mean-value theorem gives √

Δr log p(N ) = g (r) (ξ ) + O 2r exp(−c4 N ) , where N −

1 24

< ξ < N + r. Hence for r 1

Δ log p(N ) = r! r

∞ i=−1 i=0

ai

−i/2 −i/2−r ξ + (−1)r (r − 1)! ξ −r r √

+ O 2r exp(−c4 N ) .

If i 1, r 1, then i+1 i r−1 r−1 +1+j 2r −(i + 1)/2 −i/2 −1 2 + j 2 = < =1+ 3r. i i i r r j =0 j =0 +j +j 2 2 Similarly

−1

1/2

−1/2

−1

4r. r!

(r − 1)! r, (r − 1)! r!

r

r Therefore

∞ √

1/2 1/2−r

4r i

ξ Δr = c2 r!

1 + + O 2r exp(−c4 N ) O √

r c2 ξ i=1

√

1/2 1/2−r

r

ξ = c2 r!

+ O 2r exp(−c4 N ) . 1 + O √

r N

We also have

r 1/2−r ξ 1/2−r = (N + O(r))1/2−r = N 1/2−r 1 + O N r 2 = N 1/2−r 1 + O N

1/2−r =N 1 + O(N −2/3 )

by our bound for r. Finally

√

1/2 1/2−r −1

N (2N )r = exp[o( N )], 2r r!

r whence altogether

√

1/2 1/2−r

N Δr = c2 r!

1 + O(N −1/3 ) + O[exp(−c4 N /2)]

r

for all r N 1/6 , which is the lemma.

K2. Multiplicative properties of the partition function

1215

Lemma 2. Let R N 1/6 and 1 r R. Then Δr /Δr−1 R/N provided that N N0 . The N0 does not depend on R. Proof. Lemma 1 for 2 r N 1/6 supplies r− Δr = Δr−1 N

3 2

r r 3 R r < 1 + O(N −1/3 ) = − +O 4/3 N 2N N N N

if N N0 , and similarly for r = 1.

Proof of the theorem. We assume, somewhat arbitrarily, R N 1/6 . This includes for large N the range R log2 N that by an earlier remark is relevant. Consider now the numbers ω=

R−1

where xr ∈ N0 , xr

xr Δr ,

r=0

N − 1. R

They all are distinct because of R−1

R−1

xr Δr

r=s+1

N

r=s+1

R

R−1 (Δr−1 − Δr ) < Δs . − 1 Δr r=s+1

The number A of the ω’s is therefore R N R N R N 1 R R N R 1+O √ > 1− −1 = A= R R R N R N as N → ∞. On the other hand Δr = εr

r

(−1)s

s=0

r log p(N + s), s

εr = ±1,

implies that each ω is a linear combination over Z of the log p(n), N n < N + R. If q1 , . . . , qk denote the primes that make up the p(N), . . . , p(N + R − 1), p(n) =

k

a (n)

qj j

aj (n) ∈ N0 ,

,

j =1

say, then we obtain the representation ω=

k

yj log qj

j =1

with yj :=

0sr 3c2 / log 2 and a suitable N0 are chosen then for all R R0 and N N0

R 3/2 l N R R0 log 2 N R 1 2 N 1+O √ , R 3c2 R N as claimed. For each of the remaining R < R0 , formula (2) implies 3l 2R − ε, and l

therefore 3l 2R, if N is large enough, hence again 2R N 3/2 (N/R)R . (2)

3c2 3/2 2R N log 2 R

l

N R

References [1] P. Erd˝os, A. Ivić, The distribution of values of a certain class of arithmetic functions at consecutive integers. In: Number Theory, vol. I (Budapest 1987), Colloq. Math. Soc. János Bolyai 51, North-Holland, Amsterdam 1990, 45–91. [2] G. H. Hardy, S. Ramanujan, Asymptotic formulae in combinatory analysis. Proc. London Math. Soc. (2) 17 (1918), 75–115.

Originally published in New Trends in Probability and Statistics vol. 2: Analytic and Probabilistic Methods in Number Theory VSP Utrecht & TEV Vilnius 1992, 165–171


On an analytic problem considered by Sierpinski ´ and Ramanujan

Abstract. Let r(n) be the number of representations of n as a sum of two squares. An Ω-estimate for the error term in the summation formula for r(n)2 is obtained.

Let r(n) be the number of representations of n as a sum of two squares. W. Sierpiński in his doctorate thesis (see [6]), written in 1906, has proved the estimate r(n)2 = 4x log x + cx + O(x 3/4 log x), nx

where c is a certain constant. The same asymptotic formula with the better error term O(x (3/5)+ε ) was stated without proof in [4]. B. M. Wilson [8] indicated without giving the details that O(x (3/5)+ε ) can be replaced by O(x (1/2)+ε ) for every ε > 0. Recently, W. R. Recknagel [5] improved the error term to O(x 1/2 log4 x) and M. Kühleitner [2] has 1/2 (log x)11/3 (log log x)1/3 ). c improved it further to O(x As far as I know, there is no Ω-result in the literature, and it is the aim of this paper to prove such a result. Theorem. We have

r(n)2 = 4x log x + cx + Ω(x 3/8 ).

nx

The proof is based on the approach of [1]. The possibility of using this approach has been indicated to me by Prof. A. Ivić. In comparison with [1], the present paper contains no new idea, but the work is motivated by historical reasons. I thank Dr. W. G. Nowak for pointing out some obscurities in an early draft of the paper, and the Department of Mathematics in Geneva for their help in preparing the manuscript. Notation. ζ (s) is the Riemann zeta function, ζK (s) the Dedekind zeta function of the field K = Q(i), p denotes a general prime, A is a large constant, not necessarily the same at each occurrence, T is a sufficiently large real number, y = T B , where B is a sufficiently large constant; J = {T 2/3 t 2T : for any complex number z with Re z 1/3 and |Im z − t| (log T )20 we have ζ (2z) = 0};

1218


J1 = {T 2/3 − (log T )4 t 2T + (log T )4 : for any complex number z with Re z 1/3 and |Im z − t| (log T )15 we have ζ (2z) = 0}. Lemma 1. F (s) =

∞ r(n)2 n=1

16ζK (s)2 (1 + 2−s )ζ (2s)

=

ns

for

Re s > 1.

Proof. See [4]. Lemma 2. If Re s 1/3 and |Im s| 1, then

ζK (s) = O (|t| + 2)A , where s = σ + it.

Proof. See Lemma 7 in [7]. Lemma 3. If Re s 1/3 and t ∈ J1 , then 1 = O(|t| + 2). ζ (2s)

Proof. This is a consequence of Lemma 1 in [1]. Lemma 4. If Re s 1/3 and t ∈ J1 , then

F (s) = O (|t| + 2)A for a suitable A > 0.

Proof. This is a consequence of Lemmas 1–3. Lemma 5. Let

E(x) =

r(n)2 − 4x log x − cx.

nx

If

7

∞

T

E(u)2 −u/y e du log2 T , u7/4

t = Im s ∈ J , and Re s = 3/8, then we have 7 1 r(n)2

E(n + u) −(n+u)/y −n/y e + s e du + O (log T )20 F (s) = s s+1 n (n + u) 0 n>T0

nT0

for a suitable T0 , T T0 2T . Proof. We start with ∞ r(n)2 e−n/y n=1

ns

=

1 2πi

7 F (s + w)y w Γ (w) dw. Re w=2

K3. Problem considered by Sierpiński and Ramanujan

1219

Now we break off the portion |Im w| (log T )4 of the integral, with a small error, and move the line of integration to Re w = −(1/24). Now using the estimate of F (s) given in Lemma 4 (and assuming that B is large enough), and using the fact that, since t ∈ J and t + Im w ∈ J1 , we see that the value of the integral on the horizontals t +Im w = c ∈ J1 , as well as on the vertical Re w = −(1/24) is small. This proves that ∞ r(n)2 e−n/y n=1

ns

equals nearly the sum of the residues inside the contour. Thus we have F (s) =

∞ r(n)2 n=1

Now

ns

7 ∞ r(n)2 e−n/y = ns T0 n>T0 7 ∞ = T 7 0∞ = T0

e−n/y + O(T −10 ) =

nT0

+

+ O(T −10 ).

n>T0

1 −u/y 2 e d r(n) us nu

1 −u/y e d 4u log u + cu + E(u) s u 7 ∞ 1 −u/y 1 −u/y e e dE(u) 4 log u + 4 + c) du + s s u T0 u

= S1 + S 2

(say).

The first integral does not exceed up to a constant factor 7 ∞ 1 −u/y 1/24 e u du, s T0 u and hence is small by Lemma 3 of [1]. Now S2 is (after one integration by parts) 7 ∞ 7 E(u)e−u/y ∞ E(u)e−u/y 1 ∞ E(u)e−u/y + s du + du. us us+1 y T0 us T0 T0 Hence, using Lemma 4 in [1], and choosing a suitable T0 in the interval T T0 2T , we obtain 7 ∞ E(u) −u/y S2 = s e du + O(log T ) s+1 T0 u 7 1 E(n + u) =s e−(n+u)/y du + O(log T ). s+1 (n + u) 0 nT0

This proves the lemma.

1220


Lemma 6. We have

7

T

0

n

2

T + O(n) |bn |2 bn nit

dt = n

for any sequence of complex numbers bn provided the right hand side is convergent.

Proof. For the proof of this lemma we refer the reader to [3]. Lemma 7. We have

7

r(n)2 −n/y 2 dt

e

t 2 1. Re s=3/8

ns

Im s=t∈J nT0

c

Proof. It is sufficient to prove that

7 2T

r(n)2 −n/y

2 dt

e

t 2 1. n3/8 T 2/3 nT0

By Lemma 6, we have

7 2m+1

r(n)4 r(n)2 −n/y

2 m

2 e dt = + O(n)

n3/8 n3/4 2m c nT0

Hence

7

nT0

2m+1

2m

c

nT0

1/4

2 m T0

5/4

log7 T0 + T0

log7 T0 .

1/4 5/4 T0 log7 T0 T0 log7 T0 r(n)2 −n/y

2 dt e + .

t2 ns 2m 22m 2/3

Now, summing over m satisfying (1/2)T0 < 2m 2T , it follows that

7 2T

r(n)2 −n/y

2 dt

e

t 2 1. ns T 2/3 c

nT0

Lemma 8. If

7

∞

T

then we have 7

7

Re s=3/8

Im s=t∈J

1


2 7 ∞ E(n + u)e−(n+u)/y

|E(u)| −2u/y

dt du e du,

s+1 (n + u) u7/4 T

0 nT 0

for a suitable T0 , T T0 2T . Proof. This follows from Lemma 6 in the same way as in [1] Lemma 8 follows from Lemma 6.


Lemma 9. If

7

∞

T

we have

7 Re s=3/8 Im s=t∈J

1221


|F (s)|2 dt 1 + |s|2

7

∞

T

E(u)2 −2u/y e du. u7/4

Proof. This follows from Lemmas 5, 7 and 8. Lemma 10. Let f (s) =

∞ bn n=1

ns

be convergent in some half-plane and analytically continuable in σ 1/3, A t A + M. Then 7 A+M

f (3/8 + it) 2 dt M, A

provided M log A, b1 = 1, and max |f (s)| : A t A + M, Re s 3/8 eA . Proof. This is a consequence of Lemma 9 in [1] for α = 3/8.

Definition 1. ∞

f (s) =

bn ζK (s + 1/4)2 . = (1 + 2−s )ζ (2s) ns n=1

Lemma 11. If an interval [A, A + M] is contained in J and A M log A, we have 7 A+M |f (3/8 + it)|2 dt M. A

Proof. It follows from the definition of f (s) that b1 = 1, ∞ bn n=1

ns

is convergent in

Re s > 3/4

and analytically continuable in the whole plane with poles at s=

3 (2k + 1)π i , , 4 log 2

k ∈ Z,

and at the zeros of ζ (2s). If Re s 3/8 and [A, A+M] is contained in J , then by Lemma 2 we have

ζK (s + 1/4) |t| + 2 A/(2 log(2A+2)) eA/2 ,

1222 and by Lemma 3,


ζ (2s)−1 |t| + 2 2A + 2.

Hence |f (s)| eA for A large enough. The lemma follows now from Lemma 10.

Lemma 12. If [A, A + M] ⊂ J and A M log A, we have 7 A+M

F (3/8 + it) 2 dt AM. A

Proof. For Re s = 3/8, we note that |ζK (1 − s)| = |ζK (s + (1/4))|. Using the functional equation B s Γ (s)ζK (s) = B 1−s Γ (1 − s)ζK (1 − s), where B is a constant, we obtain for Re s = 3/8, |F (s)| = 16|f (s)| |ζK (s)|2 |ζK (1/4)|−2 = 16|f (s)| |ζK (s)|2 |ζK (1 − s)|−2

= 16|f (s)| |B 2−4s Γ (1 − s)2 Γ (s)−2 | f (s) |t|1/2 . Hence Lemma 12 follows from Lemma 11.

Definition 2. For every x such that T 2/3 x 2T , J (x) = J ∩ [x, 2x], N(x) is the number of zeros of ζ (2s) with t ∈ x − (log T )20 , 2x + (log T )20 and Re s 1/3. Lemma 13. We have N (x) x 49/60 . Proof. This follows from Lemma 13 in [1] for ε = 1/6.

Definition 3. J2 (x) is the portion of J (x) obtained by deleting all connected components whose length is not greater than log 2x. Lemma 14. We have

7

F (3/8 + it) 2 dt x 2 . J2 (x)

Proof. The total length of [x, 2x] \ J2 (x) is O N (x)(log T )20 = O(x 5/6 ). Hence the total length of J2 (x) is x. Now applying Lemma 12 to each connected component of J2 (x) and adding we obtain Lemma 14.

Lemma 15. We have

7 Re s=3/8 Im s=t∈J

|F (s)|2 dt log T . |s|2


1223

Proof. From Lemma 14 we have 7 |F (3/8 + it)|2 dt x 2 , J (x)

and Lemma 15 follows by integration by parts.

Proof of the Theorem. By Lemma 9 and Lemma 15 we have 7 ∞ E(u)2 −2u/y e du log T . u7/4 T If for every ε > 0 we had |E(u)| < εu3/8 for T T0 (ε), it would follow that 7 T0 (ε)B 7 ∞ 7 ∞ E(u)2 −2u/T0 (ε)B du 2 −2v dv + e du ε e 7/4 u v T0 (ε) T0 (ε) u 1

2 < ε B log T0 (ε) + 1 ,

in contrary to the above inequality.

References [1] R. Balasubramanian, K. Ramachandra, M. V. Subbarao, On the error function in the asymptotic formula for the counting function of k-full numbers. Acta Arith. 50 (1988), 107–118. 2 [2] M. Kühleitner, On a question of A. Schinzel concerning the sum nx (r(n)) . In: Österreichisch-Ungarisch-Slowakisches Kolloquium über Zahlentheorie (Maria Trost, 1992), Grazer Math. Ber. 318, Karl-Franzens-Univ. Graz, Graz 1993, 63–67. [3] H. L. Montgomery, R. C. Vaughan, Hilbert’s inequality. J. London Math. Soc. (2) 8 (1974), 73–82. [4] S. Ramanujan, Some formulae in the analytic theory of numbers. Messenger of Math. (2) 45 (1916), 81–84; Collected papers, Cambridge 1927, 113–135. [5] W. Recknagel, Varianten des Gaußschen Kreisproblems. Abh. Math. Sem. Univ. Hamburg 59 (1989), 183–189. nb [6] W. Sierpiński, Sur la sommation de la série n>a r(n)f (n), où r(n) signifie le nombre de décomposition du nombre n en une somme de deux carrés de nombres entiers. Prace Mat. Fiz. 18 (1908), 1–59 (Polish); also in: Oeuvres choisies, T. I, Varsovie 1974, 109–154. [7] W. Sta´s, Über eine Anwendung der Methode von Turán auf die Theorie des Restgliedes im Primidealsatz. Acta Arith. 5 (1959), 179–195. [8] B. M. Wilson, Proof of some formulae enunciated by Ramanujan. Proc. London Math. Soc. (2) 21 (1922), 235–255.

Originally published in Journal of Number Theory 78 (1999), 62–84


Class numbers and short sums of Kronecker symbols with J. Urbanowicz (Warszawa) and P. Van Wamelen (Baton Rouge)

1. Introduction We shall consider sums of the form S(D, q1 , q2 ) =

q1 |D| 1 we put (2.9)

ψ=

(3−η)/2 g3 gP2 gPh

h−1

gP2 i .

i=3

By Lemma 2, ψ is a primitive character of conductor f , thus the only prime satisfying (2.2) is 2. Since all the exponents on the right hand side of (2.9) divide 2, a proper choice of gP gives ψ(2)2 = 1 and that ψ is not real. Similarly to the above we obtain ψ(−1) = ε and ψ(k) = −η.

K4. Class numbers and short sums of Kronecker symbols

1231

Likewise, if Ph = 3, η = −1 and j < h − 1 we put (3+ε)/2

ψ = g3 gP1

(2.10)

gPh−1

h−2

gP2 i .

i=2

By Lemma 2, ψ is a primitive character of conductor f , thus the only prime satisfying (2.2) is 2. Since all the exponents on the right hand side of (2.10) divide 2, a proper choice of gP gives ψ(2)2 = 1 and that ψ is not real. Moreover, ψ(−1) = (−1)(3+ε)/2 = ε,

ψ(k) = 1.

If either η = −ε,

3 = P1 ,

(2.11)

j =1

and

or 3 = Ph ,

(2.12)

η = −1,

and

j =h−1

we put ψ = ψ1 ψ2 , where

x , ψ1 (x) = pi 1ih pi /| 3P

(3+εψ (−1))/2

1 ψ2 = gP . pi and P , respectively. Thus ψ is a primitive ψ1 , ψ2 are primitive characters mod pi /| 3P 2 pi P , non-real since ζϕ(P character mod ) ∈ R. We have

pi /| 3P

ψ(−1) = ψ1 (−1)ψ2 (−1) = ψ1 (−1)(−1)(3+εψ1 (−1))/2 = ε. Moreover, (2.11) implies k ≡ −1 (mod (f/3)), ψ(k) = ψ1 (−1)ψ2 (−1) = ε = η; (2.12) implies k ≡ 1 (mod (f/3)), ψ(k) = 1 = η. The only primes satisfying (2.2) are 2 and 3 and the equalities ψ(2)2 = 1, ψ(3)2 = 1 would give 24 ≡ 1 (mod P ) or 34 ≡ 1 (mod P ), P = 5, contrary to (2.7). This completes the proof of the proposition except for the last statement. That follows by inspection of the argument, where the prime 17 is avoided only because of the condition

ψ(2)2 = 1. Proposition 3. Let k, f ∈ N, (k, f ) = 1, k ≡ ±1 (mod f ), and either f /| 16 · 3 · 5 or 16 · 5 | f . For each ε = ±1 there exists a non-real primitive character ψ of conductor d, where d | f such that ψ(−1) = ε,

ψ(k) = 1,

and ψ(p)2 = 1 for all primes p,

satisfying (2.13)

p|f

and p /| d.

1232


Proof. We shall distinguish four cases: (i) (ii) (iii) (iv)

f f f f

≡ 0 (mod 4), ≡ 4 (mod 8), = 2α f1 , where α 3, f1 odd, f1 /| 15, = 2α f1 , where f1 | 15 and either α 5 or α = 4, 5 | f1 .

For the sake of brevity in each case we only define a character ψ with the required properties, leaving to the reader the actual verification. Case (i). If f is odd it suffices to take in Proposition 2 η = 1. If f ≡ 2 (mod 4) it suffices in view of Proposition 2 to consider the case 17 | f | 510. If k ≡ ±1 (mod 17) or k ≡ −1 (mod 17), ε = −1, we put (3+ε)/2

ψ = g17

.

If k ≡ −1 (mod 17), ε = 1, there is a prime p | f such that k ≡ −1 (mod p). We put ψ = gp g17 . If k ≡ 1 (mod 17), there is a prime p | f such that k ≡ 1 (mod p), p = 3 or 5. We put ⎧ (3−ε)/2 g g , if p = 3, ⎪ ⎪ ⎪ 32 17 (3+ε)/2 ⎨ , if p = 5, k ≡ ±2 (mod 5), g5 g17 ψ(x) = (7−ε)/2 ⎪ g , if p = 5, k ≡ −1 (mod 5), and either f = 170 or ε = 1, g ⎪ ⎪ ⎩ 5 17 g3 g5 g17 , otherwise. Case (ii). In Proposition 2 we replace f by f/4, ε by −ε, and η by g4 (k). If k ≡ ±1 (mod (f/4)) there exists by virtue of Proposition 2 a non-real primitive character ψ of conductor d such that d | f/4 and (2.14)

ψ (−1) = −ε,

ψ (k) = g4 (k)

and (2.15)

ψ (p)2 = 1

for all primes

p | (f/4), p /| d.

If k ≡ −1 (mod 4), k ≡ 1 (mod (f/4)), or if k ≡ 1 (mod 4), k ≡ −1 (mod (f/4)), ε = 1, the properties (2.14) and (2.15) belong to every primitive character ψ mod f/4 with ψ (−1) = −ε. In each case the character ψ = g4 ψ satisfies the condition of the proposition. In the remaining case k ≡ 1 (mod 4), k ≡ −1 (mod (f/4)), ε = −1, there exists a prime power P such that

f f

P , P, , P = 3, 5. 4 4P We put ψ = ψ1 ψ2 ,

K4. Class numbers and short sums of Kronecker symbols

where ψ1 (x) =

p|(f/4P ) p prime

x , p

(3−ψ1 (−1))/2

ψ2 = gP

1233

.

Case (iii). If k ≡ ±1 (mod f1 ) we argue as in the case α = 2 above. If k ≡ 1 (mod f1 ) we have k ≡ 1 (mod 2α ). We choose a character ψ2 mod 2α such that ψ2 (k) = 1 and then a non-real primitive character ψ1 mod f1 such that ψ1 (−1) = εψ2 (−1). Then ψ = ψ1 ψ2 has the required properties. If k ≡ −1 (mod f1 ) we have k ≡ −1 (mod 2α ). We choose a non-trivial character ψ2 mod 2α such that ψ2 (−k) = ε and then a non-real primitive character ψ1 mod f1 such that ψ1 (−1) = εψ2 (−1). The character ψ = ψ1 ψ2 has the required properties. Case (iv). If k ≡ ±1 (mod 2α−1 ) we put (3+ε)/2

ψ = g4

g2 α .

1 + 2α−1

The same formula is good if k ≡ (mod 2α ); or k ≡ −1 + 2α−1 (mod 2α ), ε = 1; α or k ≡ −1 (mod 2 ), ε = −1. If k ≡ −1 + 2α−1 (mod 2α ), ε = −1, we put ⎧ g4 g22α , if α 5, ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ if α = 4, k ≡ 1 (mod 5), ⎪ ⎨g5 g2α , ψ = g5 g2α , if α = 4, k ≡ 1 (mod 5), f = 80, ⎪ ⎪ ⎪ 2 ⎪g3 g5 , if α = 4, k ≡ 1 (mod 5), f = 240, k ≡ 1 (mod 3), ⎪ ⎪ ⎪ ⎩ g3 g4 g5 g2α , if α = 4, k ≡ 1 (mod 5), f = 240, k ≡ −1 (mod 3). If k ≡ η (mod 2α ), η = ±1, ε = 1, then f1 has a prime factor p such that k ≡ η (mod p). If α 5, or if p = 3, or if p = 5 = f1 we put ψ = g4 gp g2α . If f = 240, k ≡ η (mod 48), k ≡ η (mod 5), ε = 1 we put ψ = g3 g5 g2α . In the remaining case k ≡ 1 (mod 2α ), ε = −1, f1 has a prime factor p such that k ≡ 1 (mod p). We put gp g2α , if α 5, or if p = 3, or if p = 5 = f1 ,

ψ= g5 g22α , otherwise.

1234


3. Character sums in terms of Bernoulli numbers Let χ be a Dirichlet character mod M and let N be a multiple of M. For any integer r > 1 prime to N and natural m we have the formula from [5]: χ¯ (r) ¯ (3.1) mr m−1 χ (n)nm−1 = −Bm,χ r m−1 + ψ(−N )Bm,χψ (N ), ϕ(r) 0 0 there exists a non-zero vector n ∈ Z3 , such that for all non-zero vectors p, q ∈ Z3 and all u, v ∈ Q, n = up + v q implies ( 4 h(p)h(q) > 3 − ε h(n). Originally, in the proof of Theorem 1 some computer calculations were used which were kindly performed by Dr. T. Regińska. We thank her for the help. The proof of Theorem 1 will be based on geometry of numbers. The inner product of two vectors n, m will be denoted by nm, their exterior product by n × m, the area of a plane domain D by A(D). (i = 1, 2, 3) and M1 , M2 , M3 the three miLemma 1. Let ai , bi be real numbers a1 a2 a3 not all equal to 0. The area of the domain nors of order two of the matrix b1 b2 b3 H : |ai x + bi y| 1 (i = 1, 2, 3) equals c

2|M1 M2 | + 2|M1 M3 | + 2|M2 M3 | − M12 − M22 − M32 |M1 M2 M3 |

1250

L. Geometry of numbers

if each of the numbers |M1 |, |M2 |, |M3 | is less than the sum of the two others, and 4/ max{|M1 |, |M2 |, |M3 |} otherwise. Proof. We may assume without loss of generality that

a1 a2

a2 a3

,

> 0, |M1 | |M2 | = abs

|M1 | = abs

b1 b2

b2 b3

a1 a3

.

|M1 | |M3 | = abs

b1 b3

The affine transformation a1 x + b1 y = X, a2 x + b2 y = Y transforms the domain H into the domain

M M3

2 H : |X| 1, |Y | 1;

X− Y 1. M1 M1 c

If |M2 | + |M3 | > |M1 |, the domain H is obtained from the square |X| 1, |Y | 1 by subtracting two rectangular triangles, symmetric to each other with respect to (0, 0), with the vertices |M | − |M | M2 |M1 | − |M2 | M2 M2 1 3 ± 1, − sgn , ± . , ± 1, − sgn , − sgn M3 |M3 | M3 |M2 | M3 Hence, A(H ) = 4 −

(|M2 | + |M3 | − |M1 |)2 . |M2 | |M3 |

If |M2 |+|M3 | |M1 |, then H coincides with the square |X| 1, |Y | 1 and A(H ) = 4.

Since A(H ) = A(H )/|M1 |, the lemma follows. Lemma 2. If 0 a b < 1, then the domain D : |x| 1, |y| 1, |ax + by| 1, x 2 + y 2 + (ax + by)2 contains an ellipse E with

(

(1)

A(E) > π

3 4

.

Proof. We take E : f (x, y) = x 2 + c

ab 2 1, x + y b2 + 1

where (2)

c = max

2 3

(b2 + 1),

(b2 + 1)2 . (b2 + 1)2 − a 2 b2

In order to see that |x| 1, |y| 1 for (x, y) ∈ E, we notice that by (2) c (3) min f (x, y) = x 2 , min f (x, y) = y2 y2. a 2 b2 y x c (b2 +1)2 + 1 c

3 2

1251

L1. A decomposition of integer vectors II

Moreover, for (x, y) ∈ E we have by (2) (4) x 2 + y 2 + (ax + by)2 2 3 ab 3 2 a 2 + b2 + 1 2 2 2 3 f (x, y) . x (b x + y + + 1) 2 3 b2 + 1 3 b2 + 1 2 2 If for (x, y) ∈ E we had |ax + by| > 1, it would follow x2 + y2
n3 . In virtue of Lemma 2 the domain

n n n2

n 2 2 3 1

1 D : |X| 1, |Y | 1, X + Y 1, X2 + Y 2 + X+ Y n3 n3 n3 n3 2 √ contains an ellipse E with A(E) > π 3/4. Let a, b be a basis the existence of which is asserted by Lemma 3. The substitution X=

a1 x + b1 y , ( 4 n 3 3

Y =

a2 x + b 2 y ( 4 3 n3

transforms D into the domain D : |ai x + bi y|

(

4 3

n3

(i = 1, 2, 3),

3

(ai x + bi y)2 2n3 .

i=1

D

E

contains an ellipse with B B

−1 4 4 4

a1 a2

A(E ) = n3

(n1 , n2 , n3 ) π , A(E) > π

b b 3 3 3 1 2 √ by (8). Since the packing constant for ellipses is π/ 12, it follows that E and, hence, D contains in its interior a point (x0 , y0 ) ∈ Z2 different from (0, 0). Putting m = x0 a + y0 b, we get the assertion of the lemma.

Hence,

Lemma 5. If 0 a 1, 0 b 1 and a + b > 1, the area of the hexagon |x| 1, 1/2

. |y| 1, |ax + by| 1 is greater than 24/(a 2 + b2 + 1) Proof. In virtue of Lemma 1 the area in question equals (2ab + 2a + 2b − a 2 − b2 − 1)/ab, thus, it remains to prove that for (a, b) in the domain G : 0 a 1, 0 b 1, a + b > 1 the following inequality holds f (a, b) = (2ab + 2a + 2b − a 2 − b2 − 1)2 (a 2 + b2 + 1) − 24a 2 b2 > 0. We have ∂G = L1 ∪ L2 ∪ L3 , where L1 = {(a, 1) : 0 a 1}, L2 = {(1, b) : 0 b 1}, L3 = {(a, 1 − a) : 0 a 1}. We find f (a, 1) = a 2 (a − 1)3 (a − 5) + 3a 2 , but a 2 (a − 1)3 (a − 5) 0 for a 1, hence f (a, 1) 3a 2 0. In view of symmetry between a and b, f (1, b) 3b2 0. Moreover, f (a, 1 − a) = 8a 2 (1 − a)2 (2a − 1)2 0. Hence, for (a, b) ∈ ∂G we have f (a, b) 0 with the equality attained only if (a, b) ∈ / G. It suffices to show that in the interior of G the function f (a, b) has no local extremum.

L1. A decomposition of integer vectors II c

1253

Indeed, putting g(a, b) = 2ab + 2a + 2b − a 2 − b2 − 1, we find ∂f = 2ag 2 + 2(2b + 2 − 2a)(a 2 + b2 + 1)g − 48ab2 , ∂a ∂f = 2bg 2 + 2(2a + 2 − 2b)(a 2 + b2 + 1)g − 48a 2 b, ∂b hence

∂f ∂f −b = 2(a − b) (a + b)g + (a 2 + b2 + 1)(2 − 2a − 2b) , ∂a ∂b

∂f ∂f b −a = 4(b − a) (a + b + 1)(a 2 + b2 + 1)g − 12ab(a + b) . ∂a ∂b The equations ∂f /∂a = ∂f /∂b = 0 imply a = b or a

(13)

(a + b)g + (a 2 + b2 + 1)(2 − 2a − 2b) = 0, (a + b + 1)(a 2 + b2 + 1)g − 12ab(a + b) = 0.

Eliminating g from the above equations we obtain

2(a 2 + b2 + 1)2 (a + b)2 − 1 − 12ab(a + b)2 = 0. c (14) The left hand sides of the equations (13) and (14) are symmetric functions of a, b. Expressing them in terms of s = a + b and p = ab, then eliminating p, we get s(s − 1)(2s − 1)(4s 2 − s + 1) = 0. c

For s = a + b > 1 this is clearly impossible, there remains the possibility a = b. However, in that case ∂f

= 16a 3 − 24a 2 + 18a − 4 = 2(2a − 1)3 + 3(2a − 1) + 1 > 1. ∂a Lemma 6. For every non-zero vector n = [n1 , n2 , n3 ] ∈ Z3 there exist linearly independent vectors p, q ∈ Z3 such that pn = qn = 0, and B 2 l(n), h(p)h(q) < 3 if each of the numbers |n1 |, |n2 |, |n3 | is less than the sum of the two others; h(p)h(q) h(n),

otherwise.

Proof. We may assume without loss of generality that 0 n1 n2 n3 > 0. In virtue of Lemmata 1 and 5 the area A(K) of the domain

n n2

1 Y 1 K : |X| 1, |Y | 1, X − n3 n3 satisfies : ⎧ ⎪ 24 ⎨A(K) > n3 , if n1 + n2 > n3 , 2 (15) n1 + n22 + n23 ⎪ ⎩ A(K) = 4, otherwise.

1254


Let a, b be a basis the existence of which is asserted in Lemma 3. The affine transformation X = a1 x + b1 y, Y = a2 x + b2 y transforms the domain K into the domain K : |ai x + bi y| 1

(i = 1, 2, 3)

satisfying (n1 , n2 , n3 ) . n3 In virtue of Minkowski’s second theorem there exist two linearly independent integer vectors [x1 , y1 ] and [x2 , y2 ] such that A(K ) = A(K)

(16)

(17)

|ai xj + bi yj | λj

(i = 1, 2, 3; j = 1, 2)

and λ1 λ2 A(K ) 4.

(18)

Putting p = ax1 + by1 , q = ax2 + by2 , we infer that p, q are linearly independent, satisfy pn = qn = 0 and in virtue of (15), (18) ( < 23 l(n), if n1 + n2 > n3 ,

h(p)h(q) λ1 λ2 n3 , otherwise. Proof of Theorem 1. If n = [ε1 , ε2 , ε3 ], where εi ∈ {1, −1}, it suffices to take p = [ε1 , ε2 , 0], q = [0, 0, ε3 ]. If n = [ε1 , ε2 , ε3 ] for every choice of ε1 , ε2 , ε3 , then by Lemma 4 there exists a vector m ∈ Z3 satisfying the conditions (19) (20)

( 0 < h(m)
4t 2 + 4t.

1256


Proof. pm1 = 0 implies p1 ≡ 0 mod 6t 2 + 4t − 1, p2 ≡ 0 mod 6t 2 + 6t − 1. Hence, p1 = p2 = 0 or |p2 | 6t 2 + 6t − 1. Similarly, q1 = q2 = 0 or |q2 | 6t 2 + 6t − 1. Since

p, q are linearly independent, h(p)h(q) 6t 2 + 6t − 1 > 4t 2 + 4t. Lemma 9. If p, q ∈ Z3 are linearly independent and pm2 = qm2 = 0, then h(p)h(q) 4t 2 + 4t. Proof. The equation pm2 = (2t 2 + 2t − 1)p1 − (4t 2 + 4t)p2 + (2t 2 + 2t)p3 = 0 c

gives p1 ≡ 0 mod 2t 2 + 2t, hence, p1 = 0 or |p1 | 2t 2 + 2t. The former possibility gives |p3 | 2. Similarly, q1 = 0, |q3 | 2 or |q1 | 2t 2 + 2t. Since p, q are linearly independent, p1 = q1 = 0 is excluded, hence, h(p)h(q) min 2(2t 2 + 2t), (2t 2 + 2t)2 4t 2 + 4t.

Lemma 10. If p, q ∈ Z3 are linearly independent and pm3 = qm3 = 0, then h(p)h(q) 4t 2 + 4t. Proof. The equation pm3 = (4t 2 + 4t)p1 − (2t 2 − 1)p2 − (2t 2 + 2t)p3 = 0 gives p2 ≡ 0 mod 2t 2 + 2t, hence p2 = 0 or |p2 | 2t 2 + 2t. The further proof is similar to that of Lemma 9.

Lemma 11. If p ∈ Z3 , pm4 = 0, then either p = 0 or h(p) 2t + 1. Proof. The equation pm4 = (2t 2 + 2t + 1)p1 + (2t 2 + 4t + 1)p2 − (4t 2 + 4t)p3 = 0 gives (24)

(2t 2 + 2t)(p1 + p2 − 2p3 ) + p1 + (2t + 1)p2 = 0.

If p1 + p2 − 2p3 = 0, then p1 + (2t + 1)p2 = 0 and either p1 = 0 or |p1 | 2t + 1. If p1 + p2 − 2p3 = 0, then since by (24) p1 ≡ p2 mod 2, we obtain p1 + p2 − 2p3 = 2s, s ∈ Z \ {0},

p1 + (2t + 1)p2 = −(4t 2 + 4t)s.

Hence, p3 + tp2 = −(2t 2 + 2t + 1)s and max{|p2 |, |p3 |}

2t 2 + 2t + 1 > 2t, t +1

thus h(p) 2t + 1.

Lemma 12. If p, q ∈ Z3 are linearly independent and pm5 = qm5 = 0, then h(p)h(q) > 4t 2 + 4t (t = 1).

L1. A decomposition of integer vectors II

1257

Proof. The equation pm5 = 2p1 + (6t 2 + 8t + 1)p2 − (6t 2 + 6t)p3 = 0 gives 2p1 + (2t + 1)p2 + (6t 2 + 6t)(p2 − p3 ) = 0. If p2 = p3 , we get p1 ≡ 0 mod 2t + 1, hence, |p1 | 2t + 1. If p2 = p3 , we get (2t + 3) max{|p1 |, |p2 |} 6t 2 + 6t, hence max{|p1 |, |p2 |}

6t 2 + 6t > 3t − 2 2t + 3

and h(p) 3t − 1. Similarly, q2 = q3 and |q1 | 2t + 1 or h(q) 3t − 1. Since p, q are linearly independent, p2 = p3 , q2 = q3 is excluded and we get for t = 1 h(p)h(q) min (2t + 1)(3t − 1), (3t − 1)2 (2t + 1)(3t − 1).

Lemma 13. If p, q ∈ Z3 are linearly independent and pm6 = qm6 = 0, then h(p)h(q) 4t 2 + 4t. The equation pm6 = (6t 2 + 6t + 1)p1 + (4t + 2)p2 − (6t 2 + 6t)p3 = 0 gives (6t 2 + 6t)(p1 − p3 ) + p1 + (4t + 2)p2 = 0. If p1 − p3 = 0, we get p1 ≡ 0 mod 4t + 2, hence, |p1 | 4t + 2. If |p1 − p3 | 2, we get (4t + 3) max{|p1 |, |p2 |} 2(6t 2 + 6t), hence, max{|p1 |, |p2 |} c

12t 2 + 12t > 3t 4t + 3

and h(p) 3t + 1. If p1 − p3 = ±1, we get p1 + (4t + 2)p2 = ∓(6t 2 + 6t), hence (6t 2 + 6t) (6t 2 + 6t) or p2 = ∓ + 1. or p2 = ∓ 4t + 2 4t + 2 The last two formulae give the following possible values for ∓[p1 , p2 ]: 3t 3t + 1 3t + 2 3t + 3 3t, , t − 1, , −t − 2, , −3t − 3, . 2 2 2 2

Hence, either h(p) 3t + 2{t/2} or p1 − p3 = ±1 and p2 = (3t + 2)/2 . Similarly, either h(q) 3t + 2{t/2} or q2 − q3 = ±1 and q2 = (3t + 2)/2 . Since p, q are linearly independent it follows that t 3t + 2 h(p)h(q) 3t + 2

4t 2 + 4t. 2 2 either |p1 | 4t + 2

1258


Proof of Theorem 2. Since lim )

t→∞ c

B

4t 2 + 4t (4t 2

+ 4t)2

− (2t 2

− 1)(2t 2

+ 2t − 1)

=

4 , 3

for every ε > 0 there exist integers t such that ( 4 (25) 4t 2 + 4t > 3 − ε h(nt )

and we fix such a value of t. If nt = up + v q, u, v ∈ Q and p, q ∈ Z3 are linearly dependent, then since c (nt1 , nt2 , nt3 ) = 1, we have either p = 0 or p = snt , s ∈ Z \ {0}, thus h(p) h(nt ), and c similarly for q. It follows that for p = 0, q = 0 ( 4 h(p)h(q) h(nt )2 > 3 − ε h(nt ) . If p, q are linearly independent, then p × q = 0 and (p × q)nt = 0. On the other hand, either h(p)h(q) 4t 2 + 4t or h(p × q) 2h(p)h(q) 2(4t 2 + 4t − 1) = 8t 2 + 8t − 2. In the latter case in virtue of Lemma 7 we have p × q = mi , for an i 6. Hence, 2 c pmi = qmi = 0 and from Lemmata 8–13 we obtain h(p)h(q) 4t + 4t. In view of (25) the theorem follows.

c

Remark. There exist decompositions nt = up + v q with h(p)h(q) = 4t 2 + 4t, namely nt = (6t 2 + 4t − 1) 2t 2 + 2t, 0, −(2t 2 + 2t − 1) + (2t 2 + 2t)(6t 2 + 6t − 1)[0, 1, 2] or nt = (2t 2 + 2t)(6t 2 + 4t − 1)[1, 0, 2] + (6t 2 + 6t − 1)[0, 2t 2 + 2t, 1 − 2t 2 ].

References [1] A. Châtelet, Leçons sur la théorie des nombres. Paris 1913. [2] A. Schinzel, Reducibility of lacunary polynomials VII. Monatsh. Math. 102 (1986), 309–337; Errata, Acta Arith. 53 (1989), 95. [3] −−, A decomposition of integer vectors I. Bull. Polish Acad. Sci. Math. 35 (1987), 155–159.

Originally published in Journal of the Australian Mathematical Society Series A 51 (1991), 33–49


A decomposition of integer vectors IV In memory of Kurt Mahler Abstract. Given m linearly independent vectors n1 , . . . , nm ∈ Zk and an integer l ∈ [m, k] one proves the existence of l linearly independent vectors p 1 , . . . , pl ∈ Zk or q 1 , . . . , q l ∈ Zk of small size (suitably measured) such that the ni ’s are linear combinations of p j ’s with rational coefficients or of q j ’s with integer coefficients.

In order to generalize the results of [10] (Part III of this series) let us introduce the following notation. Given m linearly independent vectors n1 , . . . , nm ∈ Zk let H (n1 , . . . , nm ) denote the maximum of the absolute values of all minors of order m of the matrix ⎛ ⎞ n1 ⎜ .. ⎟ ⎝ . ⎠ nm and D(n1 , . . . , nm ) the greatest common divisor of these minors. Furthermore, let h(n) = H (n) for n = 0,

h(0) = 0.

Definition 1. For k l m, k > m, let l D(n1 , . . . , nm ) (k−l)/(k−m) c0 (k, l, m) = sup inf h(p i ), H (n1 , . . . , nm ) c1 (k, l, m) = sup inf

i=1

D(n1 , . . . , nm ) H (n1 , . . . , nm )

(k−l)/(k−m) l

h(q i ),

i=1

where the supremum is taken over all sets of linearly independent vectors n1 , . . . , nm ∈ Zk and the infimum is taken over all sets of linearly independent vectors p 1 , . . . , pl ∈ Zk or q 1 , . . . , q l ∈ Zk such that for all i m, ni =

l

uij pj ,

uij ∈ Q,

j =1

ni =

l

uij q j ,

uij ∈ Z.

j =1

The Bombieri–Vaaler refinement [1] of the Siegel lemma easily leads (on the lines of the proof of (8) in [10]) to the conclusion that c0 (k, l, m) is finite, first obtained by Yu. Teterin. The aim of this paper is to give bounds for c0 (k, l, m) and c1 (k, l, m) which Communicated by J. H. Loxton

1260


are independent of k. First however we shall introduce three further series of constants, this time of geometric character. Definition 2. For a given positive integer m, let κm be the volume of the unit ball in Rm , g0 (m) = sup inf

vol P , vol K

g1 (m) = sup inf

vol P κm · m, vol E (K) 2

where the suprema are taken over all m-dimensional convex bodies K situated in Rm , symmetric with respect to the origin, the infima are taken over all parallelopipeds P containing K symmetric with respect to the origin and E (K) denotes the ellipsoid of the maximum volume contained in K. (It is unique; see [7].) Clearly 2m 2m g0 (m) g1 (m). κm κm The best published result pertaining to g0 (m), g1 (m) seems to be the following inequality due to Dvoretzky and Rogers ([4], Theorem 5A): mm 1/2 . g1 (m) m! Professor A. Pełczyński who indicated to me the paper [4] has improved the above inequality by showing together with S. J. Szarek that (see [9], Proposition 2.1) m(m+1) 2 m 2 g1 (m)2 m m+1 and, on the other hand, they have proved that (ibid., Section 6) g1 (m)2

2m . m+1

For m 2 the two bounds coincide and give g1 (1) = 1,

g1 (2) =

(

4 3

.

According to Theorem 5.1 of [9], for every ε > 0, m log g1 (m) = + o(m2/3+ε ). 2 I am indebted to Professor Pełczyński also for the paradigm (for l = 2) of the proof of Lemma 1 below, which he has since proved in another way (see [9], Corollary 3.1). We shall prove Theorem 1. For all integers k, l, m satisfying k l m, k > m > 0, l/2 l! (1) c0 (k, l, m) min (l − m + 1)1/2 g1 (m)γl , g0 (m), m! l/2 l l/2 , l (l−m)/2 g1 (l)γl m where γl is the Hermite constant. For l = m 2 we have here equality.

L2. A decomposition of integer vectors IV c

1261

Theorem 2. For all integers k, l, m satisfying k l m, k > m > 0 we have l c1 (k, l, m) |δij | , f (l) = sup inf c0 (k, l, m) A U j =1

where [δij ] = UA−1 , A and U run through all lower triangular non-singular integral matrices and all lower triangular integral matrices of order l, respectively. Moreover √ 1 + 16l + 17 (l + λ + 1)! where λ = f (l) l−λ . 4 (2λ + 1)! 4 S. Chaładus and Yu. Teterin prove in the forthcoming paper [2] that the exponent (k − l)/(k − m) in the definition of c0 (k, l, m) is the correct one, that is, for any smaller exponent the corresponding supremum is infinite. Moreover they give an estimate for c0 (k, l, m) that depends on k and is better than (1) for k = o(l 2 ). Let us note that for large l the minimum on the right hand side of (1) is equal to the first term for m < c1 l/ log l, to the last term for m > c2 l, where c1 , c2 are suitable constants,

c1 > 0, c2 < 1, provided in the latter case that γl , log g0 (l)κl /2l are regularly growing functions and lim inf l→∞

log g0 (l) − l

l 2

log γl

>

1 . 2

For m = 1, (1) constitutes an improvement over Theorem 1 of [8] already for l > 50. The problem of existence of a bound for c0 (k, l, m) depending only on m remains open also for m = 1. Lemma 1. If A is a parallelohedron given by the inequalities |a i x| 1,

a i ∈ Rl (1 i k)

then for every parallelopiped P containing A, symmetric with respect to 0 and for a suitable subset S of {1, 2, . . . , k} of cardinality l we have vol P vol P0 (S), where P0 (S) is the parallelopiped |a i x| 1 (i ∈ S). Proof. We shall proceed by induction on the number n of pairs of parallel (l−1)-dimensional faces of P that do not contain (l − 1)-dimensional faces of A (in the sequel, briefly, faces). If n = 0 the assertion is true. Suppose it is true for the case of n − 1 pairs of parallel faces and consider a parallelopiped P symmetric with respect to 0 with exactly n pairs of parallel faces not containing faces of A. Let P be given by the inequalities |bi x| 1,

b i ∈ Rl

(1 i l)

and let b1 x = ±1 be the pair of hyperplanes corresponding to one of the n pairs in question. Replacing P if necessary by a smaller parallelopiped we may assume that there is x 0 ∈ A

1262


such that b1 x 0 = 1.

(2)

Let I = {i k : |a i x 0 | = 1} and let a i x 0 = εi

(3)

(i ∈ I ).

From the fact that the hyperplane b1 x = 1 is supporting A at x 0 it follows that (4)

εi a i t 0 (i ∈ I ) implies

b1 t 0 for t ∈ Rl .

Indeed, suppose for some t 0 ∈ Rl that εi a i t 0 0 and b1 t 0 > 0. Then for 1 − |a i x 0 | t0 2 t1 = min min , min i ∈I / i∈I h(a i ) lh(t 0 ) h(a i ) we have ±(x 0 + t 1 ) ∈ A, b1 (x 0 + t 1 ) > 1, b1 (−x 0 − t 1 ) < −1 < 1, and thus the hyperplane b1 x = 1 divides A. This contradiction proves (4). Hence by a theorem of Farkas ([5], page 5. I owe this reference to Professor S. Rolewicz. There is a related earlier statement in [8], page 45) we have ε i a i λi , b1 = i∈I

where λi 0

(5) and by (2) and (3)

(6)

(i ∈ I )

λi = 1.

i∈I

Therefore, (7)

−1 −l

= 2 det εi a i λi , b2 , . . . , bl

vol P i∈I

−l

λi det(εi a i , b2 , . . . , bl )

. =2

i∈I

Regarding λi as variables restricted by the conditions (5) and (6), we easily see that the right hand side of (7) takes the maximum for λi = 1 if i = i0 , λi = 0 otherwise. Hence vol P vol P1 ,

(8) where P1 is the parallelopiped |a i0 x| 1,

|bi x| 1

(2 i l).

However P1 contains A and it has only n − 1 pairs of parallel faces that do not contain faces of A. Thus by the inductive assumption there exists a set S ⊂ {1, 2, . . . , k} of cardinality l and with the property vol P1 vol P0 (S).

L2. A decomposition of integer vectors IV

1263

In view of (8) this gives vol P vol P0 (S)

and concludes the inductive argument. Lemma 2. For all linearly independent vectors c1 , . . . , cl ∈ Rk the domain C : h(c1 x1 + . . . + cl xl ) 1 satisfies vol C

2l , g0 (l)H (c1 , . . . , cl )

vol E (C)

κl . g1 (l)H (c1 , . . . , cl )

Proof. Put a i = [c1i , c2i , . . . , cli ]

(9)

(1 i k).

Then C = {x ∈ Rl : |a i x| 1 for all i k} and clearly C is a convex body symmetric with respect to 0. By Definition 2 vol C g0 (l)−1 inf vol P,

vol E (C) g1 (l)−1 2−l κl inf vol P,

where the infimum is taken over all parallelopipeds P symmetric with respect to 0 and containing C. However by Lemma 1 the infimum can be replaced by the minimum taken over the finite set of all parallelopipeds P0 (S),

|a i x| 1

(i ∈ S),

where S runs through all subsets of {1, . . . , k} of cardinality l. Since

−1

vol P0 (S) = 2l det{a i : i ∈ S}

we have by (9) that min vol P0 (S) = 2l H (c1 , . . . , cl )−1

and the lemma follows.

Lemma 3. If for all linearly independent vectors n1 , . . . , nm ∈ Zk such that D(n1 , . . . , nm ) = 1 there exist linearly independent vectors p1 , . . . , pl ∈ Zk such that ni =

l

uij pj ,

uij ∈ Q

j =1

and l j =1

then c0 (k, l, m) c.

h(p j ) cH (n1 , . . . , nm )(k−l)/(k−m)

1264


Proof. Consider m linearly independent vectors n1 , . . . , nm ∈ Zk and let N be the linear space spanned by them over R. Further, let b1 , . . . , bm be a basis of the lattice N ∩ Zk and c1 , . . . , ck−m ∈ Zk linearly independent vectors perpendicular to N . Since N ∩ Zk is the lattice of all solutions x ∈ Zk of the system ci x = 0 (1 i k − m), we have by the known theorem ([3], page 53) that D(b1 , . . . , bm ) = 1.

(10) On the other hand clearly

⎛

⎞ ⎛ ⎞ n1 b1 ⎜ .. ⎟ ⎜ .. ⎟ = A ⎝ . ⎠ ⎝ . ⎠,

(11)

nm

bm

where A is an integral square matrix of order m. It follows from (11) that D(n1 , . . . , nm ) = |det A| D(b1 , . . . , bm ), H (n1 , . . . , nm ) = |det A| H (b1 , . . . , bm ) and by (10) H (b1 , . . . , bm ) =

(12)

H (n1 , . . . , nm ) . D(n1 , . . . , nm )

By the assumption of the lemma there exist linearly independent vectors p 1 , . . . , pl ∈ Zk and a matrix U ∈ Mm,l (Q) such that ⎛ ⎞ ⎛ ⎞ p1 b1 ⎜ .. ⎟ ⎜ .. ⎟ (13) ⎝ . ⎠ = U⎝ . ⎠, bm

pl

and (14)

l

h(p j ) cH (b1 , . . . , bm )(k−l)/(k−m) .

j =1

It follows from (11) and (13) that ⎛

⎞ ⎛ ⎞ p1 n1 ⎜ .. ⎟ ⎜ .. ⎟ = AU ⎝ . ⎠, ⎝ . ⎠ nm

pl

while from (12) and (14) that l j =1

h(p j ) c

H (n1 , . . . , nm ) D(n1 , . . . , nm )

Thus, by Definition 1, c0 (k, l, m) c.

(k−l)/(k−m) .

Lemma 4. Let K be a convex domain symmetric with respect to 0 in the linear subspace L : x1 = . . . = xm = 0 of Rk , not containing in its interior any point of the lattice

1265


L ∩Zk except 0 and let K be the corresponding distance function. Let n1 , . . . , nm ∈ Zk and N be the linear space spanned by n1 , . . . , nm over R. If Δ = det(nij )i,j m = 0 and D(n1 , . . . , nm ) = 1 there exist vectors nm+1 , . . . , nk ∈ Zk such that n1 , . . . , nk are linearly independent and k A A

A(ni + N ) ∩ L A 2k−m vol K −1 |Δ|−1 . K i=m+1

Remark. Since Δ = 0 we Ahave N ∩ L = {0}, and hence (ni + N ) ∩ L consists of one A point and A(ni + N ) ∩ L AK means the distance from this point to 0 measured through K. Proof. If |Δ| = 1 the desired conclusion follows directly from Minkowski’s second theorem. Indeed, by that theorem applied to the domain K there exist linearly independent vectors nm+1 , . . . , nk ∈ L ∩ K such that k

−1

ni K 2k−m vol K .

i=m+1

Since N ∩ L = {0} we have (ni + N ) ∩ L = {ni } (m < i k) and n1 , . . . , nk are linearly independent. Therefore assume that |Δ| > 1. Let Δi (x) be the determinant of the matrix obtained from (nij )i,j m by replacing the ith row by the first m coordinates of the vector x. Let us take a real number r > 1 and consider in Rk the domain A A(k−m)r m A A A xΔ − n Δ (x) |Δ|(k−m)r . Dr (K) : max |Δμ (x)| + |Δ|r A μ μ A A 1μm

K

μ=1

Then Dr (K) is convex and symmetric with respect to 0. In order to compute its volume we make the affine transformation Δμ (x) = yμ (μ = 1, . . . , m), xμ = yμ (μ = m + 1, . . . , k). Δ(k−m)r This transformation has Jacobian equal to Δ(k−m)rm−m+1 and it transforms Dr (K) into A A(k−m)r m A A rA (k−m)r−1 A nμ y μ Δ 1, Dr (K) : max |yμ | + |Δ| A[0, ym+1 , . . . , yk ] − A 1μm

K

μ=1

where nμ is the projection of nμ on L . Clearly vol Dr (K) = |Δ|(k−m)rm−m+1 vol Dr (K) 7 (k−m)rm−m+1 vol K = |Δ|

max1μm |yμ |1

7

= 2m |Δ|((k−m)r−1)m vol K 0

1

dy1 dy2 · · · dym

mt m−1 (1 − t)1/r dt.

1 − max |yμ | |Δ|r

1/r

1266


C1

mt m−1 (1 − t)1/r dt = Ir,m . Let λi = inf λ : dim λDr (K)∩Zk i (1 i k). By Minkowski’s second theorem there exist linearly independent points m1 , . . . , mk such that

Put

0

mi ∈ λi Dr (K) ∩ Zk

(15) and (16)

k

−1 −1 λi 2k vol Dr (K)−1 = 2k−m Ir,m vol K |Δ|(1−(k−m)r)m .

i=1

We shall show that (17)

λi = |Δ|1−(k−m)r

(1 i m)

and (18)

mi ∈ N

(1 i m).

Indeed, for i m, μ m we have Δμ (ni ) = Δ if μ = i, Δni =

m

0 otherwise;

nμ Δμ (ni ),

μ=1

and hence (19)

ni ∈ |Δ|1−(k−m)r Dr (K)

(1 i m).

On the other hand, if x ∈ λDr (K) ∩ Zk and x ∈ / N we have Δx =

m

nμ Δμ (x), and

μ=1

c

m A A nμ Δμ (x)AK 1 and by the definition of thus by the assumption about K, AΔx − μ=1

Dr (K), (20)

λ(k−m)r |Δ|(k−m)r |Δ|r ;

λ |Δ|−1+1/(k−m) > |Δ|1−(k−m)r .

If x ∈ λDr (K) ∩ Zk and x ∈ N we have Δx =

m

nμ Δμ (x) and thus by the assumption

μ=1

that D(n1 , . . . , nm ) = 1 we have Δμ (x) ≡ 0 (mod Δ), and hence either x = 0 or max |Δμ (x)| |Δ|, which by the definition of Dr (K) implies 1μm

λ |Δ|1−(k−m)r .

(21)

The claims (17) and (18) follow from (19), (20) and (21). From (16) and (17) we infer that k i=m+1

−1 λi 2k−m vol K)−1 Ir,m


1267

and since by (15) A A(k−m)r x A A (k−m)r A |Δ|r A Δm − n Δ (m ) |Δ|(k−m)r λi i μ μ i A A K

μ=1

we obtain A m k A A A

−1 k−m −1 Ami − Δ−1 nμ Δμ (mi )A . vol K |Δ|−1 Ir,m A A 2

(22)

K

μ=1

i=m+1

Moreover, by (18), n1 , . . . , nm , mm+1 , . . . , mk are linearly independent. For every r > 1 there corresponds a certain choice of vectors mi ∈ Zk , however the set of values which we can obtain on the left hand side of (22) is discrete. Therefore there exist vectors ni (m < i k) such that ni (1 i k) are linearly independent and A m k A A A

−1 k−m −1 Ani − Δ−1 A n Δ (n ) . vol K |Δ|−1 lim Ir,m μ μ i A 2 A r→∞ K

μ=1

i=m+1

However

ni − Δ−1

m

nμ Δμ (ni ) = (ni + N ) ∩ L

μ=1

and

7 lim Ir,m =

r→∞

1

mt m−1 dt = 1,

0

which proves the lemma.

Lemma 5. If m < k, n1 , . . . , nm ∈ Zk , D(n1 , n2 , . . . , nm ) = 1 there exist vectors nm+1 , . . . , nk ∈ Zk such that n1 , . . . , nk are linearly independent and for each l ∈ [m, k] l

the domain D : h xi ni 1, contained in Rd , satisfies i=1

(23)

vol D

2l m! H (n1 , . . . , nm )−(k−l)/(k−m) g0 (m)l!

and (24)

vol E (D) max

−1/2 κl κl l (m−l)/2 , l g1 (m)(l − m + 1)l/2 g1 (l) m × H (n1 , . . . , nm )(k−l)/(k−m) .

Proof. Without loss of generality we may assume that H (n1 , n2 , . . . , nm ) = |Δ|, where Δ = det(nij )i,j m . By Lemma 4 applied with K = {x ∈ L : h(x) 1} there exist

1268


vectors nm+1 , . . . , nk ∈ Zk such that n1 , . . . , nk are linearly independent and k

(25)

h(ni ) |Δ|−1 ,

where {ni } = (ni + N ) ∩ L

(m < i k).

i=m+1

Permuting the vectors ni if necessary we may assume that the sequence h(ni ) is nondecreasing. Then (25) implies l

(26)

h(ni ) H (n1 , . . . , nm )−(l−m)/(k−m) .

i=m+1

In order to prove (23) let us write explicitly ni = ni −

m

aiμ nμ

(m < i l).

μ=1

Then l

x i ni =

m

l l nμ xμ + aiμ xi + xi ni

μ=1

i=1

i=m+1

i=m+1

and (27)

l

h

x i ni

m

h

l l + nμ xμ + aiμ xi |xi |h(ni ).

μ=1

i=1

i=m+1

i=m+1

It follows by a change of variables that 7 vol D

D0

m l dxm+1 · · · dxl vol x ∈ Rm : h x μ nμ 1 − |xi |h(ni ) , μ=1 l

where D0 is the domain

i=m+1

vol x ∈ R : h m

i=m+1

|xi |h(ni ) 1. However by Lemma 2,

m

xμ nμ

c

μ=1

2m c m , g0 (m)H (n1 , . . . , nm )

and hence 2m vol D = g0 (m)H (n1 , . . . , nm )

7 D0

1−

l

m

|xi |h(ni )

dxm+1 · · · dxl

i=m+1

=

l 2l m! h(ni )−1 g0 (m)l! H (n1 , . . . , nm ) i=m+1

and (23) follows from (26).

1269


In order to prove the part of (24) corresponding to the first term of the maximum on m

the right hand side, let D1 be the domain h xi ni 1. The ellipsoid E (D1 ) is given i=1

by the inequality F1 (x1 , . . . , xm ) 1, where F1 is a positive definite quadratic form. Since E (D1 ) ⊂ D1 we have for all x ∈ Rm , m ) (28) F1 (x1 , . . . , xm ) = x E (D1 ) x D1 = h xi ni . i=1

By virtue of Lemma 2, we have vol E (D1 ) κm g1 (m)−1 H (n1 , . . . , nm )−1 . However κm vol E (D1 ) = √ , d(F1 ) and thus

) d(F1 ) g1 (m)H (n1 , . . . , nm ).

(29)

Consider now the quadratic form l aiμ xi , . . . F (x1 , . . . , xl ) = (l − m + 1)F1 . . . , xμ + i=m+1

+ (l − m + 1)

l

xi2 h2 (ni ).

i=m+1

For all x ∈ Rl we have by the Cauchy inequality, by (28) and (27), that F G l l G ) F (x1 , . . . , xl ) HF1 . . . , xμ + aiμ xi , . . . + |xi |h(ni ) m

h

μ=1

i=m+1

nμ x μ +

l

i=m+1

aiμ xi

+

i=m+1

l

|xi |h(ni )

i=m+1

and thus the ellipsoid E : F (x1 , . . . , xl ) 1

c

is contained in D and by the definition of E (D), (30)

vol E (D) vol E = √

κl . d(F )

Since F is obtained from the quadratic form l xi2 h2 (ni ) (l − m + 1) F1 + i=m+1

l

h

i=1

x i ni ,

1270


by a unimodular substitution, we have l ) ) d(F ) = (l − m + 1)l/2 d(F1 ) h(ni ) i=m+1

and by (26), (29) and (30), vol E (D) κl (l − m + 1)−l/2 H (n1 , . . . , nm )(k−l)/(k−m) . In order to prove the remaining part of (24) note that H (n1 , . . . , nl ) = H (n1 , . . . , nm , nm+1 , . . . , nl ). Let M be a minor of order l of the matrix ⎛

⎞ n1 ⎜ .. ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ nm ⎟ ⎜ ⎟ ⎜n ⎟ ⎜ m+1 ⎟ ⎜ . ⎟ ⎝ .. ⎠ nl

and S the set of indices of the columns of M. Developing M according to the first m rows we obtain from the Laplace theorem (31) |M| H (n1 , . . . , nm ) |Mj1 ,...,jl−m |, where Mj1 ,...,jl−m is the minor of

⎛

⎞ nm+1 ⎜ .. ⎟ ⎝ . ⎠ nl

consisting of the columns j1 , . . . , jl−m , while {j1 , . . . , jl−m } runs through all subsets of S of cardinality l − m. By the generalized Hadamard inequality ([1], formula (2.6))

Mj21 ,...,jl−m

l i=m+1 j ∈S

l−m n2 ij l

l

h(ni )2 ,

i=m+1

and hence, by the Cauchy inequality, (32)

1/2 l l (l−m)/2 |Mj1 ,...,jl−m | l h(ni ). m i=m+1

The inequalities (26), (31) and (32) give 1/2 l |M| l (l−m)/2 H (n1 , . . . , nm )(k−l)/(k−m) , m

1271


and hence by the arbitrary choice of M 1/2 l l (l−m)/2 H (n1 , . . . , nm )(k−l)/(k−m) . H (n1 , . . . , nl ) m Now Lemma 2 applied with C = D implies −1/2 l κl l (m−l)/2 H (n1 , . . . , nm )−(k−l)/(k−m) . vol E (D) g1 (l) m

Proof of Theorem 1. Let n1 , . . . , nm ∈ Zk be linearly independent and D(n1 , . . . , nm ) = 1. Let nm+1 , . . . , nl be vectors the existence of which is asserted in Lemma 5 and consider l

the domain D : h xj nj 1. Let j =1

μi = min{μ : dim μD ∩ Zl i} (1 i l). By Minkowski’s second theorem there y i = [yi1 , . . . , yil ] (1 i l) such that

exist

linearly

independent

vectors

y i ∈ μi D ∩ Zl

(33) and l

(34)

μi 2l vol D)−1 .

i=1

By another theorem of Minkowski (see [8], §51 or [6], §18, Theorem 3), l

(35)

−1 μi Δ E (D) ,

i=1

where Δ E (D) is the critical determinant of E (D) and by the definition of the Hermite constant

−1 κl l/2 = γl (36) Δ E (D) vol E (D) (see [6], formula (37.6)). Let us put pi =

(37)

l

yij nj

(1 i l).

j =1

It follows from the definition of D and from (34)–(37) that h(pi ) = μi , hence by (34)–(37) l i=1

−1 l/2 −1 h(p i ) min 2l vol D , γl κl vol E (D)

1272


and by Lemma 5 l

h(p i ) min

i=1

l! l/2 g0 (m), (l − m + 1)l/2 g1 (m)γl , m! 1/2 l l/2 (l−m)/2 H (n1 , . . . , nm )(k−l)/(k−m) . l g1 (l)γl m

Moreover, since y 1 , . . . , y l are linearly independent the system (37) can be solved with respect to n1 , . . . , nl and we obtain ni =

l

uij pj ,

uij ∈ Q

(1 i l).

j =1

Since ni (1 i l) are linearly independent so are p j (1 j l) and we obtain from (37) and Lemma 3 that l/2 (38) c0 (k, l, m) min (l − m + 1)l/2 g1 (m)γl , 1/2 l l! l/2 (l−m)/2 , g0 (m), l g1 (l)γl m! m which proves the first part of the theorem. In order to prove the second part let us observe that if l = m = 1 the right hand side of (38) equals 1, while it immediately follows from the definition of c0 (k, l, m) that c0 (k, 1, 1) 1. If l = m = 2 the right hand side of (38) equals 43 , since B B 4 4 4 g1 (2) = , γ2 = , g0 (2) . 3 3 3 On the other hand, consider the following vectors in Zk (k 3) n1 = [2t, 4t + 1, 2t, 0, . . . , 0],

n2 = [4t − 1, 2t, −2t, 0, . . . , 0] (t ∈ N).

We have here H (n1 , n2 ) = 12t 2 + 2t,

D(n1 , n2 ) = 1.

Hence, if ni =

2

uij pj ,

uij ∈ Q, p j ∈ Zk

(1 i, j 2)

j =1

we have pj = n1 xj + n2 yj ,

[xj , yj ] ∈ Z2 \ {0} (1 j 2).

If xj = yj we have |pj 2 | > 6t, otherwise |pj 3 | 2t, and thus h(pj ) 2t (1 j 2). If for an ε > 0 we have

h(p1 )h(p 2 ) 43 − ε H (n1 , n2 ) = 43 − ε (12t 2 + 2t)


1273

then for t > t0 (ε) (39)

h(p 1 )h(p 2 ) < (16 − 10ε)t 2

and since h(p j ) 2t we obtain h(p j ) < (8 − 5ε)t 2 (1 j 2). Hence for t > t1 (ε), by consideration of the first three coordinates of pj |2xj + 4yj | 7,

|4xj + 2yi | 7,

|2xj − 2yj | 7;

|xj | 1, |yj | 1 and since [xj , yj ] = [0, 0], h(pj ) 4t − 1 (1 j 2). It follows that h(p1 )h(p 2 ) 16t 2 − 8t + 1, which for t > max{t0 (ε), t1 (ε), ε−1 } contradicts (39). This shows that c0 (k, 2, 2) = 43 and completes the proof of the theorem.

Proof of Theorem 2. The proof does not differ essentially from the proof of Theorem 2 in [10]. In formula (14) and in the fourth displayed formula on page 701 there, one has to H (n , . . . , n ) (k−l)/(k−m) 1 m replace c0 (k, l) by c0 (k, l, m) and h(n)(k−l)/(k−m) by .

D(n1 , . . . , nm ) Note added in proof. Yu. Teterin has remarked that Lemma 4 holds under a weaker assumption, namely that vol K < ∞ instead of K not containing in its interior any point k c of the lattice L ∩ Z except 0. To see this, it suffices to apply the original formulation to the body of λK for suitable λ.

References [1] E. Bombieri, J. D. Vaaler, On Siegel’s Lemma. Invent. Math. 73 (1983), 11–32; Addendum, ibid. 75 (1984), 377. [2] S. Chaładus, Yu. Teterin, Note on a decomposition of integer vectors II. Acta Arith. 57 (1991), 159–164. [3] A. Châtelet, Leçons sur la théorie des nombres. Paris 1913. [4] A. Dvoretzky, C. A. Rogers, Absolute and unconditional convergence in normed linear spaces. Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 192–197. [5] J. Farkas, Über die Theorie der einfachen Ungleichungen. J. Reine Angew. Math. 124 (1902), 1–27. [6] P. M. Gruber, C. G. Lekkerkerker, Geometry of Numbers. North–Holland, Amsterdam 1987. [7] F. John, Extremum problems with inequalities as subsidiary conditions. In: Studies and Essays Presented to R. Courant on his 60th Birthday, January 8, 1948, Interscience, New York 1948, 187–204. [8] H. Minkowski, Geometrie der Zahlen. Leipzig 1896; reprint Chelsea, New York 1953. [9] A. Pełczyński, S. J. Szarek, On parallelepipeds of minimal volume containing a convex symmetric body in Rn . Math. Proc. Cambridge Philos. Soc. 109 (1991), 125–148. [10] A. Schinzel, A decomposition of integer vectors III. Bull. Polish Acad. Sci. Math. 35 (1987), 693–703.

Originally published in Monatshefte für Mathematik 137 (2002), 239–251


A property of polynomials with an application to Siegel’s lemma

Dedicated to Professor Edmund Hlawka at the occasion of his 85th birthday

c

Abstract. It is proved that natural necessary conditions imply the existence of infinitely many integer points at which given multivariate polynomials with integer coefficients take relatively prime values. As a consequence the best constant in the simplest case of Siegel’s lemma is expressed in terms of critical determinants of suitable star bodies.

The first aim of this paper is to prove the following: Theorem 1. Let F, Fμν ∈ Z[T , T1 , . . . , Tl ] (1 μ m, 1 ν n), F = 0 and for each μ m, Fμν (1 ν n) be relatively prime. If the product Π=

m

Fμ1 (t, t1 , . . . , tl ), . . . , Fμn (t, t1 , . . . , tl )

μ=1

has no fixed prime divisor for [t, t1 , . . . , tl ] running over Zl+1 , then there exist integers t1∗ , . . . , tl∗ and an arithmetic progression P such that for t ∈ P F (t, t1∗ , . . . , tl∗ ) = 0 and for each μ m the numbers Fμν (t, t1∗ , . . . , tl∗ ) (1 ν n) are relatively prime. This implies at once Corollary 1. Let Fμν ∈ Z[T , T1 , . . . , Tl ] (1 μ m, 1 ν n) and for each μ m, Fμν (1 ν n) be relatively prime. If the numbers Fμν (t, t1 , . . . , tl ) (1 ν n) are relatively prime simultaneously (1 μ m) for at least one integer point [t, t1 , . . . , tl ], then they are relatively prime simultaneously for infinitely many integer points. The following consequence of Theorem 1 is less obvious: Theorem 2. Let f (x) and g(x) be the distance functions of two bounded star bodies in Rl+1 , both functions symmetric with respect to the coordinates of x and even with respect to each of them. Let for α ∈ Rl

Sα = {x ∈ Rl : f x, αx < 1},

L3. A property of polynomials

1275

where αx is the inner product. Then C(f, g) := lim sup c

f (x)l x∈Zl+1 \{0} g(a) inf

a∈(Z\{0})l+1 h(a)→∞

ax=0

=

sup

f (x)l '(Sα )−1 = sup , x∈Zl+1 \{0} g(a) α∈Al g(α, 1) inf

a∈(Z\{0})l+1

ax=0

Ql

c

: 0 < α1 α2 . . . αl 1} and Δ(·) is the critical where Al = {[α1 , . . . , αl ] ∈ determinant, h(a) is defined below. This in turn implies several corollaries, some of which are implicit in the literature and some are new. In order to formulate them we use the following notation. For x = [x1 , . . . , xn ] ∈ Rn and 0 < p < ∞, 1/p n |xk |p , hnp (x) = hp (x) = k=1

hn∞ (x) = h(x) = max |xk |, 1kn

c(l, p) = c

sup

hp (x)l . x∈Zl+1 \{0} hp (a) inf

a∈Zl+1 \{0}

ax=0

l/2

Corollary 2. c(l, 2) = γl , where γl is the Hermite constant. l/2

The inequality c(l, 2) γl is contained in Theorem 4D of [7]. The reverse inequality has been proved even in greater generality (see below), but not published, by Vaaler. The referee pointed out that it is a direct consequence of the results of [9] and [11] and Prof. Thunder has kindly supplied a proof. Corollary 3. For l 2, c(l, ∞) = sup Δ(Hα )−1 1, α∈Al−2

c

where Hα is a generalized hexagon in Rl given by the inequalities

l−2

|xk | 1 (1 k l),

αi xi + xl−1 + xl

1. i=1

This corollary is new. Corollary 4. c(2, ∞) = 4/3. This corollary is implicit in [5], namely the inequality c(2, ∞) 4/3 is contained in Lemma 4 of [5], while the inequality c(2, ∞) 4/3 is a consequence of Lemma 7 of [5].

1276


Corollary 5. c(3, ∞) = 27/19. The inequality c(3, ∞) 27/19 has been proved by Chaładus [4], while the inequality c(3, ∞) 27/19 has been recently proved by Aliev [1]. The proof of Theorem 2 is a generalization of arguments of Chaładus and Aliev. Corollary 6. For l 4, 1 c(l, ∞)

√ l + 1.

√ The inequality c(l, ∞) l + 1 is implicit in Theorem 1 of [2], the inequality c(l, ∞) 1 seems to be new. In order to formulate Theorem 3 we need more notation. Let for a matrix A ∈ Zm×n of rank m, D(A) be the greatest common divisor of all minors of A of order m, √ det AAT H (A) = . D(A) Combining Corollary 2 with a result of Thunder [8] we shall show Theorem 3. For all positive integers m and n, where m < n, we have h2 (x)n−m \{0} H (A)

c0 (m, n) := lim sup

inf n

=

inf n

A∈Zm×n x∈Z rank A=m Ax=0 H (A)→∞

sup

h2 (x)n−m (n−m)/2 = γn−m . \{0} H (A)

A∈Zm×n x∈Z rank A=m Ax=0

A more general form of Theorem 3, concerning algebraic number fields has been proved, but not published, by Vaaler, via geometry of numbers over adeles. The referee pointed out it is a direct consequence of the results of [9] and [11]. Proof of Theorem 1. We shall proceed by induction on l. For l = 0, since Fμν are relatively prime (1 ν n) there exist polynomials Aμν ∈ Z[t] such that (1)

n

Aμν Fμν = Rμ ∈ Z \ {0}.

ν=1

By the assumption about Π for each prime p dividing νp1 , . . . , νpm and τp ∈ Z such that for all μ m

m

Rμ there exist indices

μ=1

Fμνpμ (τp ) ≡ 0 mod p. m Now taking t ≡ τp mod p for all primes p dividing Rμ we obtain from (1) and (2) for μ=1 all μ m (2)

g.c.d. Fμν (t) = 1 1νn

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and, if t is large enough, also F (t) = 0. Now we assume that the theorem is true for polynomials in l variables and proceed to prove that it is true for polynomials in l + 1 variables. There is no loss of generality in assuming that they are all different from 0. Let F = F00 , Fμν =

(3)

rμν

T rμν −ρ Gμνρ (T1 , . . . , Tl ), where Gμν0 = 0

ρ=0

and for each μ the polynomials Gμνρ (1 ν n, 0 ρ rμν ) are relatively prime. m Let P be the product of all primes not exceeding max rμν . For each prime p μ=1 1νn

dividing P there exist indices νp1 , . . . , νpm and integers τp , τp1 , . . . , τpl such that Fμνpμ (τp , τp1 , . . . , τpl ) ≡ 0 mod p

(4)

(1 μ m).

By the Chinese remainder theorem there exist integers u∗1 , . . . , u∗l such that u∗j ≡ τpj mod p

(5)

for all p | P

(1 j l).

Since Fμν (1 ν n) are relatively prime there exist Aμν ∈ Z[T , T1 , . . . , Tl ] such that n

(6)

Aμν Fμν = Rμ ∈ Z[T1 , . . . , Tl ] \ {0} (1 μ m).

ν=1

We have G000 (P V1 + u∗1 , . . . , P Vl + u∗l )

m

Rμ (P V1 + u∗1 , . . . , P Vl + u∗l ) = 0

μ=1

and for each μ the polynomials Gμνρ (P V1 +u∗1 , . . . , P Vl +u∗l ) (1 ν n, 0 ρ rμν ) are relatively prime. Moreover m

g.c.d.

μ=1 νn, ρrμν

Gμνρ (P v1 + u∗1 , . . . , P vl + u∗l )

has no fixed prime divisor p when [v1 , . . . , vl ] runs over Zl . Indeed, suppose that such p exists. In view of (3)–(5) we have p /| P . Hence for every vector [t1 , . . . , tl ] ∈ Zl there exists a vector [v1 , . . . , vl ] ∈ Zl such that tj ≡ P vj + u∗j mod p (1 j l) and we obtain m

g.c.d.

Gμνρ (t1 , . . . , tl ) ≡ 0 mod p,

μ=1 νn, ρrμν

which by (3) gives m

g.c.d. Fμν (t, t1 , . . . , tl ) ≡ 0 mod p,

μ=1 νn

contrary to the assumption about Π .

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Therefore, we may apply the inductive assumption to polynomials G000 (P V1 + u∗1 , . . . , P Vl + u∗l )

m

Rμ (P V1 + u∗1 , . . . , P Vl + u∗l )

μ=1

and Gμνρ (P V1 + u∗1 , . . . , P Vl + u∗l ) (1 μ m, 1 ν n, 0 ρ rμν ). We obtain existence of integers v1∗ , . . . , vl∗ such that (7)

G000 (P v1∗ + u∗1 , . . . , P vl∗ + u∗l )

m

Rμ (P v1∗ + u∗1 , . . . , P vl∗ + u∗l ) = 0

μ=1

and for each μ m (8)

Gμνρ (P v1∗ + u∗1 , . . . , P vl∗ + u∗l ) = 1.

g.c.d. νn, ρrμν

Let us put tj∗ = P vj∗ + u∗j

(9)

(1 j l)

and consider polynomials in one variable F (T , t1∗ , . . . , tl∗ ) and Fμν (T , t1∗ , . . . , tl∗ ). We have F (T , t1∗ , . . . , tl∗ ) = 0, since by (7) and (9) G000 (t1∗ , . . . , tl∗ ) = 0 and for each μ m the polynomials Fμν (T , t1∗ , . . . , tl∗ ) (1 ν n) are relatively prime in view of (6), since by (7) and (9) Rμ (t1∗ , . . . , tl∗ ) = 0. Suppose that a prime p is a fixed divisor of m

g.c.d. Fμν (t, t1∗ , . . . , tl∗ )

μ=1 νn

when t runs over Z. By (4), (5) and (9) we have for each μ m Fμνpμ (τ, t1∗ , . . . , tl∗ ) ≡ 0 mod p, hence p /| P , p >

m

if p | P ,

max rμν .

μ=1 1νn

In view of (8) for each μ m there exist indices νμ n and ρμ rμν such that Gμνμ ρμ (t1∗ , . . . , tl∗ ) ≡ 0 mod p. By Lagrange’s theorem and (3) the congruence m

Fμνμ (t, t1∗ , . . . , tl∗ ) ≡ 0 mod p

μ=1

has at most

m

rμνμ < p solutions, hence it is not satisfied identically.

μ=1

The obtained contradiction shows that m g.c.d. Fμν (t, t1∗ , . . . , tl∗ ) μ=1 1νn

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has no fixed prime divisor when t runs over Z and, by the already proved case l = 0 of the theorem, there exists an arithmetic progression P such that for t ∈ P we have F (t, t1∗ , . . . , tl∗ ) = 0 and for each μ m the numbers Fμν (t, t1∗ , . . . , tl∗ ) (1 ν n) are relatively prime.

For the proof of Theorem 2 we need ⎞ a1 ⎜ a2 ⎟ ⎜ ⎟ Lemma. Let Λ be a sublattice of Zn with a basis a 1 , . . . , a m , A = ⎜ . ⎟ and let Λ⊥ , ⎝ .. ⎠ ⎛

Λ⊥⊥

be the sublattice of We have

Zn

am consisting of all vectors orthogonal to Λ, or Λ⊥ , respectively. Λ = Λ⊥⊥

if and only if D(A) = 1.

Proof. See [6], p. 336 and [3], p. 15. Proof of Theorem 2. We shall prove first that (10)

f (x)l '(Sα )−1 sup =: s. x∈Zl+1 \{0} g(a) α∈Al g(α, 1) inf

sup a∈(Z\{0})l+1

ax=0

Let a = [a1 , . . . , al+1 ] ∈ (Z \ {0})l+1 . Since f (x) and g(x) are symmetric with respect to the coordinates of x and even with respect to each of them, we may assume that 0 < a1 a2 . . . al+1 and, since g(a) is homogeneous of degree 1, we may assume that (a1 , . . . , al+1 ) = 1. We have [a1 /al+1 , . . . , al /al+1 ] ∈ Al , hence Δ(Sa1 /al+1 ,...,al /al+1 )−1 sg

a al 1 ,..., ,1 . al+1 al+1

Therefore, by the property of critical determinants, every full lattice Λ in Rl with determinant d(Λ) has a non-zero point (y1 , . . . , yl ) such that l l a ak al 1 f y1 , . . . , yl , yk sg ,..., , 1 d(Λ). al+1 al+1 al+1 k=1

Consider now the lattice Λ0 obtained as projection on the hyperplane xl+1 = 0 of the

1280


lattice Λ1 of integer vectors orthogonal to a. Let b1 , . . . , bl be a basis of Λ1 ⎛ ⎞ b1 ⎜ .. ⎟ bk = (bk1 , . . . , bk,l+1 ), B = ⎝ . ⎠ , bl and let Bk be the minor of B obtained by omitting the k-th column. Since Λ1 = Λ⊥⊥ 1 , by Lemma we have

B1 , −B2 , . . . , (−1)l Bl+1 = 1

and bk B1 , −B2 , . . . , (−1)l Bl+1 = 0 (1 k l). Since (a1 , . . . , ak+1 ) = 1 and bk a = 0 there exists an ε = ±1 such that εak = (−1)k+1 Bk (1 k l) and, in particular,

al+1 = det(bij )1i,j l = d(Λ0 ). Hence there exist integers u1 , . . . , ul not all zero such that (11) f

l

bj 1 uj , . . . ,

j =1

l

l l l ak bj k u j a k=1 l+1 j =1 a al 1 sg ,..., , 1 al+1 = sg(a1 , . . . , al+1 ). al+1 al+1

bj l uj ,

j =1

However, by the definition of Λ1 , l+1

ak bj k = 0

(1 j l),

k=1

hence l+1 ak bj k = −bj,l+1 al+1

(1 j l)

k=1

and inequality (11) takes the form f

l

l b j uj

sg(a).

j =1

Taking x =

l j =1

bj uj we find x ∈ Λ1 , hence ax = 0 and f (x)l s, g(a)

which proves (10).

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Now we shall prove that (12)

lim sup a∈(Z\{0})l+1

Δ(Sα )−1 f (x)l g(α, 1) x∈Zl+1 \{0} g(a) inf

ax=0

for all α ∈ Al . Sα as an open bounded star body has a critical lattice Λ. Let a 1 , . . . , a l be a basis of Λ. Take a positive δ < 1 and choose b1 , . . . , bl in Ql such that h(bj − a j ) < δ (1 j l),

T

det(b , . . . , bT ) − d(Λ) < δd(Λ) = δΔ(Sα ).

(13) (14)

1

l

Choose a positive integers d such that dbj ∈ Zl and dαk bj k ∈ Z for all j, k l, where bj k is the kth coordinate of bj . We shall apply Theorem 1 taking m = 1, F = 1 and taking for F1ν (1 ν l + 1) all minors of order l of the matrix M = M(T , T1 , . . . , Tl ) ⎛ db12 T ⎜db11 T + T1 ⎜ ⎜ ⎜ db22 T + T2 ⎜ db21 T =⎜ ⎜ .. .. ⎜ ⎜ . . ⎜ ⎝ dbl1 T dbl2 T

⎞ αk b1k T ⎟ ⎟ k=1 ⎟ l ⎟ ... db2l T d αk b2k T ⎟ ⎟, k=1 ⎟ .. .. .. ⎟ . ⎟ . . ⎟ l ⎠ . . . dbll T + Tl d αk blk T

...

db1l T

d

l

k=1

where T , T1 , . . . , Tl are variables. Let Mi = Mi (T , T1 , . . . , Tl ) and mi be the minor obtained by omitting the ith column in M, or in the matrix ⎛ ⎞ l b b . . . b α b 12 1l k 1k ⎟ ⎜ 11 ⎜ ⎟ k=1 ⎜ . ⎟ .. .. .. ⎜ . ⎟ , respectively. . . . ⎜ . ⎟ ⎜ ⎟ l ⎝ ⎠ bl1 bl2 . . . bll αk blk k=1

We have, by (14), (15) (16)

|ml+1 | = det(bj k ) = 0, |mi | = αi |ml+1 |

(1 i l)

and (17)

Mi = d l mi T l + polynomial of degree less than l in T ,

hence Mi is a non-zero polynomial independent of Ti (1 i l). A possible non-constant common factor of M1 , . . . , Ml+1 would have to belong to Q[T ] and, since these minors are homogeneous in T , T1 , . . . , Tl , T would be a common factor. However Ml+1 ≡ T1 T2 . . . Tl mod T ,

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hence T /| Ml+1 and Ml+1 (0, 1, . . . , 1) = 1. By Theorem 1 there exist integers t1∗ , . . . , tl∗ and an arithmetic progression P such that for t ∈ P we have

(18) M1 (t, t1∗ , . . . , tl∗ ), . . . , Ml+1 (t, t1∗ , . . . , tl∗ ) = 1. Let

a(t) = M1 (t, t1∗ , . . . , tl∗ ), . . . , (−1)l Ml+1 (t, t1∗ , . . . , tl∗ ) .

We shall show that for every ε > 0, sufficiently small δ > 0 and sufficiently large t ∈ P every integer vector x = 0 such that a(t)x = 0 satisfies (19)

f (x)l > (1 − ε)

Δ(Sα )−1 g a(t) . g(α, 1)

By (14)–(17) we have for δ < ε/2 and sufficiently large t

ε Δ(Sα ). (20) g a(t) = 1 + o(1) d l t l |ml+1 | g(α, 1) d l t l g(α, 1) 1 + 2 On the other hand, the rows of the matrix M(t, t1∗ , . . . , tl∗ ) are orthogonal to a(t) and by (18) and Lemma they form a basis for the lattice of all integer vectors with this property. Therefore, for every x = [x1 , . . . , xl+1 ] in Zl+1 \ {0} satisfying a(t)x = 0 we have (21)

xk =

l j =1

(22)

xl+1 =

l

uj (dbj k t + δj k tk∗ )

(1 k l),

l uj d αk b j k t ,

j =1

k=1

where δj k is the Kronecker delta and uj are integers not all equal to 0. Assume that, contrary to (19), (23)

f (x)l (1 − ε)

Δ(Sα )−1 g a(t) , g(α, 1)

hence, in particular, ε < 1. By (20) this gives for δ < ε/2 and sufficiently large t ε l l d t < dltl f (x)l (1 − ε) 1 + 2 and, since f is homogeneous of degree 1 and the set {x ∈ Rl+1 : f (x) < 1} is bounded, h(x) cf dt,

where cf depends only on f.

Solving the system (21) by means of Cramer’s formulae we obtain by virtue of (13)–(14) for δ < ε/2 and sufficiently large t |uj | cf,a k

(1 j l),

where cf,a k depends only on f and on a 1 , . . . , a m .

1283


From the continuity of f at the point l

uj a j ,

j =1

l

αk

l

uj aj k

j =1

k=1

we infer that for δ small enough f

l

uj b j ,

j =1

l k=1

αk

l

uj bj k

j =1

l l l ε 1− uj a j , αk uj aj k , f 2l

j =1

k=1

j =1

hence, by the choice of a j , f

l j =1

by (21) and (22)

uj b j ,

l k=1

αk

l

l uj bj k

j =1

ε l ε > 1− >1− , 2l 2

ε l l f (x)l > 1 − d t + o(t l ), 2

by (20)

2 − ε Δ(Sα )−1 f (x)l

> 1 + o(1) , 2 + ε g(α, 1) g a(t) which for t large enough contradicts (23). The obtained contradiction proves (19) and (12).

Proof of Corollary 2. For f = hl+1,2 = h2 we have 2 l l Sα = [x1 , . . . , xl ] ∈ Rl : xk2 + α k xk 1 . k=1

k=1

The matrix of the relevant quadratic form is AAT , where ⎞ ⎛ 1 0 . . . 0 α1 ⎜0 1 . . . 0 α2 ⎟ ⎟ ⎜ , A = ⎜. . . . . ... ... ⎟ ⎠ ⎝ .. .. 0

0

...

1

αk

hence −l/2

−l/2

γ γ =: l Δ(Sα ) = √ l T l det AA

k=1

and αk2 + 1

Δ(Sα )−1 l/2 = γl . h2 (α, 1)

It remains to consider a ∈ Zl with at least one coordinate 0, say a1 = 0.

1284


Then ax = 0 for x = [1, 0, . . . , 0] and h2 (x)l l/2 1 < γl . h2 (a)

Proof of Corollary 3. For f = hl+1,∞ = h, α ∈ Al we have Sα ⊃ Hα1 /αl−1 ,...,αl−2 /αl−1 . Indeed, if x ∈ Hα1 /αl−1 ,...,αl−2 /αl−1 we have c

|xk | 1

(24)

(1 k l),

l−2 αk

x + x + x k l−1 l 1,

α

(25) c

k=1

l−1

hence multiplying the last inequality (24) by αl /αl−1 − 1 and adding to (25) we obtain

l

αk

l αl

xk and

αk xk

αl 1,

αl−1 αl−1 c k=1

k=1

thus x ∈ Sα . It follows that

−1

Δ(Sα )−1 Δ Hα1 /αl−1 ,...,αl−2 /αl−1

and sup

α∈Al

Δ(Sα )−1 = sup Δ(Sα )−1 sup Δ(Hα )−1 . h(α, 1) α∈Al α∈Al−2

Since for α ∈ Al−2 , Hα = Sα,1,1 , the inequality in the opposite direction is obvious. Also Δ(Hα ) 1, since the only integer point inside Hα is 0. It remains to consider a ∈ Zl with at least one coordinate 0, say a1 = 0. Then ax = 0 for x = [1, 0, . . . , 0] and we have h(x)l 1 sup Δ(Hα )−1 . h(a) α∈Al−2

Proof of Corollary 4. By Corollary 3 we have c(2, ∞) = Δ(H )−1 , where H is the hexagon |x1 | 1, |x2 | 1, |x1 + x2 | 1. Clearly Δ(H ) =

3 vol H = . 4 4

Proof of Corollary 5. By Corollary 3 we have c(3, ∞) = sup Δ(H )−1 . α∈A1

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Now, by the result of Whitworth [12] ⎧ 3 ⎪ ⎪ ⎨ 4 Δ(Hα ) = 2 −1 ⎪ ⎪ ⎩− α + 3α − 24 + α 27

if α

1 2

if 1 α

1 . 2

Hence Δ(Hα ) takes its minimum in the interval [0, 1] at α = 1 and Δ(H1 ) = 19/27.

Remark 1. Using the equality Δ(Hα ) = 43 if α 21 and following the first part of the proof of Theorem 2 we infer that if a = [a1 , a2 , a3 , a4 ] ∈ Z4 , 0 a1 a2 a3 a4 ( and a1 21 a2 , then there exists x ∈ Z4 such that 0 < h(x) 3 43 h(a). This improves a conditional result of Chaładus [4] obtained under an unproved assumption and the stronger condition a4 −2a1 + a2 + a3 . Proof of Corollary 6. By Corollary 3 we have c(l, ∞) = sup Δ(Hα )−1 1. α∈Al−2

Now, by Minkowski’s theorem and Theorem 1 of Vaaler [10] IF J−1 G l G vol Hα 1 2 H . αk + 1 √ Δ(Hα ) 2l l +1 k=1

Proof of Theorem 3. By Corollary 1 for every positive integer l and every ε > 0 there exist infinitely many a in Zl+1 such that every y ∈ Zl+1 \ {0} with ay = 0 satisfies (26)

h2 (y)l l/2 > γl − ε. h2 (a)

Replacing if necessary a by a/D(a) we may assume that D(a) = 1. Take now l = n − m, A = (aij ) where i m, j n, ⎧ ⎪ ⎨aj , if i = 1, j l + 1, aij = 1, if i > 1, j = n − m + i, ⎪ ⎩ 0, otherwise. We have (27)

H (A) =

h2 (a) = h2 (a). D(a)

On the other hand, x ∈ Zn , Ax = 0 implies x = [y, 0, . . . , 0]T , where y ∈ Zl+1 , ay = 0, hence by (27) and (26) h2 (x)l h2 (y)l l/2 = > γl − ε, H (A) h2 (a)

1286

L. Geometry of numbers (n−m)/2

where H (A) can be arbitrarily large. This proves that c0 (m, n) γn−m . In order to prove the remaining part of the theorem consider a matrix A ∈ Zm×n of rank m. Let Λ be⎛a lattice ⎞ spanned by the rows of A, and b1 , . . . , bl (l = n − m) be a basis b1 ⎜ ⎟ of Λ⊥ and B = ⎝ ... ⎠. bl √ Since = Λ⊥ we have by Lemma D(B) = 1, hence H (B) = det BB T . On the other hand, by Theorem 1 of Thunder [8] we have Λ⊥⊥⊥

H (A) = H (B). Consider the ellipsoid E:

n l

2 bkj uk

1.

j =1 k=1

The matrix of the quadratic form on the left hand side is BB T , hence −l/2

Δ(t) = √

γl

. det BB T By the property of critical determinants there exist integers uk not all 0 such that l ) l/2 l/2 l/2 bk uk γl det BB T = γl H (B) = γl H (A). hl2 k=1

Taking x T =

l

bk uk we obtain x ∈ Zn \ {0}, Ax = 0 and

k=1

h2 (x)l l/2 γl , H (A)


c

Remark 2. Since for ellipsoids the anomaly is 1 the last argument shows in fact the existence of linearly independent vectors x 1 , . . . , x l in Zn such that Ax k = 0 (1 k l) and l

l/2

h(x k ) γl H (A).

k=1

References [1] I. Aliev, On a decomposition of integer vectors. Ph.D. Thesis. Institute of Mathematics, Polish Academy of Sciences, Warsaw 2001. [2] E. Bombieri, J. D. Vaaler, On Siegel’s Lemma. Invent. Math. 73 (1983), 11–32; Addendum, ibid. 75 (1984), 377.


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[3] J. W. S. Cassels, An Introduction to the Geometry of Numbers. Springer, Berlin 1959. [4] S. Chaładus, On the densest lattice packing of centrally symmetric octahedra. Math. Comp. 58 (1992), 341–345. [5] S. Chaładus, A. Schinzel, A decomposition of integer vectors II. Pliska Stud. Math. Bulgar. 11 (1991), 15–23; this collection: L1, 1249–1258. [6] D. R. Heath-Brown, Diophantine approximation with square-free numbers. Math. Z. 187 (1984), 335–344. [7] W. M. Schmidt, Diophantine Approximations and Diophantine Equations. Lecture Notes in Math. 1467, Springer, Berlin 1991. [8] J. L. Thunder, Asymptotic estimates for rational points of bounded height on flag varieties. Compositio Math. 88 (1993), 155–186. [9] −−, An adelic Minkowski–Hlawka theorem and an application to Siegel’s lemma. J. Reine Angew. Math. 475 (1996), 167–185. [10] J. D. Vaaler, A geometric inequality with applications to linear forms. Pacific J. Math. 83 (1979), 543–553. [11] T. Watanabe, On an analog of Hermite’s constant. J. Lie Theory 10 (2000), 33–52. [12] J. V. Whitworth, On the densest packing of sections of a cube. Ann. Math. Pura Appl. (4) 27 (1948), 29–37.

Originally published in Monatshefte für Mathematik 144 (2005), 177–191


On vectors whose span contains a given linear subspace with I. Aliev* (Wien) and W. M. Schmidt (Boulder)

Abstract. Estimates are given for the product of the lengths of integer vectors spanning a given linear subspace.

The aim of this paper is to estimate for k > l > m > 0 c(k, l, m) = sup inf H (S)(l−k)/(k−m)

(1)

l

|p i |,

i=1

where the supremum is taken over all subspaces S of Qk of dimension m and the infimum is taken over all sets of linearly independent vectors p 1 , . . . , pl in Zk , whose span contains S. Here H (S) is the determinant of the lattice S ∩ Zk and |p| is the Euclidean norm of p. Let γr,s be the generalized Hermite constant, as defined by Rankin [7], i.e. the least number such that every lattice Λ of rank r in Rr has a sublattice Γ of rank s and determinant 1/2

det Γ γr,s (det Λ)s/r . Here, γr,1 = γr,r−1 = γr is the ordinary Hermite constant. We shall prove Theorem 1. 1/2

1/2

l/2

γk−m,k−l c(k, l, m) γk−m,k−l γl . Corollary 1.

B 1/2 γk−1

c(k, 2, 1)

1/2 γk−1

4 . 3

Theorem 2. c(3, 2, 1) 6/(722)1/4 > γ2 . *

Communicated by F. Grunewald The first author was supported by FWF Austrian Science Fund, project M672.

1289

L4. Vectors whose span contains a given linear subspace

A related problem has been considered in [1], [4], [5] and [8]. Given m linearly independent vectors n1 , . . . , nm in Zk let H (n1 , . . . , nm ) denote the maximum of the absolute values of m × m-minors of the matrix (nt1 , . . . , ntm ) and D(n1 , . . . , nm ) the greatest common divisor of these minors. Furthermore, let h(n) = H (n) for n = 0. In [1], [4], [5] and [8] the following quantity has been estimated (2)

c0 (k, l, m) = sup inf

D(n1 , . . . , nm ) H (n1 , . . . , nm )

(k−l)/(k−m) l

h(p i ),

i=1

where the supremum is taken over all sets of linearly independent vectors n1 , . . . , nm in Zk and the infimum is taken over all sets of linearly independent vectors p 1 , . . . , pl in Zk such that for all i m ni =

l

uij pj ,

uij ∈ Q.

j =1

In particular, it has been proved in [8] that for fixed l, m (3)

lim sup c0 (k, l, m) < ∞, k→∞

in [1] that c0 (k, 2, 1) √ in [4] that c0 (3, 2, 1) = 2/ 3, and in [5] that (4)

c0 (k, l, m)

2 , k 1/(k−1)

1/2 γk−m,k−l

(k−l)/2(k−m) k . m

Taking for n1 , . . . , nm a basis of the lattice S ∩ Zk and using the inequalities |p| k 1/2 h(p),

H (n1 , . . . , nm ) H (S) D(n1 , . . . , nm )

one obtains from (1), (2) and (4) c(k, l, m)

1/2 γk−m,k−l

(k−l)/2(k−m) k k l/2 , m

following the proof in [5] one can omit the factor lary 1 that, in contrast to (3),

k (k−l)/2(k−m) m

. It follows from Corol-

lim c(k, 2, 1) = ∞

k→∞

and from Theorem 2 that, at least for k = 3 the lower bound given in Corollary 1 for c(k, 2, 1) is not sharp. The proof of Theorem 2 is based on the following theorem of independent interest.

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Theorem 3. If Λt is a sequence of lattices in Rl convergent to a full lattice Λ and λi (K, Λ) is the i-th minimum of Λ with respect to a centrally symmetric convex body K, then for each i l lim λi (K, Λt ) = λi (K, Λ).

t→∞

(We say, following [3], Chapter V, §3, that a sequence of lattices Λt in Rl is convergent to a lattice Λ, if there exists a linear homogeneous transformation τt such that Λt = τt Λ and τt − ι tends to 0 for t tending to infinity, where ι is the identity transformation and

τ = l max |τij |.) 1i,j l

Corollary 2. Let ft (x, y) be a sequence of positive definite quadratic forms over R, mt and mt be the first and the second minimum of ft (u, v) for (u, v) ∈ Z2 . If lim ft = f,

t→∞

where f is positive definite, then lim mt = m,

t→∞

lim mt = m,

t→∞

where m and m are the first and the second minimum of f , respectively. For i = 1 Theorem 3 has been known, see [3], Chapter V, §3.3, Remark. The proof of Theorem 1 is based on the following Proposition 1. Let S be an m-dimensional subspace of Qk . (i)

When 0 < n < m < k, there is a subspace T ⊂ S of dimension n with 1/2

H (T ) γm,n H (S)n/m .

(5) 1/2

The constant γm,n here is best possible. (ii) When m < l < k, there is a subspace T ⊃ S of dimension l in Qk with 1/2

H (T ) γk−m,k−l H (S)(k−l)/(k−m) .

(6) 1/2

The constant γk−m,k−l here is best possible. c

Proof. (i) A lattice Λ ⊂ Zk is primitive if Λ = S ∩ Zk , where S is the subspace of Qk spanned by Λ. There is a (C) correspondence of subspaces of Qk and primitive lattices. To a space S corresponds Λ = S ∩ Zk , and to a primitive lattice Λ corresponds the space S spanned by it. When S, Λ correspond to each other, dim S = rank Λ, and H (S) = det Λ. To a subspace T of S corresponds a primitive sublattice Γ of Λ. The existence of T as claimed in (i) now follows from the definition of γm,n . Km the group of matrices K = λO with Let Om be the orthogonal group in Rm , and O Km . Let λ ∈ R+ , O ∈ Om . Full lattices Λ, Λ in Rm are similar, if Λ = KΛ with K ∈ O γm,n (Λ) be the minimum such that there is a sublattice Γ of Λ of rank n with 1/2

det Γ = γm,n (Λ) (det Λ)n/m .


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Similar lattices Λ, Λ have γm,n (Λ) = γm,n (Λ ). When a 1 , . . . , a m is a basis of Λ, the matrix A = (a t1 , . . . , a tm ) ∈ GLm (R) with columns a t1 , . . . , a tm will also be called a basis. Now A can uniquely be written as A = KZ, Km and Z ∈ Hm , the generalized half-plane (see, e.g. [10], p. 38). The general with K ∈ O basis of Λ is AM with M ∈ GLm (Z) and AM = KM ZM , where again KM ∈ Om , ZM ∈ Hm . The map Z → Zm determines an action of GLm (Z) on Hm . Let Fm be Hm modulo this action, i.e., where Z, ZM with M ∈ GLm (Z) are identified. Then to Λ corresponds a unique Z = Z(Λ) ∈ Fm , and Z(Λ) = Z(Λ ) precisely when Λ, Λ are similar. There is a certain measure μ on Fk with μ(Fk ) = 1, and μ(D) > 0 for every non-empty open subset D. A lattice Λ of rank m in Rk is a full lattice in the space it spans, and Z(Λ) ∈ Fm is again well defined. In [10] it was shown that when D ⊂ Fm is open and non-empty, then the number of primitive lattices Λ of rank m with Z(Λ) ∈ D and the determinant not exceeding T is asymptotically equal to ck,m μ(D)T k , as T → ∞, where ck,m > 0. Therefore as Λ ranges through primitive lattices of rank m in Rk , the elements Z(Λ) will be dense in Fm . There is a lattice Λ1 with γm,n (Λ1 ) = γm,n . Set Z 1 = Z(Λ1 ). Given " > 0 we have (7)

γm,n (Λ) > γm,n − ",

when Λ is near Λ1 , i.e., when Λ has a basis near some fixed basis of Λ1 . There is a neighborhood D of Z 1 in Fm such that (7) holds when Z(Λ) ∈ D. By the density property enunciated above, there is a primitive lattice Λ of rank m with (7). Since " > 0 was arbitrary, and by the correspondence (C) of rational subspaces and primitive lattices, 1/2 we see that the constant γm,n in (5) is best possible. T⊥

(ii) The orthogonal complement S ⊥ of S has dimension k − m. By (i) there is a space ⊂ S ⊥ of dimension k − l with H (T ⊥ ) γk−m,k−l H (S ⊥ )(k−l)/(k−m) . 1/2

The orthogonal complement T of T ⊥ has T ⊃ S, dim T = l. Now (6) is a consequence of (8)

H (S ⊥ ) = H (S),

H (T ⊥ ) = H (T ).

(For a proof of the last formulae, see [2], pp. 27–28.) By (i), there is for any " > 0 a space S ⊥ of dimension k − m such that (9)

H (T ⊥ ) (γk−m,k−l − ")H (S ⊥ )(k−l)/(k−m) , 1/2

for any space T ⊥ ⊂ S ⊥ of dimension k − m. Let S be the orthogonal complement of S ⊥ . When T ⊃ S, dim T = l, then T ⊥ satisfies (9), hence by (8) 1/2

H (T ) (γk−m,k−l − ")H (S)(k−l)/(k−m) . This shows that the constant in (6) is best possible.

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Proposition 2. Let S be an m-dimensional subspace of Qk . When 0 < n m, there are linearly independent integer vectors p 1 , . . . , pn in S with n/2

|p 1 | · · · |pn | γm H (S)n/m .

(10) n/2

The constant γm

here is best possible.

Proof. By Minkowski’s second theorem for balls (see [3], Chapter VIII, Theorem I) in Λ = S ∩ Zk there are independent vectors p 1 , . . . , pm with m/2

|p 1 | · · · |pm | γm

m/2

det Λ = γm H (S), n/2

where |p1 | . . . |pm |, so that (10) holds. The constant γm the definition of γm .

is best possible in (10) by

Proof of Theorem 1. Let T be a space as in the part (ii) of Proposition 1. By Proposition 2 there are integer vectors p 1 , . . . , pl which span T and have l/2 1/2

l/2

|p 1 | · · · |pl | γl H (T ) γl γk−m,k−l H (S)(k−l)/(k−m) . This implies 1/2

l/2

c(k, l, m) γk−m,k−l γl . On the other hand, let p1 , . . . , pl be independent and with span T containing S. Then |p 1 | · · · |pl | H (T ) (cf. [2], formula (2.6)) and for " > 0 we shall necessarily have by Proposition 1(ii) 1/2

H (T ) > (γk−l,k−m − ")H (S)(k−l)/(k−m) . This proves 1/2

c(k, l, m) γk−l,k−m .

Proof of Corollary 1. From Theorem 1 we obtain 1/2

1/2

γk−1,k−2 c(k, 2, 1) γk−1,k−2 γ2 √ and it suffices to use γk−1,k−2 = γk−1 , γ2 = 4/3.

Proof of Theorem 3. We will use properties of convergent sequences of lattices as given in [3]. When a 1 , . . . , a l is a basis of a lattice Λ in Rl , the matrix A = (a t1 , . . . , a tl ) will also be called a basis of Λ, as in the proof of Theorem 1. There is a finite set V of non-singular integer matrices such that when M ⊂ Λ are lattices with [Λ : M] l! and B is a basis of Λ, then there is a V ∈ V such that A = BV is a basis of M. Conversely, when A is a basis of M, then there is a U ∈ V and a basis B of U with A = BU (see [3], Chapter I, §2.2, where the roles of Λ, M are reversed). Let a 1 , . . . , a l be independent elements of Λ with F (a i ) = λi (i = 1, . . . , l), where λ1 , . . . , λl are the successive minima of Λ and F is the distance function determined


1293

by K. Here a 1 , . . . , a l generate a sublattice M ⊂ Λ with [Λ : M] l! (ibid., p. 219). Say A = (a t1 , . . . , a tl ), A = BV with B a basis of Λ. Now, if Λt → Λ for lattices Λ1 , Λ2 , . . . there are bases Bt of Λt with Bt → B. Setting At = Bt V , say At = (a t1t , . . . , a tlt ), the points a 1t , . . . , a lt are independent in Λt and lim F (a it ) = F (a i ) = λi (i = 1, . . . , l). So, if λ1t , . . . , λlt are the successive minima of Λt , we have lim sup λit λi (i = 1, . . . , l).

Set now A∗t = (a ∗1t )t , . . . , (a ∗lt )t , where (a ∗1t )t , . . . , (a ∗lt )t are independent in Λt with F (a ∗it ) = λit (i = 1, . . . , l), where λ1t , . . . , λlt are the minima of Λt . We have A∗t = Bt∗ Vt , λi0 = lim inf λi0 t , as where Bt∗ is a basis of Λt and Vt ∈ V . Pick i0 , 1 i0 l, and set K c t → ∞. Pick a subsequence, where λi t → K λi0 , and a subsequence of that one, where Vt 0 is constant, so that A∗t = Bt∗ V ∗ , with V ∗ ∈ V . Since the sequence Λt is convergent, the minima λit are bounded, i.e. the F (a ∗it ) are bounded, therefore the lengths ait∗ are bounded. So taking a further subsequence, A∗t is convergent, say A∗t → A∗ = (a ∗1 , . . . , a ∗l ). Thus Bt∗ → B ∗ , where B ∗ is a basis of Λ and A∗ = B ∗ V ∗ . Thus a ∗1 , . . . , a ∗l are in Λ and linearly independent. In particular, F (a ∗1 ) λi0 = lim inf λi0 t . . . . F (a ∗i0 ), so that λi0 the i0 -th minimum of Λ, has λi0 F (a ∗i0 ) = K Since i0 was arbitrary in {1, . . . , l}, and by above inequality involving lim sup, we see that lim λit = λi

t→∞

(1 i l).

Proof of Corollary 2. If ft = at u2 + bt uv + ct v 2 it suffices to put in Theorem 3 l = 2, K = {(x, y) : x 2 + y 2 1}, ( 4at ct − bt2

√ bt Λt = Z.

at , 0 Z ⊕ √ , √ 2 at 2 at The proof of Theorem 2 is based on Corollary 2 and on four lemmas. Lemma 1. Let n = (15t 2 − t − 4, 3t 2 − 1, 3t 2 − t − 1) × (69t 2 − 5, 21t 2 − 2, 0). If t ∈ Z, t ≡ −7 mod 17, t ≡ ±3 mod 11, mn = 0 and |m| 18t 2 ,

(11)

m ∈ Z3 \ {0},

then for large t m = ±m1 , ±m2 , ±m3 , where m1 = (15t 2 − t − 4, 3t 2 − 1, 3t 2 − t − 1), m2 = (9t 2 + 4t + 11, 9t 2 + 2, −12t 2 + 4t + 4), m3 = (−6t 2 + 5t + 15, 6t 2 + 3, −15t 2 + 5t + 5). Proof. (11) implies that m = u(15t 2 − t − 4, 3t 2 − 1, 3t 2 − t − 1) + v(69t 2 − 5, 21t 2 − 2, 0),

1294


where u, v ∈ Q. Since m ∈ Z3 we have (12)

u(15t 2 − t − 4) + v(692 − 5) ∈ Z,

(13)

u(3t 2 − 1) + v(21t 2 − 2) ∈ Z,

(14)

u(3t 2 − t − 1) ∈ Z.

The relations (12) and (13) give uD ∈ Z, where D = (21t 2 − 2)(15t 2 − t − 4) − (3t 2 − 1)(69t 2 − 5) ≡ (7t + 5)(4t + 1) − t (23t + 18) ≡ 5t 2 + 9t + 5 ≡ −t 2 + 11t + 7 mod 3t 2 − t − 1, hence 3D ≡ 32t + 20 = 4(8t + 5) mod 3t 2 − t − 1. However gcd(2, 3t 2 − t − 1) = 1, 64(3t 2 − t − 1) − (8t + 5)(24t − 23) = 51, for t ≡ −7 mod 17, 8t + 5 ≡ 0 mod 17, for t ≡ −1 mod 3, 8t + 5 ≡ 0 mod 3 and for t ≡ 1 mod 3, D ≡ 1 mod 3. Thus gcd(D, 3t 2 − t − 1) = 1 and (12)–(14) imply u ∈ Z. Hence v(69t 2 − 5) ∈ Z,

(15)

v(21t 2 − 2) ∈ Z.

However (69t 2 − 5)7 − (21t 2 − 2)23 = 11, hence if t ≡ ±3 mod 11, we have gcd(69t 2 − 5, 21t 2 − 2) = 1 and (15) implies v ∈ Z. Now, if |m| 18t 2 we have |m|2 = Au2 + 2Buv + Cv 2 324t 4 ,

(16)

where for t tending to infinity A = (15t 2 − t − 4)2 + (3t 2 − 1)2 + (3t 2 − t − 1)2 = 243t 4 + O(t 3 ), B = (15t 2 − t − 4)(69t 2 − 5) + (3t 2 − 1)(21t 2 − 2) = 1098t 4 + O(t 3 ), C = (69t 2 − 5)2 + (21t 2 − 2)2 = 5202t 4 + O(t 3 ), hence AC − B 2 = 58482t 8 + O(t 7 ) = 81 · 722t 8 + O(t 7 ). The inequality (16) gives 324Ct 4 4C = < 29 + O(t −1 ), AC − B 2 722t 4 + O(t 3 ) 324At 4 4A v2 < 2 + O(t −1 ), = AC − B 2 722t 4 + O(t 3 )

u2


1295

hence for large t |u| 5,

|v| 1

and (16) implies 243u2 + 2196uv + 5202v 2 324, (27u + 112v)2 + 722v 2 972. We obtain either u = ±1, v = 0, or v = ±1, u = −4v, or u = −5v. The first case gives m = ±m1 , the second case m = ±m2 , the third case m = ±m3 .

Lemma 2. If p ∈ Z3 \ {0} and pm1 = 0, then for t tending to infinity √ |p| 18t + o(t). Proof. We easily verify that p1 m1 = q 1 m1 = 0, where p1 = (t, −4t − 1, 1 − t),

q 1 = (1, −3t − 3, 3t − 1).

Since p1 , q 1 are linearly independent, pm1 = 0, p ∈ Z3 implies p = up1 + vq 1 , where u, v ∈ Q. Now p ∈ Z3 implies ut + v ∈ Z,

(17)

u(4t + 1) + v(3t + 3) ∈ Z, u(1 − t) + v(3t − 1) ∈ Z, hence by taking determinants u(3t 2 − t − 1) ∈ Z,

(18)

u(3t 2 − 1) ∈ Z,

u(15t 2 − t − 4) ∈ Z.

However 15t 2 − t − 4 − 4(3t 2 − 1) − (3t 2 − t − 1) = 1, hence gcd(3t 2 − t − 1, 3t 2 − 1, 15t 2 − t − 4) = 1, (18) implies u ∈ Z and by (17) v ∈ Z. Now |p|2 = A1 (t)u2 + 2B1 (t)uv + C1 (t)v 2 , where A1 (t) = 18t 2 + O(t),

B1 (t) = 9t 2 + O(t),

C1 (t) = 18t 2 + O(t).

The sequence of quadratic forms A1 (t) 2 B1 (t) C1 (t) x + 2 2 xy + 2 y 2 2 t t t

1296


tends to 18x 2 + 18xy + 18y 2 and this quadratic form has minimum 18. It follows by Corollary 2 that |p|2 18t 2 + o(t 2 ),

which gives Lemma 2.

Lemma 3. If pm2 = qm2 = 0, where p, q ∈ Z3 , p, q linearly independent, t ≡ 2 mod 28, then for t tending to infinity √ |p| |q| 328t 2 + o(t 2 ). Proof. We easily verify that p2 m2 = q 2 m2 = 0, where p2 = (12t + 20, −28t − 56, −12t − 27), q 2 = (84t + 572, −84t − 1624, −761). Now, take t ≡ 2 mod 28 and r2 =

27t + 72 29 1 87t + 386 p2 + q2 = , −8t − 29, − ∈ Z3 . 112 112 7 28

If p ∈ Z3 and pm2 = 0, then p = up2 + vr 2 , where u, v ∈ Q. We assert that u, v ∈ Z. Indeed, p ∈ Z3 implies 27t + 72 v ∈ Z, 7 (28t + 56)u + (8t + 29)v ∈ Z, 87t + 386 (12t + 27)u + v ∈ Z, 28

(12t + 20)u +

hence on taking determinants D1 u, D2 u, D3 u ∈ Z, where D1 = −12t 2 + 4t + 4,

D2 = −9t 2 − 2,

D3 = −9t 2 − 4t − 11.

However, −3D1 + 7D2 − 3D3 = 7 and for t ∈ Z, 7 /| D2 , hence gcd(D1 , D2 , D3 ) = 1 and u ∈ Z. Similarly v ∈ Z. By Corollary 2 the problem reduces to finding the first two minima of the quadratic form 27 2 87 2 12u + v + (28u + 8v)2 + 12u + v . 7 28 By reduction we find that the first minimum m is obtained for u = 2, v = −7, 87 2 81 = 14.0625 =9+ m = 32 + 24 − 4 16


1297

and the second minimum m is obtained for u = 1, v = −3, 3 2 75 2 m= + 42 + = 23.358418. 7 28 Hence mm = 328.47775.

Lemma 4. If pm3 = qm3 = 0, where p, q ∈ Z3 , p, q linearly independent and t ≡ 15 mod 55, then for t tending to infinity √ |p| |q| 328t 2 + o(t 2 ). Proof. One easily verifies that p3 m3 = q 3 m3 = 0, where p3 = (10t + 25, −15t − 65, −10t − 36), q 3 = (165t + 1425, 165t − 3760, −2019). Now, take t ≡ 15 mod 55 and r3 =

75t + 228 −96t − 595 −342t − 1635 171 1 p3 + q3 = , , ∈ Z3 . 275 275 11 11 55

If pm3 = 0 and p ∈ Z3 we have p = up3 + vr 3 , where u, v ∈ Q. We assert that u, v ∈ Z. Indeed, p ∈ Z3 implies 75t + 228 v ∈ Z, 11 96t + 595 (15t + 65)u + v ∈ Z, 11 342t + 1635 (10t + 36)u + v ∈ Z, 55 hence D1 u, D2 u, D3 u ∈ Z, where (10t + 25)u +

D1 = −15t 2 + 5t + 5, D2 = −6t 2 − 3, D3 = 6t 2 − 5t − 15.

c

However 2D1 − 3D2 + 2D3 = −11 and for t ∈ Z, 11 /| D1 , hence gcd(D1 , D2 , D3 ) = 1 and u ∈ Z. Similarly v ∈ Z. By Lemma 1 the problem reduces to finding the first two minima of the quadratic form 75 2 96 2 342 2 10u + v + 15u + v + 10u + v = A3 u2 + 2B3 uv + C3 v 2 , 11 11 55 where A3 = 425,

B3 = 261.2727,

C3 = 161.3186.

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By reduction we find the first minimum 4A − 12B + 9C = 16.5948, the second minimum 9A − 30B + 25C = 19.7834,

the product 328.30.

Proof of Theorem 2. Take an " > 0 and a large integer t such that t ≡ −7 mod 17, t ≡ 2 mod 28, t ≡ 15 mod 55 and suppose that n = up + vq, n = (15t 2 − t − 4, 3t 2 − 1, 3t 2 − t − 1) × (69t 2 − 5, 21t 2 − 2, 0), where p, q ∈ Z3 ,

|p| |q|

Since

it follows that (19)

) 6 − " |n| . (722)1/4

√ |n| = 9 722t 4 + O(t 3 ), " 2 |p × q| |p| |q| 18 − t , 2

where we may assume without loss of generality that p, q are linearly independent. Therefore p × q = 0 and since (p × q)n = 0 we obtain, by Lemma 1, that p × q = ±m1 , ±m2 , ±m3 . If p × q = ±mi , then pmi = qmi = 0 implies, by Lemma i + 1, that |p| |q| (18 + o(1))t 2 , contrary to (19).

Appendix We give a proof independent of [10] of the part of Proposition 1(i). Proposition 3. If n < m < k and " > 0 there exists an m-dimensional subspace S of Qk such that for every n-dimensional subspace T of S we have (20)

1/2

H (T ) > (1 − ")γm,n H (S)n/m .

Notation. For a positive integer n we put [n] = {1, . . . , n}, for a vector a, ai is the i-th coordinate of a and for a matrix A = (aij ), AI,J = det(aij )i∈I, j ∈I , |J | is the cardinality of a set J .

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Lemma 5. If n < m < k for every three matrices U ∈ Rn×m and A, B ∈ Rm×k we have

det U AAt U t − det U BB t U t

k m max |AI,J − BI,J | max |AI,J + BI,J | det U U t , n n where the maxima are taken over all subsets I, J of [m] and [k], respectively, with |I | = |J | = n. Proof. This follows at once from the identities t t (21) det U AA U =

2

U[n],I AI,J

,

J ⊂[k], |J |=n I ⊂[m], |I |=n

(22)

det U U t =

J ⊂[m], |J |=n

2 U[n],J

and from the Cauchy–Schwarz inequality.

Lemma 6. If n m for every non-singular matrix A ∈ Rm×m , there exists a positive number c(A) such that for every matrix U ∈ Rn×m we have det U AAt U t c(A) det U U t . Proof. The right hand side of identity (21) is a quadratic form in U[n],I (I ⊂ [m], |I | = n). We shall show that for k = m it is positive definite. Otherwise there would exist xI not all equal to 0 such that xI AI,J = 0 I ⊂[m], |I |=n

for all J ⊂ [m], |J | = n. It follows hence that det B = 0 where B = (AI,J )I,J ⊂[m], |I |=|J |=n . However, by the statement 184 on p. 178 of [6], det B is a power of det A, hence det B = 0 and the obtained contradiction shows that the form in question is positive definite. Now we use the following argument, kindly supplied by M. Skałba, that is simpler than our original one. Reducing a positive definite quadratic form f ∈ R[x1 , . . . , xl ] to a diagonal form by an orthogonal transformation with matrix (cij ) we obtain that f (x1 , . . . , xl ) =

l

λj

j =1

l

2 cj i xi

.

i=1

Hence f (x1 , . . . , xl ) λ1

l l j =1

i=1

2 cj i xi

= λ1

l i=1

xi2 ,

1300


where λ1 is the least characteristic root of f . Applying this to our quadratic form and using identity (22) we obtain the lemma.

Proof of Proposition 3. It is enough to consider k = m + 1, " < 1. By the definition of γm,n there exists a full lattice Λ in Rm such that for every n-dimensional sublattice Γ of Λ we have det Γ γm,n (det Λ)n/m . Let a 1 , . . . , a m be a basis of Λ and let b1 , . . . , bm in Qm be so close to a 1 , . . . , a m , respectively, that taking ⎛ ⎞ ⎛ ⎞ b1 a1 ⎜ .. ⎟ ⎜ .. ⎟ A = ⎝ . ⎠, B = ⎝ . ⎠ am we have (23)

bm

" |det A| |det B| < 1 + 2

and

c(A)" − AI,J | min a, 2 , 6 m n a

max |BI,J

where a = max |AI,J | and the maxima are taken over all subsets I, J of [m] with |I | = |J | = n. Let d be a positive integer such that dbi ∈ Zm (1 i m) and let us consider the matrix ⎞ ⎛ db11 t + t1 . . . db1m t tm+1 ⎜ .. .. .. ⎟ . .. C = C(t, t1 , . . . , t2m ) = ⎝ . . . . ⎠ dbm1 t

. . . dbmm t + tm

t2m

We have (24)

C[m],[m] ≡ t1 · · · tm mod t,

hence C[m],[m] = 0 and also for all j m C[m],[m+1]−{j } (t, t1 , . . . , tm , db1j t, . . . , dbjj t + tj , . . . , dbmj t) = (−1)m+j C[m],[m] , hence C[m],[m+1]−{j } = 0. However C[m],[m] is independent of tm+1 , . . . , t2m and C[m],[m+1]−{j } is independent of tj (1 j m), hence the greatest common divisor D of C[m],[m+1]−{j } (1 j m + 1), which is homogeneous in t, t1 , . . . , t2m could depend only on t and by (24) D = 1. Also the m + 1 polynomials in question have no common fixed numerical divisor > 1, since by (24) C[m],[m] (0, 1, . . . , 1) = 1. ∗ and an arithmetic progression Hence, by Theorem 1 of [9], there exist integers t1∗ , . . . , t2m


1301

P ⊂ Z such that for t ∈ P (25)

∗ C[m],[m+1]−{j } (t, t1∗ , . . . , t2m ) = 1.

gcd

1j m+1

∗ ) and consider the lattice Λ and the linear Let cj be the j -th row of C(t, t1∗ , . . . , t2m 1 m subspace S of Q spanned by c1 , . . . , cm . For t ∈ P tending to infinity we have by (25) F Gm+1 G ∗ )2 C[m],[m+1]−{j } (t, t1∗ , . . . , t2m (26) H (S) = det Λ1 = H j =1

= d m t m |det B| + O(t m−1 ). Assume now that t ∈ P and T is an n-dimensional subspace of S. Since Λ1 is primitive, m the lattice T ∩ Zm+1 is generated by uij cj (1 i n), where uij ∈ Z and we put (uij )in, j m = U . Assume, contrary to (20), that

j =1

1/2

H (T ) (1 − ")γm,n H (S)n/m . Hence, by (23) and (26) we have for t large enough " 1/2 γm,n |det A|n/m . H (T ) 1 − 2 However √ H (T ) = det U CC t U t , where ∗ ), C(t) = C(t, t1∗ , . . . , t2m

thus

C(t) C(t)t t " 2 γm,n |det A|2n/m . U 1− dt dt 2 Applying Lemma 5 to the matrices U , C(t)/dt and B equal to B augmented by the (m + 1)-th column consisting of zeros we obtain " 2 γm,n |det A|2n/m + O(t −1 ) det U U t . det U BB t U t = det U B (B )t U t 1 − 2 We now apply Lemma 5 to the matrices U , A and B and obtain " " 2 γm,n |det A|2n/m + c(A) + O(t −1 ) det U U t . (27) det U AAt U t 1 − 2 2 However, by Lemma 6 the left hand side is at least c(A) det U U t . Hence for t large enough det U

c(A) det U U t γm,n |det A|2n/m and combining this with (27) we obtain for t large enough det U AAt U t < γm,n |det A|2n/m .

1302


It follows that the lattice Γ1 generated by

m j =1

uij a j (1 j n) satisfies

1/2

det Γ1 < γm,n (det Λ)n/m , contrary to the choice of Λ. The obtained contradiction proves (20).

References [1] I. Aliev, On a decomposition of integer vectors II. Acta Arith. 102 (2002), 373–391. [2] E. Bombieri, J. D. Vaaler, On Siegel’s Lemma. Invent. Math. 73 (1983), 11–32; Addendum, ibid. 75 (1984), 377. [3] J. W. S. Cassels, An Introduction to the Geometry of Numbers. Springer, Berlin 1959. [4] S. Chaładus, A. Schinzel, A decomposition of integer vectors II. Pliska Stud. Math. Bulgar. 11 (1991), 15–23; this collection: L1, 1249–1258. [5] S. Chaładus, Yu. Teterin, Note on a decomposition of integer vectors II. Acta Arith. 57 (1991), 159–164. [6] T. Muir, A Treatise on the Theory of Determinants. Dover, New York 1960. [7] R. A. Rankin, On positive definite quadratic forms. J. London Math. Soc. 28 (1953), 309–314. [8] A. Schinzel, A decomposition of integer vectors IV. J. Austral. Math. Soc. Ser. A 51 (1991), 33–49; this collection: L2, 1259–1273. [9] −−, A property of polynomials with an application to Siegel’s lemma. Monatsh. Math. 137 (2002), 239–251; this collection: L3, 1274–1287. [10] W. M. Schmidt, The distribution of sublattices of Zm . Monatsh. Math. 125 (1998), 37–81.

Part M Other papers


Commentary on M: Other papers* by Stanisław Kwapień

M1. Since the publication of the paper, it has become customary to call Gołab–Schinzel (in the sequel it is abbreviated to G–S) equation the functional equation appearing in the title of the paper. A motivation for considering the equation was an observation made by J. Aczél and S. Gołab connecting the equation with subgroups of the group A of affine transformations of R1 into itself. If we identify an affine transformation T of R1 , given by the formula T (x) = a + bx, with the pair (a, b) ∈ R1 × (R1 \ {0}) then the function f fulfils the G–S equation if and only if the part of the graph of f which is R1 × (R1 \ {0}) is a subgroup of A. Perhaps more important reason of the popularity of the G–S equation is that it is one of the simplest functional equations which combines conditions like in the Cauchy equation and of iterative type. In their paper S. Gołab and A. Schinzel do not pursue connections with the theory of groups and their subgroups. Instead, they give a treatment of the equation as elementary as possible. The main results of the article are: I.

The only differentiable solutions of the equation are the function f ≡ 0 and the functions fm (x) = 1 + mx, where m is an arbitrary, fixed real number.

II.

The only continuous solutions are the functions as in I and the functions max{fm , 0}, where fm is as in I.

III. A complete and simple description of trivial solutions of the equation is given (a function f is called trivial iff its values are in the set {−1, 0, +1}). It is proved that a trivial function is a solution of the equation if and only if it is identically equal to 0, or identically equal to 1, or it is the difference of the indicator functions of an additive subgroup of R1 and its disjoint translation. IV. A complete and simple description of nontrivial and non-microperiodic solutions of the equation is found (a function on R1 is said to be microperiodic if it has arbitrary small periods). It is proved that f is such a solution if and only if f = fm IG (fm ), where fm is as in I and IG is the indicator function of a set G which is a multiplicative subgroup of R1 \ 0, containing −1. *

The paper M2 is commented in the next article by E. Szemerédi.

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The question of a description of measurable solutions of the G–S equation is left open in the paper. The article started a steady flow of works concerning the equation and its generalizations. In a recent review paper by J. Brzdek [7], the author lists 82 publications closely related to the G–S equation. For the list and a throughout review of all the subsequent developments to the G–S equation we refer the reader to that paper. Here we will point out only some of them. A form of a general solution of the G–S equation was given independently by P. Javor [10] and S. Wołod´zko [15]. Lemma 1 from M1 is a basic tool for this goal. We will sketch their result. It is easier to describe the inverse g = f −1 , which in general is a multifunction, and we will do that. If f is not identically equal to 0, what will be assumed in the sequel, then g is defined on a multiplicative subgroup G of R\{0} (eventually G ∪ {0} if f admits values equal to 0). The set A = g(1) is an additive subgroup of R (it is the group of the periods of f ). It follows by the G–S equation that A is closed under multiplication by numbers from G, i.e., rA = A for r ∈ G. Moreover Lemma 1 implies that the values of g, restricted to G, are different cosets of A. Therefore the restriction of g to G can be treated as an usual 1–1 function from G into the quotient group R/A. The G–S equation implies that g(xy) = g(x) + xg(y) for x, y ∈ G (since rA = A for r ∈ G the multiplication by elements of G is defined on the quotient group R/A). If 0 is a value of f then g(0) = R \ g(G). Conversely, having such a multiplicative subgroup G, an additive subgroup A closed under the multiplication by numbers from G and a 1–1 function g : G → R/A which fulfils the above equation then putting g(0) = R \ g(G) if the last set is nonempty we obtain a solution f of the G–S equation such that g = f −1 . For many groups G, A we see easily that the only functions g with the above properties are of the form g(x) = xa − a for x ∈ G, where a is a fixed element of R/A. For a discussion when it is so we refer the reader to the book of K. S. Brown [5], Ch. 4, Sect. 2, p. 89, or S. Balcerzyk [3], Ch. X, Sect. 4, p. 333. It is true if there exists y ∈ G such that 1/(y − 1) is in the ring generated in R by G. In particular this is satisfied if f is continuous or has the Darboux property. The above description of solutions allows us to obtain not only the results I–IV, stated above, but also those from many other articles as well. In particular, by the very same method, we obtain a description of continuous functions fulfilling the G–S equation in the case when the function f is defined on a linear topological vector space instead on R1 . The continuous solutions of the G–S equation on a topological vector space are the same as in II, except that mx has to be understood as m(x), where m is a continuous linear functional on the topological vector space. An exposition of this result and those of P. Javor and S. Wołod´zko can be found in Chapter XI of the monograph by J. Aczél and J. Dhombres [1]. If we restrict the domain of f to R+ or R \ {0} then we cannot repeat directly the above method to find solutions of such equations. Several papers treat this problem, concentrating on a description of continuous solutions of such equations. They are called the conditional G–S equation or the G–S equation with restricted domains in those papers. In most cases the continuous solutions are similar as in the original G–S equation. An elegant paper of J. Aczél, J. Schwaiger [2] is the most representative for this approach. Another direction of generalizations considered in the papers are the G–S equations of a more general form.

Other papers

1307

The equations of the form f (f (y)k x + f (x)l y) = λf (x)f (y), and more general, were investigated in many papers. Essentially new types of continuous solutions of these generalized G–S equations were found, see e.g. J. Brzdek [6]. Of special interest is the paper by P. Kahlig, J. Matkowski [12]. In the paper the authors found continuous solutions on R+ of the equations f (x + ys(x)r ) = s(x)f (y), where s : R+ → R+ is a fixed monotone function and r is a fixed positive number. It is closely related to the G–S equation and the continuous solutions of the both equations are of the same type. What makes this paper interesting is that the authors present some applications to nonlinear processes of meteorology and fluid mechanics. The differential equations describing evaporation of cloud droplets, water discharging from reservoir, etc., exhibit some symmetries which are expressed by the G–S equation modified in the above way. We end this short review with mentioning recent results on the stability of the G–S equation. A typical result in this direction is that of J. Chudziak [8]. The result says that if for a continuous function f on R the expression f [x + yf (x)] − f (x)f (y) is bounded on R × R then either f is a bounded function or it is an unbounded solution to the G–S equation (1 ). M3. The paper treats in an elementary way a special case which can be put in a broad scheme considered in Potential Theory associated with a random walk, resp. with a Markov process. According to the general scheme for a given random walk, resp. Markov process, and a region A we can associate in a natural way a set ∂A—called the Martin boundary of A, and a kernel HA : A × (A ∪ ∂A) → R+ , in such a way that nonnegative functions which are superharmonic in A for the potential theory, associatedto random walks, HA (x, p)μ(p), resp. Markov process, are precisely those which can be represented as p∈C C resp. by C HA (x, p)dμ(p), where C = A ∪ ∂A and μ is a nonnegative function, resp. measure, on C. A subject of numerous papers in the probabilistic potential theory is a study of the behavior of the function HA (x, p), especially when x approaches ∂A. Harnack’s inequalities are estimates, independent of p, from above and below of the ratio HA (x, p)/HA (y, p). There are very few random walks (resp. Markov processes) and regions when the kernel HA can be explicitly computed as it is in the classical potential theory with A being a ball. This makes the problem quite difficult. For this probabilistic point of view we refer the reader to Chapter 3 of the monograph of F. Spitzer [14]. The paper treats simple, symmetric random walk on the plane while A is a disc. Although an explicit formula for the kernel is not given (probably there is no simple one), quite precise bounds on the kernel are proved. The paper is an outgrowth of a problem posed at one of the mathematical olympiads in Poland. It is dedicated to Stefan Straszewicz, one of the founders of The Mathematical Olympiads in Poland and their chairman through the first twenty years of their existence. The author’s deep and strong involvement in organizations of these mathematical competitions for youths is very well known and highly appreciated in Poland. (1 )

The author of this commentary would like to thank Janusz Brzdek for providing his review paper on the G–S equation and for his helpful comments also Jan Krempa for showing him connections of the G–S equation with concepts in Homological Algebra and for references on that.

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M4. The very well known Hadamard estimate of the determinant of a matrix A = (ai,j )i,j n states that |det A|

n

a i 2 ,

i=1

where, for i = 1, . . . , n, the vector a i = (ai,1 , . . . , ai,n ) is the i-th row of the matrix A and

· 2 is the standard Euclidean norm on Rn . The above inequality turns into an equality if all the vectors a i , i = 1, . . . , n, are orthogonal. Hence the Hadamard inequality is in a sense optimal. More exactly, if we want the inequality to hold for all matrices A the Euclidean norm in the inequality can not be replaced by another norm · with a a 2 for all a ∈ Rn . We can only hope for a norm which is smaller than the Euclidean norm for some norm can be found. result of the vectors a ∈ Rn . It is remarkable that such

It is the main a

aj − aj = max aj fulfils present paper that the norm a = max j :aj 0 suffisamment petit, coupe le diagramme y = f (x) dans les points x1 + η1 , x2 − η2 , où η1 , η2 > 0, η1 , η2 → 0 quand ε → 0 et f (x1 + η1 ) = f (x2 − η2 ) = 0. Les nombres x2 − x1 − (η1 + η2 ) = p − (η1 + η2 ) en vertu du lemme 1 seront des périodes de la fonction f (x), donc aussi η1 + η2 et la fonction f , comme micropério2 c dique et continue, doit être constante f (x) ≡ c. Mais en vertu de (1) c = c .

Lemme 3. Si la solution f (x) n’est pas périodique, alors elle doit être invertible dans l’ensemble A = {x : f (x) = 0}. C’est la conséquence immédiate du lemme 1. Lemme 4. Si la solution f (x) est continue et n’est pas périodique, alors elle doit avoir une des formes suivantes : a) f (x) est négative et strictement croissante dans l’intervalle (−∞, x1 ), f (x) = 0 dans l’intervalle [x1 , x2 ], f (x) est positive et strictement croissante dans l’intervalle (x2 , +∞), où −∞ x1 x2 < 0. b)

f (x) est positive et strictement décroissante dans l’intervalle (−∞, x1 ), f (x) = 0 dans l’intervalle [x1 , x2 ], f (x) est négative et strictement décroissante dans l’intervalle (x2 , +∞), où 0 < x1 x2 +∞.

Cela résulte de ce que, f (x) n’est pas constante, f (0) = 1, f (x) est continue et invertible dans l’ensemble A.

3. Lemme 5. Si la fonction f (x) est continue sur toute la droite et n’est pas constante, alors f (x) − 1 pour tout x ∈ A, x = 0, la valeur de est constante. x Démonstration. Supposons que x ∈ A, y ∈ A, xy = 0 et x + y · f (x) = y + x · f (y), et d’autre part on a

f (y) − 1 f (x) − 1 = . Donc x y

f [x + y · f (x)] = f (x) · f (y) = f [y + x · f (y)] = 0.

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Donc, en vertu du lemme 1, la fonction f est périodique, ce qui, vu les hypothèses admises, n’est pas compatible avec le lemme 2.

Lemme 6. Les hypothèses du lemme 5 étant admises, le cas −∞ < x1 < x2 < ∞ dans le lemme 4 ne peut pas avoir lieu. Démonstration. Si −∞ < x1 et x2 < ∞ alors en vertu de la continuité de la fonction f aux points x1 et x2 , nous obtenons du lemme 5 f (x1 ) − 1 f (x2 ) − 1 = x1 x2 d’où, vu que f (x1 ) = f (x2 ) = 0, x1 = x2 . En outre, dans les intervalles (−∞, x1 ) et (x2 , ∞) la fonction f doit être linéaire. Si x1 = x2 , nous obtenons les solutions obtenues déjà dans le §1. Il nous reste donc seulement le cas, où x1 = −∞ < x2 < 0 ou bien 0 < x1 < x2 = +∞.

Nous obtenons donc le Théorème 1. Les seules solutions continues de l’équation (1) sauf les solutions (2) et (3) sont les solutions : ⎧ ⎧ ⎨0 ⎪ pour x x2 (x2 < 0) ⎪ ⎪ ⎪ x a) f (x) = ⎪ ⎪ pour x x2 ⎩1 − ⎨ x2 ⎧ (10) x ⎪ ⎨1 − ⎪ pour x x1 (x1 > 0) ⎪ ⎪ x1 b) f (x) = ⎪ ⎪ ⎩ ⎩0 pour x x 1

où x1 est un nombre positif quelconque, x2 — un nombre négatif quelconque.

4. Occupons nous maintenant des solutions triviales. Nous appelons ainsi les solutions pour lesquelles les valeurs de f (x) sont comprises dans T (voir Introduction). Ici appartiennent surtout les deux solutions constantes et des autres, qui sont déjà discontinues. Nous démontrerons d’abord le Lemme 7. Si l’ensemble F des valeurs de f (x), qui est la solution de l’équation (1) n’est pas compris dans T , alors l’ensemble F est infini. Démonstration. Soit f (x0 ) = 0, 1, −1. En posant f (x0 ) = a x1 = x0 f (x0 ) + x0 xn+1 = x0 f (xn ) + xn

M1. Sur l’équation fonctionnelle f [x + y · f (x)] = f (x) · f (y)

1319

on prouve aisèment par induction que f (xn ) = a n

(n = 1, 2, . . . ).

En effet, en substituant dans (1) x = xn , y = x0 , nous avons f (xn+1 ) = f [xn + x0 · f (xn )] = f (xn ) · f (x0 ) = a n · a = a n+1 . Comme a = 0, 1, −1, l’ensemble des valeurs de a n est infini.

La lemme 7 justifie l’introduction du terme “solutions triviales”. Supposons, que f (x) est une solution triviale, ne prenant que les valeurs 0 et +1. Une de telles solutions est la suivante : f (0) = 1,

f (x) = 0,

pour x = 0.

S’il existe un x0 = 0 tel que f (x0 ) = 1, alors x0 est une période de la fonction f (x), car f (y + x0 ) = f [x0 + y · f (x0 )] = f (x0 ) · f (y) = f (y). Désignons par Ω l’ensemble de toutes les périodes. Ω forme évidemment un groupe additif. Si x ∈ / Ω, alors f (x) = 0. D’autre part si nous prenons un groupe additif quelconque Ω et posons 1 quand x ∈ Ω (11) f (x) = 0 quand x ∈ /Ω nous obtenons une solution. Le groupe Ω peut se composer des points isolés, ou former un ensemble dense. En particulier, quand Ω se compose de tous les nombres rationnels, nous obtenons une solution discontinue à chaque point. Cette fonction est nommée fonction de Dirichlet. Voici maintenant un exemple de solution non mesurable L de la forme (11), dû à M. W. Sierpiński. Soit H une base de Hamel et soit b un élément donné de H . Définissons l’ensemble Ω de façon suivante : x ∈ Ω si dans le développement de x de la forme x = a1 b1 + a2 b2 + . . . + am bm , où b1 , b2 , . . . , bm sont des éléments de la base H , et a1 , a2 , . . . , am sont des nombres rationnels, non nuls, (b1 , b2 , . . . , bm ) ne contient pas l’élément b. De l’unicité des développements considérés, il résulte sans peine, que Ω est un groupe additif, et comme Ω est non mesurable L (voir [5], p. 108), la fonction f est non mesurable. Examinons maintenant la structure des solutions triviales prenants aussi la valeur −1. Désignons Ω = {x : f (x) = 1},

Ω ∗ = {x : f (x) = −1},

B = {x : f (x) = 0}.

L’ensemble Ω forme un groupe (composé éventuellement de 0 ou des éléments isolés) dont l’ensemble Ω ∗ doit être une translation. D’autre part on démontre facilement que si nous prenons un groupe additif quelconque Ω — different de l’ensemble de tous les

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nombres réels, — si nous désignons par Ω ∗ une translation autre que Ω lui-même et par B l’ensemble (−∞, ∞) − (Ω + Ω ∗ ) et si nous définissons la fonction f (x) par la formule ⎧ ⎪ quand x ∈ Ω ⎨1 f (x) = −1 quand x ∈ Ω ∗ ⎪ ⎩ 0 quand x ∈ B nous recevrons une solution de l’équation (1). Le problème se pose, si l’ensemble B peut être vide. Or, la réponse est négative. Nous a a démontrerons que si a ∈ Ω ∗ alors ∈ B. En effet, s’il était ∈ Ω alors on aurait 2 2 a a a + = a ∈ Ω, contre l’hypothèse. S’il était ∈ Ω ∗ alors, vu que a ∈ Ω ∗ , on aurait 2 2 2 a a a a − = ∈ Ω, ce qui est impossible. Donc il doit être ∈ B. 2 2 2 A. Łomnicki a démontré dans le travail cité [4] que l’ensemble des périodes d’une fonction mesurable, non constante, est de mesure nulle. Il a démontré aussi que chaque fonction mesurable, micropériodique et non constante, possède une soi-disant valeur privilégiée, c’est-à-dire il existe un nombre p tel que l’ensemble E = {x : f (x) = p} est de mesure nulle. Or, dans le cas de notre équation, pour chaque solution mesurable, micropériodique et non-constante, ce nombre privilégié p est égal à 0. Pour le démontrer observons, que l’ensemble des périodes coïncide avec l’ensemble Ω = {x : f (x) = 1}, car comme nous avons vu, f (x0 ) = 1 entraîne f (y + x0 ) ≡ f (y) et si p est une période alors f (p) = f (0) = 1. Si le nombre q = 0 est la valeur de la fonction f (x) alors l’ensemble Q = {x : f (x) = q} est une translation de l’ensemble Ω. Car si f (x) = f (y) = q = 0, alors en vertu du lemme 1, x − y ∈ Ω. Puisque l’ensemble Ω en vertu du premier des théorèmes de Łomnicki est de mesure nulle, l’ensemble Q est aussi de mesure nulle, par là l’ensemble E = A.

5. Occupons nous maintenant avec les solutions non micropériodiques discontinues. Lemme 8. Si f (x) est une solution de l’équation (1) et s’ils existent des nombres x1 et x2 tels que : (12)

f (x1 ) = 0, 1, −1; f (x2 ) = 0, [1 − f (x2 )]x1 = x2 [1 − f (x1 )],

alors la fonction f (x) est micropériodique. Démonstration. Soit y1 = f (x1 ), y2 = f (x2 ), z0 = 0,

zn =

y1n − 1 x1 ; y1 − 1


1321

nous affirmons, que f (zn ) = y1n ,

(13)

c

pour n = 0, ±1, ±2, . . . .

En effet, pour n = 0 et n = 1 la formule est vraie (f (0) = 1, f (z1 ) = y1 ). Supposons, qu’elle soit vraie pour n 0 et substitutions à (1) x = zn , y = x1 . Nous aurons : f [zn + x1 · f (zn )] = f [zn + x1 y1n ] = f (x1 ) · f (zn ) = y1 · y1n = y1n+1 . D’autre part zn + x1 · f (zn ) = x1

yn − 1

y n+1 − 1 + y1n = x1 · 1 = zn+1 . y1 − 1 y1 − 1 1

Soit à son tour dans (1) x = zn , y = z−n ; nous obtenons : f [zn + z−n · f (zn )] = f (zn ) · f (z−n ), mais zn + z−n · f (zn ) =

y1n − 1 y −n − 1 x1 n x1 + 1 x1 · y1n = y1 − 1 + 1 − y1n = 0. y1 − 1 y1 − 1 y1 − 1

1 = y1−n et la formule (13) se trouve démontrée. Posons f (zn ) maintenant dans l’équation (1) x = zn , y = x2 , ensuite x = x2 , y = zn . Nous aurons alors De là nous avons : f (z−n ) =

f (zn + y1n · x2 ) = f (zn ) · f (x2 ) = y2 · y1n = 0 f (x2 + zn y2 ) = f (x2 ) · f (zn ) = y2 · y1n = 0. D’où, en vertu du lemme 1, chacun des nombres ωn = x2 + zn y2 − zn − x2 y1n est la période de la fonction f . Les nombres ωn+1 − ωn = y2 (zn+1 − zn ) − (zn+1 − zn ) − x2 (y1n+1 − y1n ) =

y1n+1 − y1n (y2 − 1)x1 − x2 y1n (y1 − 1) = y1n (y2 − 1)x1 − x2 (y1 − 1) y1 − 1

sont aussi des périodes. Mais d’après l’hypothèse (12) la dernière parenthèse n’est pas nulle, donc ωn+1 − ωn = 0. Quand |y1 | < 1, alors y1n → 0 pour n → ∞, quand |y1 | > 1, alors y1n → 0 pour n → −∞. Il en résulte, que f (x) a des périodes aussi petits qu’on veut, c’est-à-dire f (x) est micropériodique, ce que nous voulions démontrer.

Observons comme résultat secondaire, que selon que |y1 | < 1 ou |y1 | > 1 la suite zn respectivement z−n tend vers x1 ξ= 1 − f (x1 )

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et alors f tend vers zéro, ainsi, que, l’hypothèse de notre lemme étant admise, zéro est le point d’accumulation des valeurs de la fonction f (x) quel que soit le voisinage du point ξ . Nous avons dit plus haut, que si la solution f (x) est micropériodique, alors zéro est sa valeur privilégiée. On voit, qu’aussi pour les solutions non micropériodiques zéro est une valeur “privilégiée” dans un certain sens. Le théorème suivant le démontre : Théorème 2. Pour que la fonction non micropériodique f soit une solution non triviale de l’équation (1) il faut et il suffit qu’il existe un nombre m = 0 ainsi qu’un groupe G multiplicatif, contenant outre ±1 encore d’autres nombres et tel que 1 + mx quand (1 + mx) ∈ G (14) f (x) = 0 quand (1 + mx) ∈ / G. Démonstration. Nécessité. Il existe un x0 tel, que y0 = f (x0 ) = 0, +1, −1. Évidemment x0 = 0, puisque f (0) = 1. Comme f (x) n’est pas micropériodique, on a en vertu du lemme 8 pour chaque x l’alternative f (x) = 0

ou x0 [1 − f (x)] = x(1 − y0 ).

En posant def

m =

x0 − 1 x0

nous avons m = 0 et on peut écrire l’alternative nommée ci-dessus dans la forme f (x) = 0

ou f (x) = 1 + mx. y − 1 Il faut démontrer, que l’ensemble des y, pour lesquels f = y = 0 forme un m groupe multiplicatif. Prenons deux valeurs y1 et y2 de l’ensemble y − 1 = y = 0 . G= y:f m Nous avons f

y − 1 1 = y1 , m

f

y − 1 2 = y2 . m

S’il y avait f

y1 − 1 y2

alors il devrait être f

m

=

y1 − 1 y2 m

y1 , y2

= 0,


1323

d’où nous aurions : y − 1 y2 − 1 1 =f + y2 · f m m

y1 y2

− 1 m

y2 − 1 =f + m =f

y1 − 1 y2

− 1 y2 − 1 f m m y − 1 y1 − 1 y2 2 y1 y2

m

m

=0

y1 c’est-à-dire si y1 ∈ G et y2 ∈ G m y2 y1 aussi ∈ G d’où il résulte que G est un groupe multiplicatif. y2 Suffisance. Nous vérifions que la fonction f (x) remplit l’équation (1). Quand (1 + mx) ∈ / G, alors f (x) = 0 et et on aboutit à une contradiction. Donc f

=

f [x + y · f (x)] = f (x) = 0 = f (x) · f (y). Quand (1 + mx) ∈ G et (1 + my) ∈ G, alors f (x) · f (y) = (1 + mx) · (1 + my) 1 + m[x + y · f (x)] = 1 + m · x + m · y · (1 + mx) = (1 + mx) · (1 + my) ∈ G, donc f [x + y · f (x)] = (1 + mx) · (1 + my) = f (x) · f (y). Quand (1 + mx) ∈ G et (1 + my) ∈ / G, alors f (y) = 0, et en même temps 1 + m · [x + y · f (x)] = 1 + mx + my · (1 + mx) = (1 + mx) · (1 + my) ∈ /G d’où f [x + y · f (x)] = 0. Par là nous avons démontré, que f (x) est une solution de (1) et le théorème se trouve démontré.

Il est à remarquer, que les solutions f (x) en question sont bornées dans chaque intervalle fini. En outre chaque solution est continue au point x=−

1 . m

Pour caractériser les points de discontinuité de la fonction f (x) observons que, quel que soit le groupe G, un des cas suivantes subsiste : (15)

G se compose des nombres a n ou ±a n , où a est un nombre réel = 0, et n = 0, ±1, ±2 . . . .

(16)

G est en même temps dense et frontière sur la demidroite (0, ∞).

(17)

G contient toute la demidroite (0, ∞).

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Dans le cas (16) la fonction f est discontinue soit sur la demidroite (x, ∞) (si m > 0), soit sur la demidroite (−∞, x) (si m < 0). Dans le cas (17) une fonction f est de la forme 1 + mx (si G se compose de tous les nombres réels = 0) ou bien de la forme (10) (si G se compose de tous les nombres positifs). Puisque toutes les solutions micropériodiques sont discontinues sur toute la droite, nous pouvons renforcer le théorème 1 de façon suivante : Théorème 3. Si la solution non-triviale n’est ni de la forme (3) ni de la forme (10) elle est soit discontinue sur une demidroite, soit elle est de la forme (14) où G remplit (15). Quant aux solutions de la forme (14) partout discontinues, S. Marcus a remarqué que nous pouvons obtenir une solution non mesurable de cette forme, prenant comme G le corps non-mesurable des nombres réels, dont l’existence est démontrée dans la note posthume de M. Souslin, rédigée par C. Kuratowski ([6], p. 315).

6. Le problème se pose si l’équation (1) possède des solutions non triviales micropériodiques. Or nous démontrerons le Théorème 4. L’équation (1) possède des solutions non triviales micropériodiques. Nous allons construire deux groupes : un groupe additif Ω et un groupe multiplicatif G, tels que 1) 2) 3)

Ω contient non seulement le nombre zéro, G contient non seulement les nombres 1 et −1, y ∈ G, ω ∈ Ω =⇒ yω ∈ Ω, y1 , y2 ∈ G, (y1 − y2 ) ∈ Ω =⇒ y1 = y2 .

Des paires de tels groupes existent. √ Il suffit de classer dans Ω tous les nombres de la forme r 2, dans G tous les nombres r = 0, où r est un nombre rationnel quelconque. Posons y s’il existe ω ∈ Ω tel que y = 1 + x + ω ∈ G f (x) = 0 si un tel ω n’existe pas. La fonction f est bien définie. En effet, supposons qu’il existe ω , ω ∈ Ω et y , y ∈ G tels que 1 + x + ω = y , 1 + x + ω = y ; alors y − y = (ω − ω ) ∈ Ω, d’où y = y . Ensuite la fonction f (x) est périodique. Prenons en effet un ω ∈ Ω quelconque. Si f (x) = 0, alors f (x + ω) = 0 car s’il existait un ω tel que 1 + (x + ω) + ω ∈ G, alors il existerait aussi un ω = ω + ω tel que 1 + x + ω ∈ G en contradiction avec f (x) = 0 ; or si f (x) = 0, alors 1 + x + ω ∈ G d’où 1 + x + ω + 0 ∈ G et alors f (x + ω) = 1 + x + ω = f (x).


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Le nombre zéro est un point d’accumulation de l’ensemble Ω pour la raison suivante. Puisque le groupe G est non-trivial, 0 est son point d’accumulation. Comme yω ∈ Ω, quand ω ∈ Ω alors y ∈ G, donc il existe des périodes aussi petites que l’on veut, c’est-àdire f est micropériodique. Nous démontrerons maintenant, que f satisfait à l’équation (1). En effet, si f (x) = 0, alors l’équation (1) est satisfaite d’une façon triviale. Si f (x) = 1 + x + ω, alors nous distinguons deux cas : 1) ou bien f (y) = 0, 2) ou f (y) = 1 + y + ω1 . Dans le premier cas le second membre de l’équation est zéro. Supposons pour le c moment, que le premier membre ne soit pas zéro : f [x +y ·f (x)] = 0, donc 1+x +yf (x)+ ω2 ∈ G, c’est-à-dire 1+x +ω0 +y ·f (x)+ω2 −ω0 ∈ G, ou f (x)+y ·f (x)+ω2 −ω0 ∈ G ω2 − ω0 1 ou (comme f (x) = 0) 1 + y + ∈ G. Mais f (x) ∈ G, ∈ G, ω2 − ω0 ∈ G, f (x) f (x) donc ω2 − ω0 ω1 = ∈Ω f (x) et comme 1 + y + ω1 ∈ G, donc f (y) = 0 et on aboutit à une contradiction. Donc le premier membre de l’équation est aussi égal à zéro. Dans le second cas on a f (x) = 1 + x + ω0 ∈ G, f (y) = 1 + y + ω1 ∈ G, d’où x + y · f (x) = f (x) − 1 − ω0 + [f (y) − 1 − ω1 ] · f (x) = f (x) · f (y) − [ω0 + ω1 f (x)] − 1, c’est-à-dire 1 + x + y · f (x) = f (x) · f (y) − [ω0 + ω1 · f (x)]. Mais f (x) ∈ G, donc ω1 f (x) ∈ Ω et ω2 = ω0 +ω1 ·f (x) ∈ Ω. Ensuite f (x)·f (y) ∈ G. Alors on a 1 + x + y · f (x) + ω2 ∈ G, donc f [x + y · f (x)] = 1 + x + y · f (x) + ω2 = f (x) · f (y). Nous avons ainsi démontré que la fonction f (x) est une solution de l’équation (1). Observons enfin que le théorème 2 nous donne la structure générale de toutes les solutions non triviales et non micropériodiques, mais le théorème 4 nous donne seulement un certain ensemble des solutions non triviales et micropériodiques, n’épuisant pas nécessairement l’ensemble de toutes les solutions de ce type. Le problème de la forme de toutes les solutions mesurables reste aussi ouvert.

Bibliographie [1] J. Aczél, Beiträge zur Theorie der geometrischen Objekte III–IV. Acta Math. Acad. Sci. Hungar. 8 (1957), 19–52. [2] C. Burstin, Über eine spezielle Klasse reeller periodischer Funktionen. Monatsh. Math. Phys. 26 (1915), 229–262.

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[3] M. Hosszú, Functional equations and algebraic methods in the theory of geometric objects. Publ. Math. Debrecen 5 (1958), 294–329. [4] A. Łomnicki, O wielookresowych funkcjach jednoznacznych zmiennej rzeczywistej. Sprawozd. Tow. Nauk. Warsz. 11 (1918), 808–846. [5] W. Sierpiński, Sur la question de la mesurabilité de la base de M. Hamel. Fund. Math. 1 (1920), 105–111. [6] M. Souslin, Sur un corps non dénombrable de nombres réels. Fund. Math. 4 (1923), 311–315.

Originally published in American Journal of Mathematics LXXXVII (1965), 684–694


A combinatorial problem connected with differential equations with H. Davenport (Cambridge)

1. Let (1)

F (D)f (x) = 0

be a (homogeneous) linear differential equation with constant coefficients, of order d. Suppose that F (D) has real coefficients, and that the roots of F (λ) = 0 are all real though not necessarily distinct. As is well known, any solution of (1) is of the form (2)

f (x) = P1 (x)eλ1 x + . . . + Pk (x)eλk x ,

where λ1 , . . . , λk are the distinct roots of F (λ) = 0 and P1 (x), . . . , Pk (x) are polynomials of degrees at most m1 − 1, . . . , mk − 1, where m1 , . . . , mk are the multiplicities of the roots, so that m1 + . . . + mk = d. Let (3)

f1 (x), . . . , fn (x)

be n distinct (but not necessarily independent) solutions of (1). For each real number x, apart from a finite number of exceptions, there will be just one of the functions (3) which is greater than all the others. We can therefore dissect the real line into N intervals (−∞, x1 ), (x1 , x2 ), . . . , (xN−1 , ∞) such that inside any one of the intervals (xj −1 , xj ) a particular one of the functions (3) is the greatest, and such that this function is not the same for two consecutive intervals. It is almost obvious that N is finite, and a formal proof will be given below. The problem of finding how large N can be, for given d and given n, was proposed to one of us (in a slightly different form) by K. Malanowski. This problem can be made to depend on a purely combinatorial problem, by the following considerations. With each j = 1, 2, . . . , N there is associated the integer i = i(j ) for which fi (x) is the greatest

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of the functions (3) in the interval (xj −1 , xj ). (We write x0 = −∞ and xN = ∞ for convenience.) This defines a sequence of N terms (4)

i(1), i(2), . . . , i(N),

each term being one of 1, 2, . . . , n. This sequence has no two consecutive terms equal, which we may express by saying that it has no immediate repetition. The sequence has the further property that it contains no subsequence of the form (5)

with d + 1 terms and a = b.

a, b, a, b, . . .

For suppose that j1 < j2 < . . . < jd+1 and that i(j1 ) = a,

i(j2 ) = b,

i(j3 ) = a,

....

Then the function fa (x) − fb (x) is positive in (xj1 −1 , xj1 ), negative in (xj2 −1 , xj2 ), and so on. Hence this function has a zero between xj1 and xj2 −1 , another zero between xj2 and xj3 −1 , and so on, making at least d distinct zeros. But fa (x) − fb (x) is itself a function of the type (2), and it is known(1 ) that any such function has at most d − 1 zeros. We are therefore led to the following combinatorial problem: to find the greatest length of a sequence with no immediate repetition, each term of which is one of 1, 2, . . . , n, and which contains no subsequence of the type (5). We shall denote this greatest length (that is, greatest number of terms) by Nd (n). Any upper bound obtained for Nd (n) will be valid for the number N defined earlier in relation to the differential equation (1). We do not know whether the two problems are fully equivalent, though this appears to be the case for a few small values of d and n. The combinatorial problem is plainly equivalent to the problem of the maximum number of intervals for n functions which are continuous but not necessarily of the form (2), and which have the property that any two of them are equal for at most d − 1 values of x. An obvious upper bound for Nd (n) follows from the consideration that the pairs of integers i(j ), i(j + 1),

for j = 1, 2, . . . , N − 1,

can include any given pair i1 , i2 at most d times. Since the number of pairs i1 , i2 with 1 i1 n,

1 i2 n,

i1 = i2

is n(n − 1), it follows that (6)

Nd (n) dn(n − 1) + 1.

The problem of evaluating Nd (n) is trivial when d = 2, for then there is no subsequence a, b, a, and therefore any integer can occur only once. The longest sequences are simply the permutations of 1, 2, . . . , n, and we have (7)

N2 (n) = n. The case d = 3 is also simple. We prove:

(1 )

See, for example, [1], Section V, Problem 75.

M2. A combinatorial problem

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Theorem 1. We have N3 (n) = 2n − 1,

(8)

and two examples of maximal sequences are 1, 2, 3, . . . , n − 1, n, n − 1, . . . , 3, 2, 1; (9) 1, 2, 1, 3, . . . , 1, n − 1, 1, n, 1. For d > 3 the problem becomes much more difficult and appears to change its character. We shall concern ourselves mainly with the behaviour of Nd (n) for fixed d and large n. As regards a lower bound for Nd (n), we prove: Theorem 2. We have (10)

Nd (n) (d 2 − 4d + 3)n − C(d) if d is odd and d > 3,

(11)

Nd (n) (d 2 − 5d + 8)n − C(d) if d is even and d > 4,

where C(d) depends only on d. Also N4 (n) 5n − C. As regards upper bounds, we prove: Theorem 3. We have (12) and, for d > 4, (13)

N4 (n) < 2n(1 + log n),

Nd (n) < An exp B(log n)1/2 ,

where A, B depend only on d and (14)

B = B(d) = 10(d log d)1/2 .

2. Proof of Theorem 1 We give two proofs, based on different principles. Neither of them appears to be capable of extension to the case d > 3. In both proofs, S denotes a sequence of maximal length satisfying the conditions of the problem, that is, having no immediate repetition and containing no subsequence of the form a, b, a, b. We abbreviate N3 (n) to N (n). First proof. We can suppose without loss of generality that the first term of S is 1. We can write S as 1, S1 , 1, S2 , . . . , 1, Sk , (1), where each Sm is a sequence formed from the integers 2, 3, . . . , n, and where the final 1 may or may not occur. The sequences Sm are disjoint; for if an integer x occurred in two

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of them, there would be a subsequence 1, x, 1, x in S. Thus if nm denotes the number of distinct integers in Sm , we have n1 + n2 + . . . + nk n − 1. Since Sm is a segment of S, it satisfies the conditions of the problem, and therefore the number of terms in Sm is at most N (nm ). It follows that N (n) k + 1 + N (n1 ) + . . . + N (nk ). By induction, starting from N (1) = 1, we obtain N (n) k + 1 + (2n1 − 1) + . . . + (2nk − 1) 2n − 1. The fact that the particular sequences (9) have the desired property is obvious, and this proves (8).

Second proof. We begin with an observation, made to us by Mrs. Turán, that there is some one of the integers 1, 2, . . . , n which occurs only once in S. For if a is any integer which occurs twice in S, so that i(j1 ) = a,

i(j2 ) = a,

j1 < j2 ,

there must be some integer b which occurs between, say i(j3 ) = b,

j1 < j3 < j2 .

This integer b cannot occur as i(j ) for j < j1 or j > j2 , for then we should have a subsequence b, a, b, a or a, b, a, b. If b occurs only once we have the result, and if not we can repeat the argument with b instead of a, and this process must terminate. Now suppose, as we may without loss of generality, that n occurs only once in S. If we delete the term n from S, we obtain a sequence whose terms are formed from 1, 2, . . . , n−1 and which has no subsequence of the form a, b, a, b. This sequence may, however, have one immediate repetition, namely if the neighbours of n in S are equal: . . . , x, n , n, n , y, . . . . But this immediate repetition disappears if we delete also one of the two terms n , since x = n and y = n . Hence by deleting at most two terms from S we can obtain an admissible sequence whose terms are formed from 1, 2, . . . , n − 1. It follows that N (n) N (n − 1) + 2, and this again gives (8).


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3. Proof of Theorem 2 Suppose first that d is odd. Let A denote the sequence 1, 2, . . . , n, and let D denote the sequence n − 1, n − 2, . . . , 2. Then the (15)

A, D, A, D, . . . , A, D, 1

(which is symmetrical, in spite of its appearance) satisfies the conditions of the problem, provided each A and D is taken (d − 1)/2 times. For if a < b, the successive pairs a, b in a subsequence a, b, a, b, . . . must have their a’s in different A’s, assuming (as we may) that we take the last occurrence of each a before the corresponding b. Consequently there cannot be (d + 1)/2 such pairs. By symmetry the same holds if a > b. The sequence (15) has length (d − 1)(n − 1) + 1. If d > 3 it can be expanded into a longer sequence, which is still admissible, as follows. We replace each element in A, say the first element 1, by 1, x, 1, x, . . . , 1, x

with d − 3 terms.

Here x is an integer greater than n, and we use the same integer for all the elements of the first A in (15). We do the same with each A and D in (15), but using a different new integer for each of them, and finally we replace the last term 1 in (15) by 1, t, 1, t, . . . , 1, t

to d − 3 terms,

where t is the same new integer as that used to expand the last D. We now have a sequence with n + (d − 1) distinct terms, and of length

(d − 3) (d − 1)(n − 1) + 1 . We shall prove that this sequence satisfies the conditions of the problem, and it will follow that Nd (n + d − 1) (d − 1)(d − 3)(n − 1) + (d − 3), which gives (10). We have to prove that the expanded sequence contains no subsequence a, b, a, b, . . . with d +1 terms. No further proof is needed if a n and b n, since then the subsequence is a subsequence of (15). The result is obviously true if a > n and b > n, that is, if a and b both belong to the set x, y, . . . of additional integers, for then there is no subsequence of the form x, y, x. Thus we can suppose that either a n and b > n or a > n and b n, and it will be enough to treat the former case. We replace b by y for ease of comparison with the construction. In any subsequence (16)

a, y, a, y, . . . , a, y,

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all the occurrences of y must be in the expansion of the same A or D in (15), or possibly in that of the final D and 1. Except for the first y in (16), the a’s which precede each y are in that same A or D. The number of y’s is therefore at most 21 (d − 3) + 1. Hence the length of the subsequence (16) is at most d − 1, and this, when we allow for the possible occurrence of another a after (16), means a total length of at most d. Hence the expanded sequence has the desired property. Suppose now that d is even. We start from the sequence A, D, A, D, . . . , A,

(17)

where A occurs in (17) has length d, or indeed only d − 1 if a > b. We expand (17) by replacing each element a in the first A by

1 1 2 d times and D occurs 2 d−1 times. The longest subsequence a, b, a, b, . . .

a, x, a, x, . . . , a, x

to d − 2 terms,

where x is an integer greater than n. We apply the same treatment to the last A, using a different integer greater than n. We also expand the intermediate A’s and D’s, but here we replace each element a by a, x, a, x, . . . , a, x

to d − 4 terms,

again using a different integer x for each A and D. It can be proved, on the same lines as before, that the expanded sequence contains no subsequence a, b, a, b, . . . of d + 1 terms. The number of distinct terms in the expanded sequence is n + d − 1, and the length is > 2(d − 2)n + (d − 4)(d − 3)(n − 2) = (d 2 − 5d + 8)n − 2(d − 3)(d − 4). Hence Nd (n + d − 1) (d 2 − 5d + 8)n − 2(d − 3)(d − 4), and this gives (11). If d = 4 we do not expand the intermediate A’s and D’s, and get N4 (n) 5n − C.

4. Proof of (12). Let S be a sequence of maximal length for d = 4, this length being N4 (n). Let k(a) denote the number of times that a occurs in S, for a = 1, 2, . . . , n. Then (18)

n

k(a) = N4 (n).

a=1

If we delete a wherever it occurs in S, we obtain a sequence formed from the n − 1 integers other than a, and this sequence has no subsequence a, b, a, b, . . . of length greater than 4. But it may have immediate repetitions. To remove these, we must delete not only each occurrence of a but also one of the neighbours of a whenever these two neighbours are equal, as in the second proof of Theorem 1.


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We now prove, for any a, that there are at most two occurrences of a, namely the first and the last, which can have equal neighbours. This is immediate, for in the contrary case we should have . . . , a, . . . , x, a, x, . . . , a, . . . , containing a subsequence a, x, a, x, a of 5 terms. It follows that by deleting k(a) + 2 elements from S we can obtain an admissible sequence formed from n − 1 distinct integers. Hence N4 (n) N4 (n − 1) + k(a) + 2. Summing for a = 1, . . . , n and using (18), we obtain nN4 (n) nN4 (n − 1) + N4 (n) + 2n. This can be written N4 (n) N4 (n − 1) 2 − . n n−1 n−1 Writing down a series of such equations and adding them, and noting that N4 (2) = 4, we obtain 7 n−1 1 1 1 N4 (n) t −1 dt = 2 log(n − 1). −22 + + ... + a.

Proof. We take the first term of S to be 1, the second term to be 2, the next term other than 1 and 2 to be 3, and so on, numbering the terms in the order of their first appearance in S. Plainly (i) holds.

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Suppose S has a subsequence of the form (19), where m, a, b satisfy (20). Then before the first term, m, in (19), or possibly coinciding with it, there occurs a term b, since b m. Before this term b there occurs a term a, since a < b. But then there is a subsequence a, b, a, b, . . . to d + 1 terms, contrary to hypothesis. This proves the lemma.

We remark that (ii) implies the original hypothesis that S contains no subsequence a, b, a, b, . . . to d + 1 terms, since such a sequence always contains a sequence of d terms with the first term greater than the second, and this is excluded by (19) with m = b. Proof of (13). For any integer m with 1 < m < n we pick out the first occurrence of m in S and dissect S into S , m, S ,

(21)

so that every term in S is one of 1, 2, . . . , m − 1. We write S as (22)

S1 (1) , a1 , S1 (2) , a1 , . . . , S1 (r1 ) , a1 , S2 (1) , a2 , . . . , ak , Sk (rk ) , ak , T,

where a1 , . . . , ak are all the terms not exceeding m that occur in S , and all the terms of the sequences Si (j ) and T are integers greater than m. Note that the integers a1 , . . . , ak are not necessarily distinct, though ai = ai+1 as a consequence of our choice of notation. The sequence S consists of terms each of which is one of 1, 2, . . . , m − 1, and is an admissible sequence. Hence L(S ) Nd (m − 1),

(23)

where L(S ) denotes the length of S . The sequence a1 , a2 , . . . , ak has each term less than or equal to m and has no immediate repetition. It also contains no subsequence a, b, a, b, . . . of d terms, for this would necessarily contain a similar subsequence of d − 1 terms with the first term less than the second, and this, preceded by m, would contradict (ii) of the lemma. Hence k Nd−1 (m).

(24) The sequence

S1 (1) , S1 (2) , . . . , S1 (r1 ) , S2 (1) , . . . , S2 (r2 ) , . . . , Sk (1) , . . . , Sk (rk ) , T has all its terms greater than m, and is an admissible sequence except for possible immediate repetitions. These occur only when the last term of one of the above sequences is the same as the first term of the next. They can be removed by deleting at most ri terms at the ends of the sequences. Hence (25)

ri k i=1 j =1

L(Si (j ) ) + L(T) Nd (n − m) +

k

ri .

i=1

It remains to estimate ri . For this we consider only the sequences Si (j ) with j > 1. None of them can be empty, since otherwise there would be an immediate repetition of

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some ai in (22). We select from each of these sequences a term xi (j ) . Among the terms xi (2) , xi (3) , . . . , xi (ri ) ,

(26)

for given i, the same integer cannot occur more than 21 d times, since otherwise there would be a subsequence x, ai , x, ai , . . . , x, ai of more than d terms. It follows that the number of distinct integers among (26) is at least 2(ri − 1)/d. Let Xi be a subsequence of (26) containing si distinct terms, where si 2(ri − 1)/d. The sequence X1 , X2 , . . . , Xk

(27)

is admissible for d, except for possible immediate repetitions. Since the terms of each Xi are distinct among themselves, all immediate repetitions can be removed by deleting at most k − 1 terms. Since all the terms in (27) are greater than m, and the total number of terms is si , we have k

si Nd (n − m) + k − 1.

i=1

It follows that 2d

−1

k

(ri − 1) Nd (n − m) + k − 1,

i=1

whence (28)

k

ri < 21 dNd (n − m) + ( 21 d + 1)k.

i=1

By (23), (24), (25), (28) we have L(S) L(S ) + 1 +

k

ri +

ri k i=1 j =1

i=1

Nd (m − 1) + 1 +

L(Si (j ) ) + L(T)

k i=1

ri + Nd (n − m) +

k

ri

i=1

Nd (m − 1) + (d + 1)Nd (n − m) + (d + 2)k Nd (m − 1) + (d + 1)Nd (n − m) + (d + 2)Nd−1 (m). Taking S to be a maximal sequence, we obtain the inductive inequality (29)

Nd (n) Nd (m) + (d + 1)Nd (n − m) + (d + 2)Nd−1 (m).

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Suppose that d 5 and that

) Nd−1 (m) < A1 m exp(B1 log m)

(30) for all m, where

1/2 B1 = 10 (d − 1) log(d − 1)

(31)

and A1 depends only on d. This is a legitimate assumption when d = 4, by (12). Choose A sufficiently large to ensure that the inequality ) (32) Nd (m) < Am exp(B log m), where B = 10(d log d)1/2 ,

(33)

holds for all m n0 , where n0 = n0 (d) will be chosen later (in a manner which does not depend on the choice of A). Suppose also that A > 2(d + 2)A1 .

(34)

Now suppose that n > n0 and that (32) holds for all m < n; we have to prove that it holds for m = n. Define C = B − B1 . Let h be the integer defined by )

(35) h − 1 < n exp −C log n h. We suppose n0 chosen sufficiently large to ensure that 1 < h < n. By (29), Nd (n) Nd (n − h) + (d + 1)Nd (h) + (d + 2)Nd−1 (n − h)

)

) < A(n − h) exp B log n + (d + 1)Ah exp B log h

) + (d + 2)A1 (n − h) exp B1 log n .

√ This will be less than An exp B log n , thus giving the desired conclusion, provided that

)

)

) Ah exp B log n > (d + 1)Ah exp B log h + (d + 2)A1 n exp B1 log n .

√ Since n/ h exp (B − B1 ) log n by (35), it will suffice if ) )

A > (d + 1)A exp −B log n + B log h + (d + 2)A1 . By (34), this will hold if

) )

1 > 2(d + 1) exp −B log n + B log h .

Now, by (35),

( ) ) ) ) log n − log h > log n − log 2n − C log n > 13 C,

provided n0 is sufficiently large. Hence it suffices if BC > 3 log 2(d + 1).


1337

By (31), (33),

1/2 C = B − B1 = 10 (d log d)1/2 − (d − 1) log(d − 1) > 5d −1/2 (log d)1/2 .

Hence BC > 50 log d, and this amply suffices. This completes the proof of (13).

Note added in proof. Since the paper was written we have improved on the results of Theorems 2 and 3.

Reference [1] G. Pólya, G. Szegö, Aufgaben und Lehrsätze aus der Analysis II, zweite Auflage. Springer, Berlin 1954.

Originally published in Demonstratio Mathematica XI (1978), 47–60


An analogue of Harnack’s inequality for discrete superharmonic functions To Professor Stefan Straszewicz on his diamond scientific jubilee

Let f (p) be a harmonic function defined on the plane and positive in the disc D(o, R) = {p : |p| R}, where |p| is the Euclidean distance from the origin o. The classical Harnack’s inequality (see [2], p. 35, Th. 1.18) asserts that 2|p| 2|p| − f (o) f (o) − f (p) f (o). R − |p| R + |p| This inequality has been extended to discrete harmonic functions by S. Verblunsky [8] and R. Duffin [1]. They have proved the existence of an absolute constant A ( 50) such that every function f (p) defined on the integral lattice Z2 satisfying the equation Δf (p) = f (p + e1 ) + f (p − e1 ) + f (p + e2 ) + f (p − e2 ) − 4f (p) = 0, e1 = (1, 0), e2 = (0, 1) and positive in the disc D(o, R) satisfies the inequalities

f (ej ) − f (o) A f (o) (j = 1, 2). R An analogue of Harnack’s inequality for positive superharmonic functions is easily deduced from the well known convexity properties of subharmonic functions. Indeed, let f be a superharmonic function positive in a disc D(o, R) and let us set for r R

B(r) = sup f (o) − f (p) . |p|=r

f (o) − f (p) is a subharmonic function, hence by a well known theorem (see [2], p. 66, Th. 2.13) B(r) is a convex function of log r in the interval 1 r R, i.e. log R − log r log r B(r) B(1) + B(R). log R log R But log r 0 and B(R) f (o). Thus for |p| 1 log |p| f (o). log R The main aim of the present paper is to prove an analogue of Harnack’s inequality for discrete superharmonic functions, i.e. functions f (p) defined on Z2 and satisfying the inequality Δf (p) 0. We formulate it as f (o) − f (p) B(1) +

M3. An analogue of Harnack’s inequality

1339

Theorem 1. Let f (p) be a function on Z2 superharmonic and non-negative in the disc D(o, R). Then

f (p) − f (o) < π + o(1) f (o), if |p| = 1, R → ∞ (1) 2 log R and (2)

−

log |p| + O(1) log p + O(1) f (o) < f (o) − f (p) < f (o) log(R − |p|) − log p log R

if |p| → ∞ and R |p| or R > |p| for the left hand side and the right hand side of (2) respectively. It will be clear from Lemma 3 below that the inequalities (1) and (2) are best possible or nearly best possible. It follows from the theorem that all functions superharmonic and positive on Z2 are constants. This is known and apparently proved for the first time in a more general context by Kemeny and Snell [3]. Instead of the lattice Z2 one can consider other lattices or more generally networks. From the results on electric currents in networks due to Nash-Williams [6] one obtains the following theorem. Theorem 2. Let L be a regular lattice on the plane (triangular, square or hexagonal) with o ∈ L and the minimal distance 1. If f (p) defined on L satisfies

f (q) − f (p) 0 for all p ∈ L q∈L,|q−p|=1

and f (p) 0 for |p| R then for |p| = 1

f (p) − f (o) 2 + o(1) . log R In particular if f (p) 0 for all p ∈ L, f (p) is constant. Theorem 3 related directly to the work of Nash-Williams requires more notation and therefore, its formulation is postponed. The present paper has originated in a problem proposed at the XXVIII Polish Mathematical Olympiad, which requires a proof of the last statement of Theorem 2 with L replaced by Z. I thank Professor Z. Ciesielski, Dr K. Malanowski, Professor W. M. Schmidt, Dr M. Skwarczyński and Professor E. Wirsing for their valuable suggestions. Let G be a locally finite graph, i.e. a set of points and lines joining some of these points such that every point is joined to only finitely many others (and none is joined to itself). Let c be a function with positive real values defined on the lines of G. The pair [G, c] is called an electric network. For a set V of points of G let V = V ∪ {q ∈ G : ∃p ∈ V pq ∈ G}. V is called connected in G if for any two points p, q ∈ V there exists a sequence of points pi ∈ V such that p0 = p, pn = q and pi pi+1 ∈ G. We shall call a function f (p) defined on V c-superharmonic on V if for all p ∈ V

cpq f (q) − f (p) 0. Δc f (p) = pq∈G

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It is convenient to denote the set of points of G by V (G), the set of lines by E(G) and / G; αp = cpq . put cpq = 0 if pq ∈ q∈G

Lemma 1. If V is a finite set of points connected in G and a function h(p) is c-superharmonic on V then either V = V and min h(p) > min h(p) p∈V \V

p∈V

or h(p) is constant on V . Moreover, if h(p) 0 for all p ∈ V then for any two points p ∈ V , q ∈ V h(q) a(p, q)h(p), where a(p, q) is independent of h. Proof. Let for any two points p, q, where p ∈ V , q ∈ V , v(p, q) be the minimal length n of a sequence of points p0 , p1 , . . . , pn−1 ∈ V such that p0 = p, pn = q, pi pi+1 ∈ G.

(3) Let further

a(p, q) = min

(4)

αp0 αp1 · · · αpn−1 , cp0 p1 cp1 p2 · · · cpn−1 pn

where an empty product is 1 and the minimum is taken over all sequences satisfying (3). Finally, let m = min h(p). We shall show by induction on v(p, q) the following two p∈V

assertions, which clearly imply the lemma. A. B.

If h(p) = m, then h(q) = m. If h(V ) ⊂ [0, ∞], then h(q) a(p, q)h(p).

If v(p, q) = 0 both A and B are obvious. Assume that they are true if v(p, q) < n and let v(p, q) = n. Take any sequence pi satisfying (3). From the inductive assumption applied with q = pn−1 we get: h(p) = m

if

c

h(pn−1 ) = m;

h(V ) ⊂ [0, ∞] then h(pn−1 ) a(p, pn−1 )h(p).

if

(5)

then

Now, from the inequalities c > 0 on E(G) and

Δc h(pn−1 ) = cpn−1 r h(r) − h(pn−1 ) 0 r∈V

we get h(q) = m; αpn−1 h(V ) ⊂ [0, ∞] then h(q) h(pn−1 ). cpn−1 q if

(6)

if

h(p) = m

then

The inductive proof for A is complete, B follows by comparison of (4), (5) and (6).

1341


Lemma 2. Let V = V be a finite set of points connected in G and o ∈ V . A function f (p) c-superharmonic on V and non-negative for all p ∈ V satisfies for all these p the inequality f (p)

(7)

g(p, V ) f (o), g(o, V )

where g(p, V ) is the unique function defined on V such that g(p, V ) = 0 if p ∈ V \ V −1 if p = o Δc g(p, V ) = 0 if p ∈ V , p = o.

(8) (9) c

Proof. A function h defined on the set V can be regarded as a point in N -dimensional Euclidean space, where N is the cardinality of V . The set S of all functions h satisfying h(o) = 1, h(p) 0 for p ∈ V , Δc h(p) 0 for p ∈ V is closed. It is also bounded since by Lemma 1 we have h(p) a(o, p)h(o). Therefore S is compact and for any p0 ∈ V the functional h(p0 ) assumes in S its minimum m. The set S0 = {h ∈ S : h(p0 ) = m} is again compact hence the functional h(p) p∈V

assumes in S0 its minimum. Let h0 ∈ S0 be a function for which the minimum is assumed. We assert that it satisfies the conditions (10) (11)

h0 (p) = 0

p ∈V \V

if

Δc h0 (o) < 0, Δc h0 (p) = 0

for

p ∈ V , p = o.

Indeed, if h0 (p1 ) = 0 for a p1 ∈ V \ V , setting h0 (p) for p = p1 h1 (p) = 0 for p = p1 we find that h1 ∈ S, h1 (p0 ) h0 (p0 ) and h1 (p) < h0 (p), (12) p∈V

p∈V

contrary to the definition of h0 . Secondly if Δc h0 (p1 ) < 0 for p1 ∈ V , p1 = o, we set ⎧ ⎨h0 (p) h1 (p) = h (p ) + α −1 Δ h (p ) = α −1 c h (q) pq 0 p1 c 0 1 p1 ⎩ 0 1 q∈G

Again h1 ∈ S, h1 (p0 ) h0 (p0 ) and (12) holds.

for p = p1 for p = p1 .

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Finally if Δc h0 (o) = 0 we infer from Lemma 1 applied to −h0 (p) that max h0 (p) max h0 (p) = 0 p∈V

p∈V \V

contrary to h0 (o) = 1. It follows from (10) and (11) that the function g(p, V ) =

(13)

h0 (p) |Δc h0 (o)|

satisfies the conditions (8) and (9). If g (p) is any function satisfying the same conditions we have g (p) − g(p, V ) = 0 Δc g (p) − g(p, V ) = 0

if

p ∈ V \ V,

if

p ∈ V,

g (p) − g(p, V )

hence applying Lemma 1 to and to g(p, V ) − g (p) we get

max g (p) − g(p, V ) 0, g (p) = g(p, V ). p∈V

Thus g(p, V ) is unique and taking p = o in (11) we get independently of p0 (14)

1

Δc h0 (o) = h0 (o) = . g(o, V ) g(o, V )

If f (o) = 0 the inequality (7) is trivially satisfied. If, on the other hand, f (o) > 0 then f (p)/f (o) ∈ S and for any p ∈ V we have by (13) and (14) f (p) g(p, V ) h0 (p) = . f (o) g(o, V )

Remark. Lemma 2 and at least a part of Lemma 1 can be deduced from the Maximum Principle and the Principle of Domination of the transient potential theory for Markov chains (see [4], Corollary 8-44 and Theorem 8-45). The proof obtained in this way would be about twice shorter than one given above but far from self-contained. The existence and uniqueness of g(p, V ) has been proved first by Nash-Williams (see [6], Lemma 4 and 9). Lemma 3. Let G be the two-dimensional Euclidean lattice graph, c = 1 on E(G) and VR the set of all points of G contained in the disc D(o, R − 1). Then for any p ∈ VR \ {o} 1 1 1 (15) g(p, VR ) = . log R − log |p| + O +O 2π R |p|2 Moreover c

(16)

g(o, VR ) − g(p, VR ) =

1 4

if |p| = 1.

Proof. McCrea and Whipple [5] and independently Stöhr [7] found a function φ(p) on Z2


with the following properties: φ(o) = 0,

Δφ(p) = φ(p) =

1 0

1343

for p = o for p = o,

1 3 1 1 , log |p| + log 2 + C+O 2π 4π 2π |p|2

where C is Euler’s constant (see [7], p. 342, Theorem 1). Let us consider the function (17)

h(p) = g(p, VR ) + φ(p) −

1 3 1 log R − log 2 − C. 2π 4π 2π

For p ∈ VR we have Δh(p) = 0. If p ∈ V R \ VR we find R − 1 < |p| R and

1 3 1

1

log R − log 2 − C = O . (18) |h(p)| = φ(p) − 2π 4π 2π R 1 for all p ∈ V R . Hence by Lemma 1 applied to h(p) and to −h(p) we have h(p) = O R (15) follows now from (17) and (18). In order to prove (16) let us observe that the graph G and the set VR are symmetric with respect to the coordinate axes, hence g(p, VR ) must exhibit the same symmetry. Thus

g(p, VR ) has the same value for p = ±e1 , ±e2 and (16) follows from Δg(o, VR ) = −1. Proof of Theorem 1. Let G be the graph described in Lemma 3, c = 1 on E(G). A function f (p) superharmonic and non-negative in D(o, R) is c-superharmonic in VR , and nonnegative in V R . If |p| = 1 we get from Lemmata 2 and 3 f (o) − f (p)

π + o(1) g(o, VR ) − g(p, VR ) = f (o). g(o, VR ) 2 log R

If |p| → ∞ and R |p| we have similarly f (o) − f (p)

log |p| + O(1) log |p| + O(1) f (o) = f (o). log R + O(1) log R

In order to estimate f (o) − f (p) from below let us shift the roles of points o and p. Since D(p, R − |p|) ⊂ D(o, R) we get π + o(1) f (p) 2 log(R − 1) log |p| + O(1) f (p) − f (o) f (p) log(R − |p|) f (p) − f (o)

c

if |p| = 1, if |p| → ∞ and R − |p| |p|.

(1) and (2) follow now by simple algebraic transformations.

In order to prove Theorem 2 we need a lemma due to Nash-Williams [6]. A finite sequence Y0 , Y1 , . . . , Yn (n 1) of disjoint subsets of V (G) is called a constriction of G if Y0 ∪ Y1 ∪ . . . ∪ Yn = V (G) and p ∈ Yj , q ∈ Yk , pq ∈ G implies |j − k| 1. Using this notion we can state

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Lemma 4. Let V be a finite set of points connected in G, o ∈ V . For any constriction C = Y0 , Y1 , . . . , Yn of G such that o ∈ Y0 , V (G) \ V ⊂ Yn we have n

g(o, V )

nk (C)−1 ,

k=1

where

nk (C) =

cpq .

p∈Yk−1 q∈Yk

g(p, V ) Proof. Let S = G, c, o, V (G)\V . In the language of [6] φ = is an S-admissible g(o, V ) G-potential and fcφ (o) = g(o, V )−1 . The lemma follows now from Lemma 5 of [6].

Proof of Theorem 2. Let G be a graph consisting of all points of L and all lines between points of distance 1, let c = 1 on E(G). Take for VR the set of all points of G contained in D(o, R − 1). A function satisfying the conditions of the theorem is c-superharmonic on VR and non-negative on V R . Since G and VR are symmetric with respect to the α0 lines joining o to the nearest points (α0 = 3, 4 or 6), the function g(p, V ) must exhibit the same symmetry. Thus g(p, VR ) takes the same value for all p with |p| = 1 and Δc g(o, VR ) = −1 implies for these p g(o, VR ) − g(p, VR ) = α0−1 . Hence if |p| = 1 we obtain from Lemma 2 f (o) − f (p) α0−1 g(o, VR )f (o).

(19) c

Consider now the following constriction C of G: Yk = {p ∈ G : d(o, p) = k} (k < [R]), Y[R] = {p ∈ G : d(o, p) [R]}, where d(o, p) is the minimal number n such that for suitable pi ∈ G: p0 = o, pn = p, pi pi+1 ∈ G. An easy geometric argument shows that ⎧ ⎪ if α0 = 3, k = 1, ⎨3 nk (C) = 6k − 6 if α0 = 3, k > 1, ⎪ ⎩ α0 (2k − 1) if α0 = 4 or 6. Thus in any case nk (C) α0 (2k − 1) and 7 [R]+1 [R] −1 −1 nk (C) α0 (20) k=1

1

dt log(2[R] + 1) . = 2t − 1 2α0

It follows from (19), (20) and Lemma 4 that (21)

f (o) − f (p)

2 f (o). log(2[R] + 1)

Changing the roles of o and p and observing that D(p, R − 1) ⊂ D(o, R) we obtain 2 (22) f (p) − f (o) f (p). log(2[R] − 1)

1345


The first assertion of Theorem 2 follows from (21), (22) and Lemma 4. Passing with R to infinity we get f (p) = f (o) whenever d(o, p) = |p| = 1. By induction on d(o, p) we obtain f (p) = f (o) for all p.

Our last theorem is a counterpart of Lemma 4. Theorem 3. Let V be a finite set of points connected in G, o ∈ V . For any constriction C = Y0 , Y1 , . . . , Yn of G such that Y0 = {o}, Yn = V (G) \ V we have (23)

g(o, V )

n

m0 (C)−1 m1 (C)−1 · · · mk−1 (C)−1 ,

k=1

where

m0 (C) = α0 ,

mk (C) = min

cpq

q∈Yk+1

p∈Yk

cpq

(k 1),

q∈Yk−1

the minimum is taken over fractions with non-zero denominator and if mk (C) = 0, we take m−1 k (C) = ∞. Proof. If any of the numbers mk (C) (k < n) is zero the bound is trivial. Therefore, assume that mk = mk (C) > 0 for k < n and set s n−1 1 −1 r = 1 − 1 + (1 r n). mk s=r k=r

We have 1 > r > 0 (r < n), n = 0, moreover it is easily verified that for r < n (24)

mr (1 − r+1 ) = r−1 − 1.

Let us now define for p ∈ Yk f (p) =

k

r .

r=1

We have f (o) = 1, Δc f (o) = α0 (1 − 1) < 0 and for p ∈ Yk (n > k > 0) (25)

Δc f (p) = ap (k−1 − 1) + bp (k+1 − 1), f (p)

where ap =

q∈Yk−1

cpq ,

bp =

cpq .

q∈Yk+1

It follows from (24), (25) and the definition of mk that for k < n Δc f (p) = (1 − k+1 )(ap mk − bp ) 0. f (p)

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Thus f (p) is c-superharmonic on V and since it is also non-negative we have by Lemma 2 g(o, V ) − g(p, V ) . g(o, V ) and summing over all p adjacent to o we get f (o) − f (p)

Multiplying by cop

−Δc f (o) −Δc g(o, V )g(o, V )−1 = g(o, V )−1 , hence

−1 −1 g(o, V ) − Δc f (o) = m−1 0 (1 − 1 ) s s n−1 n−1 −1 −1 = m0 1 + mk = m−1 k . s=1 k=1

s=0 k=0

Furthermore one can show that the equality sign holds in (23) provided g(p, V ) is constant on Yk and deduce from it a formula for g(o, V ) analogous to Theorem 1 of [6]. We shall however not pursue the matter.

References [1] R. J. Duffin, Discrete potential theory. Duke Math. J. 20 (1953), 233–251. [2] W. K. Hayman, P. B. Kennedy, Subharmonic Functions, vol. I. Academic Press, London–New York 1976. [3] J. G. Kemeny, J. L. Snell, Potentials for denumerable Markov chains. J. Math. Anal. Appl. 3 (1961), 196–260. [4] J. G. Kemeny, J. L. Snell, A. W. Knapp, Denumerable Markov Chains. Springer, New York 1976. [5] W. H. McCrea, F. J. W. Whipple, Random paths in two and three dimensions. Proc. Roy. Soc. Edinburgh 60 (1940), 281–298. [6] C. St. J. A. Nash-Williams, Random walk and electric currents in networks. Proc. Cambridge Philos. Soc. 55 (1959), 181–194. [7] A. Stöhr, Über einige lineare partielle Differenzengleichungen mit konstanten Koeffizienten III. Math. Nachr. 3 (1950), 330–357. [8] S. Verblunsky, Sur les fonctions préharmoniques. Bull. Sci. Math. (2) 73 (1949), 148–152.

Originally published in Colloquium Mathematicum XXXVIII (1978), 319–321


An inequality for determinants with real entries

In memory of Bohuslav Diviš

The aim of this note is to prove the following Theorem. For every matrix A = (aij )i,j n with real entries we have the inequality n n n |det A| (1) max aij , − aij . j =1 aij >0

i=1

j =1 aij 1, then ai1 a12 = ai2 (6) ai2 − a11 and the (i − 1)-st row of B contains exactly two non-zero elements, namely the numbers of opposite signs: aiki and aili . If ki = 1 and li > 2, then the (i − 1)-st row of B contains also exactly two non-zero elements of opposite signs, namely ai2 − (ai1 /a11 )a12 and aili . Besides, by (4),

a

ai1

i1

a12 =

a12 |ai1 |. (7)

ai2 − a11 a11 Finally, if ki = 1 and li = 2, then the (i − 1)-st row of B contains only one non-zero element, namely ai2 − (ai1 /a11 )a12 . Since ai1 a12 > 0, ai2 − a11 we have (8)

a

ai1

i1

a12 < max |ai2 |,

a12 max |ai2 |, |ai1 | .

ai2 − a11 a11

By the inductive assumption, (6), (7), and (8), we have (9)

|det B|

n i=2

max |aij |, j

and (3) follows from (5) and (9). Thus (1) is true for all matrices A satisfying (2). In the general case we proceed by induction with respect to the number of non-zero elements of A.

1349

M4. An inequality for determinants with real entries

If this number is 0, then det A = 0 and (1) holds. Assume that (1) is true for all square matrices with less than N non-zero elements and consider a square matrix A with exactly N non-zero elements. If A satisfies (2), then (1) holds. If (2) is not fulfilled, then for a certain i0 there exist j1 and j2 such that j1 = j2 ,

ai0 j1 ai0 j2 > 0.

Assuming, without loss of generality, that i0 = 1, j1 = 1, and j2 = 2, we have

a11 + a12 0 . . . a1n

a21 a22 . . . a2n

a11

det A =

.. .. ..

.. a11 + a12

. . . .

an1 an2 . . . ann

0 a11 + a12 . . . a1n

a21 a22 . . . a2n

a12

+

. .. .. . .. a11 + a12 .. . . .

an1 an2 . . . ann

The inductive assumption applies to the determinants on the right hand side, since the relevant matrices contain only N − 1 non-zero elements. Hence

|det A|

n n n a11

a12

max aij , − aij

+

a11 + a12 a11 + a12 i=1

j =1 aij >0

=

n i=1

j =1 aij 0

aij , −

n

aij

j =1 aij B(1 − 6n−1/3 ).

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M5. Norms of squares of polynomials

It follows that B = lim A(n) = sup A(n). n→∞

n

The determination of B appears to be difficult. Theorem 2. 4/π B < 1.7373. A slightly better upper bound will in fact be proved. We should mention that Ben Green [1] showed in effect that (|f |1 /|f |2 )2 < 7/4 for f ∈ F, where |f |2 denotes the L2 -norm. In fact he has the slightly better bound 1.74998. . . Since |f |22 |f |1 |f |∞ , this yields B < 1.74998 . . . , which is only slightly weaker than the upper bound in Theorem 2. However, Green’s result is valid without the assumption g 0. On the other hand, Prof. Stanisław Kwapień (private communication) proved that

A(n) B 1 − 3(B/4)1/3 n−1/3 .

2. Assertions (i), (ii) of Theorem 1 When R is a polynomial or power series a0 + a1 X + . . . , set |R|∞ for the maximum modulus of its coefficients. For such R, and for a polynomial S, |RS|∞ |R|∞ |S|1 .

(2.1) When P ∈ P(n), say P =

Q2 ,

set

K = (1 + X + . . . + X l−1 )Q(X l ) Q

and

K= Q K2 . P

K l − 1 + l(n − 1) = ln − 1, so that P K ∈ P(ln). Further |Q| K 1 = l|Q|1 , Then deg Q yielding K 2 = l 2 |Q|2 = l 2 |P |1 . K|1 = |Q| |P 1 1

(2.2)

For polynomials or series R = a0 + a1 X + . . . , S = b0 + b1 X + . . . with non-negative coefficients, write R $ S if ai bi (i = 0, 1, . . . ). Then Q(Xl )2 $ |Q2 |∞ (1 + X l + X 2l + . . . ) = |P |∞ (1 + X l + X 2l + . . . ). Therefore K = (1 + X + . . . + X l−1 )2 Q(X l )2 P $ |P |∞ (1 + X l + X 2l + . . . )(1 + X + . . . + X l−1 )2 = |P |∞ (1 + X + X 2 + . . . )(1 + X + . . . + X l−1 ). K|∞ |P |∞ l. Together with (2.2) this yields n−1 |P |1 /|P |∞ Now (2.1) gives |P K|1 /|P K|∞ A(nl). Assertion (i) follows. (ln)−1 |P

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We now turn to (ii). Let P ∈ P(n) be given, say P = Q2 with Q = a0 + a1 X + . . . + an−1 X n−1 . Let g be the function with support in [0, 1) having g(x) = ai

for i/n x < (i + 1)/n

(i = 0, 1, . . . , n − 1),

i.e., for %nx& = i. Then |g|1 = n−1 |Q|1 , so that f = g ∗ g has |f |1 = n−2 |Q2 |1 = n−2 |P |1 .

(2.3)

Let x be given. The interval I = [0, 1) is the disjoint union of the intervals (possibly empty) Ii,j (x) (i = 0, 1, . . . , n − 1; j ∈ Z) consisting of numbers y with %ny& = i,

%n(x − y)& = j − i.

When y ∈ Ii,j (x) and 0 i < n, then y + (i − i)/n ∈ Ii ,j (x). Therefore Ii,j (x) has length independent of i; denote this length by Lj (x). Clearly Lj (x) = 0 unless j = %nx& or %nx − 1&. We have n−1

1=

(2.4)

Lj (x) = n

i=0 j

For y ∈ Ii,j (x) with 0 i < n,

ai aj −i 0

Therefore

7 g(y)g(x − y) dy =

(2.5) Ii,j (x)

Lj (x).

j

g(y)g(x − y) =

when j − n < i j, otherwise.

ai aj −i 0

when j − n < i j, otherwise.

Now j

ai aj −i = bj |P |∞ ,

i=0

where bj is the coefficient of X j in P . Taking the sum of (2.5) over i = 0, 1, . . . , n − 1 and j ∈ Z, and observing (2.4), we obtain 7 f (x) = g(y)g(x − y) dy |P |∞ Lj (x) = |P |∞ /n. j

Therefore |f |∞ |P |∞ /n, so that in conjunction with (2.3), n−1 |P |1 /|P |∞ |f |1 /|f |∞ B. Assertion (ii) follows.

1353


3. Assertion (iii) of Theorem 1 Pick f ∈ F with |f |1 /|f |∞ close to B. We may suppose that |f |∞ = 1 and |f |1 is close to B, in particular that |f |1 1. Say f = g ∗ g. Then for r < s, 77 7 s 2 (3.1) g(x) dx g(x)g(y) dx dy r

2rx+y2s

7

7

2s

=

dz

7 g(y)g(z − y) dy =

2r

2s

f (z) dz 2(s − r).

2r

Cy √ Setting G(y) = 0 g(y) dy, so that G(y) 2y, and using partial integration, we obtain 7 δ 7 δ 7 δ (δ − x)g(x) dx = G(y) dy (2y)1/2 dy < δ 3/2 . (3.2) 0

Similarly,

0

7

1

0

δ − (1 − x) g(x) dx < δ 3/2 .

1−δ

With c ∈ 21 Z in 1 c (n − 1)/2 to be determined later, set 7 n (i+1/2+c)/n g(x) dx (0 i < n) ai = 2c (i+1/2−c)/n and Q(X) =

n−1

ai X i .

i=0

Then |Q|1 =

n−1 i=0

ai =

n 2c

7

1

ν(x)g(x) dx 0

where ν(x) is the number of integers i, 0 i < n, having (i + 1/2 − c)/n x (i + 1/2 + c)/n. Then ν(x) is the number of integers i having max(0, nx − 1/2 − c) i min(n − 1, nx − 1/2 + c). When (c + 1/2)/n x 1 − (c + 1/2)/n, this becomes the interval nx − 1/2 − c i nx − 1/2 + c, so that ν(x) 2c, as c ∈ 21 Z. When x < (c + 1/2)/n, the interval becomes 0 i nx − 1/2 + c, and ν(x) nx + c − 1/2 = 2c − (c + 1/2 − nx). On the other hand when x > 1 − (c + 1/2)/n, then ν(x) 2c − (c + 1/2 − n(1 − x)). Therefore 7 1 7 n (c+1/2)/n g(x) dx − (c + 1/2 − nx)g(x) dx (3.3) |Q|1 n 2c 0 0 7 1 n (c + 1/2 − n(1 − x))g(x) dx. − 2c 1−(c+1/2)/n

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M. Other papers

Applying (3.2) with δ = (c + 1/2)/n we obtain 7 n (c+1/2)/n (c + 1/2 − nx)g(x) dx 2c 0
B(1 − 6n−1/3 ).

4. The lower bound in Theorem 2 Set f = g ∗ g where g(x) = x −1/2 in 0 < x < 1, and g(x) = 0 otherwise. Then f ∈ F, and |f |1 = |g|21 = 4. For 0 < z 2, 7 f (z) = (z − x)−1/2 x −1/2 dx, with the range of integration max(0, z − 1) x min(1, z). Setting x = y 2 z we obtain 7 dy , f (z) = 2 (1 − y 2 )1/2 the integration being over y 0 with 1 − 1/z y 2 min(1/z, 1). When 0 < z 1, this range is 0 y 1, so that f (z) = π , whereas in 1 < z 2 the range is smaller, and f (z) < π . We may conclude that |f |∞ = π , and B |f |1 /|f |∞ = 4/π .

5. The upper bound B 7/4 The upper bound of Theorem 2 will be established in three stages. Here we will show that B 7/4 = 1.75, and in the following stages we will prove that B 7/4 − 1/80 = 1.7375, then that B 1.7373. Our problem is invariant under translations. To exhibit symmetry, we therefore redefine F to consist of functions f = g ∗ g with g non-zero, non-negative and integrable, with support in [−1/2, 1/2], so that f has support in [−1, 1]. We will suppose throughout that f ∈ F with |f |∞ = 1, and we will give upper bounds for |f |1 . Lemma 1.

7

1

1/2

f (z)f (−z) dz 1/4.

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M. Other papers

As a consequence of this lemma, 7 7 1 7 1

|f |1 = f (z) + f (−z) dz 1 + f (z) dz = −1

7

1+

0

f (z) + f (−z) dz

1/2

1

1

1 + f (z)f (−z) dz

1/2

3 1 7 + = , 2 4 4

so that indeed B 7/4. Proof of Lemma 1.

7

f (z) = (g ∗ g)(z) =

(5.1)

7 g(x)g(z − x) dx = 2

g(x)g(y) dx.

x+y=z xy

(It is to exhibit symmetry that we write y for z − x.) Similarly 7 (5.2) f (−z) = 2 g(u)g(v) du. u+v=−z uv

Here x, y, u, v may be restricted to lie in [−1/2, 1/2]. When δ 0 and z 1/2 − δ, then x = z − y 1/2 − δ − 1/2 = −δ, also v = −u − z 1/2 − 1/2 + δ = δ, so that u v δ, We obtain 7

1

7 f (z)f (−z) dz 4

1/2−δ

−δ x y. 77

1

dz 1/2−δ

g(x)g(y)g(u)g(v) dx du. uvδ −δxy x+y=z u+v=−z

In this integral u −z/2 −1/4 + δ/2, and y z/2 1/4 − δ/2. Setting w = u + y = −x − v we have w u + 1/2 1/4 + δ/2, and in fact |w| 1/4 + δ/2. Replacing the variables x, u, z in the above integral by x, y = z − x, w = u + z − x, we obtain the bound 77 7 1/4+δ/2 dw g(x)g(y)g(u)g(v) dx dy. (5.3) 4 −1/4−δ/2

y+u=w x+v=−w −δxy uvδ x+y1/2−δ

Let us now take δ = 0. In this case 7 1 7 1/4 f (z)f (−z) dz 4 dw 1/2

−1/4

77 g(x)g(y)g(u)g(v) dx dy. x+v=−w y+u=w uv0xy

1357


Interchanging the rôles of the variables x, y, and as a result those of u, v, and replacing w by −w, we get an integral as before, except that the region u v 0 x y is replaced by the region v u 0 y x. These regions are essentially disjoint, and are contained in u 0 y, v 0 x. We therefore obtain 7 2

−1/4

7 =2

1/4

1/4

−1/4

7

dw

7

g(x)g(v) dx x+v=−w v0x

g(y)g(u) dy y+u=w u0y

dw fK(w)fK(−w)

with fK(w) =

(5.4)

7 g(y)g(u) dy. y+u=w u0y

Thus 7

1

(5.5)

7 f (z)f (−z) dz 4

1/2

1/4

fK(w)fK(−w) dw.

0

It is clear from (5.1) and (5.4) that fK(w) f (w)/2 1/2, so that we obtain 1/4, and Lemma 1 follows.

6. The upper bound B 1.7375

With f = g ∗ g as above, and ε = ±1, set 7

77

1/8

Iε =

g(εx) dx, 0

Jε =

g(y)g(u) dy du. εy>0, εu>0 ε(y+u)1/4

Lemma 2. C1 (i) 1/2 f (z)f (−z) dz 1/4 − Jε . (ii) For 0 δ 1/6, 7

7

1 δ f (z)f (−z) dz + + 4 2 1/2−δ 1

2

δ −δ

g(x) dx

.

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M. Other papers

As a consequence, 7

1

(6.1) |f |1 =

7

f (z) + f (−z) dz =

0

1/2−δ

0

7

+

1

1/2−δ

1 + f (z)f (−z) dz

1

1 − 2δ +

7

1/2−δ 7 1

3 −δ+ 2

7

7 δ f (z)f (−z) dz − + 4 2 1/2−δ

2

δ −δ

g(x) dx

.

Setting δ = 1/8 we obtain |f |1

(6.2)

27 27 + (I1 + I−1 )2 + 4M 2 16 16

with M = max(I1 , I−1 ). On the other hand, by (i), 7 1 3 7 7 (6.3) |f |1 + f (z)f (−z) dz − max Jε − M 2 . ε=±1 2 4 4 1/2 In conjunction with (6.2) this gives |f |1 7/4 − 1/80 = 1.7375, so that indeed B 1.7375. Proof of Lemma 2. When w > 0, we cannot have y + u = w and u y < 0. Therefore fK(w) as given by (5.4) is 7 7 1 1 K f (w) = g(y)g(u) dy − g(y)g(u) dy = f (w) − f5(w) 2 2 y+u=w uy

y+u=w 0uy

with

7

f5(w) =

g(y)g(u) dy. y+u=w y,u0

Now (5.5) yields 7 1 7 f (z)f (−z) dz 1/2

1/4

0

=

=

1 − 4 1 − 4

f (w) − f5(w) f (−w) dw

7

dw 0

77

1/4

1 − f5(w) dw

0

7

1/4

7

g(y)g(u) dy

y+u=w y,u0

g(y)g(u) dy du =

1 − J1 . 4

y,u0 y+u1/4

The bound 1/4 − J−1 is obtained similarly, so that assertion (i) is established.

1359


We will now suppose δ > 0, and we return to the bound (5.3). We first deal with the part where v x in the integral, so that u v x y.

(6.4)

After interchanging the rôles of x and y, and of u and v, and replacing w by −w, the integrand will be the same, but now v u y x.

(6.5)

The interiors of the domains (6.4), (6.5) are disjoint, and are contained in the region with v x and u y, so that this part of (5.3) is 7 1/4+δ/2 7 7 dw g(x)g(v) dx g(y)g(u) dy (6.6) 2 −1/4−δ/2

=

1 2

7

1/4+δ/2

−1/4−δ/2

x+v=−w vx

7

dw f (−w)f (w) =

y+u=w uy 1/4+δ/2

f (w)f (−w) dw 1/4 + δ/2.

0

It remains for us to deal with the part of (5.3) where x v in the integral, so that −δ x v δ. This part is 7 7 7 g(x)g(v) dx g(y)g(u) dy. 4 dw x+v=−w −δxvδ

y+u=w y1/2−δ−x uδ

When 0 < δ 1/6, then y 1/2 − 2δ δ u, and the last integral is 7 g(y)g(u) dy = f (w)/2 1/2. y+u=w uy

Therefore the part in question of (5.3) becomes 7 7 7 7 g(x)g(v) dx = dw 2 dw x+v=−w −δxvδ

g(x)g(v) dx =

x+v=−w −δx,vδ

Together with (6.6) this gives the asserted bound for

C1 1/2−δ

In fact we will show that B 7/4 − 1/80 − ξ < 1.7373

where ξ = 0.000200513 . . . is a root of the transcendental equation F (b(x)/a(x)) = 1/2,

2

δ −δ

f (z)f (−z) dz.

7. The upper bound 1.7373

(7.1)

7

g(x) dx

.

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M. Other papers

√ √ where a(x) = 1/10 − 2x, b(x) = ( 1/20 − x − 1/80 + x)2 /2, and ) ) √ F (x) = x 2 + x + log( x 2 + x + x). The calculation of ξ has kindly been performed by Dr. A. Pokrzywa. We will suppose that f ∈ F, |f |∞ = 1 and |f |1 > 7/4 − 1/80 − ξ,

(7.2)

and we will reach a contradiction, thereby establishing the truth of (7.1), and hence of Theorem 2. Retaining earlier notation we now set a = a(ξ ), u = I1 + I−1 ,

v = |I1 − I−1 |,

m = min(I1 , I−1 ) = (u − v)/2,

and observe that M = max(I1 , I−1 ) = (u+v)/2. Also, u0 , u1 will be the positive numbers with u20 = 1/20 − ξ = a/2,

u21 = 1/20 + 4ξ.

We may suppose that u u0 ,

(7.3)

for otherwise (6.2) yields |f |1 27/16 + u20 = 7/4 − 1/80 − ξ , against (7.2). We further may suppose that u + v u1 ,

(7.4)

for otherwise (6.3) yields |f |1 7/4 − u21 /4 = 7/4 − 1/80 − ξ , contradicting (7.2). As a consequence, 2u2 − m2 /2 = 2u2 − (u − v)2 /8 = 3u2 /2 + u(u + v)/2 − (u + v)2 /8 3u2 /2 + 3u(u + v)/8 15u21 /8 < 1/10 − 2ξ = a, so that 0 = 2u20 − a 2u2 − a < m2 /2.

(7.5) Lemma 3.

7 1 − |f |1 (u2 + v 2 ) + 4 4

7

m2 /2 ) 2u2 −a

dη (η + a)/2 − u √ . 2η

Proof. By (6.1) and (7.2), 1/80 + ξ > δ/2 −

7

2

δ

−δ

g(x) dx

for δ in 0 < δ < 1/6. Setting δ = 1/8 + η with 0 < η < 1/24, this gives 2 7 1/8+η g(x) dx > η/2 + 1/20 − ξ = (η + a)/2, −1/8−η

1361


and

7 G(η) :=

(7.6)

1/8+η

(g(x) + g(−x)) dx >

)

(η + a)/2 − u.

1/8

On the other hand, by (6.3) and (7.2), and since m2 /2 u2 /8 u21 /8 < 1/24 < 1/8, 7 1/4−x 7 1/4 1 1 2 1 2 +ξ > I1 + I−1 + 2 Jε = g(εx) dx g(εy) dy 80 2 2 0 ε=±1 ε=±1 1/8 2 7 7 2 1/4−x 1/8+m /2 1 u + v2 +2 g(εx) dx g(εy) dy 2 2 0 ε=±1 1/8 7 7 2 1/8−η m /2 1 g(ε/8 + εη) dη g(εy) dy. = (u2 + v 2 ) + 4 0 0 ε=±1

By (3.1) with r = 1/8 − η, s = 1/8, 7 7 1/8−η g(εy) dy = Iε − 0

1/8

g(εy) dy Iε −

)

2η m −

) 2η .

1/8−η

Thus 7 m2 /2 )

1 1 + ξ > (u2 + v 2 ) + g(ε/8 + εη) m − 2η dη 80 4 ε=±1 0 7 m2 /2 )

1 g(1/8 + η) + g(−1/8 − η) m − 2η dη. = (u2 + v 2 ) + 4 0 Integrating by parts we represent the last integral as 7

m2 /2

0

dη G(η) √ 2η

7

m2 /2

2u2 −a

dη G(η) √ . 2η

Since m2 /2 < 1/24 we may apply (7.6) to obtain the lemma. Lemma 4. In the domain of points (u, v) with (7.3), (7.4), v 0, the function H (u, v) =

1 2 (u + v 2 ) + 4

7

1 u−v) 2 2( 2 )

2u2 −a

) dη ( (η + a)/2 − u) √ 2η

satisfies H (u, v) H (u0 , u1 − u0 ). Proof. 2H (u, v) =

1 2 (u + v 2 ) + 2

7

1 u−v 2 2( 2 )

2u2 −a

B

) η+a dη − u(u − v) + 2u 4u2 − 2a. η

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Hence 2

(u − v)2 + 8a 1/2 v − u · (u − v)2 4 1 1/2 = v + u − (u − v)2 + 8a . 4

∂H (u, v) =v+u+ ∂v

We claim that this partial derivative is 0 in our domain. For otherwise 16(u + v)2 − ((u−v)2 +8a) > 0, or 15(u+v)2 +4uv−8a > 0. But u + v u1 and 4uv 4u(u1 −u) 4u0 (u1 − u0 ) since u u0 > u1 /2. Therefore 15u21 + 4u0 u1 − 4u20 − 8a > 0. Substituting the values for a, u0 , u1 gives 4u0 u1 1/4 − 80ξ. Squaring, we get 16(1/20 + 4ξ )(1/20 − ξ ) > (1/4 − 80ξ )2 , which is not true. Thus our claim is proven, and H (u, v) H (u, u1 − u).

(7.7) Next,

1 2H (u, u1 − u) = −u2 + u21 + 2 so that d H (u, u1 − u) = −2u + 2 du

7

1 2u−u1 2 2( 2 )

2u2 −a

(2u − u1 )2 + 8a (2u − u1 )2

B

) η+a dη + 2u 4u2 − 2a , η

1/2

2u − u1 · − 2

2u2 2u2 − a

1/2 · 4u

+ 2(4u2 − 2a)1/2 + 8u2 (4u2 − 2a)−1/2 ) 1) = −2u + (2u − u1 )2 + 8a + 2 4u2 − 2a . 2 We claim that this derivative is 0 for u0 u u1 . For otherwise 16u2 (2u−u1 )2 +8a, so that 12u2 + 4uu1 − u21 > 8a. But this entails 15u21 > 8a, i.e., 15(1/20 + 4ξ ) > 4/5 + 16ξ, which is not true. Thus our claim is correct, and H (u, u1 − u) H (u0 , u1 − u0 ),

which together with (7.7) establishes the lemma. It is now easy to arrive at the desired contradiction to (7.2). By Lemmas 3 and 4, 7/4 − |f |1 H (u0 , u1 − u0 ) 1 = (u20 + (u1 − u0 )2 ) + 4

7

1 1 2 2 (u0 − 2 u1 )

2u20 −a

B 1 η+a u0 dη. −√ 2 η 2η


1363

2 Here 2u20 − a = 0 and 21 u0 − 21 u1 = b(ξ ) = b, say, and 7 xB 7 x √ η+a dη dη = aF (x/a), √ = 2x . η 2η 0 0 Therefore 1 a (2u20 − 2u0 u1 + u21 ) + F (b/a) − u0 (u0 − u1 /2) 4 2 a 1 3 a 2 2 = −u0 /2 + u1 /4 + F (b/a) = − + ξ + F (b/a) = 1/80 + ξ, 2 80 2 2 contrary to (7.2).

7/4 − |f |1

Added in proof. Dr. Erik Bajalinov has checked that for n 26 and n = 31, 36, 41, 46, 51: A(n) < 4/π , which suggests that B = 4/π.

Addendum* The following problem equivalent to the problem considered in this paper has been proposed by L. Moser at the Institute in the Theory of Numbers (Boulder, Colorado 1959), see Report of the said Institute, p. 342, Problem 29: C1 Let f (x) 0, f (x) = 0 outside (0, 1), 0 f (x) dx = 1. Let 7 t f (x)f (t − x) dx. g(t) = 0

Find M = min max g(t). Conjecture: M = π/4. f

t

Reference [1] B. Green, The number of squares and Bh [g] sets. Acta Arith. 100 (2001), 365–390.

*

Added in 2006

Unsolved problems and unproved conjectures


Unsolved problems and unproved conjectures proposed by Andrzej Schinzel in the years 1956–2006 arranged chronologically

1 (conjecture) For every positive integer m there exist n0 (m) and n1 (m) such that for n > n0 (m) or n > n1 (m) the equation 1 1 1 m = + + n x y z is solvable in positive integers x, y, z or in integers x, y, z respectively. (formulated in [5]) 2 (problem) Does the number of integer solutions of the equation x1 +x2 +. . .+xs = x1 x2 · · · xs satisfying 1 x1 x2 . . . xs tend to infinity with s ? (formulated in [6]) 3 (conjecture) For every even k the equation ϕ(x + k) = ϕ(x) has infinitely many solutions. (paper 21) 4 (conjecture) If k is a positive integer and f1 (x), f2 (x), . . . , fk (x) irreducible polynomials with integer coefficients and the leading coefficient positive such that f1 (x)f2 (x) · · · fk (x) has no fixed divisor > 1, then there exist infinitely many positive integers x such that all numbers fi (x) (1 i k) are primes. (paper 22=J1) 5 (conjecture) For all positive integers k and n, where n > 1, k n, (k, n) = 1, (paper 22=J1) there exists at least one prime p ≡ k (mod n), p < n2 . α 6 (conjecture)

n For every integer k > 34, k = p (p prime) there exists an integer n such that n − i /| i for all i k. (paper 26=H1)

7 (conjecture) Every integer n ≡ 0, 4, 7 (mod 8), n > 130, is a sum of three positive squares. (paper 33=A4) 8 (conjecture) Every integer n ≡ 0, 4, 7 (mod 8), n > 627, is a sum of three distinct squares. (paper 33=A4) 9 (conjecture) For every positive integer k k times

σ σ · · · σ (n) < ∞. lim inf n→∞ n

([3])

1368


10 (conjecture) The product p1 p2 · · · pk−1 pk+1 , where pi is the i-th prime, is the least positive integer g(k) with the property that for every integer n sufficiently large at least one of the numbers n + 1, n + 2, . . . , n + g(k) has more than k prime divisors. ([4]) 11 (problem) Apart from 2, 3, 5 and 3, 4, 5, 7, 11 does there exist a sequence a1 < a2 < . . . < ar of positive integers such that ar < [ai , aj ]

(1 i < j r)

and

r 1 >1? ai

(paper 35)

i=1

12 (conjecture) For every k = 1 and l 0 there exists an integer m such that the equations ϕ(x) = m and σ (y) = m have exactly k and l solutions respectively. (paper 46=J2) 13 (conjecture) For every x 8 there is a prime between x and x + (log x)2 . (paper 46=J2) 14 (problem) For every pair of relatively prime integers with |a| > |b| > 0 does there exist n such that a n − bn has three primitive prime factors? (paper 53=I1) an

15 (problem) Does there exist a pair a, b (as above) with ab = ±ch (h 2) such that − bn has three primitive prime factors for infinitely many n? (paper 53=I1)

16 (problem) Does there exist a pair a, b (as above) with ab = ±2c2 , ±ch (h 2) such that the greatest prime factor of a n − bn is greater than 2n for all sufficiently large n? (paper 53=I1) 17 (problem) Does there exist a polynomial f ∈ Q[x1 , x2 , . . . , xn , y, z] such that for all integer systems (x1 , . . . , xn ) the equation f (x1 , x2 , . . . , xn , y, z) = 0

(1)

is soluble in integers y, z and for no rational functions ϕ, ψ ∈ Q(x1 , x2 , . . . , xn )

(2) the identity

f (x1 , x2 , . . . , xn , ϕ, ψ) = 0

(3) holds?

(paper 56=E1, cf. also 69=F1)

18 (problem) Does there exist a polynomial f ∈ Q[x1 , x2 , . . . , xn , y, z] such that for all rational systems (x1 , . . . , xn ) the equation (1) is soluble in rational y, z and for no rational functions ϕ, ψ satisfying (2) the identity (3) holds? (paper 56=E1) 19 Let distinct polynomials f1 , . . . , fk (k 0) satisfy the assumptions of conjecture 4. Let g be a polynomial with integral coefficients and the leading coefficient positive. (conjecture) Let n be a positive integer such that n − g(x) is irreducible and k fi (x)(n − g(x)) has no fixed divisor > 1. Denote by N (x) = N the number of i=1

1369


positive integers x such that n − g(x) > 0 and by P (n) the number of x’s such that all numbers f1 (x), f2 (x), …, fk (x) and n − g(x) are primes. Then for large n we have ω(p) N 1 −k−1 −1 (h h · · · h ) , 1 − P (n) ∼ 1 − 0 1 k p p logk+1 N p where h0 = deg g, hi = deg fi and ω(p) is the number of solutions of the congruence k fi (x)(n − g(x)) ≡ 0 (mod p). (paper 60=J3) i=1

20 (problem) Does there exist infinitely many solutions of the equation (x 2 − 1)(y 2 − 1) = (z2 − 1)2 with x even, y, z odd or at least one solution with |x|, |y|, |z| even and distinct? (paper 61) 21 (problem) Assume that f ∈ Z[x] and f (x) is representable as a sum of two integral cubes for all sufficiently large integer x. Does it follow that f (x) = u(x)3 + v(x)3 , where u, v are integer valued polynomials? (paper 66=A6) 22 (problem) Is the inequality 1 n 2

σ ϕ(n) true for all n ?

(paper 67=G5)

23 (conjecture) There exists a constant c > 0 such that for every algebraic integer α = 0 of degree n, that is not a root of unity, c α >1+ . (paper 68=C1) n 24 A factorization of a polynomial in Q[x] into a product of a constant and of coprime powers of polynomials irreducible over Q is called standard. For a given polynomial f = 0, Kf denotes the factor of f of the greatest possible degree whose no root is 0 or a root of 1 and whose leading coefficient is equal to the leading coefficient of f . If φ ∈ Q[x1±1 , . . . , xk±1 ] \ {0}, then Jφ = φ

k

− ordxi φ

xi

.

i=1

(conjecture) Let F (y1 , . . . , yk ) be a polynomial irreducible over Q which does not divide y1 · · · yk J (y1δ1 y2δ2 · · · ykδk − 1) for any integers δ1 , . . . , δk not all zero. For every system of k positive integers n1 , . . . , nk there exists an integral non-singular matrix [νij ] (1 i k, 1 j k) satisfying the following conditions: (i) 0 νij C1 (F ) (1 i k, 1 j k); k νij uj (1 i k), uj integers 0 (1 j k); (ii) ni = j =1

(iii) if JF

k j =1

ν

yj 1j ,

k j =1

ν

yj 2j , . . . ,

k j =1

r ν yj kj = const Fs (y1 , . . . , yk )es s=1

1370


is a standard factorization, then either KF (x n1 , . . . , x nk ) = const

r

KFs (x u1 , . . . , x uk )es

s=1

is a standard factorization or α1 n1 + . . . + αk nk = 0, where αi are integers not all zero and |αi | C0 (F ) (1 i k). C0 (F ) and C1 (F ) are constants independent of n1 , . . . , nk . (paper 73=D2, cf. also paper 96=D4) 25 (conjecture) In every finite covering system of congruences ai (mod mi ) (mi > 1) (paper 86=D3) we have mi | mj for at least one pair i, j with i = j . 26 Let S be the set of all polynomials with integral coefficients and the leading coefficient positive. (problem) Does there exist for every polynomial f (x) ∈ S and every ε > 0 a polynomial h(x) ∈ S of degree d such that the degree of each irreducible factor of f (h(x)) is less than εd ? (paper 87, §5=J4). 4 3 fi (x) = P x + Q, where fi ∈ Z3 [x], 27 (problem) Does there exist an identity i=1 P , Q ∈ Z3 , P = 0, Q ≡ 4 (mod 9) ? (paper 89) 28 (conjecture) Every genus of primitive binary quadratic forms with discriminant D (paper 99) represents a positive integer c(ε)|D|ε for every ε > 0. 29 (problem) To estimate the number of irreducible non-cyclotomic factors of a polynomial f ∈ Z[x] by a function of f alone, where f is the sum of squares of the coefficients of f . (paper 105=C6) 30 (problem) Let K be an algebraic number field. Does there exist a sequence {αi } of integers in K such that for every ideal q of K, integers α1 , α2 , . . . , αN(q) represent all residue classes modulo q ? (formulated in [7], earlier for K = Q(i) proposed orally by J. Browkin) 31 (problem) To improve the estimate (log n)2 / log log n for the number of non-zero coefficients of the cyclotomic polynomial with a square-free index n. (formulated in [1]) 32 (conjecture) If a polynomial P (x) with rational coefficients has at least three simple zeros, then the equation y 2 z3 = P (x) has only finitely many solutions in integers x, y, z with yz = 0. (paper 115=A8) 33 (problem) Given a, b with |a| = |b|, do there exist infinitely many quotients r such that for suitable integers m, n: m/n = r and K(ax m+n + bx m + bx n + a) is reducible? (paper 121=D7, the operation K is defined in 24 above) 34 (conjecture) Let F ∈ Z[x, y] be a form such that F (x, y) = F1 (ax + by, cx + dy) for any F1 ∈ Z[x, y] and any a, b, c, d ∈ Z

a b

= ±1. implies

c d

If f ∈ Z[t1 , . . . , tr ] has the fixed divisor equal to its content and the equation (4)

F (x, y) = f (t1 , . . . , tr )

1371


is soluble in integers x, y for all integral vectors [t1 , . . . , tr ], then there exist polynomials X, Y ∈ Z[t1 , . . . , tr ] such that identically

(5) F X(t1 , . . . , tr ), Y (t1 , . . . , tr ) = f (t1 , . . . , tr ). (paper 128=J5) 35 (conjecture) Let F ∈ Z[x, y] be any form and f ∈ Z[t1 , . . . , tr ] any polynomial. If the equation (4) is soluble in integers x, y for all integer vectors [t1 , . . . , tr ], then there (paper 128=J5) exist polynomials X, Y ∈ Q[t1 , . . . , tr ] satisfying (5). 36 (problem) Does the divisibility φ(n)+ 1 | n imply n = p or 2p, where p is a prime? (formulated in [2], p. 52) 37 (conjecture) If F ∈ Z[x, y, t] is irreducible, the highest homogeneous part F0 of F with respect to x, y is reducible over Q(t) and every arithmetic progression contains an ∗ in integers x, y, then there exist polynomials integer t ∗ such that F (x, y,

t ) = 0 is solvable X, Y ∈ Q[t] such that F X(t), Y (t), t = 0. (paper 132=A12) 38 (problem) Let a(0) = [a1 (0), . . . , an (0)] ∈ Rn and an infinite sequence a(t) = [a1 (t), . . . , an (t)] be formed by means of the formulae ai (t + 1) = |ai (t) − ai+1 (t)|, where the addition of indices is mod n. Is it true that for every n and every a(0) ∈ Rn either lim a(t) = 0 or there exists c ∈ R such that a(t) ∈ {0, c}n for all sufficiently t→∞

large t ?

(paper 155)

39 Given m linearly independent vectors n1 , . . . , nm ∈ Zk , let H (n1 , . . . , nm ) denote the maximum of the absolute values of all minors of order m of the matrix ⎛ ⎞ n1 ⎜ .. ⎟ ⎝ . ⎠ nm and D(n1 , . . . , nm ) the greatest common divisor of these minors. Furthermore, let l D(n1 , . . . , nm ) (k−l)/(k−m) c0 (k, l, m) = sup inf H (pi ), H (n1 , . . . , nm ) i=1

where the supremum is taken over all sets of linearly independent vectors n1 , . . . , nm ∈ Zk and the infimum is taken over all sets of linearly independent vectors p1 , . . . , pl ∈ Zk such that ni =

l

uij pj ,

uij ∈ Q.

j =1

(problem) Is lim sup c0 (k, l, m) finite? k,l→∞

(paper 166=L2)

1372


40 (problem) Given an integer m 3, does there exist a number K such that every polynomial in Q[x] with m non-zero coefficients has a factor irreducible over Q with at most K non-zero coefficients? (paper 168, for m = 3 already paper 56=E1) √ / Q, then 41 (conjecture) If fi (x) = ai x 2 + bi x + ci ∈ Z[x] (i = 1, 2, 3), a1 a2 a3 ∈ the number N(x) of integers x3 such that |x3 | x and the equation f3 (x3 ) = f1 (x1 )f2 (x2 ) is soluble in integers x1 , x2 , satisfies N (x) x ε for every ε > 0. (paper 170) 42 (conjecture) For every algebraic number field K there exist sets Fν,μ ⊂ K 2 (ν ∈ N, μ ∈ N) such that M M {x ν + ax μ + b} is finite ν,μ a,b ∈Fμ,ν (K)

and if x n + ax m + b, where n 2m > 0, a, b ∈ K ∗2 , is reducible over K, at least one of the following conditions is satisfied: (i) x n/(n,m) + ax m/(n,m) + b has a proper linear or quadratic factor over K, (ii) there exists an integer l such that n/ l, m/ l = 2p, p (p prime), 6, 1 , 6, 2 , 7, 1 , 8, 2 , 8, 4 , 9, 3 , 10, 2 , 10, 4 , 12, 2 , 12, 3 , 12, 4 , 15, 5 , 7, 2 , 7, 3 , 8, 1 , 9, 1 , 14, 2 , 21, 7 , (iii) there exists an integer l such that n/ l, m/ l =: ν, μ ∈ Z2 and a = uν−μ a0 (v), b = uν b0 , where a0 , b0 ∈ Fν,μ . Consequence 1. For every algebraic number field K there exists a constant C1 (K) such that if n1 > C1 (K) and a, b ∈ K ∗ then x n + ax m + b is reducible over K if and only if (i) holds. Consequence 2. For every algebraic number field K there exists a constant C2 (K) such that if a, b ∈ K then x n + ax m + b has in K[x] an irreducible factor with at most C2 (K) non-zero coefficients. Consequence 3. There are only finitely many integers b such that for some n = 2m, x n + bx m + 1 is reducible over Q. (paper 175=D10) 43 (problem) What is the least positive integer n such that all integers 2k n − 1 (k = 1, 2, . . . ) are composite? (paper 179=G6) 44 (problem) Have the integers not of the form n − ϕ(n) a positive lower density? (paper 179=G6) 45 (conjecture) Let k, m, a, b be positive integers, m > kb. There are no polynomials F1 , F2 , . . . , Fk ∈ Z[x] with the leading coefficient positive such that 1 m . = Fi (x) ax + b k

(paper 198=A15)

i=1

46 (problem) Do there exist two trinomials Ti ∈ C[x] (i = 1, 2) such that (T1 , T2 ) has more than six non-zero coefficients? (paper 199=D16)


1373

47 (conjecture) For every algebraic number field K and d = 1, 2 there exist sets d (K) ⊂ N2 × {x d + . . . + c : c , c ∈ K} such that the set Fν,μ d 1 d M M {x ν + ax μ + b} d ν,μ,F a,b,F ∈Fν,μ

is finite and if n, m ∈ N, n > m, n1 = n/(n, m), m1 = m/(n, m), a, b ∈ K ∗ , F is a monic factor of x n1 + ax m1 + b in K[x] of maximal possible degree d ∈ {1, 2}, n1 > 6, then (x n + ax m + b)F (x (n,m) )−1 is reducible over K if and only if there exists an integer l such that n/ l, m/ l =: ν, μ ∈ N2 and a = uν−μ a0 , b = uν b0 , F = u(ν,μ)d F0 (x/u(ν,μ) ), d (K). where u ∈ K ∗ , a0 , b0 , F0 ∈ Fν,μ (paper 200=D14, for d = 1 already paper 197=D13) 48 (conjecture) For every field K such that char K > d every polynomial F ∈ d fμ (αμ1 x1 + αμ2 x2 ), where fμ ∈ K[z], αμi ∈ K K[x1 , x2 ] can be represented as μ=1 (1 μ d, 1 i 2). (paper 208=E8) 49 (problem) Let K be a real quadratic field, β be a primitive integer of K, p a rational prime and M the set of Mahler measures of all algebraic numbers. √ Do the conditions β ∈ M and p splits in K imply pβ ∈ M in general, or for β = (1 + 17)/2, p = 2 in particular? (paper 211=C10) 50 (problem) Let K be a field of characteristic 0, n a positive integer. Is it true that a monic polynomial f ∈ K[x] of degree n with exactly k distinct zeros is determined up to finitely many possibilities by any k of its non-zero proper coefficients? (paper 213) 51 (problem) Let f ∈ Z[x] have the leading coefficient positive and assume that the congruence f (x) ≡ y 2 (mod m) is solvable for every positive integer m. Does there exist an k odd integer k > 0 and integers x1 , . . . , xk such that f (xi ) is a square? (paper 214=A16) i=1

52 (problem) Let f be a non-singular binary form over C. Can the existence of a non-trivial automorph of f be characterized in terms of invariants of f exclusively? (paper 219=E9) 53 (conjecture) The explicit value for the maximal order of the group of weak automorphs divided by the group of trivial automorphs of a binary form f of degree n defined over a field of characteristic π , where 0 < π n. (paper 219=E9) 54 (problem) How to compute l(P ) for cubic polynomials P over R, in particular for P (x) = 2x 3 + 3x 2 + 4 ? Here l(P ) = inf L(P G), where L(F ) is the length of F and G runs through all monic polynomials over R. (paper 221=D17) 55 (problem) Is it true that l(P ) ∈ K(P ) for all P ∈ R[x] with no zeros inside the unit circle? Here l(P ) has the meaning of Problem 54 and K(P ) is the field generated by the coefficients of P . (paper 221=D17)

1374


56 (conjecture) Let k, n and bi (1 i k) be positive integers, and let ai (1 i k) be any integers. The number N (n; a1 , b1 , . . . , ak , bk ) of solutions of the congruence k

ai xi ≡ 0 (mod n) in the box 0 xi bi

i=1

satisfies the inequality N (n; a1 , b1 , . . . , ak , bk ) 21−n

k

(bi + 1).

(paper 222)

i=1

References [1] J. H. Conway, A. J. Jones, Trigonometric Diophantine equations (On vanishing sums of roots of unity). Acta Arith. 30 (1976), 229–240. [2] R. K. Guy, Unsolved Problems in Number Theory, Springer, New York 1981. [3] A. Schinzel, Ungelöste Problem Nr. 30. Elem. Math. 14 (1959), 60–61. [4] −−, Ungelöste Problem Nr. 31. Elem. Math. 14 (1959), 82–83. [5] W. Sierpiński, Sur les décompositions de nombres rationnels en fractions primaires. Mathesis 65 (1956), 16–32; see also Oeuvres choisies, T. I, Varsovie 1974, 169–184. [6] E. Trost, Ungelöste Problem Nr. 14. Elem. Math. 11 (1956), 135. [7] R. Wasén, Remark on a problem of Schinzel. Acta Arith. 29 (1976), 425–426.

Publication list of Andrzej Schinzel


Publication list of Andrzej Schinzel

A. Research papers (papers not reviewed are marked with letters A, B, . . .) If a paper is reprinted in this collection, the number of its beginning page is put in the last column. [1]

[2] [3]

Sur la décomposition des nombres naturels en sommes de nombres triangulaires distincts, Bull.Acad. Polon. Sci. Cl. III 2 (1954), 409–410, MR0067910 (16,796g). (with W. Sierpiński) Sur quelques propriétés des fonctions ϕ(n) et σ (n), Bull. Acad. Polon. Sci. Cl. III 2 (1954), 463–466, MR0067140 (16,675f). Quelques théorèmes sur les fonctions ϕ(n) et σ (n), Bull. Acad. Polon. Sci. Cl. III 2 (1954), 467–469, MR0067141 (16,675g).

[4]

On the equation x1 x2 · · · xn = t k , Bull. Acad. Polon. Sci. Cl. III 3 (1955), 17–19, MR0069197 (16,998g).

[5]

(with W. Sierpiński) Sur l’équation x 2 +y 2 +1 = xyz, Matematiche (Catania) 10 (1955), 30–36, MR0075220 (17,711e). Carré, cube et bicarré en progression arithmétique, Mathesis 64 (1955), 31–32.

[5A] [5B] [6]

Sur l’équation x 2 + y 2 = 2z4 , Mathesis 64 (1955), 357–358. Sur une propriété du nombre de diviseurs, Publ. Math. Debrecen 3 (1954), 261–262, MR0072160 (17,238d).

[7]

Sur l’équation indeterminée x 2 +l = y 3 , Bull. Soc. Roy. Sci. Liége 24 (1955), 271–274, MR0072156 (17,237i). On functions ϕ(n) and σ (n), Bull.Acad. Polon. Sci. Cl. III 3 (1955), 415–419, MR0073625 (17,461c). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866 (with J. Browkin) Sur les nombres de Mersenne qui sont triangulaires, C. R. Acad. Sci. Paris 242 (1956), 1780–1782, MR0077546 (17,1055d). . . . . . . . . 11 Sur l’équation ϕ(x) = m, Elem. Math. 11 (1956), 75–78, MR0080114 (18,194c). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871 Sur quelques propriétés des nombres 3/N et 4/N , où N est un nombre impair, Mathesis 65 (1956), 219–222, MR0080683 (18,284a). . . . . . . . . . . . . . . . . . . 13

[8] [9] [10] [11]

1378 [12]

[13]

[14] [15] [16] [17] [18] [19]

[20]

[21] [22]

[23]

[24]

Publication list

O liczbach pierwszych, dla których suma dzielników sze´scianu jest pełnym kwadratem (On prime numbers such that the sums of the divisors of their cubes are perfect squares), Wiadom. Mat. (2) 1 (1956), 203–204 (Polish), MR0110663 (22 #1538). rozkładów na sume trzech O pewnym przypuszczeniu dotyczacym kwadratów (On a certain conjecture concerning partitions into sums of three squares), Wiadom. Mat. (2) 1 (1956), 205, MR0113856 (22 #4687). Sur l’équation x z − y t = 1, où |x − y| = 1, Ann. Polon. Math. 3 (1956), 5–6, MR0082506 (18,561d). (with W. Sierpiński) Sur l’équation x 2 +x +1 = 3y 2 , Colloq. Math. 4 (1956), 71–73, MR0077552 (17,1055j). Generalization of a theorem of B. S. K. R. Somayajulu on the Euler’s function ϕ(n), Ganita 5 (1954), 123–128, MR0083999 (18,791a). Sur un problème concernant la fonction ϕ(n), Czechoslovak Math. J. 6 (1956), 164–165, MR0084005 (18,792b). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875 (with W. Sierpiński) Sur les sommes de quatre cubes, Acta Arith. 4 (1958), 20–30, MR0095158 (20 #1664). (with Y. Wang) A note on some properties of the functions ϕ(n), σ (n) and θ (n), Ann. Polon. Math. 4 (1958), 201–213, MR0095149 (20 #1655); Announcement, Bull. Acad. Polon. Sci. Cl. III 4 (1956), 207–209, MR0079024 (18,17c); Corrigendum, Ann. Polon. Math. 19 (1967), 115, MR0209240 (35 #142). Sur l’existence d’un cercle passant par un nombre donné de points aux coordonnées entières, Enseignement Math. (2) 4 (1958), 71–72, MR0098059 (20 #4522). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Sur l’équation ϕ(x +k) = ϕ(x), Acta Arith. 4 (1958), 181–184, MR0106867 (21 #5597). (with W. Sierpiński) Sur certaines hypothèses concernant les nombres premiers, Acta Arith. 4 (1958), 185–208; Erratum 5 (1959), 259, MR0106202 (21 #4936). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1113 Sur un problème concernant le nombre de diviseurs d’un nombre naturel, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astr. Phys. 6 (1958), 165–167, MR0106203 (21 #4937). Sur l’équation diophantienne x x y y = zz , Acta Scient. Mat. Univ. Szehuensis (1958), 81–83 (Chinese), Ref. Zh. 1959 #4439.

[24A] Uwagi o zadaniu 420, Matematyka XI-3 (1958), 60–62. [25] Sur les nombres composés n qui divisent a n − a, Rend. Circ. Mat. Palermo (2), 7 (1958), 37–41, MR0106201 (21 #4935). [26] Sur un probléme de P. Erd˝os, Colloq. Math. 5 (1958), 198–204, MR0099950 (20 #6386). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 903 [27] Sur les diviseurs naturels des polynômes, Matematiche (Catania) 12 (1957), 18–22, MR0130248 (24 #A114). [28] Démonstration d’une conséquence de l’hypothèse de Goldbach, Compositio Math. 14 (1959), 74–76, MR0103870 (21 #2633).

Publication list

[29] [30] [31]

[32] [33] [34] [35] [36]

[37] [38] [39]

[40] [41] [42] [43] [44] [45] [46] [47] [48]

1379

(with S. Gołab) Sur l’équation fonctionnelle f [x + y · f (x)] = f (x) · f (y), Publ. Math. Debrecen 6 (1959), 113–125, MR0107101 (21 #5828). . . . . . . 1314 Sur l’équation indéterminée de R. Goormaghtigh, (with A. Makowski) Mathesis 68 (1959), 128–142, MR0118701 (22 #9472). liczb naturalnych (On the periodicity of O okresowo´sci pewnych ciagów certain sequences of natural numbers), Wiadom. Mat. (2) 2 (1959), 269–272, MR0117182 (22 #7965). Sur une conséquence de l’hypothèse de Goldbach, Bulgar. Akad. Nauk. Izv. Mat. Inst. 4 (1959), no. 1, 35–38, MR0141628 (25 #5026). Sur les sommes de trois carrés, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astr. Phys. 7 (1959), 307–310, MR0111728 (22 #2590). . . . . . . . . . . . . . . . . . . . . . 18 Sur quelques propositions fausses de P. Fermat, C. R. Acad. Sci. Paris 249 (1959), 1604–1605, MR0106875 (21 #5605). (with G. Szekeres) Sur un probléme de M. Paul Erd˝os, Acta Sci. Math. Szeged 20 (1959), 221–229, MR0112864 (22 #3710). (with A. Białynicki-Birula and J. Browkin) On the representation of fields as finite unions of subfields, Colloq. Math. 7 (1959), 31–32, MR0111739 (22 #2601). (with A. Wakulicz) Sur l’équation ϕ(x + k) = ϕ(x) II, Acta Arith. 5 (1959), 425–426, MR0123506 (23 #A831). (with W. Sierpiński) Sur les congruences x x ≡ c (mod m) et a x ≡ b (mod p), Collect. Math. 11 (1959), 153–164, MR0113838 (22 #4670). O równaniu diofantycznym nk=1 Ak xkϑk = 0 (On the Diophantine equation n ϑk k=1 Ak xk = 0), Prace Mat. 4 (1960), 45–51, MR0131395 (24 #A1247).; Corrigendum, Comment. Math. Prace Mat. 44 (2004), 283–284, MR2118015 (2005i:11045) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 O równaniu x 4 +ax 2 y 2 +by 4 = z2 , Prace Mat. 4 (1960), 52–56, MR0131396 (24 #A1248). (with B. Rokowska) Sur un problème de M. Erd˝os, Elem. Math. 15 (1960), 84–85, MR0117188 (22 #7970). (with J. Browkin) On the equation 2n − D = y 2 , Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 8 (1960), 311–318, MR0130215 (24 #A82). On the congruence a x ≡ b (mod p), Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 8 (1960), 307–309, MR0125070 (23 #A2377). . . . . . . . . . . 909 On some problems of the arithmetical theory of continued fractions, Acta Arith. 6 (1961), 393–413, MR0125814 (23 #A3111). . . . . . . . . . . . . . . . . . . . . 131 (with P. Erd˝os) Distributions of the values of some arithmetical functions, Acta Arith. 6 (1961), 473–485, MR0126410 (23 #A3706). . . . . . . . . . . . . . . . 877 Remarks on the paper “Sur certaines hypothèses concernant les nombres premiers”, Acta Arith. 7 (1961), 1–8, MR0130203 (24 #A70). . . . . . . . . . . . 1134 (with H. Davenport and D. J. Lewis) Equations of the form f (x) = g(y), Quart. J. Math. Oxford Ser. (2) 12 (1961), 304–312, MR0137703 (25 #1152). The intrinsic divisors of Lehmer numbers in the case of negative discriminant, Ark. Mat. 4 (1962), 413–416, MR0139567 (25 #2999).

1380 [49] [50]

[51]

[52]

[53] [54]

[55] [56] [57] [58] [59] [60]

Publication list

On the composite integers of the form c(ak + b)! ± 1, Nordisk Mat. Tidskr. 10 (1962), 8–10, MR0139565 (25 #2997). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 912 On some problems of the arithmetical theory of continued fractions II, Acta Arith. 7 (1962), 287–298, MR0139566 (25 #2998); Corrigendum 47 (1986), 295, MR0870671 (88b:11007). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 Remark on the paper of K. Prachar “Über die kleinste Primzahl einer arithmetischen Reihe”, J. Reine Angew. Math. 210 (1962), 121–122, MR0150115 (27 #118). (with W. Sierpiński) Sur les triangles rectangulaires dont les deux cotés sont des nombres triangulaires, Bull. Soc. Math. Phys. Serbie 13 (1961), 145–147, Zbl. 0131.28304. On primitive prime factors of a n −bn , Proc. Cambridge Philos. Soc. 58 (1962), 555–562, MR0143728 (26 #1280). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036 Solution d’un problème de K. Zarankiewicz sur les suites de puissances consécutives de nombres irrationnels, Colloq. Math. 9 (1962), 291–296, MR0141637 (25 #5035); Correction 12 (1964), 289, MR0174554 (30 #4755). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 n

Remarque au travail de W. Sierpiński sur les nombres a 2 + 1, Colloq. Math. 10 (1963), 137–138, MR0148601 (26 #6108). Some unsolved problems on polynomials, Neki nerešeni problemi u matematici, Matematiˇcka Biblioteka 25, Beograd 1963, 63–70, Zbl. 0122.25401. . 703 On primitive prime factors of Lehmer numbers I, Acta Arith. 8 (1963), 213–223, MR0151423 (27 #1408). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046 On primitive prime factors of Lehmer numbers II, Acta Arith. 8 (1963), 251–257, MR0151424 (27 #1409). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059 Reducibility of polynomials in several variables, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 11 (1963), 633–638, MR0159816 (28 #3032). 709 A remark on a paper of Bateman and Horn, Math. Comp. 17 (1963), 445–447, MR0153647 (27 #3609). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1142

[61]

(with W. Sierpiński) Sur l’équation diophantienne (x 2 − 1)(y 2 − 1) = 2 y−x 2 − 1 , Elem. Math. 18 (1963), 132–133, MR0163881 (29 #1180). 2

[62]

(with M. Bhaskaran) A new elementary estimation for the sum of real characters, Prace Mat. 8 (1963), 99–102, MR0182614 (32 #97).

[63]

(with A. Rotkiewicz) Sur les nombres pseudopremiers de la forme ax 2 + bxy + cy 2 , C. R. Acad. Sci. Paris 258 (1964), 3617–3620, MR0161828 (28 #5032). (with H. Davenport) Two problems concerning polynomials, J. Reine Angew. Math. 214/215 (1964), 386–391, MR0162789 (29 #93); Corrigendum 218 (1965), 220, MR0174553 (30 #4754). (with J. Mikusiński) Sur la réductibilité de certains trinômes, Acta Arith. 9 (1964), 91–95, MR0163906 (29 #1205). (with H. Davenport and D. J. Lewis) Polynomials of certain special types, Acta Arith. 9 (1964), 107–116, MR0163880 (29 #1179). . . . . . . . . . . . . . . . .

[64]

[65] [66]

27

Publication list

[67] [68] [69] [70] [71] [72] [73]

[74]

[75]

[76]

[77] [78] [79]

[80]

[81] [82] [83] [84] [85]

(withA. Makowski) On the functions ϕ(n) and σ (n), Colloq. Math. 13 (1964), 95–99, MR0173660 (30 #3870). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (with H. Zassenhaus) A refinement of two theorems of Kronecker, Michigan Math. J. 12 (1965), 81–85, MR0175882 (31 #158). . . . . . . . . . . . . . . . . . . . . . On Hilbert’s Irreducibility Theorem, Ann. Polon. Math. 16 (1965), 333–340, MR0173658 (30 #3868). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . On the composite Lehmer numbers with prime indices, Prace Mat. 9 (1965), 95–103, MR0174520 (30 #4721). Sur l’équation indéterminée de R. Goormaghtigh II, (with A. Makowski) Mathesis 70 (1965), 94–96, Zbl. 0127.01903. (with H. Davenport) A note on sequences and subsequences, Elem. Math. 20 (1965), 63–64, Zbl. 0132.25101. On the reducibility of polynomials and in particular of trinomials, Acta Arith. 11 (1965), 1–34, MR0180549 (31 #4783); Errata, ibid., 491, MR0197448 (33 #5613); Corrigenda, Acta Arith. 16 (1969), 159. . . . . . . . . . . . . . . . . . . . .

1381 890 175 839

301

(with B. J. Birch, S. Chowla and M. Hall Jr.) On the difference x 3 − y 2 , Norske Vid. Selsk. Forh. (Trondheim) 38 (1965), 65–69, MR0186620 (32 #4079). Uwaga do artykułu H. Steinhausa “Pogadanka (troche historyczna)” (A remark to the paper by H. Steinhaus “A chat (slightly historical)”), Wiadom. Mat. (2) 8 (1965), 143–144, Zbl. 0149.28202. (with S. Hartman, J. Mycielski, and S. Rolewicz) Concerning the characterization of linear spaces, Colloq. Math. 13 (1965), 199–208, MR0183808 (32 #1284). (with H. Davenport) A combinatorial problem connected with differential equations, Amer. J. Math. 87 (1965), 684–694, MR0190010 (32 #7426). . . 1327 (with W. Sierpiński) Sur les puissances propres, Bull. Soc. Roy. Sci. Liège 34 (1965), 550–554, MR0186649 (32 #4107). On a theorem of Bauer and some of its applications, Acta Arith. 11 (1966), 333–344, MR0190130 (32 #7544); Corrigendum 12 (1967), 425, MR0210684 (35 #1570). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 (with D. J. Lewis and H. Zassenhaus) An extension of the theorem of Bauer and polynomials of certain special types, Acta Arith. 11 (1966), 345–352, MR0190131 (32 #7545). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 (with H. Davenport and D. J. Lewis) Quadratic Diophantine equations with a parameter, Acta Arith. 11 (1966), 353–358, MR0184902 (32 #2373). (with H. Davenport) A note on certain arithmetical constants, Illinois J. Math. 10 (1966), 181–185, MR0188193 (32 #5632). On sums of roots of unity. Solution of two problems of R. M. Robinson, Acta Arith. 11 (1966), 419–432, MR0201418 (34 #1302). . . . . . . . . . . . . . . . . . . . . 197 (with H. Davenport) Diophantine Approximation and sums of roots of unity, Math. Ann. 169 (1967), 118–135, MR0205926 (34 #5751). Reducibility of polynomials of the form f (x)−g(y), Colloq. Math. 18 (1967), 213–218, MR0220703 (36 #3755). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715

1382 [86] [87]

[88] [89] [90]

[91]

[92] [93] [94] [95] [96] [97] [98] [99] [100] [101] [102] [103]

[104]

Publication list

Reducibility of polynomials and covering systems of congruences, Acta Arith. 13 (1967), 91–101, MR0219515 (36 #2596). . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 On two theorems of Gelfond and some of their applications, Acta Arith. 13 (1967), 177–236, MR0222034 (36 #5086); Corrigendum 16 (1969), 101, MR0246840 (40 #109); Addendum 56 (1996), 181, MR1075643. . . . . . . . . . 1145 (with A. Gru˙zewski) Sur les itérations d’une fonction arithmétique, Prace Mat. 11 (1968), 279–282, MR0224548 (37 #147). On sums of four cubes of polynomials, J. London Math. Soc. 43 (1968), 143–145, MR0223340 (36 #6388). (with L. P. Postnikova) O primitivnyh delitelyah vyraženiya a n − bn v polyah algebraiˇceskih cˇ isel (Primitive divisors of the expression a n − bn in algebraic number fields), Mat. Sb. (N.S.) 75 (1968), 171–177; Math. USSR-Sb. 4 (1968), 153–159, MR0223330 (36 #6378). On primitive prime factors of Lehmer numbers III, Acta Arith. 15 (1968), 49–70, MR0232744 (38 #1067); Corrigendum 16 (1969), 101, MR0246840 (40 #109). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066 An improvement of Runge’s theorem on Diophantine equations, Comment. Pontificia Acad. Sci. 2 (1969), No. 20, MR0276174 (43 #1922). . . . . . . . . . . 36 (with W. Narkiewicz) Ein einfacher Beweis des Dedekindschen Differentensatzes, Colloq. Math. 20 (1969), 65–66, MR0240071 (39 #1425). Remarque sur le travail précédent de T. Nagell, Acta Arith. 15 (1969), 245–246, MR0246854 (40 #123). A remark on a paper of Mordell, J. London Math. Soc. (2) 1 (1969), 765–766, MR0249358 (40 #2603). Reducibility of lacunary polynomials I, Acta Arith. 16 (1969), 123–159, MR0252362 (40 #5583). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Reducibility of lacunary polynomials II, Acta Arith. 16 (1970), 371–392, MR0265323 (42 #233). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 A refinement of a theorem of Gerst on power residues, Acta Arith. 17 (1970), 161–168, MR0284417 (44 #1644). (with A. Baker) On the least integers represented by the genera of binary quadratic forms, Acta Arith. 18 (1971), 137–144, MR0319896 (47 #8437). (with J. Wójcik) A note on the paper “Reducibility of lacunary polynomials I”, Acta Arith. 19 (1971), 195–201, MR0289463 (44 #6653). . . . . . . . . . . . . 403 (with W. Brostow) On interesting walks in a graph, J. Statist. Phys. 4 (1972), 103–110, MR0309794 (46 #8899). Integer points on conics, Comment. Math. Prace Mat. 16 (1972), 133–135, Erratum 17 (1973), 305, MR0321868 (48 #233). (with M. Fried) Reducibility of quadrinomials, Acta Arith. 21 (1972), 153–171, MR0313219 (47 #1774); Corrigendum and addendum, ibid. 99 (2001), 409–410, MR1845693. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 720 On a theorem of Bauer and some of its applications II, Acta Arith. 22 (1973), 221–231, MR0330105 (48 #8443). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

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1383

[105] On the product of the conjugates outside the unit circle of an algebraic number, Acta Arith. 24 (1973), 385–399, MR0360515 (50 #12963); Addendum 26 (1975), 329–331, MR0371853 (51 #8070). . . . . . . . . . . . . . . . . . . . . . . . . . . 221 [106] A contribution to combinatorial geometry, Demonstratio Math. 6 (1973), 339–342, MR0350607 (50 #3099). [107] On two conjectures of P. Chowla and S. Chowla concerning continued fractions, Ann. Mat. Pura Appl. (4) 98 (1974), 111–117, MR0340187 (49 #4943). 161 [108] Primitive divisors of the expression An − B n in algebraic number fields, J. Reine Angew. Math. 268/269 (1974), 27–33, MR0344221 (49 #8961). . . 1090 [109] A general irreducibility criterion, J. Indian Math. Soc. (N.S.) 37 (1973), 1–8, MR0429849 (55 #2859). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739 [110] On power residues and exponential congruences, Acta Arith. 27 (1975), 397–420, MR0379432 (52 #337). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915 [111] On linear dependence of roots, Acta Arith. 28 (1975), 161–175, MR0389835 (52 #10665). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 [112] Traces of polynomials in algebraic numbers, Norske Vid. Selsk. Skr. (Trondheim) 1975, no. 6, 3 pp., MR0412143 (54 #270) [113] (with D. M. Goldfeld) On Siegel’s zero, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 2 (1975), 571–583, MR0404213 (53 #8016). . . . . . . . . . . . . . . . . . . . . 1199 [114] On the number of irreducible factors of a polynomial, in: Topics in Number Theory, Colloq. Math. Soc. János Bolyai 13, North-Holland, Amsterdam 1976, 305–314, MR0435027 (55 #7989). [115] (with R. Tijdeman) On the equation y m = P (x), Acta Arith. 31 (1976), 199–204, MR0422150 (54 #10142). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 [116] (with J. Browkin and B. Diviš) Addition of sequences in general fields, Monatsh. Math. 82 (1976), 261–268, MR0432581 (55 #5568). [117] Abelian binomials, power residues and exponential congruences, Acta Arith. 32 (1977), 245–274, MR0429819 (55 #2829); Addendum 36 (1980), 101–104, MR0576586 (81g:12005). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939 [118] (with T. Shorey, A. J. van der Poorten and R. Tijdeman) Applications of the Gel fond-Baker method to Diophantine equations, in: Transcendence Theory: Advances and Applications (ed. by A. Baker and D. Masser), Academic Press, London 1977, 59–77, MR0472689 (57 #12383). [119] (with H. L. Montgomery) Some arithmetic properties of polynomials in several variables, ibid., 195–203, MR0472757 (57 #12447). . . . . . . . . . . . . . . . . 747 [120] An analogue of Harnack’s inequality for discrete superharmonic functions, Demonstratio Math. 11 (1978), 47–60, MR0486564 (58 #6287). . . . . . . . . . 1338 [121] Reducibility of lacunary polynomials III, Acta Arith. 34 (1978), 227–266, MR0506160 (58 #22012). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 [122] An inequality for determinants with real entries, Colloq. Math. 38 (1978), 319–321, MR0485920 (58 #5714). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1347 [123] (with G. Baron) An extension of Wilson’s theorem, C. R. Math. Rep. Acad. Sci. Canada 1 (1979), 115–118, MR0519537 (80e:05025). . . . . . . . . . . . . . . 971 [124] (with R. Perlis) Zeta function and the equivalence of integral forms, J. Reine Angew. Math. 309 (1979), 176–182, MR0542046 (80j:12008). . . . . . . . . . . . 47

1384

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[125] (with L. A. Rubel and H. Tverberg) On difference polynomials and hereditarily irreducible polynomials, J. Number Theory 12 (1980), 230–235, MR0578817 (81i:12024). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755 [126] (with D. J. Lewis) Quadratic Diophantine equations with parameters, Acta Arith. 37 (1980), 133–141, MR0598871 (81m:10030). . . . . . . . . . . . . . . . . . . 54 [127] (with H. P. Schlickewei and W. M. Schmidt) Small solutions of quadratic congruences and small fractional parts of quadratic forms, Acta Arith. 37 (1980), 241–248, MR0598879 (81m:10063). [128] On the relation between two conjectures on polynomials, Acta Arith. 38 (1980), 285–322, MR0602194 (82g:12004). . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154 [129] (with J. Browkin) On Sylow 2-subgroups of K2 OF for quadratic number fields F , J. Reine Angew. Math. 331 (1982), 104–113, MR0647375 (83g:12011). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 [130] (with K. Gy˝ory and P. Kiss) On Lucas and Lehmer sequences and their applications to Diophantine equations, Colloq. Math. 45 (1981), 75–80, MR0652603 (83g:10009). [131] (with J. W. S. Cassels) Selmer’s conjecture and families of elliptic curves, Bull. London Math. Soc. 14 (1982), 345–348, MR0663485 (84d:14028). . 62 [132] Families of curves having each an integer point, Acta Arith. 40 (1982), 399–420, MR0667049 (83k:12003). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 [132A] (with F. Laubie) Sur le théorème de Gordan-Igusa, Publ. Dept. Math. Limoges, Fasc. 4 (1982), 50–53. [133] An application of Hilbert’s irreducibility theorem to Diophantine equations, Acta Arith. 41 (1982), 203–211, MR0674833 (83k:12004). [134] (with E. Dobrowolski and W. Lawton) On a problem of Lehmer, in: Studies in Pure Mathematics. To the memory of Paul Turán (ed. by P. Erd˝os), Birkhäuser, Basel 1983, 135–144, MR0820217 (87e:11120). [134A] Généralisation d’un résultat de Smyth aux polynômes à plusieurs variables, Publ. Math. Univ. Pierre et Marie Curie 64 (1983/84), Fasc. 1, exposé no. 4. [135] On the number of irreducible factors of a polynomial II, Ann. Polon. Math. 42 (1983), 309–320, MR0728089 (86k:11056). [136] (with J. Wróblewski) On ideals in the ring of polynomials in one variable over a Dedekind domain, Studia Sci. Math. Hungar. 16 (1981), 415–425, MR0729304 (84m:13021). [137] Hasse’s principle for systems of ternary quadratic forms and for one biquadratic form, Studia Math. 77 (1983), 103–109, MR0743067 (85h:11018). 87 [138] Reducibility of lacunary polynomials IV, Acta Arith. 43 (1984), 313–315, MR0738143 (85g:11091). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 [139] Reducibility of lacunary polynomials V, Acta Arith. 43 (1984), 425–440, MR0756292 (86d:11083). [140] The number of zeros of polynomials in valuation rings of complete discretely valued fields, Fund. Math. 124 (1984), 41–97, MR0818607 (87g:12005). [141] Reducibility of polynomials in several variables II, Pacific J. Math. 118 (1985), 531–563, MR0789192 (86i:12005).

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1385

[142] The fundamental lemma of Brun’s sieve in a new setting, Rocky Mountain J. Math. 15 (1985), 573–578, MR0823268 (87h:11087). [143] Systems of exponential congruences, Demonstratio Math. 18 (1985), 377–394, MR0816042 (87c:11036). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975 [144] A non-standard metric in the group of reals, Colloq. Math. 50 (1986), 241–248, MR0857859 (88h:54057). [145] Reducibility of lacunary polynomials VI, Acta Arith. 47 (1986), 277–293, MR0870670 (88e:11104). [146] Reducibility of lacunary polynomials VII, Monatsh. Math. 102 (1986), 309–337, MR0866132 (88e:11105); Errata, Acta Arith. 53 (1989), 95, MR1045457 (91a:11049). [147] (with A. Rotkiewicz) On the Diophantine equation x p + y 2p = z2 , Colloq. Math. 53 (1987), 147–153, MR0890851 (88e:11017). [148] A decomposition of integer vectors I, Bull. Polish Acad. Sci. Math. 35 (1987), 155–159, MR0908163 (88m:11050). [149] On the number of terms of a power of a polynomial, Acta Arith. 49 (1987), 55–70, MR0913764 (89a:12007). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 [150] Second order strong divisibility sequences in an algebraic number field, Arch. Math. (Brno) 23 (1987), 181–186, MR0930320 (89c:11028). [151] Reducibility of lacunary polynomials VIII, Acta Arith. 50 (1988), 91–106, MR0945276 (89f:11144). [152] A decomposition of integer vectors III, Bull. PolishAcad. Sci. Math. 35 (1987), 693–703, MR0961707 (90m:11096). [153] (with E. Wirsing) Multiplicative properties of the partition function, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), 297–303, MR0983622 (90b:11102). 1211 [154] Reducibility of lacunary polynomials IX, in: NewAdvances in Transcendence Theory (ed. by A. Baker), Cambridge Univ. Press, Cambridge 1988, 313–336, MR0972008 (90a:11117). [155] (with M. Misiurewicz) On n numbers on a circle, Hardy-Ramanujan J. 11 (1988), 30–39, MR1011768 (90i:11024). [155A] Postscript to the paper A. Rotkiewicz and W. Złotkowski “On the Diophantine equation 1 + pα1 + . . . + p αk = y 2 ”, in: Number Theory, Colloq. Math. Soc. János Bolyai 51, North-Holland, Amsterdam 1989, 929–936, MR1058252 (91e:11032). [156] Reducibility of lacunary polynomials X, Acta Arith. 53 (1989), 47–97, MR1045456 (91e:11119). [157] An analog of Hilbert’s irreducibility theorem, Number Theory (ed. by R. A. Mollin), Walter de Gruyter, Berlin 1990, 509–514, MR1106684 (92h:12003). [158] (with J. L. Nicolas) Localisation des zéros de polynômes intervenant en théorie du signal, in: Cinquante ans de polynômes, Lecture Notes in Math. 1415, Springer, Berlin 1990, 167–179, MR1044112 (90k:30007). [159] Un critère d’irréductibilité de polynômes, ibid., 212–224, MR1044116 (91k:12002).

1386

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[160] (with P. Erd˝os) On the greatest prime factor of xk=1 f (k), Acta Arith. 55 (1990), 191–200, MR1061638 (91h:11100). [161] Postscript to the paper of A. Makowski “On Stroeker’s equation”, in: Algebra and Number Theory (ed. by A. Grytczuk), Pedagog. Univ. Zielona Góra, Zielona Góra 1990, 41–42, MR1114364 (92f:11046). [162] Special Lucas sequences, including the Fibonacci sequence modulo a prime, in: A Tribute to Paul Erd˝os (ed. by A. Baker, B. Bollobás and A. Hajnal), Cambridge Univ. Press, Cambridge 1990, 349–357, MR1117027 (92f:11029). [163] Reducibility of lacunary polynomials XI, Acta Arith. 57 (1991), 165–175, MR1092768 (92f:11139). [164] (with S. Chaładus) A decomposition of integer vectors II, Pliska Stud. Math. Bulgar. 11 (1991), 15–23, MR1106550 (92g:11065). . . . . . . . . . . . . . . . . . . . . 1249 [165] A class of polynomials, Math. Slovaca 41 (1991), 295–298, MR1126666 (92m:11024). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846 [166] A decomposition of integer vectors IV, J. Austral. Math. Soc. Ser. A 51 (1991), 33–49, MR1119686 (92f:11086). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1259 [167] (with H. Niederreiter and L. Somer) Maximal frequencies of elements in second-order linear recurring sequences over a finite field, Elem. Math. 46 (1991), 139–143, MR1119645 (92k:11017). [168] (with A. Choudhry) On the number of terms in the irreducible factors of a polynomial over Q, Glasgow Math. J. 34 (1992), 11–15, MR1145626 (92m:11022). [168A] O pewnym zadaniu Wacława Sierpińskiego (On certain problem of Wacław Sierpiński), Gradient 1 (1992), No. 2, 6–9. [169] Differences between values of a quadratic form, Acta Math. Univ. Comenian. (N.S.) 61 (1992), 91–93, MR1205863 (93m:11028). [170] (with U. Zannier) Distribution of solutions of Diophantine equations f1 (x1 )f2 (x2 ) = f3 (x3 ), where fi are polynomials, Rend. Sem. Mat. Univ. Padova 87 (1992), 39–68, MR1183901 (93h:11040). [171] (with A. Grytczuk) On Runge’s theorem about Diophantine equations, in: Sets, Graphs and Numbers, Colloq. Math. Soc. János Bolayi 60, North Holland, Amsterdam 1992, 329–356, MR1218200 (94i:11025). . . . . . . . . . . . . . 93 [172] (with J. Wójcik) On a problem in elementary number theory, Math. Proc. Cambridge Philos. Soc. 112 (1992), 225–232, MR1171159 (93e:11006). . . 987 [173] On an analytic problem considered by Sierpiński and Ramanujan, in: New Trends in Probability and Statistics, vol. 2, VSP Utrecht 1992, 165–171, MR1198499 (93j:11062). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217 [174] An extension of the theorem on primitive divisors in algebraic number fields, Math. Comp. 61 (1993), 441–444, MR1189523 (93k:11107). . . . . . . . . . . . . 1098 [175] On reducible trinomials, Dissert. Math. (Rozprawy Mat.) 329 (1993), 83 pp., MR1254093 (95d:11146); Errata, Acta Arith. 73 (1995), 399–400, MR1366047. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466 [176] (with K. Gy˝ory) On a conjecture of Posner and Rumsey, J. Number Theory 47 (1994), 63–78, MR1273456 (95k:11027). . . . . . . . . . . . . . . . . . . . . . . . . . . . 549

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1387

[177] (with U. Zannier) Distribution of solutions of Diophantine equations f1 (x1 )f2 (x2 ) = f3 (x3 ), where fi are polynomials II, Rend. Sem. Mat. Univ. Padova 92 (1994), 29–46, MR1320476 (96a:11032). [178] (with T. Šalat) Remarks on maximum and minimum exponents in factoring, Math. Slovaca 44 (1994), 505–514, MR1338424 (96f:11017a); Errata, ibid. 45 (1995), 317, MR1361826 (96f:11017b). [179] (with J. Browkin) On integers not of the form n − ϕ(n), Colloq. Math. 68 (1995), 55–58, MR1311762 (95m:11106). . . . . . . . . . . . . . . . . . . . . . . . . . . 895 [180] (with U. Zannier) The least admissible value of the parameter in Hilbert’s Irreducibility Theorem, Acta Arith. 69 (1995), 293–302, MR1316481 (96f:12002). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 [181] (with I. Z. Ruzsa) An application of Kloosterman sums, Compositio Math. 96 (1995), 323–330, MR1327149 (96a:11099). [181A] (with U. Zannier) Appendix to the paper of M. Nair and A. Perelli “A sieve fundamental lemma for polynomials in two variables”, in: Analytic Number Theory, Progr. Math. 139, Birkhäuser Boston 1996, 685–702, MR1409386 (97j:11045). [182] O pokazatel nyh sravneniyah (On exponential congruences), Diofantovy Približeniya, Mat. Zapiski 2, 121–126, MGU, Moskva 1996, Ref. Zh. 1998 #1A149. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 996 [183] (with J. Browkin, M. Filaseta and G. Greaves) Squarefree values of polynomials and the abc-conjecture, in: Sieve Methods, Exponential Sums, and their Applications in Number Theory, London Math. Soc. Lecture Notes Ser. 237, Cambridge Univ. Press, Cambridge 1996, 65–85, MR1635726 (99d:11101). [184] On the Mahler measure of polynomials in many variables, Acta Arith. 79 (1997), 77–81, MR1438118 (97k:11036). [185] Triples of positive integers with the same sum and the same product, Serdica Math. J. 22 (1996), 587–588, MR1483607 (98g:11033). [186] A class of algebraic numbers, in: Number Theory, Tatra Mt. Math. Publ. 11, Bratislava 1997, 35–42, MR1475503 (98i:11089). . . . . . . . . . . . . . . . . . . . . . . 264 [187] On homogeneous covering congruences, Rocky Mountain J. Math. 27 (1997), 335–342, MR1453107 (98b:11003). [188] On pseudosquares, New Trends in Probability and Statistics, vol. 4, VSP Utrecht 1997, 213–220, MR1653611 (99i:11071). [189] (with D. Barsky and J. P. Bézivin) Une caractérisation arithmétique de suites récurrentes linéaires, J. Reine Angew. Math. 494 (1998), 73–84, MR1604460 (99j:11011). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1001 [190] (with I. Aliev and S. Kanemitsu) On the metric theory of continued fractions, Colloq. Math. 77 (1998), 141–146, MR1622796 (99d:11087). [191] A property of the unitary convolution, Colloq. Math. 78 (1998), 93–96, MR1658143 (99k:11010). [192] On Pythagorean triangles, Ann. Math. Sil. 12 (1998), 31–33, MR1673056 (99k:11044).

1388

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[193] (with J. Urbanowicz and P. van Wamelen) Class numbers and short sums of Kronecker symbols, J. Number Theory 78 (1999), 62–84, MR1706925 (2000g:11103). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1224 [194] (with S. Chaładus) On a linear Diophantine equation, Österreich. Akad. Wiss. Math.-Natur. Kl. Sitzungsber. II 207 (1998), 95–101, MR1749915 (2001e:11026). [194A] Remark on a paper of T. W. Cusick, Biuletyn WAT 48 (1999), No. 10, 5–9. [195] Reducibility of lacunary polynomials XII, Acta Arith. 90 (1999), 273–289, MR1715532 (2001b:11099). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563 [196] (with A. Rotkiewicz) On Lucas pseudoprimes with a prescribed value of the Jacobi symbol, Bull. Polish Acad. Sci. Math. 48 (2000), 77–80, MR1751157 (2001a:11013). [197] On reducible trinomials II, Publ. Math. Debrecen 56 (2000), 575–608, MR1766001 (2001k:11207); Corrigendum, Acta Arith. 115 (2004), 403, MR2099832. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 [198] On sums of three unit fractions with polynomial denominators, Funct. Approx. Comment. Math. 28 (2000), 187–194, MR1824003 (2002a:11029). . 116 [199] On the greatest common divisor of two univariate polynomials II, Acta Arith. 98 (2001), 95–106, MR1831458 (2002k:12002); Corrigendum, Acta Arith. 115 (2004), 403, MR2099832. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646 [200] On reducible trinomials III, Period. Math. Hungar. 43 (2001), 43–69, MR1830565 (2002g:11145). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 [201] (with A. Paszkiewicz) On the least prime primitive root modulo a prime, Math. Comp. 71 (2002), 1307–1321, MR1898759 (2003d:11006). [202] On the greatest common divisor of two univariate polynomials I, in: A Panorama of Number Theory or the View from Baker’s Garden (ed. by G. Wüstholz), Cambridge Univ. Press, Cambridge 2002, 337–352, MR1975461 (2004a:11021). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 [203] On a decomposition of polynomials in several variables II, Colloq. Math. 92 (2002), 67–79, MR1899238 (2003i:11048). [204] (with W. M. Schmidt) Comparison of L1 - and L∞ -norms of squares of polynomials, Acta Arith. 104 (2002), 283–296, MR1914723 (2003f:11035). . . . 1350 [204A] Appendix to the paper “Diophantine Equations and Bernoulli Polynomials” by Yu. F. Bilu, B. Brindza, P. Kirschenhofer, Á. Pintér and R. F. Tichy, Compositio Math. 131 (2002), 185–187, MR1898434 (2003a:11025). [205] An extension of some formulae of Lerch, Acta Math. Inform. Univ. Ostraviensis 10 (2002), 111–116, MR1943030 (2003i:11050). [206] A property of polynomials with an application to Siegel’s lemma, Monatsh. Math. 137 (2002), 239–251, MR1942622 (2004h:11022). . . . . . . 1274 [207] (with A. Paszkiewicz) Numerical calculation of the density of prime numbers with a given least primitive root, Math. Comp. 71 (2002), 1781–1797, MR1933055 (2003g:11109). [208] On a decomposition of polynomials in several variables, J. Théor. Nombres Bordeaux 14 (2002), 647–666, MR2040699 (2005e:11031). . . . . . . . . . . . . . 760

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[209] (with M. Skałba) On power residues, Acta Arith. 108 (2003), 77–94, MR1971083 (2005a:11005). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1012 [210] (with M. Filaseta) On testing the divisibility of lacunary polynomials by cyclotomic polynomials, Math. Comp. 73 (2004), 957–965, MR2031418 (2004m:11207). [211] On values of the Mahler measure in a quadratic field (solution of a problem of Dixon and Dubickas), Acta Arith. 113 (2004), 401–408, MR2079812 (2005h:11242). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 [212] On the congruence un ≡ c (mod p), where un is a recurring sequence of the second order, Acta Acad. Paedagog. Agriensis Sect. Mat. (N.S.) 30 (2003), 147–165, MR2054724 (2005a:11004). [213] (with K. Gy˝ory, L. Hajdu, Á. Pintér) Polynomials determined by a few of their coefficients, Indag. Math. (N.S.) 15 (2004), 209–221, MR2071857 (2005d:11036). [214] (with M. Skałba) On equations y 2 = x n + k in a finite field, Bull. Pol. Acad. Sci. Math. 52 (2004), 223–226, MR2127058 (2005j:11046). . . . . . . . . . . . . . 124 [215] (with Th. Bolis) Identities which imply that a ring is Boolean, Bull. Greek Math. Soc. 48 (2003), 1–5, MR2230423 (2007a:16042). [216] (with I. Aliev, W. M. Schmidt) On vectors whose span contains a given linear subspace, Monatsh. Math. 144 (2005), 177–191, MR2130272 (2005m:11128). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1288 [217] Self-inversive polynomials with all zeros on the unit circle, Ramanujan J. 9 (2005), 19–23, MR2166374 (2006d:30008). [218] (with S. Kanemitsu, Y. Tanigawa) Sums involving the Hurwitz zeta-function values, in: Zeta functions, topology and quantum physics, Dev. Math. 14, Springer, New York 2005, 81–90, MR2179274 (2006g:11179). [219] On weak automorphs of binary forms over an arbitrary field, Dissert. Math. (Rozprawy Mat.) 434 (2005), MR2229645 (2007a:11051). . . . . . . . . . . . . . . 779 [220] Reducibility of symmetric polynomials, Bull. Pol. Acad. Sci. Math. 53 (2005), 251–258, MR2213603 (2007b:12005). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 828 [221] On the reduced length of a polynomial with real coefficients, Funct. Approx. Comment. Math. 35 (2006), 271–306, MR2271619. . . . . . . . . . . . . . . . . . . . . 658 [222] (with M. Zakarczemny) On a linear homogeneous congruence, Colloq. Math. 106 (2006), 283–292. [223] Reducibility of a special symmetric form, Acta Math. Univ. Ostraviensis 14 (2006), 71–74.

B.1. Lectures on own or joint work [1] Reducibility of lacunary polynomials, in: 1969 Number Theory Institute, Proc. Sympos. Pure Math. 20, Amer. Math. Soc., Providence 1971, 135–149, MR0323749 (48 #2105). [2] Reducibility of polynomials, in: Computers in Number Theory (ed. by B. J. Birch and A. O. L. Atkin), Academic Press, London 1971, 73–75, MR0314733 (47 #3285).

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[3] Reducibility of polynomials, in: Actes du Congrès International des Mathèmaticiens (Nice 1970), t. I, Gauthier-Villars, Paris 1971, 491–496, MR0424768 (54 #12726). [4] Privodimost’ cˇ etyrehˇclenov, Trudy Mat. Inst. Steklov. 132 (1973), 143–144, MR0332707 (48 #11033); English version (Reducibility of quadrinomials): Proc. Steklov Inst. Math. 132 (1973), 163–165, MR0369228 (51 #5463). [5] Les extensions pures et les résidus des puissances, Astérisque 24–25 (1975), 69–74, MR0379433 (52 #338). [6] Les résidus de puissances et les congruences exponentielles, Astérisque 41–42 (1977), 103–109, MR0447173 (56 #5488). [7] Diophantine equations with parameters, in: Journées Arithmétiques 1980, London Math. Soc. Lecture Notes 56, Cambridge 1982, 211–217, Zbl. 0539.10017. [8] Le nombre de zéros des polynômes dans les anneaux de valuation des corps complets valués discrètement, Groupe d’étude d’Analyse Ultramétrique (Y. Amice, G. Christol, P. Robba), 9e année 1981/82, exposé no. 6, Inst. Henri Poincaré, Paris, 1983, MR0720554 (85c:11118). [9] Reducible lacunary polynomials, Séminaire de Théorie des Nombres Univ. de Bordeaux, Année 1983–1984, exposé no 29, Univ. Bordeaux I, Talence, 1984, MR0784075. [10] On reducible trinomials, Publ. Math. Univ. Pierre et Marie Curie no. 108 (1993), exposé no. 10.

B.2. Other papers [1] (with W. Sierpiński) O równaniu x 2 − 2y 2 = k, Wiadom. Mat. (2) 7 (1964), 229–232, MR0184904 (32 #2375). [2] (with S. Bergman et al.) Kazimierz Zarankiewicz, Colloq. Math. 12 (1964), 277–288, MR0174454 (30 #4658); Polish version — Wiadom. Mat. (2) 9 (1967), 175–185, MR0209106 (35 #10). [3] Liczb teoria, Wielka Encyklopedia Powszechna PWN 6, Warszawa 1965, 501–503. [4] O ró˙znych działach teorii liczb, Wiadom. Mat. (2) 9 (1967), 187–197, MR0227081 (37 #2666); Bulgarian version — Fiz.-Mat. Spis. Bulgar. Akad. Nauk. 13 (46) (1970), 130–138, MR0396377 (53 #244). [5] O pracach Michała Kaleckiego z teorii liczb, Ekonomista 1970, 1021–1022. [6] Równania diofantyczne, Wiadom. Mat. (2) 12 (1971), 227–232, MR0453627 (56 #11889); Serbian version — Uvodenje mladih u nauˇcni rad VI, Matematiˇcka Biblioteka 41, Beograd 1965, 29–34. ˙ [7] Zyciorys Wacława Sierpińskiego, Wiadom. Mat. (2) 12 (1971), 303–308, MR0396189 (53 #57). [8] Wacław Sierpiński’s papers on the theory of numbers, Acta Arith. 21 (1972), 7–13, MR0300846 (46 #9b); French version—W. Sierpiński “Oeuvres choisies”, t. I, Warszawa 1973, 65–72, MR0414302 (54 #2405); Polish version—Wiadom. Mat. (2) 26 (1984), 24–31. [9] (with J. Oderfeld) O pracach matematycznych Michała Kaleckiego, Wiadom. Mat. (2) 16 (1973), 71–73, MR0453459 (56 #11722).

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1391

[10] Postep w teorii liczb w latach 1966–1978, Wiadom. Mat. (2) 22 (1979), 1–11, MR0571455 (83e:10002). [11] O pracach H. Iwańca dotyczacych metody sita, Wiadom. Mat. 22 (1979), 13–16, MR0571456 (82f:01140). [12] Ułamki łańcuchowe, Delta nr 5 (65) (1979), 1–3. [13] Wacław Sierpiński a szkoła s´rednia, Matematyka 33 (1980), 68–69. [14] Commentary to the Bulgarian edition of Dirichlet’s “Vorlesungen über Zahlentheorie”, Sofia 1980, 567–578. [15] Wacław Sierpiński, Matematyka 35 (1982), 126–128. [16] Paul Turán’s work in number theory, in: Topics in Classical Number Theory, Colloq. Math. Soc. János Bolyai 34, North-Holland 1984, 31–48, MR0781134 (86e:11002). [17] Zwiazki miedzy własno´sciami lokalnymi a globalnymi w teorii równań diofantycznych, Wiadom. Mat. (2) 25 (1984), 169–175, MR0786112 (87b:11027). [18] Rola Wacława Sierpińskiego w historii matematyki polskiej, Wiadom. Mat. (2) 26 (1984), 1–9, MR0778895 (86i:01054a). [19] Teoria rozmieszczenia liczb pierwszych, Delta nr 4 (136) 1985, 12. [20] Propozycje terminologiczne, Wiadom. Mat. (2) 27 (1986), 148–149, MR0888159 (88f:00021). [21] Sprostowanie, Matematyka 40 (1987), 271. [22] 50 tomów Acta Arithmetica, Wiadom. Mat. 28 (1988), 81–83, MR0986063 (90c:01046). [23] Rozwój teorii liczb pierwszych w XIX w., Matematyka XIX wieku (ed. by S. Fudali), Szczecin 1988, 29–33. [24] Postep w teorii liczb w latach 1978–1988, Wiadom. Mat. (2) 29 (1990), 3–10, Addendum, ibid. 276, MR1111891 (92k:11003). [25] Rekordy i otwarte problemy w teorii liczb, Delta nr 3 (202), 1991, 1–3. [26] Wacław Sierpiński, Matematyka przełomu XIX i XX wieku, Prace Naukowe Uniwer´ askiego nr 1253, Katowice 1992, 9–15, MR1196940 (93k:01078). sytetu Sl [27] Progress in number theory during the years 1989–1992, Discuss. Math. 13 (1993), 75–80, MR1249152 (94m:11003); Polish version – Gradient 1 (1992), No. 7, 1–7. [28] Historia teorii liczb w Polsce w latach 1851–1950, Wiadom. Mat. 30 (1993), 19–50, MR1281604. [29] Prawda i istnienie w matematyce, Nauka – Religia – Dzieje, VII Seminarium Interdyscyplinarne w Castel Gandolfo, Kraków 1994, 65–76. [30] Solved and unsolved problems on polynomials, in: Panoramas of Mathematics, Banach Center Publ. 34, Warsaw 1995, 149–159, MR1374345 (97a:12001). [31] O liczbach Fermata, Gradient 5 (1996), 92. [32] Arithmetical properties of polynomials, The Mathematics of Paul Erd˝os, I, Algorithms Combin. 13, Springer, Berlin 1997, 151–154, MR1425182 (97k:11035). [33] Teoria liczb w Polsce w latach 1851–1950, Matematyka Polska w Stuleciu 1851–1950, Uniwersytet Szczeciński – Materiały Konferencyjne nr 16, Szczecin 1995, 61–67. [34] Sierpiński Wacław, Polski Słownik Biograficzny, tom 37/3, Warszawa – Kraków 1997, 356–359. [35] Pál Erd˝os 1913–1996, Nauka 1997, Nr 3, 253–255.

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[36] O liczbach doskonałych parzystych, Matematyka 50 (1997), 270. [37] Sto lat twierdzenia o liczbach pierwszych, Wiadom. Mat. (2) 33 (1997), 91–98, MR1615772 (99b:11001). [38] Jan Wójcik i jego prace z teorii liczb, ibid., 199–204, MR1615831 (99m:01087). [39] IX Problem Hilberta, Problemy Hilberta w pie´ s´mierci ich twórcy, cdziesieciolecie Warszawa 1997, 119–122, MR1632441. [40] XII Problem Hilberta, ibid. 147–151, MR1632444. [41] The Warsaw period of Voronoi’s creative work, Voronoi’s Impact on Modern Science, Book I, Proc. Institute of Mathematics of the National Academy of Sciences of Ukraine 21, Kyiv 1998, 29–33; Ukrainian version: Vpliv naukovogo dorobku G. Voronogo na suqasnu nauku, Kyiv 2003, 43–47. [42] Reducibility of polynomials over Kroneckerian fields, Number Theory. Diophantine, Computational and Algebraic Aspects (ed. by K. Gy˝ory, A. Peth˝o and V. T. Sós), Walter de Gruyter, Berlin 1998, 473–477, MR1628863 (99i:11097). [43] O tak zwanym twierdzeniu Chińczyków, Gradient 7 (1998), 214–215. [44] Georgij Woronoj – mistrz Wacława Sierpińskiego, XII Szkoła Historii Matematyki, Krynica 19–25 maja 1998, Kraków 1999, 155–161. [45] The Mahler measure of polynomials, in: Number Theory and its Applications, Lecture Notes in Pure and Appl. Math. 204, Marcel Dekker 1999, 171–183, MR1661667 (99k:11041). [46] 50 lat Olimpiady Matematycznej, Matematyka 52 (1999), 350–352. [47] Exponential congruences, Number Theory and its Applications, Kluwer Academic Publishers, Dordrecht 1999, 303–308, MR1738825 (2001e:11031). [48] Pie´ cdziesiat lat Olimpiady Matematycznej, Gradient 8 (1999), 323–332; Wiadom. Mat. (2) 36 (2000), 155–161. [49] Remark to the article of G. N. Sakovich “U sviti matematiki”, 6 (2000), no. 3, 46 (Ukrainian). [50] Teoria liczb w “Disquisitiones Arithmeticae”, Matematyka czasów Gaussa, 137–141, Zielona Góra 2001, MR1847691. [51] Erd˝os’s Work on Finite Sums of Unit Fractions, Paul Erd˝os and his Mathematics I (ed. V. T. Sós), vol. I, Budapest 2002, 629–636, MR1954717 (2003k:11055). [52] Teoria liczb w pracach Kummera, Kroneckera, Dedekinda i Webera, Matematyka czasów Weierstrassa, Szczecin 2002, 85–93. [53] Reducibility of polynomials in one variable over the rationals, IV International Conference “Modern Problems of Number Theory and its Applications”, dedicated to 180th anniversary of P. C. Chebysheff and 110 anniversary of I. M. Vinogradov, Current Problems, Part II, Moskva 2002, 143–150. [54] Monadologia E. Kählera, Nauka – Religia – Dzieje, XI Seminarium w Castel Gandolfo 7–9 sierpnia 2001, Kraków 2002, 119–122. [55] Niektóre osiagni ecia teorii liczb w XX wieku, Matematyka 55 (2002), 68–70. [56] Przeglad e´ osiagni c teorii liczb w XX wieku, Wiadom. Mat. (2) 38 (2002), 179–188, MR1985661 (2004b:11002). [57] Sto tomów Acta Arithmetica, Wiadom. Mat. (2) 38 (2002), 189–192.

Publication list

1393

[58] Wacław Sierpiński, Słownik Biograficzny Matematyków Polskich, Tarnobrzeg 2003, 214–216. [59] Algorytm Euklidesa i jego uogólnienia, Zeszyty Naukowe Uniwersytetu Opolskiego, Matematyka 31, Opole 2003, 163–169. [60] Nieskończono´sc´ w matematyce, Nauka – Religia – Dzieje, XII Seminarium w Castel Gandolfo 5–7 sierpnia 2003, Kraków 2004, 109–116. [61] Progress in the number-theoretic problems considered by Bouniakowsky, Viktor koviq Bunkovski (do 200 riqqa z dn narodeni), Proc. Institute of Mathematics of the National Academy of Sciences of Ukraine, vol. 53, Kyiv 2004, 94–100. [62] Całki pseudoeliptyczne i równanie Pella dla wielomianów, Matematyka abelowa w dwóchsetlecie urodzin Nielsa HenrikaAbela (1802–1829), Nowy Sacz 2004, 45–48. algebry Netta i Webera, Sławne dzieła matematyczne i rocznice, Białystok [63] Podreczniki 2005, 153–157. [64] (with W. M. Schmidt) The mathematical work of Eduard Wirsing, Funct. Approx. Comment. Math. 35 (2006), 7–18; Corrigendum, ibid. 36 (2006), 133.

C. Books [1] Wacław Sierpiński, Warszawa 1976, 1–50. [2] Selected Topics on Polynomials, XXII+250 pp., University of Michigan Press, Ann Arbor 1982, MR0649775 (84k:12010). [3] Polynomials with Special Regard to Reducibility, X+558 pp., Encyclopaedia of Mathematics and its Applications 77, Cambridge Univ. Press, Cambridge 2000, MR1770638 (2001h:11135).

Selecta

Kvant Selecta: Combinatorics, I

Obra selecta

Masatoshi Fukushima: Selecta

Masatoshi Fukushima: Selecta

Masatoshi Fukushima: Selecta

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Anticonceptie (Capita Selecta)

Selecta expository writing

KVANT selecta: combinatorics 1

Masatoshi Fukushima: Selecta

Opuscula Selecta Neerlandicorum De Arte

Kvant Selecta: Algebra and Analysis

Kvant Selecta: Algebra and Analysis

Opuscula Selecta Neerlandicorum De Arte

KVANT selecta: algebra and analysis, 1

Inequalities: selecta of Elliott H. Lieb

KVANT selecta: algebra and analysis, 2

Andrzej Schinzel, Selecta (Heritage of European Mathematics)

Selecta. 2 Vols: Diophantine problems and polynomials

Inequalities: Selecta of Elliott H. Lieb

Hans Freudenthal: Selecta (Heritage of European Mathematics)

Milestones in matrix computation. Selecta of G.H. Golub

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The Stability of Matter: From Atoms to Stars: Selecta of Elliot H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

Selecta

Kvant Selecta: Combinatorics, I

Obra selecta

Masatoshi Fukushima: Selecta

Masatoshi Fukushima: Selecta

Masatoshi Fukushima: Selecta

Selecta: Expository Writing

Anticonceptie (Capita Selecta)

Selecta expository writing

KVANT selecta: combinatorics 1

Masatoshi Fukushima: Selecta

Opuscula Selecta Neerlandicorum De Arte

Kvant Selecta: Algebra and Analysis

Kvant Selecta: Algebra and Analysis

Opuscula Selecta Neerlandicorum De Arte

KVANT selecta: algebra and analysis, 1

Inequalities: selecta of Elliott H. Lieb

KVANT selecta: algebra and analysis, 2

Andrzej Schinzel, Selecta (Heritage of European Mathematics)

Selecta. 2 Vols: Diophantine problems and polynomials

Inequalities: Selecta of Elliott H. Lieb

Hans Freudenthal: Selecta (Heritage of European Mathematics)

Milestones in matrix computation. Selecta of G.H. Golub

Semigroups and Automata: SELECTA Uno Kaljulaid (1941- 1999) (Stand Alone)

Selecta. 2 Vols: Elementary, analytic and geometric number theory: I,II (2 Vols.)

The Stability of Matter: From Atoms to Stars: Selecta of Elliot H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

The Stability of Matter: From Atoms to Stars: Selecta of Elliott H. Lieb

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