Selecta. 2 Vols: Elementary, analytic and geometric number theory: I,II (2 Vols.)

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schinzel_vol_2_titelei

5.3.2007

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

1031

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 . . . . . . . . .

1103

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

1033 1036 1046 1059 1066 1090 1098

. 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.

Diophantine equations and integral forms

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 . . . . . . . . . . . . . . . . . . . . . . . . On the Diophantine equation nk=1 Ak xkϑk = 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

. .

283

. . . . .

. . . . .

295 301 333 344 381

. . . . .

. . . . .

403 409 447 450 466

. . . . . . .

. . . . . . .

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 . . . .

. . . .

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

Part G Arithmetic functions

Andrzej Schinzel Selecta

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 Makowski and Schinzel examined in G5 the lim sup and lim inf of the functions 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 true for all n ? This remains open, the best result to date being σ (ϕ(n))/n 39.4 and due 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.

Arithmetic functions

865

[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 (i) σ (Ai,k ) σ (mk + i) 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 = p 6k+1 (p − 1) et p1 = p6k+1 (p − 1) + 1 > p 2 + 1 > p2 − p + 1, et comme, d’autre part p1 = p 6k+2 − p 6k+1 + 1 = p 6k (p 2 − p + 1) − (p 6k − 1), p6 − 1 | p 6k − 1,

p 6 − 1 = (p 3 − 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 ϕ(xi ) = p1 1 p2 2 · · · pi−1 (p1 − 1) · · · (pi−1 − 1)(pi+1 − 1) · · · (pk − 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

Let g = [2l / l 2 ]. Given distinct primes q1 , q2 , . . . , qg greater than x 1/2 with q1 q2 · · · qg x + 1, the number of integers n x satisfying n ≡ a (mod b) and x q1 · · · qg | n + 1 is at most + 1. Thus, by theorems of Chebyshev and Mertens, aq1 · · · qg

880


the number of integers of the second class is less than 1 g x 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

Q = h! N 2 P h! N 2

x+hq>(x+h)/p

x+hq>(x+h)/p

μ q

μ q

x+hq

∞

pk, p/| N

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 1 1 x +O 1+ 1 . + h 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 let (1/p) be divergent,

f (p) /p convergent, then the distribution function of 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 p 1/20 . 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

(1 − ε)

>

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 lim sup 1 , · 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 σ ϕ(234 − 4) 233 − 1 1 1 = = + 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 pα−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 c(σ ) a in L. For each σ ∈ G, one has σ (α)/α = ζσ ∈ ζn , and σ (ζn ) = ζn with c(σ ) ∈ Z. c(σ ) = σ (τ (α)/α) = τ σ (α)/σ (α); hence For any σ , τ ∈ G one has τ (α c(σ ) )/α c(σ ) = ζτ c(σ ) α /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 c(σ )−1 if and only if d divides all numbers c(σ )−1, if and only if all elements σ (ζd )/ζd = ζd 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, p α ) = 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 (30 − i) = 1 : valeurs envisagées dans la table suivante, pour lesquels on a n − i, 30 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 + 1, a(p ¯ j ) = a(pj ) − pj . αj = 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 ¯ j0 ) (mod pj0 ), d’où, D’après la condition 1◦ il existe un j0 l tel que i0 ≡ a(p 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

donc i(t) a(p ¯ j0 ) + [k/pj0 ]pj0 − pj0 , d’où 0 < −a(p ¯ j0 ) + pj0 + i(t) k < pj0 0 . 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 = p α , 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 γ 2 · · · 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. As l | α, l /| β for any rational integer x: l /| αx − β. Hence, for all p ∈ P , for all x: c 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 τ the greatest integer such that ζ2τ + ζ2−1 ∈ K. Let n1 , . . . , nk , n be positive inteτ 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 αi i i = γ n ; (i) β 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 γ βα u = ζ2τ + ζ2−1 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, , 0, . . . , 0 (1 i q = min(r, s)), ⎪ ⎪ (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]. / K then Proof. Set ζ4 = i, ϑ = α + βi, α, β ∈ K. If i ∈ K we have (i); if i ∈ (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 ϑ =β = β (−1) 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 ϑ n = αn ν = α n (−1)ν 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 = + ∈K ζ2μ−1 + ζ2−1 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 if μ = τ + 1, ν ≡ 1 mod 2, then ϑ n = − ζ2τ + ζ2−1 γ and n ≡ 2τ mod 2τ +1 ; τ +2

n/2 if μ = τ + 1, ν ≡ 0 mod 2, then ϑ n = ζ2τ + ζ2−1 γ n, 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 mod p x 2 ≡ ζ2τ + ζ2−1 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

αi i

= (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 Here γ = δ means that γ /δ is an nth power in K and ω = ζ2τ +ζ2−1 . Moreover, in 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

The proof is complete.

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

= γp ; (i) α p (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 (v) p = 2, ω = 1, ν τ + 2, α = − ζ2τ + ζ2τ + 2 γ .

Here ω is the greatest integer such that ζpω ∈ 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. ( If the binomial in question is irreducible (iv) implies − ζ2τ + ζ2−1 / K, λ = 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 = p ν 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 α ζ w i c0 = ζ(w,n) , = ζnci . p n p n If n = pν , 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 / K and ζ4 ∈ K ∗ ξ . ηp ∈ K ∗ ξ and η ∈ K(ξ ) then either η ∈ K ∗ ξ or p = 2, ζ4 ∈ ∗ ∗ 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 ϑ 2 = − ζ2τ + ζ2−1 γ ; τ +2

2ν−1 2ν −1 2ν ν > τ, ϑ = ζ2τ + ζ2τ + 2 γ .

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

ζ2τ + ζ2τ + 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 or ζ2τ + ζ2−1 γ , γ ∈ K, −1 = ζ22τ +2 = γ 2 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. / K and ζ2τ is conjugate over K to ζ2−1 If ζ2τ +1 ∈ K(ζ4 ), then ζ4 ∈ 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

ζpκ = 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(ζ j

: K] | p − 1, it follows that ζp ∈ K. By Lemma 4 we

p)

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 NK(ζ4 )/K x 2 c − η2 + 2δx 2 + 2δ 2 | x 2 + 4δ 4 , = x2 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 β = −γ 2 .

−1 = δ 2 β j , 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 α = ± ζ2τ + ζ2−1 γ , 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 − α = α Xp μ , 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 ,

hence for τ 3, α = −γ 4 , γ ∈ K. Thus (iv) implies τ 3, − ζ2τ + ζ2−1 / K, 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

(p ω + 1)j − 1 ≡ 0 mod p ν . 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 pλ . 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 − ζpω γ

j =0

ν−ω

ω −1 p

μ=λ+1

j =0 (j,p)=1

μ

j

x p − ζpω 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 p μ (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 − ζ2 τ 1 γ 2

j =1 j ≡1 mod 4 ν−τ +1

j

ν−τ

If fj (x) = x 2 1 − ζ2τ1 γ 2 1 were reducible over K(ζ4 ), then since ζ2τ1 +1 ∈ / K(ζ4 ) we should have ν = τ1 = τ + 1 and j

ζ2 τ 1 γ = ϑ 2 , 2τ

c

2τ +1

ϑ ∈ K(ζ4 ),

whence −γ = ϑ contrary to Lemma 2. Thus fj (x) is irreducible over K(ζ4 ) and 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 → ζ2 ν

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)

c ord2 (1 + 2τ )r − 1 = ord2 r + τ 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

T S(ξ ) = T (ζ2ν )T (ξ ) = ζ

−j (1+2τ ) 2ν

j

ζ2ν−τ +1 ξ = ζ

ξ,

−j (1+2τ −1 ) 2ν

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 S1 ζ2τ +1 + ζ2−1 + ε 1 ζ2 ν = 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 Hence S1 S2 ξ = S2 S1 ξ and the group in question is not abelian. Therefore, if n = piνi 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 piνi , w i = wpνi . and let again n = 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 / K in which case the group groups are cyclic of order piνi unless piνi ≡ 0 mod 4 and ζ4 ∈


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 p = 2, ω = 1, ν = τ, α = − ζ2τ + ζ2−1 γ ; τ +2 ν−1

ν 2 p = 2, ω = 1, ν > τ, α = ζ2τ + ζ2−1 γ2 , 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

μp max ordp [K(ζq s ) : K]. 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 pν n, the binomial x p − α is abelian over K and by Theorem 2 c

(30)

pν−λ

α = ζpκ γp

;

κ λ min(ν, ω),

γp ∈ K.

where ω has the meaning of Theorem 1. Suppose first p ν ≡ 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 p ν ≡ 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 ) ) )

G1 = Gal K(ζn , n ζw , n α1 , . . . , n αk /K(ζn ) c 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 di = K(ζn )∗ ξ1 , . . . , ξk : K(ζn )∗ ξ1 , . . . , ξi−1 , ξi+1 , . . . , ξk c 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 , ξ0 → ξ0 ,

ξj → ξj

ξj → ξj

(1 j k);

(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 n (w, cr+1 , . . . , ck ) n cs ≡ − bh0 t0 + dhs ts + t0 + dhs ηs mod n. 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

dhs ps , = 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 t k 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 p s 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) for r = q − 1, q; n(1) r = nr 1 (0) = nq−1 + (n(0) − p + 1), n(1) q = p − 1, p q


965

verify (51) and find again (1)

(0)

(0) f (n1 , . . . , n(1) s ) < f (n1 , . . . , ns ),

which is impossible. (0) (0) Since n1 + . . . + 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 / K, n ≡ 0 mod 2τ +1 , with the property asserted in Theorem 3. In particular if ζ4 ∈ (

−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 ≡ x1 + . . . + xp−1 mod p,

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 | al . Moreover we set and A(Tj ) = 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 sequences ai satisfying ai = n − 1 and consider a sequence ai with ai = n 2. In 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 (8) σν = μ < κ μ . 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 condition on x ν + ανj x ν −j that 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 a contradiction. Therefore cs = 0 for all s > k and taking m = 2wek , tj = − for 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 sequence π1 , π2 , . . . that any non-zero element of K is represented uniquely as ζ πsxs , 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 p γp +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 (5) a l − b ≡ a − b mod l, p lαp ≡ p αp 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 p ep p lαp p min{0,βp } p βp s, a l − b = sgn a 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.

H8. On a problem in elementary number theory

991

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

− b (where n μ). Consider now the number a l 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 b = bq2 ξ −G(n)η ≡ bξ a2 mod p2 ,

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 :

)a ≡ 0 mod l det(A1 ) br+1 − (B1 A−1 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 c En outre, pour tout t ∈ Z, on a (at + b, l) = 1. Comme l est sans facteur carré, si l /| a, 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) T α1 , . . . , αr ≡ 0 (mod P)


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) T β1 c , . . . , βs ≡ 0 (mod P) donc

T

ce qui implique

β β1 s ,..., p wm p wm

≡ 0 (mod p)

β β1 s N T ≡ 0 (mod N (p)). ,..., 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 . e b1 Posons = , avec a1 , b1 ∈ Z, a1 > 0 et (a1 , b1 ) = 1. d a1 a1 (d,e) β Comme = 1, on a β a1 = ζwc1 α b1 . 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 Φl (β n ) = Φl (α kλn/ l ) = Φld (α kn/(lτ ) ) = Φkd/τ (α n ). 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 Np 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} with gj = g and |A| odd equals the corresponding number with |A| even, hence there j ∈A is an involution σg of the family of subsets A of {1, . . . , l} with gj = g such that 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 g= gj .

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 /J such that exists χ1 ∈ J 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 prime ideals in question has a positive Dirichlet density. Hence for every χ1 ∈ J /J n there 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 (Np − 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 (Np0 − 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 + Np0 + . . . + 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 +...+Np0 0

= ζni

n −1 bj (1+Np0 +...+Np0 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 ) subgroup of K ∗ generated by αi (1 i k) and K(ζm )∗m , resp. 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 ∗ , (42) (−1)|A| + (−1)|A| = 0, 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 βj20 γj−2 if j0 ∈ A0 , 0 βj = (47) −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 ,

hence β3−j ∈ which gives (10) with i = 3 − 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

cj dj c3−j d3−j

σm =

if q | m, if q /| m.

If (12) holds, we take

if 4 | m, if 4 /| m.

cj dj c3−j d3−j

σm = 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 (ζ2τ ε0 ∈ 1, (−1)2

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

1037

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 = ±(p α + 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 xQn,m (x 2 ) (4) Pn,m (x 2 ) − (−1|m)m = (x − ζns ) (x + ζnt ) = ψn,m (x) (m odd), (5)

s

1/2

Pn,m (−x ) − i (−1|m)m 2

t 2

xQn,m (−x ) (x + iζns ) (x − iζnt ) = ψ2n,m (x) = 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) Am,m (x) and Am,m (x) are integers of the field generated by {(−1|m)m}1/2 , and corresponding 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)). (ε)

It follows that the coefficients of An,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.


1039

(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 φn (x 2 ) = Pn,1 (x 2 ) − x 2 Qn,1 (x 2 ),

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).

1040


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.


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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 (1) (−1) we deduce that any common prime factor of φn (a, b) and φn (a, b) must divide 2νB. (1) By (23) and Zsigmondy’s theorem, quoted in §1, each prime factor of either φn (a, b) (−1) or φn (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 φν (A, B) A − ζνr 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 (δ) (δ) φ (ε) = φν Aq(l) , B q(l) /φν (A, B), where δ = ±1 depends on ε and on the residue classes of k(ab) and of q(l) (mod 4). Using the inequalities

(δ)

(x 1/2 − 1)φ(ν ) φν (x) (x 1/2 + 1)φ(ν )

(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) − B 1/2 , ( 18 A + 18 B)1/2 , |Q(ε) n (α, β)| > max A )

ϕ(ν −1/2 |Q(ε) (A − B)Aq(l)/2−3/2 (l 3, ν = ν/q(l)). 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) (−1) then the numbers δQn (α, β) and δQn (α, β) are coprime rational integers (2 ). 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.

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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 L { 1 ϕ(ν)} A + B 1 ϕ(ν) 1 1 4 2 (α + β)2[ 4 ϕ(ν)] δR(A, B) = (A + B)− 2 ϕ(ν) R(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 δQ(ε) (α, β) = δR(A, B) ± κ(α + β)2 αβ δ n α + β αβ A+B and the numbers AB 1/2 A+B , = ±(αβ)(n−ν)/2ν , α+β αβ

κ(α + β)2 αβ

1/2

=κ

LM 1/2 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 (ε) the numbers r −1/2 Qn (α, β) are algebraic integers. First, since κ = r 1/2 , it is clear from 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) (ε) r −1/2 Qn (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) (−1) factor of δQn (α, β) and δQn (α, β) must divide 2νM. By Lemma 1 and (10) each (ε) prime factor of δQn (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 |x − ζmr | < exp(2m1/2 log2 m). Π= 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 N ζ log m (13) 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=1

k−1

i

i

−1 i

|cos i | + |sin i | cot π

π + = π

sin

i sin k k k 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 (α − β) 1 |Q(ε) 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, (ε) |Qn (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 |Q(ε) )L. 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).

(ε)

On the other hand, by (5) and (6), Qn (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

Let S(χ) = max

χ (n) . We shall need the following lemmas. 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 2 1/2 1/6 dβ dβ k + k 1/2 3 1/2k β 1/6 β 2 1/2 k 2 1 = 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 S(χ1 ) < 1 + k 1/2 log k + log log k + 1 + . 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 S(χ ) S(χ1 ) |μ(d)χ1 (d)|. 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 (A1) |μ(d)χ1 (d)| < . 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 (A2) |μ(d)χ1 (d)| |μ(d)| = 2l0 +l , 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 if n has a primitive root, 1 mod n 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 = p μ , p odd, if n = 2p μ , 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 ν−2

ν−2

U (x, y) = y 2 T (x, y) = x 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 χ (−1)f . Moreover, for n odd ωn (χ ) = ωn (χ )2χ(2) , thus Φn(ε) (χ; x 2 , y 2 ) = ωnε (χ )

n

(x 2 − ζn2r y 2 ) = ωn2εχ(2)

r=1 χ(r)=εχ(2)

n

(x − ζnr y)

r=1 χ(r)=εχ(2)

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 , )

√ (25) Q(ε) (ε = ±1) 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) (ε) defined by formula (6) there coincide with Qn (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 (ε) also δ1 Qn (α, β) are rational. (ε) They are also obviously algebraic integers, thus they are rational integers. δ12 Qn (α, β)2 are perfect squares and since they are divisible by a square-free number δ12 they are di(ε) visible by its square δ14 . Thus δ1−1 Qn (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 Qn (α, β) and δ1−1 Qn (α, β) 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) otherwise; Φn (χ2 ; α, β) 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 δ2 Q(ε) c − εζ4 β 2 ) 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 (ε) δ2 Qn (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 (30) max |Q(−ε) (α, β)|, |Q(−ε) (α, β)|, |Q(−ε) (α, β)| < |α|ϕ(n)/2 exp(4n1/2 log2 n). 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 (α, β)Q(−θ) (α, β)Q(−εθ) (α, β) (δδ1 δ2 )−1 Q(−ε) 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)

) Q(ε,θ (α, β) = 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)

) Q(ε,θ (α, β) = 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 ,θ )

(ε ,θ ))

the greatest common factor of Qn 1 1 (α, β) and Qn 2 2 (α, β) for ε1 , θ1 = ε2 , θ2 c divides at least two of the numbers

(1) (1) Qn (α, β)2 , Q(−1) (α, β)2 , (α, β)2 , Qn (α, β)2 , Q(−1) n n

(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 (ε,θ) {1, 1, 1} according to whether n = q ν , 2q ν or otherwise. It follows that Qn (α, β) are 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

are relatively prime in pairs. In virtue of Lemma 2 of [8] there primes of Q(ζe ) and 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) (ε) Qm (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 (ε) (θ) Qm (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 (ε) (em)em . Since (ω, ω) = 1 it follows by (36) that any common prime factor of Qm (ω, ω) (θ) and Qm (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

⎧ ⎪ if K ≡ 0 mod 8, ⎨α, n ω, m = ζ4 α, n/2 if L ≡ 0 mod 8, ⎪ ⎩ ζ4 α 2 , n/4 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 ωe Ω r ω s (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 −ν 1/2 9 u/σ , ζ ν Δ1/2 β u/σ if u 0, Δ α ζ 2e 2e 1 1 Ω(ω − ω ), Ω (ω − ω) = 8 −ν 1/2 9 ν Δ1/2 α |u|/σ ζ2e Δ1 β |u|/σ , ζ2e if u < 0. 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

Nij ζe < (2n)1/2 log 2n (1 h < e, 1 i kd ). (A5)

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) e ϕ(n) − >l> , e 2 2e hence we may choose an integer m prime to e/2 such that

(A9)

ϕ(n) ϕ(n) e 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), n e−1

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 A Ad 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 i1 − 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

exp (m − l) log

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

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

1091

A α A Let Q = K0 , = , 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 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 N d−1 (|β σ | + |α σ |) 2d ew(α/β) . 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 N d2|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 p d/(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

max σ , 1 log 1 + = log > . w β β 52d log 6d 52d log 6d + 1 σ ∈S

α If is not an integer, then (β) = d and β α log Nβ − log N d log 2. w 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 as the numerator and the denominator of in its reduced form. Set d = d1 d2 Q 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.

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 Pn (α, β) was proved for n > n0 (L, M) (given explicitly). Since for n > 2, Φn (α, β) ∈ Q 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 1 if KL ≡ 0 mod 27, η3 = 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 max(|a|, |b|, |c|) c(ε) p . 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

exp (1 + ε)ϑ q(f (x)) p ,

(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

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[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. x k=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.

[16] P. Erd˝os, On the greatest prime factor of

[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 X 2 + 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.

1120

J. Prime numbers

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

1122

J. Prime numbers

(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 ap2 + 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 (p 2 − 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 ap 2 + (d − a) = dq, d’où

(p 2 − 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.

1124

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 f1 (x) ≡ 3 (mod 8), d’où = −1 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


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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.


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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.

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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


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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.


1133

[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

1135

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: p α , 2pα , α 4p , p α q β , 2p α 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. On a conjecture of Murphy. An. Soc. Paran. Mat. (2) 3 (1960), 13. [12] A. Makowski, [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): N ω(p) 1 −k −1 (1) P (N) ∼ 1 − h · · · h ) , 1 − (h 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.

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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 (165) p exp δ1 q(d) + o(q(d)) 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.

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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.

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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 , β= . α= √ √ |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 Gk (x) = x d F − − k = a0 1 − x − xHk (x) , x (d − 1)a0 (d − 1)a0 a1 where Hk (x) is a polynomial, Hk (0) = dk and if F − − k = 0, Hk (x) is of (d − 1)a0 degree d − 1 with the leading coefficient a1 −1 −a0 F − −k . (d − 1)a0 Clearly

Gk (x) 1 a1 (168) F Hk (x) − k ≡ F − + − −k 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) lim P (ν) 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), where G1 , G2 are cubic polynomials with integer coefficients. Applying again

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 product as (l.c., formula (7.10)). Hence Theorem 15 takes the form: 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) lim 41 21 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 p 2 or P (t0 + p) ≡ 0 mod p 2 . 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 ax + y = (a1 x + b1 y)ω1 + (a2 x + b2 y)ω2 , 2

hence F (x, y) = F1 (a1 x + b1 y, a2 x + b2 y) and by (1)

a1

a2

b1

= ±1. b2

√ b+ Δ It follows that a, is itself a basis for a and by a well known result 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: Np | 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|Np 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 The solubility of the equation N (ω) = f (t) for all t ∈ Zr implies, by Theorem 1 A of [14], the existence of a polynomial ω(t) ∈ K[t] such that F (x, y) =

(10)

Na N ω(t) = f (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,

1160

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 F X1 (t), Y1 (t) = N ω1 (t) = N γ1 N ω(t) = N c · f (t) = f (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 Nα A = 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 +

√ √

( 21 ω4 − ω)t 2 + ω − 8 .

−23 −23

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.

1164

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 ) t 2 ∈ Nr such that t 2 ≡ t 1 mod m, all the ideals are prime in K, distinct and do 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

1166

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 (21)

Δ,

n

fν (τ 0 ) = 1.

ν=1

Let fν (τ 0 ) ≡ ν mod Δ, ν > 0 (ν k). Without loss of generality we may assume that Δ Δ = 1 (1 ν j ), = −1 (j < ν k). (22) 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 = −1 (27) 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 By (25) and (26) the discriminant of fi (t, τ ) equals D(τ ) ≡ 0 mod pν . Since 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

m, Δ

ν=k+1

n N ψν (t 1 ) =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/j cx 1/n − 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

G(x, y), xy = 1. F (x, y) = x k y l G(x, y), k 1, l 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 F ξ ,η = F ,1 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 .

1172

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 ax0 + y0 b + √Δ 2 N = |f (μt 2 + τ )|, where a = a, . 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 (50) ax0 + , b, y0 a = b = 1. 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.

1174

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, ζ |μ

|f0 |

(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) (58)

ai x + bi y = ci

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 Ds (t 2 ) = 0, (60) (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 Ds (t 1 ) = 0 f (μt 1 + τ ) 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 e(a1 x + b1 y) + , (a2 x + b2 y) g = b 2 C(ψν ) ν=1

g + √Δ , where g = e, 2 aNb = (B),

(62)

αν + βν + βν = eν

(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 (70) Dσ s (t 1 ) ≡ 0 mod m 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 (72) Es (t 1 , εσ ) ≡ 0 mod m, 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 f (μt 1 + τ ) Es (t 1 , εσ ) = 0 s∈S0 r is soluble in integers σ . Thus for every t 1 ∈ Z either f (μt 1 + τ ) = 0 or f (μt 1 + τ ) = 0 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

a0 b0 Ψ0s (t 1 )

√ √

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,

b0 a0

√ √

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 + 2 + 1 . (76) 2T + 1 > |f |(|S0 | + 1) 3|f | 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 (2T + 1)r 3|f | + 2 + 1 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 f (μt + τ ) Es (t, εσ0 ) = 0 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 F (x, y) = (a0 x + b0 y)N e(a1 x + b1 y) + (a2 x + b2 y) e−1 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].

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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)

Since, by Lemma 5, n

w v

f0 (μt + τ ) =

e n ψν (t) ν . B v C(ψν )eν

w

ν=1

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 ) and, by Lemma 5, H implies the existence of a t 2 ≡ t 1 mod m such that (ν n) 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 ) = N b

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 p n ,

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

It is easily seen that f (t) ∈ Z[t] and pn (109) |f | = p(n − 1) c

v(t) = cu(t) + p n−1 .

if α > 0, if α = 0.

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) = pn−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) ≡ p n−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 h(d) = − kχd (k) = χd (k), |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 ων = , ω0 = ω, 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 = (−1)ν Qν /2. αA2ν−1 + βAν−1 Bν−1 + γ Bν−1

(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 Bl−1 ω + = Bl−1 . + (17) ε0 = 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 √ max (xi + 1) = √ M |a| d d

where maximum is taken over all non-decreasing sequences of at most l real numbers satisfying √ xi ε0 . 2 xi 21 d − 1 = D, 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) if 4θ < D, ⎪ ⎨2 √ 1 M = 2 e d + 2θ + 4 if 2θ < D 4θ, ⎪ ⎪ ⎩1 √ if D 2θ. 2e d + θ + 7 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−

1 √ log ε0 log θ log 4 log 4θ =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 1 − > − η √ √ . 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 1 h0 log ε0 + 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 ω p(n) m(N ) (1 − ε) 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 − log x g(x) = c2 x + log + log 1 − √ 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)/2 r

i+1 i r−1 r−1 +1+j −i/2 −1 2 + j 2r 2 = < =1+ 3r. i i r i j =0 j =0 +j +j 2 2

Similarly

−1/2

(r − 1)!−1 r, r!

r

−1

1/2

(r − 1)! r!

4r. r

Therefore

∞ √

1/2 1/2−r

4r i

ξ 1 + + O 2r exp(−c4 N ) O Δr = c2 r!

√

r c2 ξ i=1

√

1/2 1/2−r

r

ξ = c2 r!

1 + O + O 2r exp(−c4 N ) . √

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 1 + O(N −1/3 ) + O[exp(−c4 N /2)] Δr = c2 r!

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

xr ∈ N0 , xr

where

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 1 N R 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 + S2

(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) e−(n+u)/y du + O(log T ). =s 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

2m+1

2m

c

nT0

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 3 = P1 ,

(2.11)

η = −ε,

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 character mod pi P , non-real since ζϕ(P ) ∈ 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 , if p = 3, g3 g17 ⎪ ⎪ ⎪ ⎨g 2 g (3+ε)/2 , if p = 5, k ≡ ±2 (mod 5), 5 17 ψ(x) = (7−ε)/2 ⎪ g g , if p = 5, k ≡ −1 (mod 5), and either f = 170 or ε = 1, ⎪ ⎪ ⎩ 5 17 otherwise. g3 g5 g17 , 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, ⎪ ⎪ ⎪ ⎪ ⎪g g 2 , ⎪ if α = 4, k ≡ 1 (mod 5), ⎪ ⎨ 5 2α 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) ¯ χ (n)nm−1 = −Bm,χ r m−1 + ψ(−N )Bm,χψ (N ), (3.1) mr m−1 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

,

|M1 | = abs

> 0, |M1 | |M2 | = 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 X− Y 1. H : |X| 1, |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 M2 M2 |M1 | − |M2 | 1 3 , ± . , ± 1, − sgn , − sgn ± 1, − 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 y2 y2. (3) min f (x, y) = x 2 , min f (x, y) = 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 + b 1 y , ( 4 n 3 3

Y =

a2 x + b2 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 h(p)h(q) < l(n), 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 K : |X| 1, |Y | 1, X − Y 1 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 otherwise. n3 , 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 pj ’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 , . . . , p l ∈ 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 b1 = ε i a i λi , i∈I

where λi 0

(5) and by (2) and (3)

(6)

(i ∈ I )

λi = 1.

i∈I

Therefore, (7)

vol P

−1

= 2 det εi a i λi , b2 , . . . , bl

i∈I

−l

=2

λi det(εi a i , b2 , . . . , bl )

. −l

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 p 1 , . . . , p l ∈ 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 p1 , . . . , p l ∈ Zk and a matrix U ∈ Mm,l (Q) such that ⎛ ⎞ ⎛ ⎞ b1 p1 ⎜ .. ⎟ ⎜ .. ⎟ (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 ⎛

⎞ ⎛ ⎞ n1 p1 ⎜ .. ⎟ ⎜ .. ⎟ = 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 Δni =

if μ = i, m

0 otherwise;

nμ Δμ (ni ),

μ=1

and hence (19)

ni ∈ |Δ|1−(k−m)r Dr (K)

(1 i m).

/ N we have Δx = On the other hand, if x ∈ λDr (K) ∩ Zk and x ∈

m

nμ Δμ (x), and

μ=1

c

m A A thus by the assumption about K, AΔx − nμ Δμ (x)AK 1 and by the definition of 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 k A m 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 k A m 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

1

lim Ir,m =

r→∞

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 l κl (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

(m < i l).

aiμ nμ

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)

h

l

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

|xi |h(ni ) 1. However by Lemma 2,

m

vol x ∈ R : h m

i=m+1

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

)

(29)

d(F1 ) g1 (m)H (n1 , . . . , nm ).

Consider now the quadratic form l F (x1 , . . . , xl ) = (l − m + 1)F1 . . . , xμ + aiμ xi , . . . 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 ) i=m+1

i=m+1

m l l l h + nμ x μ + aiμ xi |xi |h(ni ) h x i ni , μ=1

i=m+1

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 (l − m + 1) F1 + xi2 h2 (ni ) i=m+1

i=1

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 κl l vol E (D) l (m−l)/2 H (n1 , . . . , nm )−(k−l)/(k−m) . 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} By Minkowski’s second theorem there y i = [yi1 , . . . , yil ] (1 i l) such that

exist

(1 i l). 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 (36) Δ E (D) = γl 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 pj (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, pj ∈ 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(p j ) 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

where Al = {[α1 , . . . , αl ] ∈ : 0 < α1 α2 . . . αl 1} and Δ(·) is the critical determinant, h(a) is defined below. Ql

c

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

αi xi + xl−1 + xl

1. |xk | 1 (1 k l),

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

1277


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.

1278


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

1279


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 yk sg ,..., , 1 d(Λ). f y1 , . . . , yl , 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).

1281


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 α b 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 ,

1282


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 Δ(Sα ) = √ l =: 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 ) =

vol H 3 = . 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 Δ(Hα ) αk + 1 √ . 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)

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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)

which completes the proof.

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 , . . . , p l 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.

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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 , . . . , p l in Zk such that for all i m ni =

l

uij ∈ Q.

uij pj ,

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 , . . . , p n in S with n/2

|p1 | · · · |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 , . . . , p m with m/2

|p1 | · · · |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 , . . . , p l which span T and have l/2 1/2

l/2

|p1 | · · · |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 , . . . , p l 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


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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 ) . . . F (a ∗i0 ), so that λi0 the i0 -th minimum of Λ, has λi0 F (a ∗i0 ) = K λi0 = lim inf λi0 t . 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),

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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


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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 p 1 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

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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


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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 t , |p × q| |p| |q| 18 − 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 det U AA U = (21)

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 ,

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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 ⎛ ⎞ ⎛ ⎞ a1 b1 ⎜ .. ⎟ ⎜ .. ⎟ A = ⎝ . ⎠, B = ⎝ . ⎠ am we have (23)

bm

" |det B| < 1 + |det A| 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 (26) H (S) = det Λ1 = H C[m],[m+1]−{j } (t, t1∗ , . . . , t2m 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 H (T ) 1 − γm,n |det A|n/m . 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 det U BB t U t = det U B (B )t U t 1 − γm,n |det A|2n/m + O(t −1 ) det U U t . 2 We now apply Lemma 5 to the matrices U , A and B and obtain " " 2 (27) det U AAt U t 1 − γm,n |det A|2n/m + c(A) + O(t −1 ) det U U t . 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 .

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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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M. Other papers

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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M. Other papers

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 (x) − 1 f (y) − 1 = . Donc x y

f [x + y · f (x)] = f (x) · f (y) = f [y + x · f (y)] = 0.

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M. Other papers

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


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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

f (x) = 1 + mx. y − 1 = y = 0 forme un Il faut démontrer, que l’ensemble des y, pour lesquels f m groupe multiplicatif. Prenons deux valeurs y1 et y2 de l’ensemble y − 1 = y = 0 . G= y:f m Nous avons f

ou

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,


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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

1335


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 put cpq = 0 if pq ∈ / G; αp = cpq . 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: if h(p) = m if

(5) c

h(pn−1 ) = m;

then

h(V ) ⊂ [0, ∞] then

h(pn−1 ) a(p, pn−1 )h(p).

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 if (6)

h(p) = m

h(q) = m; αpn−1 h(q) h(pn−1 ). cpn−1 q

then

if h(V ) ⊂ [0, ∞] then

The inductive proof for A is complete, B follows by comparison of (4), (5) and (6).


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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) for p = p1 h1 (p) = h (p ) + α −1 Δ h (p ) = α −1 c h (q) for p = p . pq 0 1 p1 c 0 1 p1 ⎩ 0 1 q∈G

Again h1 ∈ S, h1 (p0 ) h0 (p0 ) and (12) holds.

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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) g(p, V ) f (p) 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 . log R − log |p| + O +O (15) g(p, VR ) = 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 1 3 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 1 3 1

(18) |h(p)| = φ(p) − log R − log 2 − C = O . 2π 4π 2π R 1 Hence by Lemma 1 applied to h(p) and to −h(p) we have h(p) = O for all p ∈ V R . 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)

g(o, VR ) − g(p, VR ) π + o(1) = 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 (20) nk (C) α0 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 n−1 s 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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M. Other papers

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| max aij , − aij . (1) 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.

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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 ).

1351

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|1 = |Q| K 2 = l 2 |Q|2 = l 2 |P |1 . |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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M. Other papers

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 Lj (x) = |P |∞ /n. f (x) = g(y)g(x − y) dy |P |∞ 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 (3.2) (δ − x)g(x) dx = G(y) dy (2y)1/2 dy < δ 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

1/2−δ

f (z) + f (−z) dz =

0

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 g(y)g(u) dy − g(y)g(u) dy = f (w) − f5(w) f (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

f (w) − f5(w) f (−w) dw

0

=

=

1 − 4 1 − 4

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 (6.6) 2 dw g(x)g(v) dx g(y)g(u) dy −1/4−δ/2

=

1 2

7

1/4+δ/2 −1/4−δ/2

x+v=−w vx

y+u=w uy

7

1/4+δ/2

dw f (−w)f (w) =

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 4 dw g(x)g(v) dx g(y)g(u) dy. 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 2 dw g(x)g(v) dx = 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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√ √ 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.

1 7 − |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 7 1/8+η 2 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 1 2 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 = (u2 + v 2 ) + g(ε/8 + εη) dη g(εy) dy. 4 0 0 ε=±1

By (3.1) with r = 1/8 − η, s = 1/8, 7 1/8−η 7 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

1362

M. Other papers

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 2 H (u, u1 − u) = −2u + 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 g(t) = f (x)f (t − x) dx. 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 m 1 1 1 = + + 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, there exists at least one prime p ≡ k (mod n), p < n2 . (paper 22=J1) 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 (paper 53=I1) exist n such that a n − bn has three primitive prime factors? 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 ) , 1 − P (n) ∼ 1 − (h 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 (ii) ni = νij uj (1 i k), uj integers 0 (1 j k); 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 27 (problem) Does there exist an identity fi (x) = P x + Q, where fi ∈ Z3 [x], 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 represents a positive integer c(ε)|D|ε for every ε > 0. (paper 99) 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 exist polynomials X, Y ∈ Q[t1 , . . . , tr ] satisfying (5). (paper 128=J5) 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) H (pi ), c0 (k, l, m) = sup inf 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 , . . . , p l ∈ 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) √ 41 (conjecture) If fi (x) = ai x 2 + bi x + ci ∈ Z[x] (i = 1, 2, 3), a1 a2 a3 ∈ / Q, then 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). O pewnym przypuszczeniu dotyczacym rozkładów na sume trzech 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). O okresowo´sci pewnych ciagów liczb naturalnych (On the periodicity of 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) = y−x 2 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. byA. 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). “On Stroeker’s equation”, in: Algebra [161] Postscript to the paper of A. Makowski 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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[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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[10] Postep w teorii liczb w latach 1966–1978, Wiadom. Mat. (2) 22 (1979), 1–11, MR0571455 (83e:10002). metody sita, Wiadom. Mat. 22 (1979), [11] O pracach H. Iwańca dotyczacych 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). miedzy własno´sciami lokalnymi a globalnymi w teorii równań diofantycznych, [17] Zwiazki 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.

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[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).

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Analytic Number Theory

Analytic Number Theory

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Analytic Number Theory

Analytic Number Theory

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Selecta. 2 Vols: Elementary, analytic and geometric number theory: I,II (2 Vols.)

Selecta. 2 Vols: Diophantine problems and polynomials

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Field's Virology [2 vols]

The LispKit manual, vols.1 and 2

History of Indian Philosophy, 2 Vols

Fretboard Logic Vols 1 2 & 3

Indogermanisches Etymologisches Woerterbuch Set 2 vols

Polymer Blends Handbook 2 vols set

Bluethendiagramme: 2 vols. Leipzig 1875-1878

Vols. A and B

Number theory vol.2. Analytic and modern tools

Number theory vol.2. Analytic and modern tools

Analytic number theory

A Course in Applied Mathematics, Vols 1 & 2

Bretherick's Handbook of Reactive Chemical Hazards 2 vols set

Baker's Biographical Dictionary of Popular Musicians Since 1990 (2 vols)

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