Jiri Herman Radan Kucera Jaromir Simsa v
Equations and Inequalities Elementary Problems and Theorems in Algebra and Num...
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Jiri Herman Radan Kucera Jaromir Simsa v
Equations and Inequalities Elementary Problems and Theorems in Algebra and Number Theory Translated by Karl Dilcher
��'z._ � :7
Canad1an Mathematical Society
Societe mathematique du Canada
CMS Books in Mathematics Ouvrages de mathematiques de Ia SMC 1
2 3
4
HE RMAN/KutERA/SIMSA Equations and Inequalities ARNOLD Abelian Groups and Representations of Finite Partially Ordered BORWEIN/LEWIS Convex Analysis and Nonlinear Optimization LEVIN/LUBINSKY Orthogonal Polynomials for Exponential Weights
5 KANE Reflection Groups and Invariant Theory
Sets
Jifi Herman Radan Kucera Jaromir Simsa v
Equations and Inequalities Elententary Problents and Theorents in Algebra and NUinber Theory Translated by Karl Dilcher
Springer
Jiff Herman
Radan Kucera Department of Mathematics Faculty of Science, Masaryk University
Brno Grammar School tf. Kpt. Jarose 14 658 70 Brno Czech Republic
Janackovo nam. 2a 662 95 Brno Czech Republic
Translator
Jaromfr Simsa Mathematical Institute Academy of Sciences of the Czech Republic ZiZk.ova 22 616 62 Brno Czech Republic
Karl Dilcher Department of Mathematics and Statistics Dalhousie University Halifax, Nova Scotia B3H 3J5 Canada
Editors-in-Chief Redacteurs-en-chef
Jonathan Borwein Peter Borwein Centre for Experimental and Constructive Mathematics Department of Mathematics and Statistics Simon Fraser University Burnaby, British Columbia VSA 1S6 Canada
Mathematics Subject Classification (2000): 16G20, 20Kxx Library of Congress Cataloging-in-Publication Data
Herman, Jiff.
[Metody resenf matematiclcych uloh I. English]
Equations and inequalities : elementary problems and theorems in algebra and number
theory I Jiff Herman, Radan Kueera, Jaromir Simsa; translated by Karl Dilcher. p. em.- (CMS books in mathematics) Includes bibliographical references and index.
ISBN 0-387-98942-0
(alk.
paper)
1. Equations-Numerical solutions.
n. SimSa, Jaromir. 2000 512.9'4--dc21
Radan.
QA218.H4713
m. Title.
2. Inequalities.
IV. Series.
3. Problem solving.
I. Kucera
99-047384
Printed on acid-free paper. © 2000 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Allan Abrams; manufacturing supervised by Jerome Basma. Photocomposed copy prepared from the translator's files.
UT:EX
Printed and bound by R.R. Donnelley and Sons, Harrisonburg, VA. Printed in the United States of America. 9 8 7 6 5 4 3 2 1
ISBN 0-387-98942-0 Springer-Verlag New York Berlin Heidelberg
SPIN 10745911
Preface
This book is intended as a text for a problem-solving course at the first or second-year university level, as a text for enrichment classes for talented high-school students, or for mathematics competition training. It can also be
used
as a source of supplementary material for any course dealing with
algebraic equations or inequalities, or to supplement a standard elementary number theory course. There are already many excellent books on the market that can be used for a problem-solving course. However, some are merely collections of prob lems from a variety of fields and lack cohesion. Others present problems according to topic, but provide little or no theoretical background. Most problem books have a limited number of rather challenging problems. While these problems tend to be quite beautiful, they can appear forbidding and discouraging to a beginning student, even with well-motivated and carefully written solutions.
As
a consequence, students may decide that problem
solving is only for the few high performers in their class, and abandon this important part of their mathematical, and indeed overall, education. One of the reasons why problem solving is often found to be difficult is
the fact that in recent decades there has been less emphasis on technical
skills in North American high-school mathematics. Furthermore, such skills
are rarely taught at university, where most courses are quite theoretical or structure-oriented.
As
a result, a lack of ''mathematical fluency" is often
evident even in upper years at university; this reduces the enjoyment of the subject and impairs progress and success. A second reason is that most students are not used to more complex or multi-layered problems. It would certainly be wrong to give the impres-
vi
Preface
sion that all mathematical problems should succumb to a straightforward approach. Indeed, much of the attractiveness of mathematics lies in the satisfaction derived from solving difficult problems after much effort and several futile attempts. On the other hand, being "stuck" too often and for too long can be very discouraging and counterproductive, and is ulti mately a waste of time that could be better spent learning and practicing new techniques. This book attempts to address these issues and offers a partial remedy. This is done by emphasizing basic algebraic operations and other techni cal skills that are reinforced in numerous examples and exercises. However, even the easiest ones require a small twist. We therefore hope that this pro cess of practicing does not become purely rote, but will retain the reader's interest, raise his or her level of confidence, and encourage attempts to solve some of the more challenging problems. Another aim of this book is to familiarize the reader with methods for solving problems in elementary mathematics, accessible to beginning uni versity and advanced high-school students. This can be done in different ways; for instance, the authors of some books introduce several general methods (e.g., induction, analogy, or the pigeonhole principle) and illus trate each one with concrete problems from different areas of mathematics and with varying degrees of difficulty. Our approach, however, is different. We present a relatively self-contained overview over some parts of elementary mathematics that do not receive much attention in high-school and university education. We give only enough theoretical background to make these topics self-contained and rig orous, and concentrate on solving particular problems in the three main areas corresponding to the first three chapters of this book. These chapters are fairly independent of each other, with only a limited number of cross-references. Within each chapter, clusters of sections and subsections are tied together either by topic, or by the methods needed to solve the examples and exercises. The problem-book character of this text is underlined by the large number of exercises; they can be solved by using a method or methods previously introduced. We suggest that the reader first carefully study any relevant examples, before attempting to solve any of the exercises. The individual problems (divided into approximately
760
330
examples and
exercises) are of varying degrees of difficulty, from completely straight
forward, where the use of a method under consideration will immediately lead to a solution, to much more difficult problems whose solutions will sometimes require considerable effort. The more demanding exercises are marked with an asterisk
(*).
Answers to
all
exercises can be found in the
final chapter, where additional hints and instructions to the more difficult ones are given as well. The problems were taken from a variety of mainly Eastern European sources, such as Mathematical Olympiads and other competitions. Many
Preface
vii
ofreader. themAnare important therefore notcriterion otherwise easily access i ble to the English-speaking for the selection of problems that their We solutions should, in principle, be access i ble to high-school students. believe that even with this limitation one can successfully stimulate creative workOneinofmathematics and illustrate its diversity and richness. our objecti v es has been to stress alternative approaches to solving aplace, givenwhile problem. Sometimes we provide several different solutions in one at other times we return to a problem in later parts of the book. This book is a translation of the second Czech edition of Metody resen-t matematick'gch uloh I ( "Methods for solvirig mathematical problems I" ) published in 1996 at Masaryk University in Brno. Apart from the correction ofshortsomesections minorwith misprints, the main material has been left unchanged. Three "competition-type" problems at the ends of the chapters were deleted, but some of these problems were incorporated into the main body of the book. alphabetical index was added. The Czech editions of this book have been used by the authors in spe cial enrichment class e s for secondary-school students and for Mathematical Olympiad training in the Czech Republic. It has also been used on a regular basis at Masaryk University in courses for future mathematics tea.C h ers at secondary schools. A preliminary English version of the first two chapters successfully tried out by the translator in a newly created problem was solving course in the Fall of 1998 at Dalhousie Uni v ersity, in Halifax, Canada. It was felt that a course partly based on this book worked qui t e well even in a class with a wide range of mathematical backgrounds and 's structure, the worked examples, and the range in level abilities. The book ofassdifficulty of the exercises make it particularly well suited as a source for iThegnedtranslator readings also and homework exercises. feels that, following the Czech example, this book may be suitable for a course in a teachers' education program for high school mathematics teachers. Finally, we hope that the reader will find this book a rich source of useful identities, equations, and inequalities. Jifi Herman Brno and Halifax Radan Kucera July 1999 Jaromir Sim8a Karl Dilcher was
An
Contents
Preface
v
Symbols
xi
1
2
1
Algebraic Identities and Equations
12 34 5 67
Formulas for Powers . . . . . . Finite S . . . . . . . . . Polynomials . . . . . . . . . . . . . . . Symmetric Polynomials Systems ofEquations Equations . .. . . . . . .. . .. . . . . Irrational Some Applications of Complex Numbers .
ums
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1 523 38 46 607 5
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Algebraic Inequalities
12 34 56 7 8 9
. . . . . . . . . . . Definitions and Properties Basic Methods . . . . . . . . . . . . . The Use of Algebraic Formulas . . . . . . . . The Method of Squares . . . . . . . . . . The Discriminant and Cauchy's Inequali. ty. .. . The Induction Principle . . . 's Inequality . . . . . . . . . . Chebyshev Inequali t ies Between Means . . . . . Appendix on Irrational Numbers . .
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89
9094 106 113 124 13145 5 150 169
Contents
x
3
4
Number Theory
.. . . . . . . . . Concepts Basic . . . . . Numbers Prime . Congruences . . . . . . . . . . . . . . . .. . . Variable One in Congruences . . . . . . . Equations Diophantine . Equations Diophantine of Solvability . . . . . Part Fractional and Part Integer . . . . . Representations Base .... . . . . Principle Dirichlet's Polynomials . . . . . Hints and Answers 1 1 . . . . Chapter to Answers and Hints 2 2 Chapter to Answers and Hints 3 Hints and Answers to Chapter 3 .
1 2 3 4 5 6 7 8 9 10
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173
174 183 189 201 217 238 251 258 268 274 283
283 294 312
Bibliography
337
Index
341
Symbols
to
Inof numbers addition willusual mathematical notation, the foll o wing symbols for sets be used: N No Z
Q Q+ R R+ R-
JRt
IRQ
C
the set {1, 2, 3, } of all natural numbers, or positive integers the set of all nonnegative integers the set of all integers the set of all rational numbers the set of all positive rational nwnbers the set of all real numbers the set of all positive real numbers the set of all negative real numbers the set of all nonnegative real numbers the set of all nonpositive real nwnbers the set of all complex numbers .
.
.
1 Algebraic Identities and Equations
Inrestricted contrasttotoonethetopic; other rather, chaptersit ofis this book, this first chapter will not be devoted to several relatively disparate . The emphasis lies on finite sums, polynomials, and solutions of matters systems of equations. The final section deals with several types of problems that can be solved by using complex numbers. 1
Formulas for Powers
This is a short preparatory section. We introduce several identities that will be used throughout this text. 1 . 1 Combinatorial Numbers
For an arbitrary number n E No we define the number n! read factorial ) as follows: O! 1 , and for n > 0 let n! be the product of natural numbers not exceeding the number n: Definition. ( n
=
all
n! =
1 · 2 · · · (n- 1) · n.
we
Factorials help us define combinatorial numbers (�) (read n choose k) by the relation (nk) - k!(nn!- k)!' where n, k E No, k < n.
2
1 . Algebraic Identities and Equations
From this definition we obtain directly the identities
< k < n; where 0 -
(1)
-
in the case where 0 < k < n, we have furthermore
( )+( )= n k
n k- 1
=
+
n! n! (k - 1)!(n - k 1)! k!(n - k)! n! [(n k + 1) k ] = k!(n - k 1)f -
+ + (n+k 1)
+
(2)
·
These identities enable us to show that combinatorial numbers are the coefficients in the expansion of the nth power of a binomial expression. Therefore they are often called binomial coefficients. In the following theo + 2A B = we rem, which generalizes the known formula can take A, to be numbers or expressions.
( A + B ) 2 A2
B
+ B2 ,
1. 2 The Binomial Theorem Theorem. For an arbitmry natuml number n we have
(A + B) n = (�) An + (�) An- l B + (;) An-2B2 + . . . + (:) Bn . PROOF. Since (�) = G) 1, the theorem is certainly true for n = 1. We assume that it is true for some fixed n > 1; we will prove that it holds for n + 1. If we add the two identities A · (A + B)nn (�) An+l + (�) AnB + · · . + (:)ABn , B · (A + B) (�)AnB · · · + (n �1)ABn + (:) Bn+l , we obtain on the left-hand side the power (A + B ) n+l , while the sum on the right is, according to (2 ) , equal to (�) An+l + (n il) AnB + (n �l)An- 1 B2 + . . . + (n �l) ABn + (:) Bn+l . Since (�) = 1 (n 61 ) and (:) = 1 = (:tD , the statement of the theorem 0 holds also for n + 1, which completes the proof by induction. =
=
=
+
=
1. 3 Examples
B = 1, resp. A = 1, B -1, we obtain interesting A = identities between combinatorial numbers, valid for all n E N: {i) If we set
=
(3)
1 Formulas for Powers
3
+ v2) 100•
(ii) We will determine the largest summand in the expansion of (1 To do this, we determine the quotient of the (i 1 )th and the ith summand (0 < i < 100) of the expansion, namely
+
- i)! . h C�0)( ( vi2)i 1 = i! (100100!- i)! . (i - 1)!(101 100! (J��) vf2)i= 101 . - i . h. �
This quotient is bigger than 1 exactly when ( 101 - i ) v'2 > i, that is, exactly when 101 i< 101(2 - J2)
""59.1644, + �2 = and for i < 59 the ( i + 1 )th summand is bigger than the ith. Similarly, we convince ourselves that for i > 60 the (i + 1 )th summand is less than the ith. Therefore, the largest summand is (�09°) . ( v'2) 59 1
so
_
1.4 l?xercises Show that (i)-(iii) hold for arbitrary natural numbers n:
(�) + (�) + (�) + . . . = 2n-1. (�) + (;) + (:) + . . . = 2n-1 (the sums in (i), (ii) end with ( :) or ( �1) ). n (iii) en:1) + ( 2n i 1 ) + en:1) + . . . + en:1) = 4n . {i) {ii)
(iv) Find the largest summand in the expansion of ( v'2 (v) Determine all natural numbers m, n satisfying
n+11) = (nm+1) = §.3 (mn+1 ( m+ -1)
+ vf3) 50.
·
The binomial theorem is going to be used extensively in what follows, in particular in Sections 2.9 and 2.10, where we find identities similar to (3) and Also, in Section 3 . 7 of Chapter 2 we show how the binomial theorem can be used to prove ip.equalities. Now, however, we introduce a generalization of the known identity As in may denote numbers or expressions. 1.2, the letters
(4).
A, B
A2 - B2 = (A - B) · (A + B).
1. 5 A Factorization Theorem. For an arbitrary natural number n we have
An - Bn = (A - B)· (An-1 + An-2B + · · · + ABn-2 + Bn- 1 ). (5)
1 . Algebraic Identities and Equations
4
PROOF.
By direct calculation: If we subtract the identities
A . (Ann-11 + Ann-2B + . . . + Bnn- 11 ) = An + Ann- 11 B + . . . + AABnn--11 ' Bn , B . (A - + A -2B + . . . + B - ) = A - B + . . . + B +
we
0
(5).
immediately obtain
1.6 Example Find the coefficients of the polynomial
where r, s are integers,
0 < r < s.
(1 + x)r, we obtain F(x) = (1 + xr · [1 (1 + x) + · · · + (1 + x)s-r]. 1 in (5), we get Replacing A = 1 + x, B = 1, = s (1 + x)s-r+1 - 1 = X · [(1 + x) s-r + (1 + x)s-r- 1 + · · · + (1 + x) + 1], SOLUTION.
If we factor out
+
n
r
+
hence
F(x) = -X1 · (1 + x) r [(1 + x) s-r+ I - 1] = -X1 · [(1 + x) 8+1 - (1 + x) r] . According to the binomial theorem 1.2, � 1 ) x•- l _t (:) xi- 1 , (s F(x) = E �=1 �=1 and therefore the coefficient of xi in F(x) is (j!D - (i�1) for 0 < j < and
(;!D
for
r
< j < s.
r
o
1. 7 Exercises Show that for an arbitrary natural number
(i) (ii)
m
we have
A2m_B2m = (A+ B)(A2m- 1 _A2m-2B + . . . _B2m- 1 ), A2m- 1 + B2m- 1 = (A + B)(A2m-2 _A2m-3B + . . . + B2m-2 ) .
(6) (7)
2 Finite Sums
2
5
Finite Sums
In this section we give several methods for obtaining formulas for some interesting finite sums, that is, sums with finitely many summands. We point out that we avoid on purpose the use of mathematical induction, since in many cases it is quite difficult to obtain the necessary hypothesis on how the result should look. This does not mean, however, that this method could not be used for certain problems·. Indeed, the many examples in this section would be very useful for practicing this important method of proof. For the remainder of this sectiqn, let k, m, n be natural numbers.
2. 1
Example
Determine
S1 (n) = 1 + 2 + 3 + · · · + n.
We write out the sum S1 ( n) twice, with the second one written in opposite order; then we add the two expressions: SOLUTION.
S1 (n) = 1 + 2 + 3 + · · · + (n - 1) + n S1 (n) = n + (n - 1) + (n - 2) + · · + 2 + 1 2S1 (n) = (n + 1) + (n + 1) + (n + 1) + · · · + (n + 1) + (n + 1). ·
That is,
2S1 (n) = n(n + 1), and therefore sl ( n)
= 1 + 2 + 3 + . . . + n = n(n2+ 1) .
(8)
We were able to determin� this sum very easily, simply by changing the order of the summands. It is now clear that this method can be used for summing finitely many terms of any arithmetic progression. If an arithmetic progression is determined by its first term a 1 and difference d, then we get for the sum SA of its first n terms,
+ · · · + [a 1 + ( n - 1 ) d] + a1 SA = [a 1 + d] a1 SA = [a 1 + (n - 1) d] + [a 1 + (n - 2) d] + · . . + 2SA = [2a1 + (n - 1) d] + [2a 1 + (n - 1) d] + · · · + [2a1 + (n - 1) d] , that is,
where we have used the fact that
an = a 1 + (n - 1) d .
D
1 . Algebraic Identities and Equations
6
2. 2 Remark If we change the above procedure a little, we can also determine the sum of the first n terms of a geometric progression. If a geometric progression is determined by its first term a 1 and quotient q 1= 1 , we obtain a formula for the sum So of its first n terms as follows:
So = a1 + a 1 q + a 1 q2 + · + a 1 qn- 1 q · So = a1 q + a 1 q2 + + a1 qn - 1 + a 1 qn q · So - So = -a 1 + Since q 1= 1, we obtain n-1 q (10) So = a 1 · q - 1 . The trick that led to the expression (10) is based on an appropriate multiplication of the original expression for So, followed by a subtmction ·
·
·
·
·
of the two equations, so that on the resulting right-hand side all but two terms disappear. We note that if we begin by factoring out a 1 in the sum So, then formula (10) can be obtained directly from (5) by setting A = 1,
B = q.
2. 3 Exercises Evaluate the following sums:
(i) S = 1 + 3 + 5 + · · · + ( 2n - 1). (ii) S = 1 - 2 + 3 - 4 + · · · + ( -1 ) n+ 1 n. (iii) S = n + (n + 3) + (n + 6) + · . . + 4n. (iv) S = ( -31) + ( -27) + ( -23) + · · · + 29 + 33. ( ) S = 2 + 22 + 23 + · · · + 2n . (vi) S = 1 -! + � - � + . . + (-1) n - 1 · 2 n':.1• (vii) S = 2 + 5 + 11 + . . · + (3 · 2n- 1 - 1). (viii) S = (x + �)2 + (x2 + �)2 + + (xn + x�)2 , where x E 1R, x 1= 0. V
·
·
2.4
·
·
Some Combinatorial Identities
The methods of changing the order of summands or appropriate multipli cation of the sum can also be applied to certain combinatorial sums, that is, sums that contain factorials or binomial coefficients. In some cases both methods can be combined in an appropriate way.
2 Finite Sums
7
2. 5 Example
Determine S = 1 · (�) + 2 (�) 3 · (�) + · + n (:) the fact that k (�) = k · ( �k) , we can (i)write the given sum in aIfdiffweeuse n rent way, S n (�) + (n - 1) (�) + (n - 2) · (�) + + 1 · (n�1). We then add both expressions of the sum 1 · (�) + 2 · (�) + . + (n - 1) · (n�1) + n · (:) S= S = n·(�) + (n - 1) · (�) + (n - 2) · (�) + . . + 1·(n�1) +
·
·
·
·
FIRST SOLUTION.
as
=
·
·
.
·
·
·
·
S:
. ·
·
Hence, according to (3) , S = �n [ (�) + (�) + (�) + · + (:)] = � n 2n = n 2n-1. 0 (ii) Each term in the sum S is of the form k (�) (for k = 1, 2, .. . , n) and we have (n)!-(1)-! k)! - n (nk-1) . k . k!( n! k)! - k . k·(kn-1 -1 n n Then S = 1 · (�) + 2 (�) + · · + (n - 1) (n�1) + n (:) = n (no1) + n (n11) + . . . + n . 2 and alsoS = 1 for n = 1. ·
·
·
.
SOLUTION .
as
·
0
so,
_
us
·
_
o
8
1. Algebraic Identities and Equations
2. 7 Example
Determine
= f(�) + �(�) + !(;) + · · · + n�1 (:). immediatelbyy ). We ( Here each summand is of the form � · k.,!. 1 reach our goal if we multiply each summand (and thus the whole sum) the number n + 1 , since +1). 1 (nk) = k+1 . k!(nn-! k)! - (k+(1n)!+1(n)!-k)! - (kn+1 ( n + 1) k+1 Then we have (n + 1) . S = nt1 (�) + nt1 (�) + nt1 (;) + . . . + nt1 (n�1) + :t� (:) = (nt1) + (nt1) + (nt1) + . . . + (n�1) + (:!D = [(nt1) + (nt1) + (nt1) + . . . + (:!D J (nt1) = 2n+1 1. 0 From this we obtain S = 2:�11 . S
SOLUTION.
!!:±!
.
_
_
_
_
2. 8 Exercises
Determine the following sums: (i) S = 1 · (�) + 2 (�) + 3 · (;) + · · · + (n + 1) · (:). (ii) S = 1 · (;) 2 · (;) + 3 · (:) - 4 (�) + · · + ( 1 )n (n- 1 ) · (:) for n > 2. (iii) S = 2 · (;) + 6 · (;) + 12 (:) + · · + n( n- 1) (:) for n > 2. *(iv) S = 12 · (�) + 22 . (;) + 32 . (;) + . . . + n2 (:). (v) S = �(�)- �(;) + i(;)- �(:) + . . . + ( -��n1+1 (:). (vi) S = �(�) + i(�) + i(;) + · · · + n..!-2 (:) . *(vii) S = 222 (n1) + 233 (n2 ) + 244 (n3) + . . . + 2nn++1l (nn) · ( ...) 1!(2n1-1)1 + 3!(2n1-3)! + 51(2n1-5)! + · · + (2n-11 )!1! ·
-
·
·
-
·
·
·
•
Vlll
s-
·
·
2. 9 Applications of the Binomial Theorem
The binomial theorem 1 . 2 can be used successfully for deriving sev eral combinatorial identities We consider the binomial expansion of the . expression
9
2 Finite Sums
which is a polynomial of degree k in the variable x. If we succeed in express ing this polynomial in a different form for certain k, then by equating the coefficients of equal powers xi we can obtain interesting formulas between binomial coefficients. For example, let us prove the identity Here then appropriate exponent is k = 2n since clearly (1 + x)2n (1 + x) (1 + x)n . After expanding both sides of this equation accord ing to the binomial theorem, we obtain the required identity by equating the coefficients of xn: (1 + x)2n = e;) + ef) x + ...+ (�) xn + ... + (�:) x2n, (1 + x)n · (1 + x)n = ((�) + ( �) x + · · · + (n�1) xn-1 + (:) xn) ·
X
((�) + ( �)x + · · · + (n�1) xn-1 + (:) xn)
= (�) (�) + [(�)(�) + (�)(�)]X+ ...
+ [(�)(:) + (�)(n�1) + ...+ (:) (�)J xn + . . .
+ (:) (:) . x2n.
Therefore, we have e:) = (�)(:) + (�)(n�1) + ...+ (n�1) (�) + (:) (�) = (�)2 + (�)2 + ...+ (n�1)2 + (:)2, where in the last equality we have again used (1). In general, the two expressions of this polynomial form the basis of the method of genemting functions. With the help of this method one can obtain a large number of interesting identities. The interested reader is encouraged to consult, e.g., the books [9] and [6]. 2. 1 0 Exercises
Using the method of 2.9, prove ( i) and (ii), where and < n: r
p, r
E N,
p
1 · tn2±11) (ii) S - 1·32 + 21·33 + 337 ·4 + · · · + n3n(nt 2t2nt1 (iii) S = 1·2 + 21·33 + 325· 4 + ... + 2nn(nt 1) · (i)
s
·
_
...§_
·
2.43 Example Let us determine
12 + 22 + 32 + ··· + n2 8 =1·3 3·5 5·7 (2n - 1) (2n + 1)" �-�;-----:-
0
23
3 Polynomials
(2 i_1){2 i+ 1) = 4 i -1 . If
i2 In this sum each term is of the form we consider the sum 48, we can write each term in the form
S OLUTION .
·2
�
4i 2 4i2 - 1 + 1 1 1 + + = = 1 1 4i 2 - 1 4i 2 - 1 4i2 - 1 (2i - 1 )(2i + 1) ' Therefore, we have, by Exercise 2.40.(ii), 4S
S0 finally, 8 = 2.44
· ( (2n- 1)1(2n+1) ) = n + 2n�1 ·
= (1 + /3 ) + (1 + 3�5 ) + ·· + 1 +
n(n+ (2 2n+11))
·
Exercises
Evaluate
1
0
1
2
3
n -----(2n - 1 ) (2n + 1 ) (2n + 3) '
3 5 3 · 5 · 5·7 9 1 4 24 34 n4 . ( ) 8 = - + - + - + ·· · + 1·3 3 · 5 5 · (2n - 1) · (2n + 1)
* ( .) 8 =
1·
·
+
7
.. u
3
+
·
+ ··· +
7
Polynomials
The search for zeros of polynomials was for many centuries the basic prob lem of all of algebra. Even when algebra in its development departed to a great extent from this problem, polynomials continued to play an impor tant role in algebra as well as in a number of other areas (among the more remarkable occurrences are the Taylor polynomial for the approximate eval uation of functions, and the use of polynomials in interpolation questions). In this section we recall some basic terms connected with polynomials, and also several methods for finding their zeros . However, we are not con cerned with providing a well-rounded theory of polynomials; instead, we limit ourselves to some properties that receive relatively little attention in basic algebra courses but that are useful for solving certain problems. Although we have already used polynomials in the preceding sections, we will first recall some basic terms from their theory.
3. 1
Definitions
A polynomial in the variable form
x is an
expression that can be written in the
F(x) = anXn + an- 1 Xn- 1 + ·· · + a1 x + a , o
(21)
24
1. Algebraic Identities and Equations
No ai F(x), a0 ai
aixi
where n E and ( i = 0, 1 , . . . , n) , called coefficients, are numbers from are called terms o� some number domain (Z, Q, IR, C). The expressions the polynomial and is called the constant term. Those terms whose coefficients are zero can be disregarded in (21). A polynomial all of whose coefficients are zero = 0) is called the zero polynomial. After dropping possible zero terms, any nonzero polynomial can be written in the form 2 ) with the additional condition 1= 0. In this case the number n is called degree of the polynomial denoted by n = deg The degree of the zero polynomial will remain undefined. Polynomials of degree 1 are called linear, those of degree 2 quadratic, and of degree 3 cubic. We remark that the polynomial is uniquely determined by its coefficients, although it can be expressed in a variety of different forms.
(1
(F(x)
a F(x), n F(x)
aix�
F(x) .
3. 2 Example Show that the polynomial
F(x) = (1 - x + x2 - . . . - xgg + xioo) . (1 + x + x2 + . . . + xgg + xloo) has zero coefficients for all odd powers of x. SOLUTION. Using (5) and (7) with A = 1 and B = x, we get 1 x 1o1 1 x l O l 1 x2o2 F(x) = 1 + x 1 - x = 1 - x2 ' from which, using (5) with A = 1, B = x2 and = 10 1, we get F( x) = 1 + x2 + x4 + . . . + x19B + x2oo . Thus we have proved the statement of this example, and at the same time we also found the coefficients of the even powers of x in F(x) . Let us add that in general, the following is true: The coefficients of the odd powers of 0 x in F(x) are zero exactly when F(-x) is identical with F(x) . +
_
_
·
n
3. 3 Exercises For arbitrary k E N find the coefficients of the polynomials (i) = (1 + + + (ii) = + · + · + + . * (iii} Decide which of the two polynomials = + +
x)2k (1 - x) 2k - (1 x2 ) k , x) (1 x2) (1 x4 ) . . . (1 x2k - l ) (1 x2 x3 ) 10oo , G(x) (1 - x2 x3 ) 1ooo has the larger coefficient of x20 • (iv) For each k E N consider the polynomial Fk (x) = (x2 - x + 1)(x4 - x2 + 1) (x2k - x2,._ 1 + 1) . Determine the coefficients of all polynomials (x2 + x + 1)Fk (x). F(x) G(x) (1 F(x)
_
=
·
·
·
3 Polynomials
3.4
Zeros of Polynomials
25
x ax 1 ao
F(x) anxn an- l xn- l
We can substitute any number for the variable and carry out the required operations. If = + is a polynomial + + ···+ and a number, then the number
c
0, c F(c) . F( x)
F(x)
= is called the value of the polynomial at the number If then the number is called a zero of the polynomial (or a root of the polynomial equation = The following paragraphs will be devoted to the problem of finding zeros of polynomials, which is, as we already mentioned, one of the basic problems of algebra. Finding the (single) zero of a linear polynomial is very simple. We also have a formula for determining the zeros of a quadratic polynomial: The polynomial and discriminant + bx + with real coefficients ± .Ji5) D = has the zeros which are real if and only if D > Formulas for determining the zeros of polynomials of degree and do exist, but their application is so cumbersome that they are al most useless. For n > 5 the young Norwegian mathematician Niels Henrik Abel showed in that there does not exist a formula that would allow one to evaluate the zeros of an arbitrary polynomial of de gree n from its coefficients with the help of algebraic operations, including roots of arbitrary degree. This is why the approximate evaluation of zeros of polynomials belongs to the basic problems of numerical mathematics.
c F(x) 0).
2 ax b20.- 4ac
4 (1802-1829)
3. 5 Examples
b, c a, j2a,
c (b -
3
1823
a�, a2 , a3 , b�, b2 , b3 have to (a 1 + b1 x)2 + (a2 + b2x) 2 + (a3 + b3x) 2
(i) Which conditions will the real numbers satisfy so that the polynomial
is the square of some linear polynomial with real coefficients? SoLUTION .
that
We ass ume that there exist real numbers c, d with d
=I= 0 such
If the two polynomials are equal, then their values for any number must also be equal, in particular for 2 Therefore, -
(a 1
�
·
b1 �) 2 + (a2 b2 �) 2 + (aa •
�
•
�
ba
·
�) 2
=
0,
26
1. Algebraic Identities and Equations
from which in view of the fact that the numbers in parentheses are real, it follows that
(23) where � - Fu rthermore, if we equate the coefficients of x2 in (22) , we obtain (24) b� + b� + b� d2 1= 0. Conversely, if (24) and (23) are satisfied, then (22) holds for d. The required condition for the numbers a 1 , a2 , a3 , b1, b2 , b3 is therefore b� + b� + 0 b� f= 0 and the existence of a real number satisfying (23). q =
=
c = q
·
q
(ii) Find a real number p such that the difference between the real zeros of px + p 9 is equal to
x2 -
-
x 1 , x2
6.
x1 p+
= fD SOLUTION. The zeros of the polynomial are given by Since in this notation we have and = p-fD, where D = r - p + > the condition of the problem can be written as
4 36. 6 Xl -X2 (p + v75) -2 (p - v75) VD, which holds if and only if D 36. From the equation 2 4 + 36 36 0 and 4, and therefore obtain the polynomials we find the values 2 9 (x + 3) (x - 3) and x2 - 4x + 5 (x + 1) (x - 5) with the sets of xzeros 0 { -3, 3} and { -1, 5}. X 2 x 1 x2 ,
=
=
=
=
p =
-
p
-
=
p
p =
=
=
3. 6 Exercises (i} Find the sum S of the coefficients of the polynomial
3x + 3x2 ) 1988
3x - 3x2 ) 1989 . (ii) Show that for arbitrary real numbers a, with a f=. 0 and f=. the equation 1 1 1 + X X - a2 has two distinct real roots. * (iii) Decide whether there exist real numbers a, b, such that for each E :JR+ the polynomial ax2 has exactly two distinct F(x)
=
(1 -
·
(1 +
p, q
-
--
-p
A
positive real zeros. * (iv) Determine the numbers
a, b,
q
+ bx +
p, q
p
E
c
+A
lR such that
c
q,
27
3 Polynomials
+ a - lX - l + + + a
In analogy to Section 3.4, where we substituted a number for the variable a 1 x o , we can also substitute a x in F(x) = an xn n n polynomial G(x) for x . We thus obtain the polynomial ·
·
·
3. 7 Example Find those nonzero polynomials F(x) for which
F(x2 )
is equal to
(F(x)) 2 .
We will try to find the polynomial F(x) in the form (21), where n 1= 0. We assume that at least one of the coefficients ao is different from 0 and we denote by k < n the largest 0. Then the condition F(x2 ) = (F(x)) 2 can be written as k for which SOLUTION.
a an- b an-2 , .a. . ,=I=a�, k
n+k , we obtain 0 2anak , which x 0 contradicts the assumption. Therefore, an -1 a a 1 o and F(x) anxn . The condition F(x) (F(x)) 2 thus takes the form anx2n a;x2n , which holds if and only if an a;, that is, an 1. The required polynomials are therefore exactly the polynomials xn , where
If we compare the coefficients of
=
=
=
=
=
=
=
=
n
· · ·
=
=
D
E No .
3. 8 Exercises Find all polynomials F(x) that satisfy:
+
(i} (ii) (iii) *(iv) *{v)
F(x + 3) x2 + 7x 12. F(x2 1) x 1 1 - 8x7 6x5 - 4 . F(x - 1 ) + 2F(x + 1) 3x2 - 7x. F(x2 - 2x) (F(x - 2)) 2 • F(F(x)) (F(x)) n for a given n E N (deg F(x) > 0).
3. 9
Polynomial Division with Remainder
+
=
=
+
=
=
=
Let F(x), G(x) be polynomials, with G(x) nonzero. Then there exists a unique pair of polynomials H(x), R(x) such that F(x) = H(x) G(x) + R(x), (25) where R(x) 0 or deg R(x) deg G(x). The coefficients of the polyno mials H(x) , R(x) belong to the same number domain , or C as the Q coefficients of F(x), G(x). Theorem.
·
=
0 the equation ask+2 + bsk+ l + csk = 0 is satisfied.
Example
Show that if the polynomial F(x) three real zeros, then p < 0. SoLUTION.
0
=
Since q 1= p < 0.
=
x3 + px + q, where p, q
E
lR, q 1= 0, has
By (29) the zeros xb x2 , X3 of the polynomial F(x) satisfy (x1 + X2 + X3 ) 2 = X� + X� + X� + 2(X 1 X2 + X 1 X 3 + X2 X3 ) = X� + X� + X� + 2p.
0,
we have X 1 X2 X3
1= 0 and therefore x� + x� + x� > 0;
hence D
3. 22 Exercises
{i)
Suppose that the sum of two zeros of the polynomial F(x)
{ ii)
=
is equal to 1. Determine A and the zeros of F(x) . Suppose that the zeros Xb x2 , X3 of the polynomial x3 +px + q satisfy
(iv} (v)
=
1
1
+ -. X 1 X2 Which condition do the coefficients p, q satisfy? Find a relation satisfied by the coefficients of the polynomial x3 + px2 qx + r .if one of its zeros is equal to the sum of the other two. Let the polynomial F(x) = x3 +px2 +qx + r have the zeros Xb x2 , X3 . Find a polynomial G(x) that has zeros x 1 x2 , x 1 x3 , x2 x3 , and a polynomial H(x) which has zeros x 1 + x 2 , x 1 + X3 , x2 + X3 . Express the coefficient r of the polynomial x3 + px2 + qx + r in terms of the remaining coefficients if it is known that one of the zeros of this polynomials is the arithmetic mean of the remaining two. X3
(iii)
2x 3 - x2 - 7x + A
+
-
1. Algebraic Identities and Equations
34
(vi) Show that for arbitrary nonzero numbers a, b E 1R the zeros X l , x 2 , X3 of the polynomial ax3 - ax 2 + bx + b satisfy the condition
(
)
1 +1 +1 = -1. (x 1 + x2 + x3 ) · X 1 X2 X3 * (vii)Let p, q, r, b, c be five different nonzero real numbers. Determine the sum x + y + z , provided that the numbers x, y, z satisfy the system y X Z + p p - b p - c 1, X + y + Z = 1, q q b q-c Z y X = 1. + -+ r r-b r-c
-+
=
-
3.23 Reduction of the Degree by Way of a Known Zero If we know a zero x 1 of a polynomial F( x), we can use this fact for finding its remaining zeros x2 , x3 , . . . , xn . By 3. 10, F(x) is divisible by the linear polynomial (x - x 1 ), and according to 3.14, F(x)j(x - x 1 ) has exactly the zeros x2 , X 3 , . . . , Xn · 3. 24 Example Show that
(31) SOLUTION. Let A denote the left-hand side of theorem 1.2 we have
(31) .
By the binomial
A3 = 16 - 3 · A · \j(3 v'2i + 8) (3 v'2i - 8) = 16 - 15A, and therefore A is a zero of the polynomial F(x) = x3 + 15x - 16. Since F(1) = 0, the number 1 is a zero of F(x). By dividing, we obtain the identity F(x)/(x - 1) = x2 + x + 16. This last trinomial has no real zeros, since its discriminant is -63. The only real zero of F(x) is therefore the D number 1. Since A is a real zero of F(x) , we have A = 1 .
3. 25 Exercises Prove: (i) (ii)
{/20 - 14v'2 + {/20 + 14J2 = 4. {//5 + 2 + \1-15 - 2 = -/5.
3 Polynomials
35
3.26 Rational Zeros of Polynomials with Integer Coefficients There exists an easy algorithm for finding rational zeros of a polynomial with integer coefficients (compare this with the general case described in 3.4) . Suppose we are given the polynomial (32) with integer coefficients ai, where an 1= 0. We wish to :find its rational zeros. If ao = 0, then one of the zeros is 0, and using the method described in 3.23, we may reduce the degree of F(x) and search for the zeros of F(x) lx. After a finite number of steps we always obtain a polynomial with nonzero constant term. We may therefore assume in addition that F(x) in (32) satisfies ao 1= 0. If a rational number a is a zero of F( x) , then we have F(a) = 0, and if we express a = rls, where r E Z, s E N and r, s are relatively prime, then n- 1 n F = an · + an - 1 + · · · + a 1 · + ao = 0,
(:)
(�)
(:)
from which upon multiplying by sn we obtain
�
1 1 an r n + an - 1 rn - s + · · · + a 1 · r · sn - + aosn = 0 . Since the integer r divides the first n terms on the left of this last equation, it also divides the last term aosn , which implies that r divides ao, since r and s are relatively prime. In complete analogy we find that s divides an . Since ao and an have only finitely many divisors, the rational zero a can take on only finitely many values. By successively substituting these values into F(x) , we find all rational zeros of F(x) . We remark that it is not necessary to substitute all possible values of a into F(x) . Indeed, for each integer k it is true that if rls is a zero of the polynomial (32) and r and s are relatively prime, then the integer F(k) has to be divisible by (r - ks) . This result will be proved in Section 10.11 in the third chapter. Using this criterion will save work with the substitutions: To begin with, it suffices to calculate F ( k) for suitably chosen k E Z (the usual choice is k = ± 1), then for each "'suspected" value a = r Is to check whether F(k) is divisible by (r - ks) . An example for this approach is given in 3.27. (ii) .
3. 27 Examples (i) Find the rational zeros of the polynomial F(x) = x4 - 4x 2 + x + 2. SOLUTION. By 3.26 we expect the rational zeros of F(x) to be of the form r Is, where r E { -2, -1, 1, 2} and s = 1 . Therefore, all rational zeros are in
1. Algebraic Identities and Equations
36
fact integers (this is true for any polynomial with integer coefficients that has 1 as leading coefficient). Substituting, we get F( -2) = 0, F( - 1) = - 2, F(1) = 0, F(2) = 4, and therefore -2 and 1 are the only ration al zeros of F(x) . We can now apply 3.23 and determine all zeros of F(x) : We divide 2 F(x) by the polynomial (x + 2) (x - 1) and get F(x) = (x + 2)(x - 1 ) (x x - 1). The quadratic polynomial x2 - x - 1 has zeros ( 1 ± .J5) /2, and from D this we have that the zeros of F(x) are -2, 1, (1 + v'5) /2, (1 - -15) /2.
(ii)
Find the rational zeros of G(x)
=
3x4 + 5x3 + x2 + 5x - 2.
SOLUTION. If the rational number rjs, with r, s relatively prime integers and s > 0, is a zero of G(x), then r E { 1, -1, 2, -2}, s E { 1, 3}, and there fore r/s E 1 , -1, }, -!, 2, -2, i, -i}- To avoid having to check whether �!loch of these eight values is a zero of G(x), we use the criterion from the end of Section 3.26: We set k = 1 (resp. k = -1 ) , so G ( 1 ) = 1 2 (resp. G( -1 ) = - 8 ); therefore, 1 and - 1 are not zeros of G(x). Now we set up a table with all six remaining triples of the values of rjs, r - s and r + s.
{
1:.
3
1
-3
1
2
-2
2
s
3
-3
r-s
-2
-4
1
-3
-1
-5
G(1 ) = 12
r+s
4
2
3
-1
5
1
G (- 1 ) = - 8
2
Since -5 is not a divisor of 12, and 3, 5 are not divisors of -8, it remains only to decide for r/s E -}, -2} whether G(r/s) 0. We verify by direct calculation that G (3) = G( -2) = 0, G( -}) = -100 / 27, which D means that G(x) has exactly two rational zeros: } and -2.
{�,
=
(iii)
Decide whether there exist rational numbers a, � such that the polynomials x2 + ax + b, x5 - x � 1 have a common zero.
SOLUTION . Then
We assume that some number a is a zero of both polynomials. a
5
= a
+1
and
a
2
=
-aa - b.
From the equation a2 -aa - b we first find an expression with coefficients c, d depending on a, b: =
a
5
=
5
= ro
2 (aa + b) = a[a2 ( -aa - b) + 2aba + b2 ] 2 ( -aa - b) (2ab - a3 ) + (b - a2 b)a (a4 - 3a2 b + b2 ) a + (a3 b - 2ab2 ) .
= a · =
( a2 ) 2
a
= a·
+d
3 Polynomials
37
From the equation a5 = a + 1 it follows that
=
(a4 - 3a2 b + b2 ) a + ( a3 b - 2ab2 )
a ( a4 - 3a2 b + b2 -
1) = -a3 b + 2ab2 + 1 .
Since the polynomial x5 - x 1 has no rational zeros (according to 3.26 the only rational zeros would be ±1 , which obviously does not occur), a is not a rational number. If a4 - 3a2 b + b2 - 1 =/:. 0, we would have -a3 b + 2ab2 + 1 E ' aa4 ...... 3a2 b + b2 - 1 which is a contradiction. Therefore, -
Q
( 33) ( 34)
a4 - 3a2 b + b2 - 1 = 0, -a3 b + 2ab2 + 1 = 0.
We will now eliminate b from equations ( 33 ) and ( 34 ) . If we subtract the identity ( 34) from ( 33 ) multiplied by 2a, we get
(35)
and thus
4a 10 - 8a6 - 4a5 + 4a2 + 4a + 1 2a5 - 2a - 1 2 . b= and b = 25a6 5a3 If we substitute these expressions for b and b2 into ( 34 ) , we obtain after simplification the following equation for the coefficient a: a 10 + 3a6 - 1 1a 5 - 4a2 - 4a - 1 0. =
This equation, however, has no rational roots (by 3.26 the only candidates would be ±1 which, however, is impossible) . This means that there are no D rational numbers a, b satisfying the required condition.
3. 28 Exercises Find the zeros of the polynomials in (i) -(iii): (i) F (x ) = x4 + 2x3 - 2x2 - 6x + 5. (ii) F (x ) = 2x4 + 7x3 - 12x2 - 38x + 2 1 . (iii) F (x ) = 3x6 + 17x5 + 22x4 - 4x3 - 23x2 - 13x - 2 . *(iv) Let a , b, p, q E and suppose that the two polynomials F (x) x3 + px + q, G( x ) = x2 + ax + b have a common irrational zero. Show that G ( x ) divides F ( x) . (v) The parentheses in the equation ( ) x4 + ( ) x3 + ( ) x2 + ( ) x + ( ) = 0 are replaced in an arbitrary order by the numbers 1 , -2, 3, 4, and -6. Show that each such equation has at least one rational root.
Q
38
4
1. Algebraic Identities and Equations
Symmetric Polynomials
For the solution of systems of algebraic equations in Section 5 we will require the concept of a polynomial in several variables. In this section we will study such polynomials and will be especially interested in symmetric polynomials and their connection with Vieta ' s relations from Section 3.18.
4 . 1 Definitions
polynomial monomials, (36) ( k 1 ' k2 , ' kn E ) a . xkl l x2k2 xnkn where a a number (depending on the n-tuple k 1 , k2 , . . . , kn), called a coefficient. If some ki is zero, we just leave out x�. Similarly, we drop
A in n variables x 1 , x2 , . . . , Xn is a sum of finitely many expressions, called of the form ·
·
·
·
·
·
·
1M .l"qQ '
i$
the terms with zero coefficients. Since monomials with the same exponent n-tuple can always be added together (by adding their coefficients) , we will for the remainder of this section assume that a given polynomial is a sum of terms (36) with distinct n-tuples of exponents k1 , k2 , . . . , kn · A polynomial all of whose coefficients are zero is called the The sum k 1 + k2 + · · · + kn is called the degree of the monomial (36). will be defined as the largest degree The of the monomials occurring in the given polynomial. If in a polynomial all monomials with nonzero coefficients have degree r , then we talk about a
degree of a nonzero polynomial homogeneous polynomial of degree is called symmetric A polynomial
zero polynomial.
r.
if for an arbitrary F(x 1 , x2 , . . . , Xn ) order Yb Y2 , . . . , Yn of the variables x 1 , x2 , . . . , Xn we have F (y�, Y2 , . . . , Yn ) = F(x�, X2 , . . . , Xn ) · It is clear that in order to check whether F(x 1 7 x2 , . . . , xn ) is symmet ric, it suffices to verify that F(x� , x2 , . . . , xn ) does not change when x 1 is interchanged with Xi for arbitrary = 2, 3, For example, the polynomial x� + 2x 1 x 2 + 3x� is homogeneous of second degree in the two variables x 1 , x2 , but is not symmetric (by interchanging x 1 and x2 we obtain 3x� + 2x1 x2 + x�, which is a different polynomial). On the other hand, the fourth-degree polynomial x� + x� x� + x� in the two variables x 1 , x2 is not homogeneous, but it is sy:rpmetric. However, this
i
. . . , n.
last polynomial can also be considered as a polynomial in three variables x 1 , x2 , X3 (where X3 occurs with exponent zero in all terms) ; then this polynomial is not symmetric (for example, by interchanging x 1 and x3 we obtain x� + x�x� + x�). Now, we could begin constructing a general theory of symmetric polyno mials in n variables. However, since in applications we most often come across polynomials in two or three variables, we will basically restrict ourselves to these cases.
4 Symmetric Polynomials
39
4 . 2 Elementary Symmetric Polynomials in Two Variables It is easy to convince ourselves that the polynomials
O'l (Xb X2 ) = X l + X2 , (37) 0'2 (x1 , X2 ) = Xl X2 are symmetric in the two variables x 1 , x2 • More generally, the following is true: If we choose an arbitrary polynomial F(y1 , Y2 ) and substitute Yl = 0'1 (x�, x2 ) and Y2 = 0'2 (x� , x2 ), then we obtain the polynomial F(x1 + x2 , x 1 x2 ), which is symmetric in the variables x 1 , x2 . The question ·
arises whether this simple method of constructing symmetric polynomi als is sufficiently general, that is, whether one can obtain any symmetric polynomial in this way.
4 . 3 Representation of Symmetric Polynomials in Two Variables Theorem. For an arbitrary symmetric polynomial F(x 1 , x2 ) in two vari ables there exists a unique polynomial H (Yl , y2 ) in two variables such that F(x�, x2 ) = H(x 1 + x2 , x 1 · x2 ). A proof of a more general statement can be found in [13] . By Theorem 4.3 it is therefore possible to use the polynomials (37) to construct arbitrary symmetric polynomials. Therefore, 0'1 and 0'2 are called elementary symmetric polynomials. We will now describe a method for expressing a given symmetric polynomial F(x 1 , x2 ) in terms of these. The terms of F(x�, x2 ) of the form aii xix� will be change4 to aii o:� , and we collect the remaining terms in pairs of the form aijXi� + a3ix{x�, where i > j. The symmetry of F(x�, x2 ) implies the equation aij = a3i, and thus we can factor out ai3 (x 1 x2 )i = aij � · It therefore remains to find j expressions for the sums of powers Si -j = xi- + x;- j . 4-4 Sums of Powers Using the binomial theorem
1.2,
we obtain the following expressions for
S k = xt + x�: S 1 = X 1 + X2 = 0'1 , S2 = X� + X� = (x 1 + X2 ) 2 - 2X 1 X2 = 0'� - 20'2 , s3 = x� + x� = (x 1 + x2 ) 3 - 3x 1 x2 (x 1 + x2 ) = 0'� - 30'10'2 , S4 = xi + x� = (x 1 + x2 ) 4 - 4x l x2 (x� + x�) - 6x�x� = O't - 40'2 S 2 - 6 0'� = O't - 40'2 (0'� - 20'2 ) - 60'� = O't - 40"� 0'2 + 20'� '
40
1. Algebraic Identities and Equations
85 = x� + x� = (x 1 + x2 ) 5 - 5x 1 x2 (x� + x�) - lOx�x�(x 1 + x 2 ) = err - 5er2 83 - 10er�er1 = err - 5er2 (er� - 3er1 er2 ) - 10er�er1 = er51 - 5er31 er2 + 5 er1 er22 .
We could continue in this way; however, in this text we will not require formulas for 8i (i > 5). Another way of deriving these formulas is based on the recurrence relation 8i+2 = er1 8i + 1 - er2 8i, of whose validity we can easily convince ourselves.
4 . 5 Examples To illustrate the method described above we will express the given poly nomials F(x�, x2 ), G(x�, x2 ) , and H(x� , x2 ) by way of the elementary symmetric polynomials er 1 and er2 introduced in (37).
(i) F(x 1 , X2 ) = X�X2 + xfx� + X� X� + X 1 X� + X�X2 + x 1 x�. SoLUTION . We divide the terms of the polynomials F(x�, x2 ) into pairs of the form xi� + x{x�, and factor out and substitute the sums of powers from Section 4.4: F(x�, x2 ) = (x�x2 + x 1 x�) + (xfx� + x�x� ) + (x�x2 + x 1 x�) = x 1 x2 (xi + x� ) + (x 1 x2 ) 2 (x� + x�) + x 1 x2 (x1 + x2 ) = er2 8 4 + er�8 2 + er2 er1 = er2 (erf - 4er�er2 + 2er�) + er�(er� - 2er2 ) + er2 er1 o = erier2 - 3er�er� + er 1 er2 . SOLUTION .
We have
G(x� , x2 ) = (x 1 - x2 ) 2ooo = [(x 1 - x2 ) 2pooo = (x� + x� - 2x 1 x2 ) 1 ooo = [(x 1 + x2 ) 2 - 4x 1 x2 pooo = (er� 40"2 ) 1000 . _
SOLUTION .
0
By combining factors in an appropriate way, we obtain
H(x�, x2 ) = (x 1 + 4x2 )(4x 1 + x 2 )(2x 1 + 3x2 )(3x 1 + 2x2 ) = [4(x� + x�) + 1 1x 1 x2] (6(x� + x�) + 13x 1 x2] = [4(er� - 2er2 ) + 1 1er2] [6(er� - 2er2 ) + 13a2] = (4er� + 9er2 )(6a� + er2 ).
o
4 Symmetric Polynomials
41
If we compare (37) and Vieta's relations (28), we notice that the right hand sides are identical. More exactly, the right-hand sides in (28) are the values of the elementary symmetric polynomials 0'1 and 0'2 evaluated at the zeros of the single-variable polynomial a2 x 2 + a 1 x + a0• Theorem 4.3 has therefore the following important consequence: An arbitrary symmetric
polynomial in two variables evaluated at the zeros of a given quadratic poly nomial F ( x) in one variable x can be expressed in terms of the coefficients of F(x). (A similar statement is true for polynomials of higher degrees. )
4 . 6 Examples (i) Compute the number c = x� + x�, where x 1 , x2 are the zeros of F(x) = x2 + x SoLUTION . One possibility would be to find the zeros of F(x) and simply
- 19.
substitute. However, it will be easier to express c in terms of the coefficients of F(x). Since by (28), 0'1 (x� , x2 ) = x 1 + x2 = and 0'2 (x� , x2 ) = x 1 x2 = we have
-1
-19,
c = s s = O'r 50'� 0'2 + 50'1 0'� = 5. +5.
-
=
(- 1)5 - ( -1)3 • (-19) -1901 .
(-1) . (-19)2
0
(ii) Determine the coefficient c in the polynomial x2 + x + c if its zeros x 1 , x2 satisfy the equation 2x� + 2x� = 2 + X2 2 + X 1 SOLUTION. We rewrite this equation into the equivalent form (x 1 =/:. -2, X2 =/:. -2) 2x� (2 + x 1 ) + 2x�(2 + x2 ) + (2 + x 1 )(2 + x2 ) = 4(x� + x�) + 2(xi + x� ) + 4 + 2(x 1 + x2 ) + x 1 x2 = Using the relations for sums of powers from the table in 4.4 we express the left-hand side of this last equation in terms of 0' 1 = x 1 + x2 and 0'2 = x 1 x 2 ,
-1.
0, 0.
and get
40'� - 120'1 0'2 + 20'{ - 80'� 0'2 + 40"� + 4 + 20'1 + 0'2 = Since by (28) we have 0' 1 = x 1 + x2 = 0'2 = x 1 x2 = c, we obtain by substitution a quadratic equation for the unknown coefficient c, namely 4c2 + 5c =
0.
-1, 0,
whose roots
c1 =
0,
C2 =
-� are the solutions to our problem.
0
42
1 . Algebraic Identities and Equations
4. 7 Exercises Let x 1 , x2 be the zeros of the polynomial x 2 + px + q. In (i),
(ii) determine
the quadratic polynomial with the zeros
( 1.) x3l , x32 .
(ii) X� + X2 , X 1 + X�. (iii) Find the real coefficient b of the polynomial x2 + bx - 1 x 1 , x2 satisfy the equation
if its zeros
4 . 8 Example Find the quadratic polynomial with zeros
X l + X2 = 1. SOLUTION. Then
x 1 , x2 , given xf + x� = 31
Let the polynomial in question be of the form
and
x 2 + px + q.
p
-
= X l + X2 = 0'1 , q = X 1 X2 = 0'2 .
By the identity for the power sum
ss
in
4.4 we have
Since p =
-0'1 = -(x 1 + x 2 ) = -1, we obtain for q the quadratic equation 2 5q - 5q - 30 = 0 which has the roots q1 = 3, q2 = -2. The problem is therefore solved by the two polynomials x2 - x + 3 and x2 - x - 2 (note 0
that one of them has complex zeros) .
4. 9 Exercises In problems (i)-(iii) find a quadratic polynomial with zeros x 1 , x 2 satisfying
(i) 4(x l + x2 ) = 3x l x2 , x 1 + x2 + x� + x� = 26. (ii) X� + X 1 X2 + X� = 49, xf + X�X� + X� = 931. ( iii) X� + X� = 7 + X 1 X2 , x¥ + X� = 6X 1 X2 - 1. *( iv) Let a, b be two distinct roots of the equation x4 +x3 = 1. Show that their product ab is a root of the equation x 6 x 4 + x3 - x 2 - 1 = o.
+
4 Symmetric Polynomials
43
4. 1 0 Elementary Symmetric Polynomials in Three Variables To express symmetric polynomials in three variables we need the three elementary symmetric polynomials
0'1 (x�, x2 , x3 ) = X 1 + x2 + x3 , 0'2 (X 1 , X2 , X3 ) = X 1 X2 + X 1X3 + X2X3 , 0'3 (X 1 , X2 , X3 ) = X 1 X2 X3 .
(38)
H we substitute them into an arbitrary (not necessarily symmetric) poly nomial in three variables 0' 1 , 0'2 , and 0'3 , we obtain a symmetric polynomial in the variables x�, x2 , and x3 •
4 . 1 1 Representation of Symmetric Polynomials in Three Variables Theorem. For an arbitrary symmetric polynomial F(x�, x 2 , x3 ) in three variables there exists a unique polynomial H(y1 , y2 , y3 ) in three variables such that F(x�, x2 , x3 ) = H(x 1 + x2 + x3 , x 1 x2 + x 1 x3 + x2x3 , x 1x2 x3 ). Once again, a proof of a �ore general result, which also includes Theorem 4.3, can be found in (13].
4 . 12 Table Finding the polynomial H ( Y1 , Y2 , Y3 ) of the previous theorem without deeper knowledge of the theory can be difficult. Therefore, we give at least a few such expressions in the following table; with its help we can continue our work in this and the following section.
S2 X� + X� + X� = 0'� - 20'2 , S3 = X� + X� + X� = 0'� - 30'1 0'2 + 30'3 , S4 = xi + X� + X� = O'i - 40'�0'2 + 20'� + 40"1 0'3 , ss = x f + x� + xg = O"f - 50'� 0'2 + 50'1 0'� + 50'�0'3 - 50'2 0'3 , sa = x� + x� + xg = 0'� - 60'{0'2 + 90'�0'� - 20'� + 60'�0'3 - 120'1 0'2 0'3 + 30'�, x�x� + x�x� + x� x� = 0'� - 20'1 0'3 , x�x� + ��x� + x�x� = 0'� + 30'� - 30'1 0'2 0'3 , X�X2 + X 1 X� + X�X3 + X 1 X� + X�X3 + X2X� = 0'1 0'2 - 30'3 , X�X2 + X 1 X� + X�X3 + X 1 X� + X�X3 + X2X� = 0'�0'2 - 20'� - 0' 1 0'3 , xix� + x�x� + x{x� + x�x� + x�x� + x�x� = 0'� 0'� - 20'� - 20'� 0'3 + 40'1 0'2 0'3 - 30"�. =
1. Algebraic Identities and Equations
44
4. 13 Examples (i) Find all values of the parameter
x3 - 6x2 + ax + a satisfy
a E 1R such that the zeros Xb x2 , X3 of
(k = 2, 3). x� 8 4.12 xt = x� + + k a, = 0' 6, , x I, = 0' X X 1 3 2 2 0'3 = -a, 81 0'1 = 6, 82 = O"� - 20'2 = 36 - 2a, 83 = 0'� - 30'10'2 + 30'3 = 216 - 21a. for SOLUTION. We use the formulas in Since are zeros of the given polynomial, we have and therefore =
Using the binomial theorem we then obtain
= (x1 - 3)3 + (x2 - 3)3 + (x3 - 3) 3 = 83 - 382 · 3 + 38 1 32 - 3 · 33 = 216 - 21a - 9(36 - 2a) + 27 · 6 - 81 = -27 0 and therefore a = -9. 0
3a,
·
(ii) Prove that the number
F(x) = x3 + �x2 - 1. SOLUTION. We set a 1 = \jf, a 2 = A, a3 {ji. Clearly, a31 + a32 + a33 = 1 2 + = 1 a1 · a2 · a3 = \j! · (- i) · ( t) = - i' a1 a2 + a1a3 + a2a3 = H_ + -\[j; + � = A ( \ff + A+ iff) = A · c.
is a zero of
=
9 - 9
9 4
3'
a� + a� + a� = (a1 + a2 + a3) 3 - 3(a 1 + a2 + a3 )(a 1 a2 + a 1a3 + a2a3 ) + 3a1 a2a3 , and therefore l = e - 3c· A · c - i , which implies that the number c is a zero of F( x) x3 + � x 2 - 1. =
o
4 Symmetric Polynomials
45
4. 1 4 Exercises (i) Construct a cubic polynomial whose zeros are squares of the zeros of x3 - 2x2 + x - 12. *(ii) Suppose that the three distinct numbers a, b, c E 1R satisfy a(a2 + p) = b(b2 + p) = c(c2 + p) E :JR. Show that in this case a + b + c = 0. (iii} Show that if a, b, c E 1R and a + b + c = 0, then for an appropriate p
(iv) (v)
Construct a cubic polynomial with zeros x 1 + x2 , x 1 where x 1 , x2 , X3 are the zeros of x3 + px2 + qx + r . Prove that
+ x3 , x 2 + x3 ,
if the numbers a, b, c E 1R satisfy
1 1 1 = + + a b c _a_+_b_+-c ' 1
then for each odd
n
E N,
Elementary symmetric polynomials can also be successfully used to find decompositions of symmetric polynomials (into symmetric factors) .
4. 15 Example Decompose the polynomial
into a product of symmetric factors. SOLUTION.
By 4.12 we have
and thus
F (x 1 7 X2 , X3 ) = (x l + X2 + X3 ) [(x 1 + X2 + X3 ) 2 - 3(X1 X2 + XtX3 + X2 X3 )) = (x 1 + X2 + X3 ) · (x� + X� + X� - X 1 X2 - X 1 X3 - X2 X3 ). 0
46
1 . Algebraic Identities and Equations
4 . 1 6 Exercises Write the polynomials ( i)-( iv) as products of symmetric factors:
(i) (x 1 + X2 )(x 1 + X3 )(x2 + X3 ) + X 1 X2X3 . (ii) 2(x� +x� + x�) + x�x2 + x1 x� + x�x3 + x1 x� + x�x3 + x2 x� - 3x1 x2 x3 . (iii) (x� + x� + x� + x 1 x2 + x1 x3 + x2 x3 ) 2 - (x 1 + x2 + x3 ) 2 (x� + x� + x� ). *(iv) (x 1 + x2 + x3 ) 3 - (x 1 + x2 - x3 ) 3 - (x 1 - x2 +x3 ) 3 - ( -x 1 + x2 + x3 ) 3 . 5
Systems of Equations
The problem of solving systems of equations belongs to the basic ques tions of classical algebra; it receives considerable attention both in secondary-school mathematics and in introductory university algebra courses. Unfortunately, attention is mainly concentrated on systems of lin ear equations, while as a rule much less attention is paid to systems of equations of higher degree, and to methods of their solutions. We will there fore try to partially fill this gap in what follows. The main emphasis in this section will be on Sections 5.7-5. 14 (especially the method of symmetric polynomials) . In the initial parts we will introduce several nonstandard approaches to solving systems of linear equations (5.1-5.6) .
5. 1 Systems of Linear Equations As we already mentioned, we assume that the reader is acquainted with algorithmic methods for solving systems of linear equations ( Gaussian elimination, Cramer's rule) . We will therefore restrict ourselves to those problems for which it may be advantageous to solve them -differently, especially by introducing new auxiliary variables. We point out that we will solve systems of equations with real coefficients and look for their real solutions.
5. 2 Example Solve the following system of n equations in n variables:
X 1 + 2x2 + 3x3 + · + ( n - 1 )xn- 1 + nxn X2 + 2x3 + 3x4 + · + (n - 1) xn + nx1 X3 + 2x4 + 3xs + + (n - 1 )x 1 + nx2 · ·
· ·
· · ·
Xn + 2x 1 + 3x2 +
· ·
·
= 1'
2, = 3' =
+ (n - 1)Xn- 2 + nxn- 1 = n.
5 Systems of Equations
47
SOLUTION.
It turns out to be very useful to introduce the new variable s = x 1 + x2 + · · · + Xn , since by adding all equations we now obtain
+
s + 2s + 3s + · · · + (n - 1 ) s + ns = 1 + 2 3 + ·
·
· + (n -
1) + n,
and thus s = 1. Next we subtract the second equation from the first, the third from the second, etc., and finally the first equation from the nth, thus obtaining the system
(1 - n )x 1 + x2 + · · · + x1 + ( 1 - n )x2 + · · · +
Xn- 1 + Xn- 1 +
Xn = -1, Xn = -1,
X2 + · · · + ( 1 - n )Xn- 1 + Xn = -1, Xn- 1 + (1 - n)xn = n - 1, X2 + · · · + which can be rewritten as
s - nx 1 s - nx2
-1, = -1, =
s - nXn- 1 = -1, s - nxn = n - 1
.
2- n . X 1 = X2 = · · · = Xn- 1 = n2 , Xn = n-
From this we get
We realize that the last two systems are only consequences of ( as opposed to being equivalent to) the original system; we convince ourselves by a test that the n-tuple obtained above is indeed a solution of that system: For k = 1, 2, . . . , n the kth equation of the original system has the form
Xk + 2 · Xk + 1 + · · · + (n - k + 1)xn + (n - k + 2)x 1 + · · · + n · Xk - 1 = k. By substituting the left-hand sides we then obtain
+ ( n - k) · ; + ( n - k + 1) . 2�n 2 2 +( n - k + 2) · -n + + n · -n = ; (1 + 2 + · · · + (n - k + 1) + · · · + n) + (n - k + 1) ( - :) l = ; . n(n+ ) - ( n - k + 1) = n + 1 - n + k - 1 = k. 2
;+2 ;+ ·
·
··
·
·
·
·
Therefore, the system has a unique solution.
·
0
48
I . Algebraic Identities and Equations
5. 3 Exercises Solve the following systems of equations:
(i) 2X 1 + X2 + X3 + · · + X 1 + 2X2 + X3 + · · + X1 + X2 + 2X3 + + ·
·
· · ·
Xn = 1, Xn = 2, Xn = 3,
X 1 + X2 + X3 + · · + 2xn n. (ii) X2 + x3 + · · · + Xn- 1 + xn = 0, + X3 + · · + Xn- 1 + Xn = 1, X1 + + Xn- 1 + Xn = 2, ·
=
·
· · ·
= n - 1. X 1 + X2 + X3 + · · · + Xn- 1 (iii} X 1 + 2x2 + 3x3 + · · · + ( n - 1 ) Xn- 1 + nxn = n, + 3x3 + · · + ( n - 1 ) Xn- 1 + nxn = n - 2, X1 + + ( n - 1 ) xn- 1 + nxn = n - 3, X 1 + 2x2 ·
· · ·
X 1 + 2x2 + 3x3 + · + ( n - 1 ) Xn- 1 ·
·
= 0.
5.4 Systems of Linear Equations with Parameters These systems actually represent a whole (normally infinite) set of systems, one each for a fixed choice of values of the parameters. For solving them, it is often necessary to discuss the form of the solutions ( possibly also their number) depending on the values of the parameters. In the following examples we will explain several methods of solution of systems of linear equations with parameters. Here, Xi are variables and ai are parameters.
(i) x 1 + x 2 + x3 = a�, + X4 = a2 , X 1 + X2 + x3 + x4 = a3 , x1 x2 + x3 + x4 = a4 .
( 39)
Introducing the auxiliary variable 8 = x 1 + x2 + X3 + X4 , we can rewrite the system ( 39) as
x 1 8 - a4 , x 2 = 8 - a3 , X 3 = 8 - a2 , X4 = 8 - a 1 . =
5 Systems of Equations
49
Summing the equations of the last system, we obtain 3s a t + a2 + a3 + a4 , hence s = (a t +a2 +a3 +a4 )/3, and the system (39) has the unique solution =
x2 = (a t + a2 - 2a3 + a4 )/3, Xt = (a t + a2 + a3 - 2a4 )/3, X4 = ( -2a t + a2 + a3 + a4 )/3, X3 = (at - 2a2 + a3 + a4 )/3, which must be verified by substituting. (ii) Xt - X2 - X3 - · · · - Xn- t Xn = 2a, -Xt + 3x2 - X3 - · · · - Xn- t Xn = 22 a, -Xt - X2 + 7X3 - · · · - Xn- t Xn = 23 a, (40) -Xt - X2 - X3 - · · · - Xn- t + (2n - 1 )Xn = 2n a . We rewrite the system ( 40) in the equivalent form 2Xt - (Xt + X2 + X3 + · · · + Xn ) = 2a, 22 x2 - (x t + X2 + X3 + · · · + Xn ) = 22 a, 23X3 - (Xt + X2 + X3 + · · + Xn ) = 23 a, ·
2n xn - (xt + X2 + X3 + · · · + Xn ) = 2n a; then, again setting s = Xt + x2 + X3 + · · · + Xn , we have Xt - � = a, x2 - -[2 = a, x3 - � = a, Xn - 2� = a, and by adding the equations of the last system we get
s - s (! + 2\ + � + · · + 2� ) = · a, from which by formula (10), s = · 2n · a. The system (40) has therefore a unique solution: Xk = a + s/2k = a ( 1 + · 2n- k ) for k = 1, 2, which n
·
n
n
. . . , n,
has to be verified by substituting.
(iii) 2xt - x2 -Xt + 2X2 - X3
= at , = 0, = 0, -Xn-2 + 2Xn - t - Xn = 0, - Xn- t + 2Xn = a2 .
(41)
50
1. Algebraic Identities and Equations
We rewrite the equations of the system ( 41) from the second to the second to-last one in the form x 1 - x2 = x2 - X3 , x2 - X3 = X3 - X4 , . · Xn - 2 Xn- 1 = Xn- 1 - Xn· The ''trick" of elimination relies on the introduction of the new variable t = x1 - x2 , and in view of the preceding equations we have .
,
t = X 1 - X2 = X2 X3 = X3 - X4 = · · = Xn- 1 - Xn· ·
-
The first and the last equations in the system (41) can also be expressed in terms of the variable t:
Thus we obtain the system
a 1 - t, a 1 - 2t, a 1 - 3t,
X1 = X2 = X 1 - t X3 = X2 - t
(42) Xn- 1 = Xn-2 - t = Xn = Xn- 1 - t = Xn =
a 1 - (n - 1)t, a 1 - nt,
From the last two equations of ( 42) we then determine t = ( a 1 -a2 ) / ( n + 1). Substituting this into the first n equations of ( 42), we find that Xk = a 1 - k · (a 1 - a2 )/(n + 1) for each k = 1, 2, . . . , n. To complete the solution it remains only to substitute into ( 41).
5. 5 Example Solve the system with parameters a 1 , a2 , . . . , an E JR.:
X 1 + X2 = a 1 , X2 + X3 = a2 , X3 + X4 = a3 , Xn- 1 + Xn an - 1 , Xn + x 1 = an . =
(43)
5 Systems of Equations
51
(i) FIRST SOLUTION. We begin with step-by-step elimination of the variables in the individual equations in ( 43):
X2 = a1 - x 1 , X3 = a2 - x2 = a2 - a 1 + x 1 , X4 = a3 - X3 = a3 - a2 + a 1 - x�,
Xn = an- 1 - Xn- 1 = an- 1 - an- 2 + · · + ( - l ) n+ l X! , X 1 = an - Xn = an - an- 1 + · · + (-l) nXl · ·
·
From the last two equations in ( 44) it is apparent that to proceed further we have to distinguish between the cases n even and n odd. (a) n = 2k. From the last equation in ( 44) it is clear that this system, and therefore also ( 43), has a solution if and only if a 1 + a3 + · · · + a2 k- 1 = a2 + a4 + · · · + a2k . In this case, ( 43) has infinitely many solutions, which can be expressed, by ( 44), in terms of the free variable x 1 . (b) n = 2k + 1 . In this case it follows from the last equation in ( 44) that 2x 1 = a2k + l - a2k + · · · + a 1 ; hence x 1 = (a 1 - a2 + · · · - a2k + a2k+ l )/2. If we write the original system as
X2 + X3 = a2 , X3 + X4 = a3 , Xn + X1 = an , X 1 + X2 = a 1 , then by repeating the above approach we obtain
a +a a X2 = a2 - 3 + 2- 2 k+ 1 1 , · · ·
and in complete analogy,
ai- 2 + ai- l a -a + Xi = i i+ l 2 · · · -
0
(ii) SECOND SOLUTION. We distinguish between the parities of n from the beginning, and introduce the auxiliary variable s = x 1 + X2 + · · · + Xn . (a) n = 2k. By adding the odd equations in the system (43) we find that 8
=
+
X 1 + X2 + · + X2k- 1 X2k = a 1 + a3 + · • · + a2k- 1 , · ·
52
1. Algebraic Identities and Equations
and adding the even ones,
therefore, a necessary condition for the solvability of the system is the equation a 1 + a3 + · · · + a2k- l = a2 + a4 + · · · + a2k . If this condition is satisfied, then ( 43) has infinitely many solutions, which can be expressed, as in 5.5.(i)(a), in terms of the free variable x 1 ; in the opposite case (43) has no solutions. (b) n = 2k + 1. H we add all the equations in ( 43), then upon dividing by 2 we get s
= x 1 + x2 +
· · ·
+ X2 k + l
= (a 1 + a2 +
· · ·
+ a2k + l )/2 .
To evaluate the variable x�, we add all odd equations in ( 43) , and obtain hence x 1 = ( a 1 - a2 + a3 - • · · a2k + a 2k + l ) /2. This method can also be used to find Xi , i = 2, 3, . . . , 2k + 1; we only have to change the order of the equations in (43) and rewrite that system in the form -
Xi + Xi+ l = ai, Xi+ l + Xi+2 = ai + l , (45) Then by adding all odd equations in ( 45) we get 0
5. 6 Exercises Solve the systems (i)-(v) , where p and ai are real parameters:
(i) X l + X2 + X3 + X4 = 2ab X 1 + X2 - X3 - X4 = 2a2 , X 1 - X2 + X3 - X4 = 2a3 , X 1 - X2 - X3 + X4 = 2a4 . (ii) (1 + p) xl + X3 = 1, X2 + X3 = 1, X 1 + ( 1 + p)x2 + X1 + X2 + (1 + p)x3 = 1.
5 Systems of Equations
(iii) x 1 + x 2 x 1 + X3 x 1 + X4 x2 + X 3 x2 + X 4 X3 + X 4
=
=
=
=
=
=
a 1 a2 , a 1 a3 , a 1 a4 , a2 a3 , a2 a4 , a3 a4 .
In (iv) and (v) , assume that a 1 + a2 + a3 1= 0: (iv) ( a2 + a3 )( x2 + x3 ) - a 1 x1 a2 - a3 , ( a 1 + a3 )( x1 + x3 ) - a2x2 a3 - a� , ( a 1 + a2 )( x 1 + x2 ) - a3x3 a 1 - a2 . (v) a 1 x 1 + a2 x2 + a3 x3 a 1 + a2 + a3 , a2x 1 + a3 x2 + a 1 x3 a1 + a2 + a3 , a3x 1 + a1 x2 + a2 x3 a 1 + a2 + a3 . (vi) Solve the following system with the parameter a E lR: 3 a, X + + X X X + + 1 3 n 2 2 2 a 5 Xn 4 , X 1 + 4 X2 + X 3 + + =
=
=
=
=
=
9 X l + X 2 + 8X 3 +
· · ·
=
· · ·
=
· · ·
+
Xn
=
a
8'
(vii) Solve the following system with parameters a�, a2 , a3 , a4 : a1 , 2x 1 - x2 a2 , -x1 + 2x2 - X3 -x2 + 2x3 - X4 = a3 , -x3 + 2x4 a4 . (viii) Show that if X 1 + X2 + X3 0, X2 + X3 + X4 0, =
=
=
=
=
Xgg + X lQQ + X1 0, X100 + X 1 + X2 0, Xgg X100 0. =
=
then X l
=
X2
=
X3
=
· · ·
=
=
=
53
54
1. Algebraic Identities and Equations
5. 7 Systems of Equations of Higher Degree As opposed to systems of linear equations, there is no universal method for solving systems of higher-degree equations. However, there are several methods that can be used in special cases. The simplest one is the elimina tion method, which with the help of one equation of the system eliminates several variables in the remaining equations. The disadvantage, however, is that if we use a nonlinear equation, then often the degree of the remaining equations will increase.
5. 8 Examples We will illustrate the advantages and shortcomings of the elimination method by solving three systems of nonlinear equations: (i) x 1 + 2x2 + x3 = 8,
x 1 + 3x2 + 2x3 = 12, X� + X� + X� = 1 9 .
SOLUTION . If we subtract the first equation from the second, we obtain x2 = 4 - X3 . Substituting this into the first and third equations, we obtain
a system of two equations in two variables:
X1 - X3 = 0 , x� + 2x� - 8x3 = 3. If we now substitute the first equation into the second, we get 3x� - 8x3 -3 = 0, which has the two solutions X3 = 3 and X3 - l Hence the system also has two solutions: x 1 = 3, x2 = 1, X3 = 3 and X1 = l , x2 = 13/3, 0 X3 = - l · =
·
-
(ii) X1 (X 1 + X2 ) = 9 ,
X2 (X 1 + X2 ) = 16.
SOLUTION. Adding the two equations, we obtain (x 1 +x2 ) 2 = 25 , and thus x 1 + x2 = ±5. The system therefore has two solutions, which we obtain by substituting x1 + x2 into the original equations: x 1 = �' x2 = 16/5 and 0 X1 = -�, X2 = -16/5.
( ) x21 + x22 = 37 ' � + x� = 3 . .1 ·11 ·
SOLUTION.
theorem,
We will try to eliminate the variable x 2 . By the binomial
4 x62 - (x22 )3 - (13 x21 ) 3 - 343 27 39 x21 + 7x41 - x61 ' x� = (x�) 2 = (3 - x� ) 2 = 9 - 6x� + x�, _
_
_
_
_
5 Systems of Equations
55
and therefore
2x� - 7xf - 6x� + � x� - �� = 0; this, however, is an equation of degree 6 that we cannot solve. The only possibility, namely reducing its degree by using 3.23, does not work, because
we have no root of this equation at all. Trying out all possibilities according to 3.26 (after multiplying by 27 so that we get a polynomial with integer coefficients) , we find that this polynomial has no rational roots. One method that leads to the solution of this problem will be described 0 in 5.10, and example 5.8.(iii) will then be solved in 5.1 1 . (ii).
5. 9 Exercises Solve the following systems of equations: (i) x� + x� + 6x 1 + 2x2 = 0, (ii) x 1 + x2 = -8.
(iii} (v) *(vii)
x� + 2x� = 9, (x 1 + 1) 2 + 2(x2 + 1) 2 = 22. X� - X 1 - 5 = 0, 1- 1 - 1 1 = 1 •
X2
X2 +
Xt
.!.1.
� X2 + X t
=
163
'
x1 + x2 = 5. (iv) 5x 1 + 5x2 + 2x1 x2 = -19, 3x 1 x2 + x 1 + x2 = -35. (vi) X� - X1 X2 = 28, X� - X1 X2 = - 1 2 . *(viii) 2x� + 3X1 X2 + X� = 1 2 , 2(X1 + X2 ) 2 - X� = 14.
X� - X 1 X2 + X� = 21, X� - 2X 1 X2 + 15 = 0. (ix) X� - X 1 X2 - X� + 3X 1 + 7X2 + 3 = 0, 2x� + x1 x2 - x� = 0. *(x) 3x� + X1 X2 - X1 + 4x2 - 7 = 0, X� + 2X 1 X2 - 2X1 - 2X2 + 1 = 0. x 1 + x2 = 2 , *(xt.} X 1 X3 + X2 X4 = - 8, X 1 X� + X2 X� = -4, XIX� + X2 X� = - 1 28 .
5. 10 The Method of Symmetric Polynomials The method of symmetric polynomials can be applied to the solution of sys tems of equations in which the unknowns occur as variables of symmetric polynomials. By the results of Section 4 we know that an arbitrary symmet ric polynomial can be expressed as a polynomial in elementary symmetric polynomials. Therefore, the whole system can be rewritten as a system in which elementary symmetric polynomials of the original variables occur as
56
1 . Algebraic Identities and Equations
new variables. The advantage of this method lies in the fact that as a rule, the new system is of lower degree than the original system. If we find the values of the elementary symmetric polynomials, then the values of the origial variables can be obtained as solutions of equations constructed with the help of Vieta's relations 3.18, which have degrees equal to the number of variables.
5. 1 1 Examples (i) Let us solve the system of equations x� + x� = 464, x 1 + x2 = 4. SOLUTION. It is true that by substituting the second equation into the first one we could obtain an equation in one variable, but we would be unable to solve it. Therefore, we use the method of symmetric polynomials, in the notation of (37) . By the power sum relationship 85
= X� + X� = O'r
-
50'� 0'2 + 50'1 0'�
from Section 4.4 we can rewrite our system as 0'� - 50'� 0'2 + 50'1 0'� = 464, 0'1 = 4, which, after simplification, gives 0'� - 160'2 + 28 = 0, and therefore 0'2 = 2 or 0'2 = 14. Thus we get two systems of equations in the original variables x1 , x2 :
x 1 + x2 = 4, X 1 X2 = 2,
x 1 + x2 = 4, X1X2 = 14, from which by 3.18 it follows that x 1 , x2 are zeros of the polynomials x2 4x + 2, resp. x2 - 4x + 14. The first one of these has the real zeros 2 ± J2, resp.
while the second one has no real zeros. Our system has therefore the two soutions x 1 = 2 + J2, x2 = 2 - J2 and x 1 = 2 - J2, x2 = 2 + J2 in real numbers. (If we wanted to solve the system in the field of complex numbers, we would obtain the additional two solutions x 1 = 2+iVTii, x2 = 2 - i VIO, resp. x 1 = 2 - iVTii, x2 2 + iVlO.) D =
(ii) Let us now solve 5.8.(iii) using the method of symmetric polynomials.
5 Systems of Equations
57
SoLUTION. Since by Section 4.4 we have the identities x� + x� = u� - 20'2 and x� + x� = 0'� - 30'1 0'2 , we can rewrite the original system as 21 - 2 2 = 7 0' 3' 0'� - 30' 1 0'2 = 3. 0'
By substituting 0'2 from the first equation into the second one, we obtain
which is an equation of degree three and has zeros 1, 2, and -3 (by 3.26). If 0'1 = 1, we get 0'2 = -�; if 0'1 = 2, then 0'2 = �' and for 0'1 = -3, we get 0'2 = 10/3. The original variables x1 , x2 are then the zeros of the polynomial x2 - x - i , resp. x2 - 2x + �' resp. x2 + 3x + 10/3. Thus we obtain the four real solutions (x1, x2 ) E { ( (3 + v'33) /6; (3 - v'33) /6) , ((3 - v'33) /6; (3 + v'33) /6), ( 1 + ¥'6/6; 1 - ¥'6/6) , ( 1 - ¥'6/6; 1 + ¥'6/6)} . If we wanted to solve the system over the complex numbers, we would obtain the additional two solutions x1 = ( -9 + iv'39) /6, x2 = ( -9 - iv'39) /6 and 0 X 1 = ( -9 - i v'39) /6, X2 = ( - 9 + iv'39) /6.
5. 12 Exercises Solve the following systems of equations; in (ix) , a E 1R is a parameter. (i)
X� + X� = 35, x1 + x2 = 5.
X� + X� + X1 + X2 = 32, 12(X 1 + X2 ) = 7X 1 X2· x21 x22 (v) - + 12 ' X2 X1 1 1 1 - + - = -. X 1 X2 3 * (iiv) X� - X� 19 (X 1 - X2 ) , X� + X� = 7 (x 1 + X2 ) · (i:x) X1 + X2 = a, x41 + x42 = a4 . (iii }
=
* (xi)
x1 + x2 - x3 = 7, X� + X� - X� = 37, X� + X� - X� = 1.
(ii)
X1 + X2 = 7, X X2 25 -1 + - = - . X2 X 1 12 (iv) X� - X 1X2 + X� = 19, x 1 - x 1 x2 + x2 = 7. x� + x� 31 * (vi) x� + x� - 7, x� + x 1 x2 + x� = 3. *(viii) (x)
x� + x2 = 5, x� + x� = 65.
X 1 + X2 = 3, x� + x� = 33.
58
1 . Algebraic Identities and Equations
5. 13 Examples Solve the following systems of equations: (i)
X l + X2 + X3 = 2, X� + X� + X� = 6, X� + X� + X� = 8.
SOLUTION. We express the polynomials on the left sides in terms of the elementary symmetric polynomials (38) , using Table 4.12, and we obtain the system
0'1 = 2, 0'� - 20'2 = 6, O"� - 30'1 0'2 + 30"a = 8, from which it follows that 0'1 = 2 , 0'2 = -1, 0'3 = -2. By 3 .1 8 , x 1 , x2 , X3 are zeros of the polynomial x3 - 2x2 - x + 2. By 3.26, the rational zeros of this polynomial can be only 1, 2, -1, -2. By substituting we find that the zeros are -1, 1, 2. The system has therefore six solutions, one of which is x1 = -1, x2 = 1, X3 = 2, and the others are obtained by changing the 0 order of the zeros. (ii)
X 1 + X2 + X3 = 6, X 1 X2 + X 1 X3 + X2 X3 = 11, (x 1 - X2 )(X2 - X3 )(X3 - X 1 ) = 2.
SOLUTION. While the first and second equations have symmetric polyno mials on the left-hand sides, the polynomial L = (x 1 -x2 )(x2 -x3 )(x3 -x 1 ) is not symmetric, since upon interchanging x 1 and x2 the sign will change. However, if we square it, it becomes symmetric, and we have
L2 = [(x 1 - x2 )(x2 - x3 )(x3 - x 1 )] 2 = (x� - 2x 1 x2 + x�)(x� - 2x2x3 + x�) (x� - 2x 1 x3 + x�) = [s2 - (2x 1 x2 + x�)] [s 2 - (2x2 x3 + x�)] [s2 - (2x 1 x3 + x�)], where 82 = x� + x� + x�, consistent with the notation in 4.12. We further set a = 2x 1 x2 + x�, b = 2x2x3 + x�, c = 2x 1 x3 + x�. Then L2 = (82 - a)(82 - b)(s2 - c) = s� - (a + b + c)s� + (ab + be + ca)s2 - abc. We express a + b + c, ab + be + ca, abc according to 4.12: a + b + C = 2X 1 X2 + x� + 2X2 X3 + X� + 2X 1 X3 + X� = 82 + 20'2 .
5 Systems of Equations
59
Since
+
ab = (2x 1 x2 + x�)(2x2x3 + x�) = x�x� 4x 1X�X3 + 2(x�x2 + x2 x�), we obtain by symmetry
ab + be + ca = (x�x� + x�x� + x� x�) + 4x 1 x2x3 (x 1 + x2 + x3 ) + 2(x�X2 + X 1 X� + X� X3 + X 1 X� + X�X3 + X2 X�) = 20'� 0'2 - 30'� ' and finally
abc = (ab)c = [x�x� + 4x1 X�X3 + 2(x�x2 + x2 x�)] (2x 1x3 + x� ) = 9x�x�x� + 2(x�x� + x�x� + x�x�) + 4x 1 x2 x3 (x� + x� + x�) = 270'� + 20'� - 180'1 0'2 0'3 + 40'r0'3 . Therefore, we have
L2 = s� - ( s2 + 2D"2 )s� + (20'�0'2 - 30"�) s2 - (270'� + 20'� - 180'1 0'2 0'3 + 40'r0'3 ) = -20'2 (0'� - 20'2 )2 + (20'�0'2 - 30'�)(0'� - 20'2 ) - 270'� - 20'� + 1 80'1 0'2 0'3 - 40'� 0'3 = 0'�0'� - 40'� - 40"r0'3 + 180'1 0'2 0'3 - 270'�. If we substitute 0' 1 = 6, 0'2 = 11 from the first two equations into the equation L2 = 4, we obtain 0'� - 120'3 + 36 = 0, and thus 0'3 = 6. The solutions of the original system are therefore the
zeros of the polynomial
x3 - 6x2 + 11x - 6 = (x - 1)(x - 2)(x - 3). Since in the course of this solution we carried out some irreversible steps (we took the square of an equation ) , not all six permutations of the numbers 1, 2, 3 may be solutions of our system. Indeed, by checking we find that the system has exactly three solutions: x 1 = 1, x2 2, X3 = 3 and x 1 = 2, 0 X2 = 3, Xa = 1 and X I = 3, X2 = 1, X3 = 2. =
5. 14 Exercises Solve the following systems of equations, where in (vi) a is a real parameter.
(i)
X 1 + X2 + X3 = 9, 1 1 1 -1 ' Xt + X2 + X3 X 1 X2 X 1 X3 + X2 X3 = 27.
+
(ii)
X 1 + X2 + X3 '-; , 13 ...!.. + ...!.. + ...!.. Xt X2 X3 = 3 . X 1 X2X3 = 1. =
·
60
1. Algebraic Identities and Equations
(x 1 +x2 )(x2 +x3 )+(x2 +x3 )(x3 +x l )+(x3 +x l )(x 1 +x2 ) = 1, X�(X2 + X3 ) + X� (X3 + Xt) + X�(Xt + X2 ) = -6, x 1 + x2 + x3 = 2. (iv) XtX2 (Xt + X2 ) + X2 X3 (X2 + X3 ) + XtX3 (X t + X3 ) = 12, XtX2 + X 1 X3 + X2X3 = 0, x1 x2 (x� + x�) + x2 x3 (x� + x�) + x 1 x3 (x� + x�) = 12. (vi) x 1 + x2 + x3 = a, (v) Xt + x2 + x3 = 0, x21 + x22 + x23 = a2 , X� + X� + X� = X� + X� + X�, X� + X� + X� = a3 • X t X2 X3 2. X� + X� + X� = 9, (vii) ..!.. + _!_ + _!_ = Xt + X2 + X3 , X l X2 X3 XtX2 + XtX3 + X2 X3 = -4. *(viii) X� + 8x� - X� = 3, 2XtX2 - X1X3 - 2X2 X3 = Xt + 2x2 - X3 , X 1 X2 X3 = - 21 · *(ix) X� + X� + xg - X� = 210, X� + X� + X� - X� = 18, x� + x� + x� - x� = 6, Xt + X2 + X3 - X4 = 0. *(x) 3XtX2X3 - X� - X� - X� = - 9, Xt + X2 + X3 = 3, X� + X� - X� = 5. (iii)
=
6
Irrational Equations
In this section we introduce several methods that can be used to solve irrational equations, namely equations that contain a variable underneath a radical sign. We will solve these equations in IR; therefore, we first recall the definition of a root of a real number: If n is an even natural number, we define the nth root of a nonnegative number a as the (unique) nonnegative number b for which bn = a (written b = y'a); for a < 0 the symbol yla is not defined. If n is an odd natural number, we define the nth root of a real number a as the (unique) real number b for which bn = a (again written b = y'a). We point out that the numbers a and y'a have the same sign.
6 Irrational Equations
61
6. 1 The Implication Method Irrational equations will be solved, as a rule, by using what we will call the implication method. ·we consider as part of this not only the usual re versible manipulations of equations (interchanging the sides of an equation, adding the same number or expression to both sides of an equation, mul tiplying both sides by the same nonzero number or expression) , but also several irreversible manipulations, especially those that are derived from the following theorem.
Theorem. Let L( x), R( x ) be expressions in the variable x defined on the set D C 1R. Then for each x E D we have
(i) L (x) R(x) ===} [L(x)] n = [R(x)] n for each (ii ) L(x) = R(x) ===} L(x) · G(x) R(x) · G(x), =
n
E N,
=
if the expression
G (x)
is defined for each x E D.
Remark. The implications (i) and Let now
(ii) can in general not be reversed.
Lo(x) = Ro(x)
(46)
be some equation in the variable x E lR, which we will solve by way of the implication method; that is, we construct the chain of implications
Lo(x) = Ro(x) ===} L 1 (x) = R1 (x) ===} • • • ===} Ln (x) = Rn (x)
(47)
such that we can easily determine the roots of the equation Ln (x) = Rn ( x); in fact, L n (x) and Rn ( x) are most often polynomials. If for i = 0, 1 , . . . , n we denote by /Ci the set of roots of the equation Li ( x) = R;, (x), then by ( 47) we obviously have /Co C /C1 C · · · C ICn , and therefore /Co C ICn· The set ICn certainly contains all the roots of the equation ( 46) (if they exist at all) and possibly also several other numbers that we drop when we carry out a check by substituting all numbers of the set ICn into ( 46). If we solve an irrational equation ( 46) by the implication method, then this check must be part of the solution. Therefore, this method is also called the method of analyzing and checking (by analyzing we mean the chain of implications (47)). Using this method for solving irrational equations therefore frees us from the necessity to keep track of whether for individual manipulations of an equation its domain of definition increases, or whether upon taking even powers the expressions on both sides of the equation have the same sign. The members of the set ICn \ /Co (which are sometimes called ''false roots" ) will be determined by checking.
1. Algebraic Identities and Equations
62
6. 2 Examples We now solve several simple irrational equations.
(i) yf1 + 3x = 1 - x. If we square both sides, we obtain the quadratic equation 1 + 3x = 1 -2x + x2 , that is, x2 - 5x = 0, which has the roots x 1 = 0, x2 = 5. Substituting these into the original equation, we obtain L(O) = 1 = R(O), L(5) = 4 1= -4 R(5). Therefore, the equation v'1 + 3x = 1 - x has exactly the one solution x = 0. (ii) yf1 + 3x = x + 1. Again by squaring we obtain the quadratic equation 1 + 3x = x2 + 2x + 1, that is, x2 - x = 0, with roots x 1 = 0, x2 = 1. Since L(O) = 1 R(O), L(1) = v'4 = 2 = R( 1) (having substituted into the original equation) , the =
=
given equation has the two roots 0 and 1 .
(iii) J2 + x2 = x - 1.
By squaring we obtain 2 + x2 = x2 - 2x + 1, thus 2x -1 and x = � By substituting we see that the original equation has no root at all, since =
-
.
L(-�) = li = J ;/= -J = R(- !).
6. 3 Exercises Solve the following equations:
Jx2 + 7 = 2x + 2 . vf31 + x - x2 = 5 - x. 2x - 6 = J6x - x2 - 5 . 2 (iv) J5x +- 1x = 1.
(i) (ii) (iii)
6. 4 Remark Even though by using the implication method one need not worry about the equivalence of the manipulations, one can sometimes significantly accelerate the solution of a problem by determining the domains of definition of the equation or checking the signs of its sides. Thus, for example, in problems (i)-(iii) below one can easily decide that the equation has no solution in lR, and in problem (iv) it is easy to "guess" the solution. (i) The domain of definition of the equation yf4 - x = v'x - 6 is the empty set, since the inequalities x < 4 and x > 6 cannot be satisfied at the same time. (ii) The equation v'x + 2 = -2 has no real root, for while its left-hand side is nonnegative, on the right we have the negative number - 2. �-
6 Irrational Equations
63
(iii) The left-hand side of the equation v'2x + 3 + Jx + 3 = 0 is a sum of a nonnegative and a positive number (indeed, for x > - � we have v'2x + 3 > 0 and Jx + 3 > 0, while for x < - � the equation makes no sense ) , and is thus a positive number. Therefore, the equation has no solution . (iv ) Let us now solve the equation
v'11x + 3 - v'2 - x - v'9x + 7 + Jx - 2 = 0. We easily convince ourselves that the domain of definition of this equation is D = { 2 } . H we substitute x = 2, we see that L(2) = v'25 - 0 - v'25+0 = 0 = R(2); the given equation therefore has exactly the one solution x = 2.
6.5 Exercises Show that the following equations have no roots in lR:
(i) (ii) (iii) (iv)
Jx - 6 + V3 - x = 4x - 3x2 + 1 . v'4x + 7 + v'3 - 4x + x2 + 2 = 0. v'x + 9 + y'x = 2. 2 · v'x3 - 4x2 + 1 + Jx2 · (x + 1) 3 = -5.
6. 6 Examples We solve now two equations in a standard way.
(i)
v'2 - x +
4 = 2. v'2 - x + 3
SOLUTION. If we multiply the whole equation by the expression 3, we obtain
v'2 - x +
2 - X + 3 · V2 - X + 4 = 2 · V2 - X + 6, which implies v'2 - x = x, and thus x2 + x - 2 = 0; this quadratic equation has roots x 1 = 1 , x 2 = -2. By trial we find that the original equation has O the unique root x = 1. v'x + 5 - v'X = 1. (ii) (48) SOLUTION .
By squaring equation
( 48) we obtain
x + 5 - 2 · .../x (x + 5) + x = 1, and thus x + 2 = .jx(x + 5). By squaring once more we get x2 + 4x + 4 = x2 + 5x, so x = 4. Since in ( 48) we have L( 4) = 1 = R( 4), the equation has O the root x = 4.
64
1. Algebraic Identities and Equations
6. 7 Remark It is often advantageous to begin solving an irrational equation by trans forming it appropriately. This can be done, for example, by separating one of the roots from the remaining terms and putting it on one side of the equation before taking powers . The necessary calculations are often sim pler this way. Let us convince ourselves of this by solving the following three equations:
( i) v'x + 5 - y'X = 1.
We have already solved this equation in 6 . 6 . ( ii ) . Here we transform it into Jx + 5 = 1 + y'X, then x + 5 = 1 + 2y'i + x, and thus y'X = 2 and x 4. This approach for solving (48) requires less work than the one used in 6.6 . (ii ) . =
(ii) 1 +
V1 - v'x4 - x2 = x.
We put the equation in the form .../1 - vx4 - x2 = x - 1, thus 1 Jx4 - x2 = x2 - 2x + 1, and by separating the root from the remain ing terms we obtain the equivalent equation 2x - x2 = vx4 - x 2 , which upon squaring gives x2 (2 - x) 2 = x2 (x2 - 1). Therefore, x 2 (5 - 4x) = 0, which means x 1 = 0, x2 = �. By checking we convince ourselves that while L (O) = 2 ;/= 0 = R(O), we have L(�) = £ = R( £ ), and the original equation has a unique root .
(iii) Jx + 1 + v'4x + 13 = ..j3x + 12.
Here we rewrite the equation as ..jr=3x-+""""1-="' 2 - Jx + 1 = v'4x + upon squaring we obtain the easiest possible consequence
13,
so that
3x + 12 - 2 V(3x + 12)(x + 1) + x + 1 = 4x + 13. J(3x + 12)(x + 1) = 0, which implies x 1 = -1, x2 = -4. While for x2 = -4 the expression v'x2 + 1 is not defined, x = -1 is a solution of the Thus
original equation.
6. 8 Exercises Solve the following equations:
(i) (ii) (iii) (iv) (v) (vi) (vii)
Jx - 1 + v2x + 6 = 6. Jx2 + 1 + Jx2 - 8 = 3. VX + 2 - V2X - 3 = VrX ----1.-.,.x2 + v'x2 + 20 = 22. v5x + 1 - v6x - 2 - vx + 6 + v'2x + 3 = 0. v2x - 4 - v3x - 11 = Jx - 3. + 3-= = 0. ..j6x + 1 + ..j4x + 2 - J8X - v'�2x---
6 Irrational Equations
65
6. 9 Multiplication by a Conjugate Expression
For solving certain irrational equations of the form ..Jf(X5 + ..j9(X} = cp(x ) , it may be useful to multiply by an expression of the form ..Jf(X5- ..j9(X}, which is called conjugate to the expression ..Jf[X5 + ..fi(X). The same is true for equations of the form ..Jf(X5 ..fi(X) = cp(x); in both cases use of the formula -
( ..Jf[X5 + v"9W) ( ..Jf[X5 - v"9W)
=
f(x) - g(x)
may simplify the equation, especially in cases where f(x) and g(x) are "similar." After this it is useful to consider the following pair of equations together:
..Jf[X5 ± v"9W cp(X ) ' =
f(x) - g(x) = cp(x) · ( J7(X) =t= v"9W) .
We illustrate the use of this method by solving the equation
.../2x2 + 3x + 5 + .../2x2 - 3x + 5 = 3x. If we multiply simplification,
( 49)
(49)
by the difference of the square roots, we obtain after
2x = x · ( .../2x2 + 3x + 5 - .../2x2 - 3x + 5), and since
x = 0 is not a solution of ( 49), we proceed to solve the system
.../2x2 + 3x + 5 + .../2x2 - 3x + 5 = 3x, .../2x2 + 3x + 5 - .../2x2 - 3x + 5 = 2. By adding the two equations we get 2 J2x2 + 3x + 5 = 3x + 2, and after squaring and simplification, x 2 - 1 6 = 0; thus x 1 = -4, x 2 = 4. Since the left-hand side of ( 49) is nonnegative, the right-hand side must also be nonnegative, and the number x 1 = -4 cannot be a root of ( 49). By substituting into (49) we verify that x = 4 is the ( unique) root of (49). ·
6. 10 Exercises Use the method of 6.9 to solve the following equations:
(i) (ii) (iii)
v3x2 + 5x + 8 - J3x 2 + 5x + 1 = 1. .Jx3 + x2 - 1 + .../x3 + x2 + 2 = 3. .../2x2 + 5x - 2 - J2x 2 + 5x - 9 = 1 .
66
1 . Algebraic Identities and Equations
6. 1 1 Method of Substitution A very efficient method for solving some irrational equations substitution. We illustrate this with several examples.
( i) In the equation
�+3
V 2+X
·
J
2+x 3-x
is
the use of
=4
we set J(3 - x)/(2 + x) t > 0; then the equation becomes t + � 1, t 2 3. If t 1 which gives a quadratic equation with roots h J(3 - x 1 )/(2 + x 1 ) 1, then x 1 !, and if t2 = J(3 - x 2 )/(2 + x2 ) 3, then x 2 - � . By checking we verify that both numbers are indeed roots of the original equation.
=
4,
=
=
=
=
=
=
(ii} x2 + Jx2 + 2x + 8 12 - 2x. We begin by writing this equation in the form x 2 +2x+8+ v'x 2 + 2x + 8 20. If we then set .../x2 + 2x + 8 z > 0, we obtain the quadratic equation z 2 + z-20 0 with roots z 1 z2 -5. In view of the condition z > 0, only the root z 1 makes sense, and the equation Jx2 + 2x + 8 which gives the numbers x 1 2 and x 2
=
= 4, = -4,
=
=
=
4 are also roots of the given equation.
=
=
=
(iii) To solve the equation
Vx - Jx - 2 + Vx + .../x - 2 = 3, in addition to the substitution Jx - 2 t 2:: 0 we use also the method of Section 6.9 . We have .../t 2 - t + 2 + v't 2 + t + 2 3, and by multiplying by the conjugate of the left-hand side of this equation we obtain the following system: =
=
v't 2 - t + 2 + v't2 + t + 2 3, 3 . ( v't2 - t + 2 - v't2 + t + 2) -2t. =
=
This implies 6 · v't2 - t + 2 9 - 2t, and thus 32t2 Since x t2 + 2, the root of the original equation is x we can convince ourselves by checking that L(73/32) 2
=
=
0, t2 9/32. 73/32, of which 9.
9
=
=
=
=
(iv) If the equation
is given, it is convenient to use the substitution y vfx + Jx + 2. We then have y2 x + 2 · Jx(x + 2) +x + 2 2(x + 1 + vx 2 + 2x), and the original equation becomes y2 /2 - 1 + y 3 , which has the number 2 as its unique nonnegative root. As solution of the equation Vx + .../x + 2 2 we find x i, which can be easily checked to be the root of the original equation.
=
=
= =
=
=
67
6 Irrational Equations
6. 12 Exercises Use an appropriate substitution to solve the following equations.
� - 2 . Iii£ = 1 . (ii) � - � = �. (iii) x2 - 4x + 6 = v'� 2x--:2::--_ : �8x-+--:1-::-2. (iv) 2x2 + 6 - 2 V2x2 - 3X + 2 = 3( X + 4). (v) Jx3 + 8 + {/x3 + 8 = 6. (vi) 3x2 + 15x + 2 Jx2 + 5x + 1 = 2. (i)
·
·
X · � - 1 - � - 1 - 4. �+ 1 �- 1 'x====l 5. 6.v x-+--=8=------=(viii) v'x + 3 + 4 . v'x - 1 + v' (ix) v'x + 2vx - 1 - v'x - 2vx - 1 = 2. 2x--:2::-: +� 2x-+ �9. (x) v'x2 + x + 4 + Jx2 + x + 1 = v'� (xi) v'x2 + x + 7 + vlx2 + x + 2 = vf3x2 + 3x + 19. (xii) v'x + v'x + ll + v'x - Jx + 11 = 4. (xiii) v'x - 1 + Jx + 3 + 2 · J(x - l)(x + 3) = 4 - 2x. ��
=
6. 13 Examples We solve now two equations that are more complicated; again it will be convenient to use substitutions.
(50)
(i) SOLUTION. If we set y of equations
= �2x - 1, then y3 = 2x-1. We obtain the system x3 + 1 = 2y, y3 + 1 = 2x.
=
By subtracting the second equation from the first, we get x3 -y3 2(y-x), thus (x -y)(x 2 +xy+y2 + 2) 0, and so either x y or x 2 +xy+y2 -2. If x y, then x3 + 1 2x, which has roots x 1 1 , x2 = -(1 + .../5) / 2, x3 -(1 - J5)/2; all three numbers also satisfy equation (50), which is easy to verify. The equation x2 + xy + y2 -2 has no real roots, since for each pair x, y of real numbers we have x2 +xy+y2 (x+y/2) 2 +3y2 /4 > 0. Equation (50) therefore has the three roots 1 , -(1 + J5)/2, -(1 - -/5)/2. D
=
(ii)
=
=
=
=
=
=
=
=
4(vfl + X - l)(vfl - X + 1) = X.
(51)
1. Algebraic Identities and Equations
68
v'1 + x, v = v'1 - x. Therefore, = 1 + x, v2 = 1 - x, and thus u2 + v 2 2 and u2 - v2 = 2x, where 0 < u, v < v'2. By substituting into (51) we obtain after simplification 8 · [uv + ( u - v) - 1 ] = u2 - v2 . In this equation we substitute the expression uv: Since (u - v) 2 = u2 - 2uv + v 2 , and using u2 + v2 = 2, we obtain 2uv = 2 - (u - v) 2 . Hence 8 - 4(u - v) 2 + 8(u - v) - 8 - (u - v)(u + v) = 0, (u - v)(8 - 5u + 3v) = 0. The condition u - v = 0 implies x = 0, which is also a solution of (51). The equation 5u - 3v = 8 has no solution, since from the conditions 0 < u, v < 0 v'2 it follows that 5u - 3v < 5u < 5v'2 < 8. SoLUTION. First we note that x E [-1, 1] , and we set u
u2
=
=
6. 14 Exercises Solve the following equations:
(i)
{1f;_- �
=
1.
y =:;:s = 1-12. y+-=2-+-=3=-vJ;r2= (n) Jy - 2 + J2y - 5 + v,.-
2 (iii) Jx + 8x + v'x + 7 = 7 Jx + 1 Jx + 1 *(iv) -x1 + 1 = 35 . v'1 - x2 12 -
6. 15 Equations of the Form y'a + cp (x) + Jb - cp (x)
= c
The method of ''paired" substitution, which we have used to solve example 6.13. ii , can be used in a standard way to solve equations of the type n a + cp(x)+ \jb - cp(x) = c, where n E N, a, b, c E lR, cp(x) is an expression in the variable x E lR, or equations of a similar type . If we set u = \1a + cp( x), v = \jb - cp( x), then simultaneously
u + v = c, un + v"' = a + b. Systems like this can be solved, for �ance, by using the method of symmetric polynomials from Section 4. Let us again consider two examples.
(i) The equation
�32 - X + �1 + X = 3, with the substitutions u = �32 - x, v = -¢/1 + x, leads to the system u + v = 3, u5 + v5 = 33,
6 Irrational Equations
69
which we have solved in Exercise 5.12.(x) ; hence, u 1 = 1, v1 = 2, or u 2 = 2, V2 = 1 . From the equation �32 - x = 1 we obtain the first root x 1 = 31 of the original equation, and from �1 + x = 1 we get the second root x 2 = 0, of which we can easily convince ourselves.
(ii) To solve the equations �8 + X + �8 - X = 1, we
use
the substitutions
u = {/8 + x, v = {/8 - x and solve the system u + v = 1, u3 + v3 = 1 6
If we set o-1
=
u1 (u, v) = u + v, o-2 = u2 (u, v) = u · v, then 1 = Ut ,
16 and thus o-1
.
1 and o-2
=
0"� - 30"1 0"2 ,
-5 . Therefore, u, v aie roots of the polynomial F(t ) = t 2 -u1 t+u2 = t2 - t-5, that is, {u , v} = { ( 1 + v'2I)/2, ( 1 - v'2I) /2} . Since X = u3 - 8, we obtain X 1 = 3 . v'2I, X 2 = -3 . m. We remark that the solution of these problems could be accelerated: Indeed, if u2 ( u, v) = u v = -5, then -5 = {/8 + x · .V8 - x . Therefore, {/64 - x2 = -5, which gives again x1 , 2 = ±3v'2f. By substituting we verify that both numbers are roots of the original equation; to do this, we can use the result of Example 3.24. =
=
·
6. 1 6 Exercises Use the method of 6.15 to solve the following equations:
(i) {/1 - X + {/15 + X = 2. (ii) {/2x - 2 + {/6 - 2x = vf2. (iii) �1 + 2x + �1 - 2x = �. (iv) \1'1 + vfx = 2 - \1'1 - JX. (v) 2 · {/(6 - x)(x - 2) + v'6 - x + v'x - 2 = 2.
6. 1 7 Equations of the Type .ifl(X} + \I9[X) = h(x) By using the formula
(u + v) 3 = u3 + v3 + 3uv(u + v),
70
1 . Algebraic Identities and Equations
we can modify the method introduced in Section 6.15 for n = 3 to in clude the solution of more complicated equations . Indeed, if in the equation ifJ(X) + \I9(X) = h(x) we set u = VJ{x), v = {19(X), then according to the formula above we have
ifJ(X) + \I9(X) = h(x)
====>
f ( x) + g(x) + 3h(x) · Vf(x) g(x) = h3 (x). (52) ·
We illustrate this approach with the following examples .
(i) The equation
�8 + X + �8 - X = 1 was already solved in 6.15. ii) . Now, using (52), we rewrite it in the form 8 + x + 8 - x + 3 . 1 . 3 ( 8 + x)(8 - x) = 1, which implies �64 - x2 = -5; hence again x 1 ,2 = ± 3 J21 . ·
(ii) Let us now solve the equation
�X - 1 + �X + 1 = X · -e-'2. Rewriting it according to (52), we get x - 1 + x + 1 + 3x {/2( x 2 - 1) 2x3 , and thus x · [3{/2(x2 - 1) - 2 (x2 - 1)] = 0, so x 1 0 or 3 {/2(x2 - 1) = 2 (x2 - 1); this last equation has roots ·
·
=
=
·
X2 = 1, X3 = -1, X 4,5 = ± V( 2 + 3vl3)/2.
The roots of the original equation are therefore some of the numbers Xi, 1 < i < 5. While it is easy to check each of the first three numbers (by simple substitutions we verify that each one is indeed a solution of the original equation) , we carry out this check for the remaining numbers X4 , Xs with the help of the following trick: We set a = �x4 - 1 , b = �x 4 + 1, c = X4 � and show that a + b = c. Indeed, an easy calculation shows that the numbers a, b, c satisfy the relation a3 + b3 + 3abc = c3 ; the desired equation a + b = c then follows from the result of Example 3. 16. (iii) . This works similarly for x5 . •
6. 18 Exercises Solve the following equations:
(i) �12 - x + ?'14 + x = 2 . (ii) ?'x + 7 + �28 - x = 5 . (iii) �2x - 1 + ?'x - 1 = 1. (iv) ?'x + 1 + ?'x + 2 + �r--x-+-=3 = o . (v) � + ?'2x - 3 = {/12(x - 1). * (vi) � + �x - 16 = ?'x - 8.
6 Irrational Equations
71
6. 19 Irrational Equations with a Parameter We now turn our attention to solving several types of irrational equa tions with parameters. We will use all the methods described above, and pay special attention to those problems that often cause considerable complications. (i) Jx2 + a = a - x, where x is a variable and a is a real parameter. We first square the equation: x2 + a = a 2 - 2ax + x 2 , and thus 2ax = a( a - 1 ) . For a = 0 we get the equation 0 = 0, whose roots are all numbers x E lR; for a 1= 0 it is x = (a - 1)/2. We now do the necessary checking. ( a) a = 0: By substituting into the original equation we get ..JX2 = -x, which is equivalent to lxl = -x, and holds if and only if x < 0. For a = 0 we thus have an infinite set K. of roots, with K. = RO . (b) For a 1= 0 we have
( )
( )
2 L a -2 1 = (a -4 1 ) + a = I a +2 1 I '. R a -2 1 = a +2 1 ' so that L = R if and only if Ia + 1 1 = a + 1, that is, a > - 1 . Therefore,
J
the given equation has the unique root x = (a - 1)/2 for each a E [-1 , 0) U (O, oo) , while it has no root for a E (-oo, - 1) . (ii) The equation
a · Jx + 1 + b a - b · Jx + 1
a+b a-b
with real parameters a, b makes sense only when a 1= b. We substitute y = Jx + 1 and rewrite it as (ay + b) (a - b) = (a + b)(a - by); then after some manipulations we obtain ( a2 + b2 ) y = a2 + b2 • Thus y = 1 and x = 0 for each pair of parameters a, b, where a 1= b. The checking in this case is easy. •
6.20 Example Solve the following equation with parameter a E lR:
x2 - Ja - x = a.
(53) ( i) FIRST SOLUTION. By separating the square root we obtain Ja - x = x2 - a, where a - x > 0 and at the same time x2 - a > 0, hence x < a < x2 • After these observations, by squaring we obtain
(54) which is an equation of degree four in the variable x, where none of its zeros are easy to �'guess. We will therefore use the following trick: Let us for a while consider equation (54) as an equation with variable a and parameter '1
72
1. Algebraic Identities and Equations
x; for then we have a quadratic equation a 2 - (2x2 + 1)a + (x4 + x) = 0 with discriminant D = ( 2x - 1) 2 . Therefore, we have either 2 + 1 + ( 2x - 1) 2x = x2 + x, a= 2
or
2 + 1 - (2x - 1) 2x = x2 _ x + 1. a= 2
Now we return to the original variable and parameter, and distinguish between two cases: ( a) a = x2 +x. Since a < x2 , we have x < 0. Then the quadratic equation x2 + x - a = 0 has discriminant D = 4a + 1 (which means a 2:: --}) and roots x 1 , 2 = ( -1 ± v'4a + 1) /2. The root x1 = (- 1 + v'4a + 1) /2 makes sense only for a E [-� , 0] , since for a > 0 we have X1 > 0. The root x2 ( -1 - v'4a + 1) /2 is negative for all a > --! . ( b ) a = x 2 -x+l. Since a < x2 again, we have ( -x+1) < 0, that is, x > 1, from which we furthermore obtain the condition a > 1. We will now solve the equation x2 -x+(1-a) = 0 only for parameter values a > 1, and we find X3,4 (1 ± ../4a - 3) /2. Since x > 1, the number X4 = (1 - v'4a - 3) /2 is not a solution. On the other hand, X3 = ( 1 + .J4a - 3) /2 works, under the conditions a > 1 and x > 1. In summary, we have that for a < --} the equation (53) has no solutions; for a = - � it has the unique root - �; for a E (--!, 0] it has the two roots x 1 , x2 ; for a E ( 0, 1) it has the unique root x2 ; and finally, for a > 1 it has D the two roots x2 , x3 . =
=
Note: The roots of equation (53) are "unpleasant," and checking them
by substituting would be difficult. Therefore, in the course of solving the problem above we made sure that all manipulations carried out were equivalences; we carefully followed the domains of definition of expres sions, squaring only nonnegative expressions. We will follow this cautious approach also in the second solution of (53) . (ii) SECOND SOLUTION. We use the substitution y = .Ja - x; then y > 0, a > x, and we solve the system
_ a + y, x2 y2 = a - x. Subtracting the equations, we obtain x 2 - y 2 = y + x; hence (x + y)(x y - 1) = 0. Now we distinguish again between two cases: ( a) x + y = 0; then x = -.Ja - x, that is, x2 + x - a = 0 under the condition that x < 0. We then continue as in 6.20. (i) , case ( a) . ( b ) x - y - 1 = 0; then x - 1 = .Ja - x, and thus x > 1. For a > i the equation x2 - x + (1 - a) = 0 has the roots x3,4 = (1 ± v'4a - 3) /2.
6 Irrational Equations
(1 -
73
Since X4 = ..j4a - 3) /2 < !, the number x4 is not a root of (53) for any value of the parameter a; the number x3 = 1 + J4a 3) /2 is a root exactly when X3 > 1 , that is, when a > 1 . A summary of these results can be found at the end of the first 0 solution.
(
-
6. 21 Exercises Solve the following equations, where parameter: (i)
(ii) (iii) (iv) *(v)
x
is a variable and
a is
a real
..,Jx2 + a2 = x + a. 1 + y'x = ..,jx - a. ..jx - a = a · JX. J(x + a) 2 + 4a + a = x. Ja + ..,ja + x = x.
6.22 Examples At the end of this section we will solve a few systems of irrational equations. We limit ourselves to systems of two equations in two variables, and we apply some of the tricks that were used earlier.
(i) vfx + 2V'Y = 9,
x - 4y = 9. SOLUTION . Since x > 0, y > 0, the second equation of the system can be rewritten in the form ( JX + 2..,JY) ( JX - 2..,JY) = 9, from which by the first equation it follows that JX - 2..,JY = 1. The system vfx + 2 JY = 9, vfx - 2 JY = 1, has the solution JX = 5, ..,JY = 2; hence x = 25, y = 4. By substituting we convince ourselves that the ordered pair (x, y) = (25, 4) satisfies both equations of the system and is therefore its unique solution .
(ii) � +
w = 35,
rx + {IY = 5.
SOLUTION . We use substitution; if we set y = v5 , we obtain
u3 + v3 = 35, u + v = 5,
0
u = rx, v = flY, i.e., X = u4 ,
1. Algebraic Identities and Equations
74
which we solve using the method of symmetric polynomials (see Section 4) If 0'1 = 0'1 ( u, v ) = u + v, 0'2 = 0'2 ( u, v ) = u · v, then we have : 0'1 = 5 and 35 = O" r - 30'10'2 , and thus o-2 = 6 . The numbers u, v are zeros of F(t) = t2 - 5t + 6, so either u = 2 and v = 3, or u = 3 and v = 2. In the first case we have x 1 = 16, y1 = 243, and in the second one, x 2 = 81 and y2 = 32. By substituting we convince ourselves that both pairs (16, 243) 0 and (81, 32) are indeed solutions of the original system . ·
[2X=T + fY+2 = 2 V Y+2 V 2i=l '
(iii)
(55)
X + y = 12.
SOLUTION. If we use the substitution
� = t > 0 for the first equation
of the system, then this equation appears in the form t + f = 2; this is a quadratic equation with a unique (double) root t = 1 . The system (55) then takes the form of the following system of two linear equations in two variables:
2x - 1 y + 2, X + y = 12, =
with the unique solution x of the system (55).
= 5, y = 7. The pair (5, 7) is then also a solution 0
6.23 Exercises Solve the following systems of equations, where in (viii) the numbers and b are real parameters:
(i) (iii) (v)
{IX + w = 4, xy = 27. y'x2 + 4xy - 3y2 = x + 1, x - y = l. 10 · � + 3x - 3y = 58, x - y = 6.
(ii)
[t + � = � ,
(iv) (vi)
* (vii)
x+yx 2 - y2 x - yx2- y2 1 7 x-y'x2 - y2 + x+y'x2- y2 - 4 '
x(x + y) + y'x2 + xy + 4 = 52.
*(viii)
X + y = 10. x + xy + y = 12, {IX + � + � = o . y - _§Q_ .V!iy + V!Vi ...fXY ' oiJY x + y = 20.
a
7 Some Applications of Complex Nwnbers
7
75
Some Applications of Complex Numbers
Now, at the end of the first chapter, we will concentrate on problems that thematically belong to the preceding sections but that can be conveniently solved by use of complex numbers, although in the statements of the prob lems only real numbers appear . We stress that the use of complex numbers for solving these problems is not imperative; however, it replaces a number of artificial tricks required in the corresponding "real" solutions, and thus makes for clearer solutions. We assume that the reader is familiar with the definition of complex numbers, knows how to work with their algebmic (a+bi) and trigonometric (r(cos � + i sin �)) forms and knows de Moivre 's theorem
(cos � + i sin �) t = cos t� + i sin t�, which holds for arbitrary numbers
� E 1R and t E Z.
7. 1 Sums of Binomial Coefficients Already in Section 1 . 3 . ( i) we saw that by appropriate application of the binomial theorem we can obtain the sums of certain binomial coefficients. If in 1 . 2 we now set A = 1 and B = i, we get
We evaluate the left-hand side by way of de Moivre's theorem:
Comparing the real and imaginary parts on both sides, we obtain
(;) (;) + (:) (�) - (;) + (;) -
-
· · · =
-
· · · =
(ht ros n; ,
(56)
( h t sin
(57)
n; .
Here, as in the following equations, the dots denote the terms in an obyious pattern, continuing as long as the binomial coefficients involved are defined (we have defined (�) only for k = 0, 1 , . . , n) . By adding and subtract ing (56), resp. (57) , to/ from 1 .4 . ( i ) , resp . 1 . 4 . ( ii ) , we obtain the following .
76
1 . Algebraic Identities and Equations
formulas:
n ;, (�) (�) (;) n + ( ) + (;) + . . . = 2n- 2 + (V2r - 2 sin ; , (�) ; n1r ' n( ) + (n) + ( n ) + = 2n-2 - ( vrn2) n-2 4 +
2
+ . . . = 2" - 2 + ( V2r- 2 k = en+l ) = 3 . 1 k= l n
7 Some Applications of Complex Numbers
81
0
This proves equality ( 64) .
For the sake of interest we will now show how this result can easily be used to derive the sum of the important infinite series
w2 . 1 + 212 + 312 + 412 + . . . + n12 + . . . = 6
To achieve this, we begin with the fact that for each a, 0 < a < � , we have O < sin a < o < tan a. (Justify this inequality geometrically by comparing the areas of three shapes: an isosceles triangle with length of the equal sides and angle a between them, a circular sector with radius and angle a, and a right angled triangle with sides and tan a making the right angle). From these inequalities we get cot2 a < 2 < + cot2 a.
1
1
1
1 1 If we add over all a = 2!� 1 (k = 1, 2, . . . , n) , we obtain with (64) , n(2n - 1) � ( 2n + 1 ) 2 n(2n - 1) < LJ < n+ Q
3
k=l
3
kw
·
2n�l ) 2 , we reach the conclusion that for 1 212 + 312 + . . . + n12 , for each n E N we have the bounds 1( - 1 ) (1 - 2 ) w2 < S(n) < (1 - 1 ) ( 1 1 ) w2 From this, after multiplying by ( the partial sums S(n) = +
2n + 1 6 which confirms that S(n) --+ w 2 /6 for n --+ oo . 2n + 1
2n + 1
+
2n + 1 6 '
(v) Prove the identity
3C4i 3� _3 / 1 · (5 - 3 v3r,:;7) , cos 7 = y cos 7 + y 7+y y3C2; 2 c�
where (as in Section 6) the third root is considered to be a real function, that is, a bijection on the set :JR.
12
The equation x6 + x 5 + · · · + x + = 0 has the complex units e:, . . . , c-6 as roots, where c = cos + i sin These six complex numbers, upon substituting y = x + � (a standard method of solution for these so-called reciprocal equations), turn into the three real numbers SOLUTION .
c-2 ,
2;
Yl = 2cos 72w , Y2 = 2cos -4w7 ,
;.
82
1 . Algebraic Identities and Equations
which are solutions of the equation y3 + y2 - 2y - 1 = 0. This last equation can be derived by first dividing the original equation of degree 6 by x3 and then using the expressions
1 = y3 - 3y. X3 + 3 X
x + -X1 = y, Hence we have to verify the equality
{IYl + {IY2 + W3 =
\/5 - 3{/7,
where Yi are real roots of an equation whose coefficients we know. Let us therefore formulate the following more general problem: Given the cubic equation y3 - ay2 +by - c = 0 with real roots Yl , Y2 , Y3 (in our case a = -1, b = -2, c = 1) , construct an equation z3 - Az2 + Bz - C = 0 with real roots .vYl, {IY2 WS"· From Vieta's relations
Y l + Y2 + Y3 = a, Y1 Y2 + Y2 Y3 + Y3 Yl = b, Y1Y2 Y3 = c , it is immediately clear that from Table 4. 12:
{IYl + {/y2 + W3 = A,
� + � + � = B, �Y1 Y2 Y3 = C, C = {/C.
Next we use the following identity
(m + p + q) 3 = m3 + p3 + q3 + 3(m + p + q)(mp + pq + qm) - 3mpq. If we substitute the triple WI, {fY2, {(Y3, we obtain A3 = a + 3AB - 3C, and by substituting � , W2fj3, �, we get B3 = b + 3BCA - 3C2 • In the case of the coefficients a = - 1 , b = -2, c = 1 we therefore conclude that C = 1 and that A and B are the solutions of the pair of equations A3 = 3AB - 4 and B3 = 3AB - 5. Multiplying these two equations we obtain, for the new variable w = AB, the equation w 3 = (3w - 4)(3w - 5), or ( w - 3) 3 = -7. Hence w = 3 - {/7, and so A3 = 3w - 4 = 5 - 3 {/7 and B3 = 3w - 5 = 4 - 3{/7. Since A = .vY1 + f(ii2 + {(Y3, the proof is now D
complete .
7. 5 Exercises (i) Express tan 6a in terms of tan a for any a :JR. {ii) For each n prove the identity 211" 47r + cos 61r . . + cos 2n7r . cos + cos + 2n + 1 2n + 1 2n + 1 2n + 1 {iii) Show that for arbitrary n a 1R we have
E
EN
E N, E
(�) cos na + (�) cos(n - 2)a + (�) cos(n - 4)a + . . . + (:) cos(n - 2n)a = 2n
·
cosn a.
1
2
7 Some Applications of Complex Numbers a:,
83
For arbitrary n E N, {3 E find the sums 81 = (�) cos {3 + (�) cos( f3+o:) + (�) cos({3+2o:) + · · + (:) cos({3+no:) , 82 = (�) sin {3 + (�) sin({3+o:) + (�) sin({3+2o:) + · + (:) sin( f3+no: ). * (v) Let r, cp E lR, sin o: =F 0 and n E N. Prove the identitites sin cp + r · sin(cp+o:) + r2 · sin(cp+n2o:l ) + · · + rn sin(cp+nn o:2) sin cp - r sin(cp - ) r + sin(cp+ (n+1) r + sin( cp+no:) ) + o: + r2 1 2r cos cos cp + r cos(cp+o:) + r2 cos(cp+2n o:1 ) + · + rn cos(cp+nno:+)2 r cos( cp+no:) ) cos cp - r · cos( ) - r + cos(cp+ (n+1) + o: 1 - 2r cos + r2 * (vi) For each n E N verify the identity tan2 -4n1r +tan2 -37r4n +tan2 -4n57r + · · · +tan2 (2n 4n- 1)1r = n(2n - 1) . (vii) Let us return to the polynomial identity xn- 1 + xn-2 + · · · + x + 1 = (x - c:)(x - c:2 ) · (x - c:n- 1 ) (65) (c: = cos(21r/n)+i sin(27r/n)) from Example 7. 4 .(iii), where we havE:! substituted x = 1 . In the case of an odd n = 2m + 1 it is worth also substituting x = -1 . Do this, and thus derive the identity (for m E N) cos 2m1r+ 1 cos 2m + 1 cos 2m + 1 . . . cos 2m + 1 = 21m . (66) Let us note that the corresponding formula for even n = 2m, 21r- cos -37r- . . . cos (m - 1)7r = -Vm cos -2m1r- cos -2m 2m 2m""--�- 1 2m > 2), follows directly from (63) with the help of the (m E N, m identities k1r = sm. (m - k)1r = sln. (m + k)1r cos 2m 2m 2m used for k = 1, 2, . . . , m - 1. * (viii) Implement another method for deriving formulas (63) and (66) , namely one based on the multiple-angle formulas for sine obtained in(a)7.4.(i): First determine the polynomial A ( x) of degree m with zeros m 2 Xk = sin 2,!�1 (1 < k < m), Bm ( Y) of degree m with zeros (b) the polynomial 2 Yk = sin 2,!�2 (1 < k < m). the help of Vieta's relations then obtain again formulas (63) With and (66) . *(iv)
R.
·
··
a: ,
·
_
a: -
·
_
·
cp - a:
·
·
a:
··
a:
··
�
k
�
1 . Algebraic Identities and Equations
84
Using approach similar to 7.4.(v), prove the identity �/ 1 · ( 3v3/n9 - 6) 3L&i 3� 3C'ii + cos cos + = cos · y -g y -g y -g y 2 (x) Prove the following equalities: cos � = 1 cos 2;- -1�
* (ix)
an
+4v'S ,
=
v'5
.
7. 6 Polynomials I
s polynomial involving problems in used be also often can numbers Complex the of way by divisibility for testing in stance n i for coefficients, real with criterion introduced in 3. 15. (ii) . We will illustrate this with the following examples. (i) Suppose that the integer n > 1 is divisible neither by 2 nor by 3. Show that the polynomial '
is divisible by G(x) = x3 + 2x2 + 2x + 1 . G(x) . We easily polynomial the of roots the find first us Let G( -1) = 0, and thus G( x) is divisible by x + 1. By that ourselves convince dividing we obtain G(x) = (x + 1) (x2 + x + 1), x2 + x + 1 are ( -1 ± ivfa)/2. By 3. 15. (ii) we are done of zeros the where weBy can F(x) has -1, (-1 ivfa)/2, and ( - 1 - ivfa)/2 as zeros. that show the binomial theorem, we have for each x E lR, x 1= 0, 1 [(x + 1)n - xn - 1] , F (x) = F( -1) - [on - ( -1)n - 1] = 0, since get we substituting by which from . The values of the polynomial F(x) at the numbers ( - 1 + iVJ)/2 nandis odd ( -1 - ivfa) /2 can be computed together: SoLUTION.
if
+
X
•
=
F
( - l � iV3) = - l � ivl3 [ c ± i� r - (-� ± i �r 1
-1 .
By de Moivre's theorem we have c ± i�r - ( � ± i�r = (cos (± ;) + isin (± ;))" - (cos (± 2; ) + isin (± 2; ) r . . n1r - cos 3 n1r ± �s1n 2n7r . 2n1r . =F�sm = cos 3 3. 3 -
85
7 Some Applications of Complex Numbers
Ink Eview of the fact that n is divisible neither by 2 nor by 3, there exists a N such that n 6k ± 1, and thus sin n31r = sin (2k7r ;) = sin (± ; ) 2n7r , = sm. (± 321r ) = sm. ·(4k7r ± 321r ) sin 3 n1r cos (2k1r ± 1r ) cos (± 1r ) 1 , cos 3 2 3 3 2n7r (4k7r± 321r ) (± 21r ) = - 1 . s 3 2 By substituting these values we get ( - 1 ±2 iv'3) 2 cos 3 ± i sin 3 -1- cos± iv'32mr3 ± i sin 2n31r - 1 0 This completes the proof of the statement. (ii) Show that for arbitrary numbers n E N and a E 1R satisfying the conditions n > 1 and sin a 1= 0, the polynomial sin a - x sin na + sin(n - 1)a P(x) is divisible by the polynomial Q(x) x2 - 2xcosa + 1 . SOLUTION. It is easy to convince ourselves that Q(x) has the zeros x1 ,2 cos a ± i sin a. Thus, P(x 1 ,2 ) (cosa ± isina) n sina - (cosa±isina)sinna+ sin(n - 1)a. Ifuponwe simplification the expression (cos a ± isina)n cosna ± isinna, we obtain P(x 1 ) P(x2 ) cosnasina - cosasinna + sin(n - 1)a. expression on the right vanishes thanks to the the formula sin( n 1 )a cosasinna - cosnasina, which we can derive by applying de Moivre's theorem to both sides of the equality (cos a + i sin a) n-1 (cos a + i sin a) n (cos a + i sin a) - 1 have thus come the con and then equating, the imaginary parts. We clusion that P(x1 2 ) 0. By 3.15.(ii) this means that Q(x) divides0 =
±
=
=
COS
=
= COS
F
= COS
1!2!:
=
=
=
.!m.
=
o.
:l"
=
=
=
use
=
=
=
=
=
=
P(x).
·
to
1 . Algebraic Identities and Equations
86
7. 7 Exercises
Show that for arbitrary E No the polynomial (x + 1) 2n+ 1 + xn+2 is divisible by x2 + x + 1 . (ii) Show that the polynomial ( x sin + cos ) n - x sin - cos divisible by x2 + 1 for any E 1R and any E N,4a > 41. 1 4c+2 + (iii) Show that for any a, b, d E No the polynomial x + x b+ + x x4d+3 is divisible by x3 + x2 + x + 1. * (iv) Find all real numbers a E ( -2, 2) satisfying the following condition: polynomial The x1 54 - ax71 + 1 is a multiple of the polynomial x1 4 - ax7 + 1 . To concerning problem more one with deal we section this conclude polynomial with real coefficients, where again we apply complex numbers. (i)
n
na
a
a
n
a
na
n
is
c,
a
7. 8 Example X1 , x2 , . . . , Xn Xn + P1 Xn- 1 + P2 Xn -2 + · · · + Pn- 1 X + Pn P1 , P2 , . . . , Pn . (x� + · · · (x; + P2 + P4 - · · · ) 2 + (p1 - P3 + Ps - · · · ) 2
Let
be the zeros of the polynomial
Show that with real coefficients 1) 1) (1 SOLUTION . By 3. 14 we have =
Xn + P1 Xn- 1 + · · · + Pn- l X + Pn (x - Xl )(x - X2 ) · · · (x - Xn ) · x i, x -i (i - Xk )( -i Xk ) x� + (x� + ) · · · (x; + ) (in + P1 i n- 1 + · · · + Pn - 1 i + Pn ) n 1 n X (( -i) + P1 ( - i ) - + · · · + Pn - 1 ( -i) + Pn ) · a = - P2 + P4 - · · · , b = P1 - P3 + Ps - · · · , (x� + · · · (x; + ) = (ina + in- 1 b)(( -i) na + ( -i) n- 1 b) a2 + b2 , =
expressions two the multiply and this into We substitute thus obtained. Since 1, we have 1 1 then If we set 1 1) 1 also holds identity the that follows it (Furthermore, shown. be to was which in the case where the coefficients P1 , . . . , Pn are complex numbers.) =
=
-
=
=
=
D
7. 9 Exercises (i)
F(x) xn + an_ 1 xn- l + . . + a 1 x+ao polynomial the that Suppose has zeros Xt, x2 , . . . , Xn . Show that (x� + X1 + 1 ) · (x� + X2 + 1 ) (x; + Xn 1 ) 1 = 2 · [(A - B ) 2 + ( A - C) 2 + ( B - C) 2 ] , ·
=
· · ·
+
7 Some Applications of Complex Numbers as +
87
with A + + B + + C + · where and where the sums are taken over indices not exceeding we understand that 1 . *(ii) Let be the roots of the polynomial = ao
a3
· · · ,
= a1
a4
· · · ,
= a2
n,
an =
Xt , x2 , X3 , X4
Show that
· · ,
2 Algebraic Inequalities
Inequalities are essential tools in many areas of mathematics. Orderings of numbers, which are obtained by way of inequalities, find applications even outside of mathematics in various theoretical and practical fields. Secondary-school curricula have therefore traditionally devoted a certain amount of attention to linear and quadratic inequalities, and to inequalities involving absolute values. Already less emphasis is given to the •'calculus'> of operations with inequalities and to methods and approaches for deriv ing estimates for algebraic expressions, and for finding proofs. The present chapter is devoted to exactly these questions. First we de The necessary basic results are summarized in Section scribe a way of introducing inequalities between real numbers, and then we derive some basic rules of arithmetic with inequalities. FUrther theorems and important results, connected with various methods, will be considered throughout the appropriate sections. The proofs in two particular sections and require a more exact concept of irrational numbers. Although in the text itself we restrict ourselves to the case of rational numbers, the reader can find in Section 9 the necessary supplementary remarks on the general case. In the solutions to some examples, the letters L and R denote the left, resp. the right, sides of the inequalities under consideration.
1.
(1.10 8.6)
90
1
2. Algebraic Inequalities
Definitions and Properties
Already in early childhood, when our first ideas of the numbers 1, 2, 3, · · . are formed, we know about "bigger'' and "smaller'' ; this obviously occurs before we learn how to add or subtract these first numbers. Later we become aware of the fact that the difference a-b makes sense only when �'a is greater than b." We overcome this limitation by enlarging the collection of the original (positive) numbers 1, 2, 3, . . . by new numbers, namely zero and the negative numbers - 1 , -2, -3, . . (as numbers �'opposite" to the positive numbers) . Now we generalize this process: We construct the required theory of inequalities on the basis of dividing the integers into positive and negative numbers. In the following paragraphs we express the known properties of positive and negative numbers by way of the "rules of sign." .
1 . 1 Positive and Negative Numbers
1= 0
Definition. Every real number x is either positive or negative. The set 1R of all real numbers is therefore divided into three parts: the set JR+ of all positive numbers, the set :JR- of all negative numbers, and the singleton How is this subdivision preserved under arithmetic operations? We give only those rules that will be required later:
{0} .
(i) (ii) (iii) (iv) (v)
a E JR+ A b E JR+ ==> (a + b) E JR+ , a E JR+ A b E :JR+ ==> ab E :JR+ , a E lll- � (-a) E :JR+ , a E JR.+ A b E :IR- ==> ab E :IR- , a E :JR+ � -!- E :JR+ .
A real number x ¢ 1R- is called nonnegative; the set of all nonnegative numbers is denoted by :JRt , so that JRt = :JR+ U We will often require a generalization of (i) to the case of a sum of several numbers from :JRt . We therefore list this as an additional rule:
{0} .
(vi) If a1 , a2 , . . . , an are nonnegative numbers, then s = a1 + a2 + · · · + an is also nonnegative; furthermore, s = if and only if a 1 = a2 = · · · = an = o.
0
It is easy to derive (vi) from (i): The value of the sum we leave out all summands ai that are equal to zero.
s
does not change if
1. 2 Introducing Numerical Inequalities Definition. We say that the number a is greater (resp. less) than the number b (written a > b, resp. a < b) if and only if a - b is positive ( resp.
1 Definitions and Properties
91
negative) :
a>b a -F=>
(a - b) E :JR+ , ( a b) E lR- .
(1)
-
Since the numbers (a - b) and (b - a) are mutually "opposite," it follows from l . l . (iii) that a > b is the same as b < a. From the decomposition 1R = JR+ U U lR- we immediately obtain the trichotomy law: For an arbitrary pair of numbers a, b E lR, one and only one of the relations a > b, a = b, a < b is true. In the case that a > b (resp. a < b) does not hold, we write a < b ( resp. a > b). Inequalities with < and > ( resp. < and > ) are called strict (resp. weak). Thus, the weak inequality a > b means that
{0}
(a - b) E JRt.
H we represent the real numbers in the usual way by points on the num
ber line oriented from left to right, then the inequality a > b acquires geometrical meaning: the image of the number a lies to the right of the image of the number b. Thus, for example, we have -1 > -2, even though one sometimes inaccurately says "-2 is a larger negative number than -1." It is better to avoid this last expression. Finally, we note that according to (1) with b = a number is positive ( resp. negative) if and only if it is greater ( resp. less) than zero. Thus the notations x E JR+ , y E JR.t , and z E R- for real x , y, z also stand for :i; > 0, y > and z < Now we use definition (1) and the rules l . l . (i) - (vi) to derive all the basic properties of inequalities between numbers.
0
0,
0.
1. 3 Transitivity Theorem If a > b and b > c, then a > c. More generally: If a 1 > a2 , a2 > a3 , . . . ' an - 1 > an , then a 1 > an , where a1 = an if and only if a1 = a2 = . . . = an . PROOF. If a > b and b > c, that is, (a - b) E :JR+ and (b - c) E :JR+ , then by l . l . (i) we have (a - b) + (b - c) E :JR+ , that is, a > c. The more general statement follows from l . l . (vi) and the equality a 1 - an = (a 1 - a2 ) + (a 2 0 aa) + · · · + (an - 1 - an ) · We remark that we can write the pair of inequalities a > b, b > c more concisely as a > b > c; similarly, we often write the system of inequalities a1 > a2 , a2 > aa , . . . , an- 1 > an as the chain a1 > a2 > · · · > an . However, the notation x < y > z is not customary. ..
1. 4 Adding a Number Theorem. If a > b, then a + c > b + c for each c.
92
2. Algebraic Inequalities
PROOF.
b+�
If a
> b, then (a + c) - (b + c) = (a - b)
E :JR+ , that
is, a + c > 0
We note that upon replacing the number c by -c we obtain the rule of �'subtracting a number" : If a > b, then a - c > b - c for any c.
1. 5 Adding Inequalities Theorem. If a > b and c > d, then a + c > b + d. More generally: If a 1 > b�, a2 > � ' . . , an > bn , then .
(2)
where we have equality in (2} if and only if a 1 = b�, a2 = b2 , . . . , an = bn .
a > b and :> d, then by 1 .4 we have a + c > b + c and b + c > b + d, which by 1 . 3 gives a + c > b + d. The more general assertion PROOF.
If
c
follows from l . l . (vi) and the equation
(a 1 + a2 + . . · + an ) - (b1 + b2 + . . · + bn ) = (a 1 - b1 ) + (a2 - b2 ) + · · · + (an - bn ) ·
0
1. 6 Multiplying by a Number Theorem. If a > b, then ac > be for any c > 0, and ac < be for any c < 0. If a > b, then (a - b) E :JR+ . By l . l . ( ii) we have (a - b)c E JR+ whenever c E :JR+ , and by l.l. ( iv ) we have (a - b)c E lR- whenever c E lR- . Since (a - b) c = ac - be, in the first case we get ac > be, while in the second 0 case it is ac < be. PROOF .
We note that upon replacing the number c by 1 / c we obtain, in view of l . l . ( v ) , the rule of "dividing by a number" : If a > b, then afc > b/c for c > 0, and afc < b/c for c < 0.
1. 7 Subtracting Inequalities Theorem. If a > b and c > d, then a d > b - c. More generally: If a > b and c > d, then a - d > b - c, where equality a - d = b - c occurs if and only if a = b and c = d. -
By 1 .6 the inequality c > d holds if and only if -c < -d. Hence 0 1 .7 follows from 1.5 for the pair of inequalities a > b and -d > -c.
PROOF.
1 Definitions and Properties
93
1. 8 Multiplying Inequalities Theorem. If a > b > 0 and c > d > 0, then ac > bd. More generally: If a1 > b1 > 0, a2 > b2 > 0, o o o , an > bn > 0, then (3) with equality if and only if a 1 = b1 , a2 = b2 ,
.
o
o
, an
=
bn .
From a 1 > b 1 > 0 and a2 > b2 > 0 it follows by 1.6 that a 1 a2 > b 1 a2 and b1 a2 > b1b2 . Hence 1.3 implies a 1 a2 > b 1 b2 o Since b1 b2 > 0 by 1.1o(ii), we can repeat this argument for the case n > 30 Thus we obtaln. a1 a2 a3 > b1 b2 b3, etc. , and finally (3), where the last step is the chain of inequalities PROOF
o
(4) H equality occurs in
(3),
then by
(4)(sincewea have a1 a2 · 0 0 an - 1 an a o a f. 0) and by
a1 a2 · · · an - 1 bn . This implies an = bn 1 2 · n- 1 a1 a2 · · · an- 1 = b1 b2 0 bn - 1 · Repeating this argument, an - 1 = bn- b etc . , and finally a 1 = b 1 o
(4),
•
°
•
we obtain
0
1. 9 Dividing Inequalities Theorem. If a > b > 0 and c > d > 0, then � > �0 More generally: If a > b > 0 and c > d > 0, then � > �' where equality occurs if and only if a = b and c = d. In particular, for a = b = 1 we get that if c > d > 0, then 1 > -c1 · "d Since cd > 0, by Hence 1.9 follows from 1.8. PROOF.
1.6 we have a/d > b/c if and only if ac > bd. 0
1 . 10 Exponentiating Theorem. If a > b > 0, then am > bm and y'a > v'b for each integer m > 2 . More generally: If a > b > 0, then ar > br for each r > 0, and a7- < br for each r < 0. a > b > 0. Using 1.8 with m identical inequalities a > b, we obtain am > bm . Let us assume that y'a < v'b; then by the above, ( y'a) m < ( V'b) m , that is, a < b, which is a contradiction. If r > 0 and r E Q (the case r E :JR+ \ Q will be discussed in Section 9), then r = m/n ( m , n E N), and by the above we obtain am > bm; hence '\/(iiii > \fbiii, that is, ar > br . If finally r < 0, then by the above we have a-r > b- r (since 0 -r > 0) , which by 1.9 means that 1/a- r < 1/b-r , that is, ar < br . PROOF.
Let
94
2. Algebraic Inequalities
1 . 1 1 Exponentiating Inequalities Theorem. If a > b and 0 < c < 1 < d, then ca < cb and da > db · PROOF. Let 0 < c < 1 < d and r = a - b > 0. Then by 1.10 we have cT < 1 T < dT , that is, ca -b < 1 and da- b > 1. If we multiply the last two D inequalities by cb , resp. db , then by 1.6 we obtain ca < cb and da > db . To conclude this section, the properties we have thus far obtained summarized in the following table.
are
a > b A b > c =====* a > c a > b ===* a + c > b + c l\ a - c > b - c a > b A c > d =====* a + c > b + d a > b A c > 0 ==> ac > be A -ac > -bc a > b A c < 0 ==> ac < be A -ac < -b c a > b A c > d =====* a - d > b - c a > b > 0 1\ c > d > 0 =====* ac > bd A da > �b a > b > O ===* -a1 < -b1 a > b > 0 1\ r > 0 ==> aT > bT a > b > 0 1\ r < 0 ==> aT < bT a > b 1\ 0 < c < 1 ==> ca < cb a > b A d > 1 ==> da > db 2
Basic Methods
We will now show how the basic rules obtained in Section solve some easy problems involving inequalities.
1 can be used to
2. 1 Equivalent Transformations We begin by considering five problems that will be solved as follows: The inequality that is to be proved will be transformed in a succession of steps until we reach an inequality that is obviously true. In doing so it is im portant to make sure that each step is equivalent, that is, the validity or failure of the inequality is not changed by the operation. According to Sec tion 1, such operations include, for example, adding the same expression to both sides of the inequality, multiplying the inequality by a positive ex pression, or raising an inequality between positive terms to the rth power
2 Basic Methods
95
( r > 0) . Also, we must not forget algebraic transformations of both sides
of an inequality. As a '�emplate" for a proof we often choose the opposite direction: We transform the final, "obvious," inequality in a succession of steps until we arrive at the original inequality that was to be proved.
(i) Let us show that -ersT < W. Since both numbers are positive, by 1.10 it suffices to show that ( .W) 72 < ( W) 72 , that is, (8!) 9 < (9!) 8 . Since 9! = 9·8!, we can transform this inequality to (8 ! ) 9 < 98 · (8 ! ) 8 . Upon division by (8!)8 (using 1.6) we obtain 8! < 98 . This last inequality holds by 1.8, since it is the product of the eight inequalities 1 < 9, 2 < 9, . . . , 8 < 9. (ii) Let us decide which one of the three numbers x
= (a + b) (c + d), y = (a + c) (b + d) ,
is largest, under the assumption that difference y - x: y
z
= (a + d) (b + c)
a < b < c < d. We evaluate the
- x = (ab + cb + ad + cd) - (ac + be + ad + bd) = d( c - b) - a( c - b) = ( d - a) ( c - b) > 0,
since both numbers ( d - a) and ( c- b) are positive. Hence y > x. Similarly, we convince ourselves that z > y . Therefore, z is the largest of the numbers.
(iii) We now show that for each a > 1 we have the inequality 1 Va < v'a + 1 - v'a - 1 .
(5)
Since a + 1 > a - 1 > 0, by 1 .10 we have v'a + 1 > v'a - 1 . Both sides of are therefore positive; hence by 1.10 we can compare their squares:
(5)
After multiplying out and simplifying we get
2a - ! > 2 Va2 - 1 . a Both sides are again positive (2a > 2 > 1 > 1/a); by squaring we obtain 1 4a2 - 4 + 2 > 4( a2 - 1) , a and after subtracting the expression 4(a2 - 1) we get the correct inequality 1/a2 > 0. (iv) Let the numbers a, b, r, s E JR+ satisfy a > b and r > s . We will prove
the inequality
2. Algebraic Inequalities
96
Multiplying out both sides and subtracting (as+ r - bs+ r ), we obtain arbs as br > as br - arbs , that is, 2arbs > 2as br , and after dividing by 2as bs we finally get ar- s > br - s . But this holds by 1.10, since a > b and r - s > 0.
(v) We will show that for arbitrary a, b, r, s E :JR+ we have if a
1= b.
The difference of the left and the right sides is
L - R = ar (as - bs ) - br (a5 - bs )
=
(ar - br )(as - bs ) .
= a7 - br and What can we say about the signs of the two numbers B = as - bs ? By 1.10 we know that if a > b (resp. a < b), then both A and B are positive (resp. negative) . Therefore, in the case a 1= b we always have L - R > 0, that is, L > R. We remark that similar "discussions" of the values of parameters are often used in proofs of inequalities (see also 2.3. (iv) below) .
A
2. 2 (i) (ii) (iii)
Exercises Generalize the result of 2 1 . (i) Which of the numbers 2700 and Which number is larger: .
101997 + 1 101998 + 1
.
5300 is larger? or
101 998 + 1 10 1999 + 1
?
Prove the inequalities (iv) - (viii) : b - a2 (iv) Vb > a +
(v) (vi) (vii) (viii)
(O < a < Vb < a + 1). 2a + 1 1 2(v'a + 1 - Va) > (a > 0). Va c+a > c+b ( 0 < b < a, v'ab < c) . v'c2 + a2 v'c2 + b2 (a, b E :JR+ ). o/OJ + 1W > ifci{lb + {t'a� (n! ) 2 < k!(2n - k)! (n, k E N, n > k ). ::: ::= ::::;: : ---;:::::
2.3 Irreversible Transformations We will now deal with problems in which the course of solution from an original "obvious" inequality to the desired inequality cannot be reversed. Thus, for example, the inequality L > R can be proved by finding decom positions L = L 1 + L2 + · · · + Ln and R = R1 + R2 + · · · + Rn such that
2 Basic Methods
L k > Rk for each k = 1, 2,
. . . , n.
97
( This may be a tough nut to crack even
in the case n = 2). A similar approach is provided by the multiplication law 1.8. Let us now consider the following eight problems.
(i} Prove the inequality
a+b < a + b 1+a+b 1+a 1+b
(a, b E &.+ ).
SOLUTION.
The fraction on the left has the largest denominator. We can therefore write
L = 1 +a +a +b b = 1 + aa + b + 1 + ab + b < 1 +a a + 1 +b b = R.
0
(ii} Let c be the hypotenuse of a right triangle, and a, b the other two sides. Show that for every integer k > 2 we have ak + bk < ck . By the Pythagorean theorem we have t? = a2 + b2 • Multiply ing by ck - 2 , we obtain ck = a2 ck - 2 + b2 ck - 2 . Since a < c and k > 2, by 1.10 we have ak - 2 < ck- 2 ; hence ak = a2 ak- 2 < a2 ck- 2 . Similarly, we get bk < �ck- 2 , and by adding these two inequalities, SOLUTION .
0
a > b > 0 and E N. Determine which of the numbers 1 + a + a2 + · · ----� 1 + b + b2 + · · ----� · + bn · + an and B = ------�---A = ------�---1 + b + b2 + . . . + bn - 1 1 + a + a2 + · · + an- 1
(iii) Let
n
·
is larger. SOLUTION .
We rewrite both numbers according to the identity
n 2 1 +x+x + ··· +x -
"" x ----::---:::---1 + 1 + x + x2 + ----n 1 + x ··· 1 =1+ 1 + 1 + ••+ 1 " · z
1 + x + x· · + x :- = ------: :- +------: 2 n 1 ·
:z:n
From a we get
:z:n - 1
> b > 0 it follows by 1.10 that a- k < b-k (1 < k < )
n .
1 1 1 < -1 + -1 + · · · + 1, + · · · + -a + bn an b b2 a2 which leads to the conclusion that A > B. (iv) Show that
By adding
0
x6 + 2 > x4 + 2x for all x E :JR.
The given inequality is an equality for x = 1; hence the poly nomial L- R is a multiple of (x - 1) ( see Chapter 1, Section 3) . By dividing SOLUTION .
98
2. Algebraic Inequalities
find that (x6 + 2) - (x4 + 2x) = (x - 1)Q(x), where Q(x) = x 5 + x4 - 2. Therefore, it remains to show that Q(x) > 0 for x > 1 and Q(x) < 0 for x < 1. Since Q(1) = 0, one could repeat the division of the polynomial by (x - 1); however, in view of the form of Q ( x), the following discussion suggests itself: we
x > 1 ===> x5 > 1 A x4 > 1 ==> Q(x) > 1 + 1 - 2 = 0, -1 < x < 1 ==> x5 < 1 A x4 < 1 ==> Q(x) < 1 + 1 - 2 = 0, x < - 1 ==> x4 > 0 A 1 + x < 0 ==> Q(x) = x4 (1 + x) - 2 < -2. 0
This completes the proof.
We remark that the search for linear factors of a polynomial is a basic method for solving problems involving polynomial inequalities. We will use this approach also for solving the following problem.
(v) Let a, b, c E :JR+ be such that c > a + b. Show that
a3 + b3 + c3 + 3abc > 2(a + b) 2 c. SOLUTION . With the notation F(c) = L-R = c3 +3abc-2(a+b) 2 c+a3 +b3 we will show that F(x) > 0 for all x > a + b. Since F(a + b) = (a + b) 3 + 3ab(a + b) - 2(a + b) 3 + a3 + b3 0, =
F(x) is divisible by (x - a - b); by dividing we obtain F(x) = (x - a - b) (x2 + (a + b)x - a2 + ab - b2] .
the polynomial
Hence it suffices to show that the term in square brackets is positive for x > a + b. But this is easy: For such x and positive a, b we have
x2 + (a + b)x - a2 + ab - b2 > (a + b) 2 + (a + b) 2 - a2 + ab - b2 = a2 + 5ab + b2 > 0.
0
(vi) Show that for each n E N and each a E :JR+ , a 1= 1, we have the
inequality
(6) SOLUTION .
Since both sides of (6) are sums of 2n numbers (on the left we have n-times the number a2n+ l and n-times the number 1), we first check whether (6) is a sum of 2n valid inequalities. For a > 1 we have the chain a2n+ l > a2n > a2n - l > · · · > a2 > a > 1 (for positive a < 1 the chain goes in the opposite direction) ; this, however, does not imply ( 6) . Therefore, we try another possibility, namely the sum of n inequalities a2n+ l + 1 > Rk (1 < k < n) , where R1 , R2 , . . . , Rn are appropriate sums
2 Bas�c Methods
99
of two terms each from the right-hand side R of the inequality (6), with R R1 + R2 + · · · + Rn,. Obviously, it is best to combine �'symmetric" terms: the smallest with the largest, etc. , that is, to set Rk = ak + a2n+ 1 - k for k = 1, 2 , . . , n. This attempt is successful, since =
.
a2n+ I + 1 Rk _
=
=
a2n+ 1 + 1 (ak + a2n+ I - k ) (a2n- k+ 1 - 1)( ak - 1) 0. _
>
0
(vii) Show that if 0 < a < 1, then (1 + a) 1 - a . (1 - a) 1+a < 1.
b (1 - a)/(1 + a), then (1 + a) 1 -a · (1 - a) 1+a (1 + a)(1 - a)ba = (1 - a2 )ba < ba , since 0 < 1 - a2 < 1 if 0 < a < 1 . Next, since 0 < b < 1 , 1.11 implies 0 ba < 1. SOLUTION.
If we set
=
=
(viii} Show that for arbitrary numbers a, b, c E :JR+ we have
>
(aa . bb . cc ) 2 ab+c . bc+a . ca+b . SOLUTION. First we note that for any a, b E :JR+ we have a" · bb ab · ba . Indeed, this inequality can be written in the equivalent form aa -b ba -b ; now it suffices to distinguish between the cases a b, a b, a < b and to use rule 1.10. By multiplying the three valid inequalities
>
=
0
we then obtain the desired inequality.
2.4
>>
l?xercises
Prove the inequalities (i)-(iii):
(i) (1•1.)
(1•1•1.)
(iv)
>
3 1 1 + 1 (a, b, c E :JR+ ). + a+b b+c c+a a+b+c X 1 + X2 + • · · + X6 < X 1 + X2 + · · · + X 10 (x 1 < · · · < x 10 ). 10 6 (1 + a)(1 + b) < 1 + a + b ( a, b E 11]) + ) 2 2+a+b Show that if the numbers a, b, c are the side lengths of some trian gle, then the same is true for the triples ..ja, v'b, ...jC and (a + b) - 1 , (b + c) - 1 , (c + a) - 1 . �
Prove the inequalities (v )-( ix) :
(v) x8 - x5 + x2 - x + 1
>0
(x E lR) .
·
100
2. Algebraic Inequalities
2 n +, n > 2) . (vi) a 1++a2a++aa3 + · · · ++ an- 1 >- n + 1 :JR (a + ·· a n-1 (k, *(vii) + + �) · · · + 4k2I ) > J <m-:n�k)! * (viii) a3 + b3 + c:3 + 3abc > ab( a + b) + be( b + c) + ac(a + c) (a, b, c :JR+ ). (ix} (1 + na)n+ 1 > [1 + (n + 1)a] n (a :JR+ , n
E
·
(m �) (m
(m
E
(Hint: Use the identity
E N, m E N). E E N). n
1 + na 1 + 2a 1 + a + 1 )a ----:1 + (n + 1 ) a = 1 +1 (+ na · 1 ' 1 + ( n - 1 )a 1 + a and change the product of the n + 1 terms on the right to a power n
of the smallest one of these terms. ) (x) Which one of the numbers 31 11 and
171 4 is larger? 1 *(xi) Which one of the numbers 2 31 00 and 32 50 is larger? (xii) Using a similar trick as in example 2.3.(iii) , show that the equation
(xiii}
with nonnegative coefficients aj cannot have two distinct positive roots. Use the method of Example 2.3.(v) to show that if a, b, c :JR+ and c < a + b, then
E
a3 + b3 + c3 + 3abc > 2(a + b)�. * (xiv) Let a, b, c, d :JR+ . Show that of the three inequalities
E
a + b < c + d, (a + b)(c + d) < ab + cd, (a + b) cd < ab(c + d), at least one is false.
A, B, C, a, b, c :JR+ an a + b + c. Show that the numbers A+B+a+b B +C+b+ c A+C+a+c A+B+s ' B+C+s ' A+C+s are the side lengths of some triangle. (Hint: Use the fact that if 0 < x < y, then xjy < (x + p)/(y + p) for any p > 0.) * (xvi} For fixed positive numbers a 1 , a2 , . . . , an set A(r) = a� + a2 + · · · + a� ( r lR). *(xv)
Let
E
E
Show that if r < s < t < u and r + u = s + t, then A(r)A(u) A(s) A (t), with the exceptional case a1 = a2 = · · · = an .
>
101
2 Basic Methods
2. 5
The Estimation Method
In mathematics it is often required to estimate the value of some "com plicated" expression Q. This can be done, for example, by replacing it with a simpler expression (with an error as small as possible). The num bers or, more generally, the expressions L and U satisfying the inequality L < Q < U are called the lower and upper bound of the expression Q; the lower bound L1 is called sharper than the bound L if L < L1 < Q. Similarly, we define a sharper upper bound. The following seven examples will illustrate some simple approaches to obtaining bounds (especially of finite sums) .
(i)
Let us find bounds for the expression
d a b c + :JR+ ) . b, c, d E (a, + + d a + b + d a + b + c b + c+ d a + c+ It is clear that 0 < R < 4; by numerical experimentation we are unable to find concrete values a, b, c, d such that R < 1 or R > 2 . In fact, such values do not exist, for we derive the bounds R > 1 and R < 2: a b c d 1, R> + + + a + b + c + d a +b + c + d a + b + c + d a + b + c+ d a + b + c + d R< a + b a + b c + d c + d 2. R
=
---
=
=
By approprate choices of numbers sharper bounds (see [ 7], p. 165).
a, b, c, d one can show that there are no
(ii) The easiest approach to obtaining bounds for the sum a 1 + a2 + · · + an is as follows: We find the smallest and the largest among the values ab a2 , . . . , an ; these, multiplied by n, are then bounds for the given sum. ·
Thus, for the sum
S(n)
=
1 + ··+1 1 +1+·
v'2 v'3
we obtain S(n) > n · (1/fo) for S(n) will be obtained in 2
= .
Vn and S(n) < n · 1
5 ( iv) and 2.6 (iv) .
(n > 1)
Vn
.
=
below.
n.
Sharper bounds
(iii) The method of the previous example can sometimes be improved by first subdividing the terms ai of the sum a1 + a2 + · · · + an into smaller
groups (for instance pairs) ; then we find bounds for the sums of these · groupings, and finally we add these bounds. We use this method to prove the inequality
1 + 1 1 1 1 we pair the summands as follows:
(n � l +
2n
� l ) + ( n � 2 + 2n � 2 ) + ( n � 3 + 2n � 3 ) + · · · ·
To avoid having to distinguish between even and odd original inequality by two and write it in the form
n- l
n, we multiply the
)
1 3 1 1 L n + k + 2n - k < -2 - -n · k=l
(
(7)
1 < k < n - 1 we have 1 1 3n 3n - 3 < + 2n · n + k 2n - k 2n2 + k( n - k ) 2n2 + 0 Hence the sum on the left-hand side of (7) is less than _!_ 3n - 3 3n - 2 .!_ (n 1) 2n = 2n < 2n = 32 n . For
--
_
_
(iv) If we sum the inequalities 1 < Jk + 1 - Jk - 1 Jk
obtain a new upper bound for the sum S(n ) in 2.5. (ii) that is considerably sharper: S(n) < Jn + 1 + .Jii - J2. It follows from 2.6. ( iv ) below that the error of this estimate for n = 10 6 is less than 1.
(see 2.1.(iii)) ,
we
(v) We show that for arbitrary n, k E N we have 1 .!_ 1 1 ...+ 1 < (8) + + (n + 1)! (n + 2)! (n + k)! - n n! (n + k ) ! Indeed, if 1 < j < k, then 1 n < n+j - 1 = 1 . (n + j) ! - (n + j)! (n + j - 1)! (n + j)! Adding these estimates and dividing by n, we obtain (8) . (vi) A number of estimates for finite sums can be derived from results of Chapter 1 , Section 2. Thus, for example, we obtain the bound 1 +1 +1 + +1 < -7 - -1 T(n) = · · (n > 2) · 1 2 2 2 32 n2 4 n as follows: We keep the first two terms in T( n) without change and estimate the remaining summands using the inequality 1/ k 2 < 1/ ( k 2 -k) , 3 < k < n; finally, we apply the result of Exercise 2.40. ( i ) in Chapter 1 for k = 2.
[..!..
].
2 Basic Methods
103
(vii) Finally we mention an example of an estimate for a finite product. For the expression Q(n)
=
1 3 5 · ·6 2 4
2n - 1
(n > 2)
· · · -2n-
we will prove the bounds 1 -2y'n n Since ! < we have Q (n)
1
(9)
< Q ( n) < -v�2n= :::;: +1
i and more generally (2k - 1) /2k < 2k/(2k + 1) for each k > 1 , =
1 3 5 · ·6 2 4
···
2n - 1 2n
2 4
2n
< 3 · 5 · · · 2n + 1
1 (2n + 1)Q (n) "
-
This implies (2n+ 1) [Q( n)] 2 < 1 , which gives the right-hand side of (9). If we use the inequalities ! > i and more generally (2k- 1)/2k > (2k-2)/ (2k-1) for each k > 2, we get 2Q(n)
=
3 5 ·6 4
···
2n - 1 2 4 > · 2n 3 5
2n - 2
· · · 2n - 1
=
1 2nQ(n) ;
hence 4n[Q (n)]2 > 1 , which is the left-hand part of (9) .
2. 6 Exercises Prove the inequalities (i)-(iii) for integers n > 2:
(i) (n - 1 ) v'2 < v'2 + J3 + (ii) (n!)2 > nn.
2 n n 1 (iii) > E E k k=2 k k=n+l 1
and
· · + y'1i < (n - 1)y'n. ·
n2 - l 1 n - 1 1 . <Ek E k k=n k=l
(iv) Using the result of 2.2. (v) , find a lower bound for the sum S(n) in Example 2.5.(ii) . Then determine the integer N for which the estimate N < 8(106) < N + 1 holds. (v) With the help of the estimate in 2.5. (vi) prove, for each integer n > 2, the inequality 1 1 1 3 1 1 · · + < + + · + n(2n + 1) 8 2n . 2·5 3·7 4 9 ·
(vi) Using the result of Exercise 2.40. (v) in Chapter 1 , prove the following estimate for each integer n > 2: 1 1 1 1 1 + + · · · + < --23 33 n3 4 2(n2 + n) ·
104
2. Algebraic Inequalities
in Chapter 1, prove the 2.40.(vi) n > 2: 1 + 1 1 + · · · + 1 < 1 - 4n2- 1 1 n(2n - 1) 2 2(4n - )" 2.3 3·5 4·7 (viii) Show that for each integer n > 1 we have the bounds + 1) < 2n(n+1). 5·7 + · · · + (2n - 1)(2n n(2n+1) < -1-· 3 + -23 -5 + g n (ix) For each integer n > 1, prove the inequality 3 · 7 · 11 · · · 4n - 1 < n4n + 1 V � 5 9 13 (x) Let p E N. Show that for the expression Q(n) from 2 . 5 . (vii) we have Q(p) v[Pn < Q(n) < Q(p) J2n2p++ 11 for each n > p.
(vii)
Using the result of Exercise following estimate for each integer +
1
·
2. 7 Symmetry and Homogeneity
We will now mention two important properties of inequalities, which will be used in the following sections. The first two of the inequalities
a + b + c > abc, -ab + -bc + -ac > 1, a + 2b + c > bac are ( as opposed to the third one ) symmetric in the variables a, b, c, that is, they do not change if the order of the triple a, b, c is changed. (Note that the third inequality is symmetric in the variables a, c. ) Symmetries in the variables x 1 , x2 , . . . , Xn are sometimes used as follows: We may assume that the variables x�, x2 , , are appropriately ordered, for instance x 1 > X n an assumption can considerably shorten the arguments xin2 the> · course · · > Xnof. Such a proof. We illustrate this by the example of the inequality (10) which, as we will holds for arbitrary b, c E JR.+ . If a > b > c, then a + b + c < a + a + a 3a, and thus (a + b + c) lOO < (3a) 100 < 31 00 (alOO + blOO + ClOO ) . This completes the proof. It may be interesting to note that, excepting the case a b c, a variant of (10), still with a strict inequality, holds even when the number 3 1 00 on the right is replaced by 399 ( see 7.8. (v ) below) . .
see ,
=
=
=
.
•
a,
2 Basic Methods
105
homogeneous inequality by way of the example as + bs + cs > �a + �b + � a3 b3c3 . c We find out how both sides of this inequality change if we replace a, b, c by ta, tb, tc for an arbitrary t E JR.+ : L' = (ta) 8 + (tb) 8 + ( tc) 8 = t8 L, 3 (tb) 3 (tc) 3 = t8 R. (ta) +t + R! = t t We explain the term
(
)
( � � �)
This leads to an equivalent transformation: Both sides change by a factor of t8 , thus L > R holds if and only if L' > R'. In general, the inequality Q(x�, x2 , . . . , Xn ) > 0 is called homogeneous (in the variables x�, x2 , . . . , Xn ) if it is equivalent to Q(tx 1 , tx2 , . . . , txn ) > 0 for each t E JR.+ , especially if
Q(tx� , tx2 , . . . , txn ) = t r Q(x� , x2 , . . . , Xn ) for some r E :JR. ( The inequality L > R can be written as Q = L R > 0.) Through the choice of the number t E :JR+ we can appropriately •'normalize" the variables x�, x 2 , . . . , Xn in a homogeneous inequality such that, for example, we get Xn = 1 ( and thus reduce the number of unknowns by 1 ) , or x 1 +x 2 + · · + xn = 1 , etc. As an illustration we prove the following assertion: H 0 < r < s, then for arbitrary numbers a, b E :JR+ we have (as + bs ) lf s < (ar + br ) llr . ( 11 ) This inequality is homogeneous in the variables a, b; through an appropriate choice of t E :JR+ (the reader should determine this value) we may assume that a1· + br = 1 . Then ar < 1 , which by 1 . 1 1 gives a5 = (ar) s fr < ar , since sfr > 1. Similarly, we have b8 < br, and thus a:s + b8 < ar + br = 1, that is, (a8 + b8 ) 1 1 s < 1 , which is ( 1 1 ) under the assumption that a.,. + br = 1.
-
·
This completes the proof. A similar homogenization approach leads to the so-called
( n a ) 1/s ( n a ) 1/r
inequality
L � k= l
2). From this we derive the fol lowing interesting property: If n > 2 and p, a 1 , a2 , . . . , an E 1R+, p =/:. 1 , then
(a 1 + a2 +
· · ·
+ an ) P > af + a� + · · · + a� , if p > 1,
(12)
respectively
( a 1 + a2 + · · · + an ) P < ai + a� + · · · + a�, if 0 < p < 1.
( 13)
In closing we remark that symmetric and homogeneous inequalities will be very prominent throughout this chapter.
106
2. Algebraic Inequalities
2. 8 Exercises
a, b, c E :JR+ prove the symmetric inequalities (i) a( a - b)(a - c) + b(b - c) (b - a) + c(c - a)(c - b) > 0, (ii) (a - b) 2 (a + b - c) + (a - c) 2 (a + c - b) + (b - c) 2 (b + c - a) > 0. Show that the inequality (ii) can be transformed into a4 + b4 + c4 + abc( a + b + c) > 2(a2 b2 + a2 c2 + b2 c2 ) by multiplication by a + b + c and further equivalent transformations.
For arbitrary numbers
(iii)
It is very difficult to prove this last interesting inequality "directly'/ (that is, without recognizing the above connection ) ; the authors know of no other approach than the use of (ii) . Prove the homogeneous inequality
a + b - vr; (a - b) 2 (a + 3b)(b + 3a) > b ao - B(a + b)(a2 + 6ab + b2 ) 2
--
�-��-�-___,.-:"--
which provides a nonnegative lower bound for the difference between the arithmetic and the geometric means of the positive numbers a and b. (Hint: Setting b = 1 and a = t 2 , where t > 0, you get a polynomial inequality that can be proved by way of factorization. )
3
The Use of Algebraic Formulas
We now add formulas for An - Bn and (A + B ) n , which we derived in Chapter 1, to our collection of tools for proving inequalities. For the entire section we assume that n, k E N.
3. 1
The Decomposition of An - Bn
In the following three examples we illustrate the use of the formula
An - Bn = ( A - B)(An- 1 + A - 2 B + An- a B2 + . . . + ABn- 2 + Bn- 1 ) n
(14)
for algebraic transformations of expressions that occur in the given inequalities.
(i)
Show that
SOLUTION.
if a, b E JR+ , a 1= b, then an+ 1 + nbn+ 1 > (n + 1 )abn .
By
(14) we get
a( an - bn ) - nbn (a - b) = a(a - b)(an- l + an- 2 b + + abn-2 + bn- 1 ) - nbn (a - b) = (a - b)(an + an- 1 b + + a2 bn- 2 + abn- 1 - nbn )
L-R=
·
·
·
·
·
·
.
3 The Use of Algebraic Formulas
Upon adding a k bn - k > bn ( if a > b), resp. a k bn- k < bn ( if a < k = 1, 2, . . . , n we obtain in both cases L - R > 0, that is, L > R.
107
b),
for
0
(ii) Suppose the numbers a, b, c E :JR+ have the following property: For each n > 1 there exists a triangle Tn with sides an , bn , and en . Show that all triangles Tn are equilateral. SOLUTION. Let a > b > c > 0. The triangle inequality for Tn means that c"' > an - bn , that is, The last term in parentheses is at least ncn - 1 , since a > c and b > c implies an - 1 - k bk > cn - 1 - k ck = cn - 1 , 0 < k < n - 1 . We thus obtain c1" > (a - b)ncn- 1 , that is, c > (a - b)n. Since this last inequality holds for all n > 1 and furthermore we have a > b, it follows that a = b. ( In the case 0 a > b we would obtain a contradiction when n > a�b . )
(iii) Show that if a > b > 0, then n+ 1 - bn+ 1 n + 1 n+1 a n a > an - bn > n b. SOLUTION. By the identity (14) we have
(15)
an+ 1 - bn+ 1 an _ bn
(a - b) (an + an- 1 b + . . . + abn- 1 + bn ) (a _ b)(an- 1 + an- 2 b + . . . + abn-2 + bn- 1 ) an + an - 1 b + . . . + abn- 1 + bn - an - 1 + an- 2 b + . . . + abn - 2 + bn- 1 · We note that if we multiply the denominator D of the last term by the number a (resp. b), we get the first ( resp. last ) n terms of its numerator.
'rhus,
where
D = an- 1 + an-2 b + . . . + abn - 2 + bn- 1 . Since the "outer" expressions in (15) are a+aln, resp. b+bln, it suffices to �how that aln > bn I D and bin < an I D, that is, aD > nbn and bD < nan . But this is easy: a > b implies and similarly we obtain
bD < nan .
0
108
2. Algebraic Inequalities
3. 2 Exercises Prove the assertions (i) -(iv) :
1 - an 1 - an+ l (i) 0 < a < 1 A n > 1 => n > --n+1 (ii) a > b > 0 A c > 0 A n > 2 ==? y'an + c - ytbn + c < a - b. 3 1 2 (1.1·1·) a > 1 ==? 1 - 1 > 2 - 2 > a - 3 > · · · · a -a a -a a -a *(iv) a > v'2 A b > v'2 => a4 - a3 b + a2 b2 - ab3 + b4 > a2 + b2 . (v) Suppose that a, b E JR+ satisfy a3 + b3 = a - b. Show that a2 + b2 < 1. * (vi) Show that for arbitrary a, b, c E JR+ , 3 3 3 a + b+ c . c b a > + + �---3 a2 + ab + b2 b2 + be + c2 c2 + ca + a2 -
3. 3 Estimates for An - Bn The proofs of a number of inequalities rely on the estimates n(A - B)Bn- 1 < An - Bn < n(A - B) An -1 ,
(16) ( A > B > 0, n > 2) , which are easy to obtain from formula (14) : Just as in 3 . 1 . ( i) it is clear that nBn - 1 < An-1 + An-2 B + . . . + ABn- 2 + Bn-1 < nAn -1 . We give three examples. (i) Show that if 0 < a < 1, then n + (1 + a) n < na + 2n for n > 2. SOLUTION. If 1 < 1 + a < 2, then by the left part of (16) with A = 2 and B = 1 + a we have 2n - (1 + a) n > n(1 - a) (1 + a) n- 1 > n(1 - a) · 1 n- 1 , 0 and after an easy simplification we get the statement.
(ii) Show that the sums Sk (n) (studied in Chapter 1 , Section 2) satisfy the
lower bound
1 . k+ 1 Sk (n) = 1 k + 2k + · · · + nk > n . k+1 SOLUTION. By the right part of (16) with A = j, B = j - 1, and n = k + 1 we have for each j > 1 the inequality j k+ 1 - (j-1) k+ 1 < (k+1)jk . Summing over j 1, 2, . . . , n we obtain nk+ 1 = (1 k+ 1 - ok + 1 ) + (2k + 1 - 1 k+ 1 ) + . . . + [nk + 1 - (n - 1) k+ 1 ] < (k + 1)(1 k + 2 k + · · · + n k ) , =
3 The Use of Algebraic Formulas
109
0
from which upon division by k + 1 we obtain the desired bound.
(iii) Prove the so-called Bernoulli inequality (1 + x) n > 1 + nx (x > -1, n > 2), with equality if and only if x = 0.
(17)
SOLUTION. H x > 0, then (17) follows from the left inequality of (16) with A = 1 + x and B 1. If -1 < x < 0, then (17) follows from the right part of (16) with A = 1 and B = 1 + x. In both cases the inequality (17) 0 is strict, and for x = 0 we get equality in (17). =
3.4 Exercises Show that the inequalities (i) -(iii) hold for n > 2:
(i) (ii) (iii) (iv)
2(a + 1 ) n + n2n > (na + 2)2n (a > 1 ) . n + ( 1 + a) n < na2 + 2n (0 < a < 1 ) . ( 1 - a) n + (2n - 1 ) an > nan - l (0 < a < 1 ) . The arithmetic means A 1 7 A2 , , AN of the real numbers ai are •
.
•
defined as follows:
( 1 < n < N) .
Show that if 0 < a 1 < a2 < · · · < aN, then for each n = 2, 3, . . . , N we have the inequality ( An ) n > an ( An- 1 ) n- 1 , unless a 1 = a2 = · · = an . (Hint: Use ( 16 ) with A = An and B = An- 1 ·) ·
Using ( 17) , show that for each k > 2 the numbers Ck = ( 1 + i) k satisfy (v) and (vi):
(v) Ck > 2. Ck+ 1 < 1 + 1 . * (vi) 1 < ---; ;:k2 + 2k
(Further properties of the numbers Ck will be derived in 3.8.(iv) .) (vii) Using (17), show that for a, b E JR+ and n > 2 we have
(1 8) (Hint: Set x = abr - 1 in ( 17) for an appropriate r.) When is (1 8) strict? Remark. The inequality (1 8 ) follows easily from the so-called AM GM inequality, which will be proved later, in Section 8. Indeed, we can apply the inequality (84) to the n-tuple an , b, b, . . . , b.
1 10
2. Algebraic Inequalities
(viii) Show that *(vi) follows from (18) with a = 1 and appropriate b, n. *(ix) Let the largest one of the nonnegative numbers d 1 , d2 , . . . , dk be equal to D. Show that for each n > 2 we have
where an = (n - 1) n1�n . 3. 5 Estimates Using Geometric Series From the formula (14) with A = 1 and B = q it follows that
1 + q + q2 + · · · + qm =
1 - q +1 1-q m
ai such that bi + l < qbi for each j, where 0 < q < 1. Using this approach, we will prove the bound q
·
·
n+ 2 . 1 (20) n + 1 (n + 1 ) r Indeed, from the inequality (n + j)! > ( n + 2 ) ! ( n + 2)i 2 for j > 3 and from ( 19) with q = 1 / (n + 2) it follows that 1 1 1 +...+ + (n + 1)! ( n + 2) ! ( n + k) !
0, then +
see
=
=
c2
c2
Since (27) implies
( ab ) 2 + ( be ) 2 + ( ca ) 2 > ab . be + abe . cab + be . ca e
a
e a2 a b = a2 + b + � = 1, we obtain the bound V2 > 3, that is, V > v'3. Equality V = J3 occurs in the case a = b = e VJ/3; hence we have found J3 as the smallest value of V. =
b
117
4 The Method of Squares
4 . 2 Exe rcises
Prove (i)-(:xxviii) for arbitrary a, b, c E JR+ and x, y, z E IR: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
(ix) (x) * (xi)
(xii ) (xiii) (xiv) (xv ) (xvi) (xvii) (xviii )
(xix ) (xx)
(xxi )
(xxii) * (xxiii) (xxiv) (xxv)
.
* (XXVI) (xxvii) * (xxviii)
2xyz < x2 + y2 z2 . (x2 y2 ) 2 > 4xy(x y) 2 . x4 + y4 > x3 y + xy3 . > .,;a + v'b. + x2 < 1 1 + x4 2 · x2 (1 + y4 ) + y2 (1 + x4 ) < (1 + x4 ) (1 + y4 ). x2 + 4y2 + 3z2 + 14 > 2x + 12y + 6z. x2 + 2y2 + 2xy + y + 1 > 0. 2 + x2 (1 + y2 ) > 2x(1 + y). x4 + y4 + z2 + 1 > 2x(xy2 - x + z + 1). .../a2 + b2 > a + b - (2 - .../2) v'£ib. 2x2 + 4y2 + z2 > 4xy + 2xz. 2(x4 + y4 ) - 12xy + 10 > 0. 2{ � + {/b + {'C) < 3 + vfa + � + .[C. a(1 + b) + b(1 + c) + c(1 + a) > 6.J(ibc. 2(a2 + b2 ) + a + b > 2(ab + av'b + bvfa,). a4 + 2a3 b + 2ab3 + b4 > 6a2 b2 : 4x2 y2 + ( Z + + y) ( Z + - y) ( Z - + y) ( Z - - y) > 0. x4 - x2 - 3x + 4 > 0. a4 + a3 - 8a2 + 4a + 4 > 0. x; + y2 + z2 > xy - xz + 2yz. a + b + c > 2(fo + v'bC - v'iib) . a3 > 36. a; + b2 + t? > ab + be + ac, abc = 1 (1 + x + y) 2 > 3(x + y + xy). ab + be + ca > J3abc(a + b + c). a8 + b8 + cB > !. + !. + !. a3 b3 c3 - a b c · xy + yz + zx > -4, x2 + y2 + z2 = 1. 4x(x + y) (x + z)(x + y + z) + y2 z2 > 0. pz - 2qy + rx = 0 p, q, r, x, y, z E IR 2 2 pr - q > 0, xz - y < 0. F(x) = x4 + ax3 + bx2 + ex + d t? + a2 d < 4bd. a, b, c, d E IR, d > 0, _
_
.:; .:a
�
X
X
X
if
X
and
if
Assuming thatpr ve the inequality satisfy � (xxx } Show that the polynomial and no real zeros, where
(xxix)
and has
1 18
2. Algebraic Inequalities
*(xxxi) Show that for arbitrary positive numbers a, b, c we have (p E R.ri ) . This is a generalization of (27) , which we obtain by setting P * (xxxii) For arbitrary a, b, c E JR+ decide which of the two numbers
(
) (
=
0.
)
A = 3 + (a + b + c) + a1 + b1 + c1 + ab + cb + ac ' + 1) (c + 1) B = 3(a + 1)(b abc + 1 is larger. * (xxxiii) Find the largest p E 1R such that the inequality
x4 + y4 + z4 + xyz(x + y + z) > p(xy + xz + yz) 2
holds for all x, y, z E JR. *(xxxlv} For each p E {1, �, ! }, determine the inequality
n
> 2 for which the
X� + X� + · · · + x! > p(X1X2 + X2X3 + · · · + Xn- l Xn ) holds for all x 1 , x2 , . . . , Xn
E R.
4 . 3 A Lower Bound for A + B Using (25) with X VA and Y = VB, we obtain =
A + B > 2v'JiB (A, B E JRt) , (29) with equality only when A = B. We use this lower bound for the sum A + B
to solve the following seven problems . (i) Show that for any a, b, c E JR+ the inequality ab + � > 2y'aC holds. SoLUTION. The result follows from (29) with A = ab and B = cjb.
(ii) Show that if a, b E JR+ , then ab(12 - 2a - 5b)
2v'fib. Multiply ing these together, we obtain (2a + 5b) (l + ab) > 4v'Wab > 12ab, since 0 v'W > 3. SoLUTION .
4 The Method of Squares
1 19
(iii) Show that if a, b, c, d E JRt, then a + b + c + d > 4 � . SOLUTION . Since by (29) we have a + b > 2Val) and c + d > 2...;cd, it ·
suffices to show that Vab + ...;cd > 2 · � . But this is just (29) with o A = Val) and B = ...;cd.
(iv) Under the assumption that the numbers a 1 , a2 , b 1 , b2 , C1 , c2 E Jil sat
isfy a 1 c1 > lJt and a2 c2 > � ' prove the inequality (a 1 + a2 ) (c1 + c2 ) > (bl + b2 ) 2 . SOLUTION. We rewrite the desired inequality as
(a 1 c1 - b� ) + (a 2 c2 - b� ) + (a 1 c2 + a2 c1 - 2bl �) > 0. Thus it remains to show that a 1 c2 + a2 c1 - 2b1 b2 > 0, and indeed, by (29) with A = a 1 c2 and B = a2 c1 we have
a 1 c2 + a2 c1 > 2 y'a 1 c2 a2 c1 = 2 vfaiC1 · .Jli2C2 > 2b1 b2 .
0
(v) Show that if a, b E JR+ and ab < 1, then (1 + �)(1 + i) > 4.
SOLUTION .
By (29) we have 1 + � >
(1 + .!.a ) (1 + !b ) -> 2_ va,
·
}a and 1 + i > J£;, and thus
2_ = .....!__ >4 v'b Val) '
< 1. since ab -
0
(vi) Show that if a, b, c E JR+, then (a + b) (b + c) (a + c) > Babe.
By (29) we have a+b > 2v'ab, b+c > 2v'bC, and a+c > 2vfaC; multiplying these three inequalities together, we obtain 0 (a + b) (b + c) (a + c) > Bv'a2 b2 c2 = Babe. SOLUTION.
(vii) Suppose that the real numbers x 1 , x2 , . . . , Xn satisfy the condition x1 x2 • • • Xk > 1 for each index k = 1 , 2, . . . , n. Prove the inequality
n 2 1 +...+ 2 + (1 + X 1 )(1 + X2 ) . . (1 + Xn ) < • (30) 1 + X1 (1 + X 1 )(1 + X2 ) 1 , 2, . . . , n. Hence by (29) with SOLUTION . Obviously, Xk > 0 for each k A = 1 and B = Xk we have 1 + Xk > 2vfxk, and thus •
=
k � < < (1 + X 1 ) (1 + X2 ) . . (1 + Xk ) 2 k ..jX 1 X2 " Xk 2k for each k = 1, 2, . . . , n. Therefore, the left-hand side of (30) does not exceed n+2 n 2 -1 + · · + -n = 2 - n < 2 + · 2 2 2 22 (see Chapter 1, Example 2.37) . o k
•
•
120
2. Algebraic Inequalities
4.4 l?xercises Show that the inequalities ( i) -(viii) hold for all a, b, c, d E JR.+ :
(i) (ii) (iii} (iv) (v) . ( ) VI
(vii) (viii) *(ix)
a2 b +
� > 2a.
6a + (c + 2) (2c + 3)b > 6(c + 2) v'ab. a4 + a3 b - 4a2 b + ab + b2 > 0. (a + 1)(b + 1)(a + c) (b + c) � 16abc. J(a + b)(c + d) + J(a + c) (b + d) + J(a + d) (b + c) > 6 1 1 8 + > ab cd (a + b) (c + d) ( VO, + Vb) 2 > 2 V2(a + b) v'ab. 2(a + b) 2 + (a + b) > 4(a Vb + by'a) . Show that for arbitrary a 1 , a2 , . . . , an E JR.+ we have
( ) i=l j=i+l
(
n- 1 n
n- 1 n
1 1 � � � � > 4 L-i L-i 2 L-i L..i a,·a, + ! a · + a3· i=! J=i
n
·
)
•
·
{IOi)Cii.
2 .
4 - 5 Upper Bound for A - B Rewriting (29) , we obtain
(A , B E JRt),
(31)
with equality occurring only if A = B . Using the bound (31), we now solve the following five problems. (i) Show that if a, b E JR.+ and a + b = 1 , then (1 + �)(1 + i) > 9. SOLUTION. By (31) with a + b = 1 we get ab < -! · Hence
a+ +1 � ( 1 + :b = 1 + :b > 1 + � � = 1 + ) ) ( 1+
=
9.
D
(ii} Let bb b2 , . . . , bn be an arbitrary ordering of the numbers a 1 , a2 , . . . , an , with 0 < ak < 1 for k = 1, 2, Show that at least one of the numbers . . . , n.
does not exceed
-!.
4 The Method of Squares
SOLUTION . k = 1 , 2, . . .
Suppose to the contrary that , n. Then
C)
n
ak (1 - bk ) > �
1 21
for all indices
< a 1 ( l - b1 )a2 (l - ba) · · · an (l - b,) a 1 ( 1 - a I ) a2 ( 1 - a2 ) · · · an ( 1 - an ) ,
=
which means that ak (1 - ak ) > � for some k 1 , 2 , . . . , n. But this is a 0 contradiction, since by (31) we have a( 1 - a) < i whenever 0 < a < 1 .
=
(iii}
Show that
if 0 < b < a, then a + 1/(ab - b2 ) > 3.
By (31) with A a - b and B b we have the inequality 2 (a-b)b < a /4, which implies a+1/(ab-IJ2) > a+4/a2 . Therefore, it suffices to show that a+ 4/a2 > 3. But this is easy to rewrite as (a + 1) (a - 2) 2 > 0.
=
=
SOLUTION.
FUrthermore, we see that the original inequality becomes an equality only 0 if a - b = b and a - 2 0, that is, a 2 and b 1.
=
=
=
(iv) Show that if a > 0, then 4 vfci(./(i - 2) (1 + 2 vfci) < (a + 1 ) 2 . SOLUTION.
vfci( vfci - 2 )
It suffices to examine the case vfci > a - 2 vfci and B 1 + 2 vfci we get
=
=
y'a (y'a - 2) (1 + 2 v'a)
4. + c C �) (iv) 4ab(3 - a) - 4a(1 + b2 ) < b. (v) a2 + b + via+ v'ab (a v'b - 4v!a) > 0. (ii)
5
The Discriminant and Cauchy's Inequality
Let us consider the quadratic polynomial (a,
F(x) = ax2 + bx + c
(33) a :;f 0) . Completing the square ( the introduction to Section 4), we obtain the identity 2 ( ) b F(x) = a x + -2a - -4aD where D b2 - 4ac. (34) The number D is called the discriminant of the polynomial F(x). The expression (34) implies not only the known formula for the roots x 1,2 of the equation F(x) = 0, but also the following result . b, c
E
IR,
see
=
,
5. 1
Values of a Quadratic Polynomial Theorem. (33} = 2! . D as {34}. E lR, :;f (i) > D 0, E 1R (ii) >0 >0 (iii) D 0, E IR;
Let F(x) be the polynomial with positive leading coeffi cient a and discriminant in Put xo Then x x , F(x) for any x F(x ) o o i f and only i f for all x F(x) < ionly f i=f x then F(x) for all x in this case F(x) = 0 if and x , o (iv) if D > 0, the equation F(x) = 0 has two different roots, say x 1 > x2 ; in this case .F(x) > 0 when x < x2 or x > x1 1 and F(x) < 0 when X2 < X < X1 (The reader should think about how ( i) - ( iv ) change in the case a < 0.) PROOF. If a > 0, then by (34) we have F(x) > - 4Da (x lR) , where equality occurs only for x x0 . This implies the assertions (i) ( iii) ; the roots x 1 , x2 in part (iv) are given by the well-known quadratic formula, and the last assertion about the sign of F(x) follows easily from D the factorization F(x) a(x - x 1 )(x - x2 ) (see 3. 1 4 in Chapter 1). =
E
=
=
5 The Discriminant and Cauchy's Inequality
5. 2
125
Examples
In the following seven examples we will practice the use of the properties of the discriminant stated and proved in 5. 1.
a, b, (35) (a2 b2 + c2 ) 2 > 2(a4 + b4 + c4). The above inequality is equivalent to a4 - 2(b2 + c2 )a2 (b2 -c2 ) 2 < The equation x2 -2(b2 +c2 )x+(b2 -c2 ) 2 has the two roots x 1 (b+c) 2 and x2 (b c) 2 . By 5.l. (iv), the inequality (35) holds and only if 2 < a2 < (b c) 2 , that is, if and only if l b - cl < a < b + c, which is (bc) the known criterion for side lengths of a triangle. 1R for (ii) Let x 1 , x 2 , . . . , Xn be given real numbers. Determine the x which the sum S (x - x 1 ) 2 + (x - x 2 ) 2 · · · + (x -Xn ) 2 is minimal. We rewrite S = nx 2 +bx+c, where b -2(x 1 +x2 +· · · +xn ) · Then by part (i) of Theorem 5. 1 , S attains its smallest value for b X1 + X2 · · Xn 2n n (iii) Let x, y, z JR. Show that the smallest of the numbers (x-y) 2 , (y-z) 2 , and (z - x) 2 is greater than or equal to ( x2 + y2 + z2 ) f2. In view of the symmetry we may assume without loss of gen > > x y b z erality that x < y < z, and we set a y Then 2 + y2 + z2 (y - a) 2 + y2 + (y b) 2 . By 5.2.(ii) this sum is (for fixed xvalues b)/3 = (a - b)/3. Hence of a, b) smallest when y = (a + x2 y2 + z2 > ( � - a) 2 ( � ) 2 + (� + b) 2 2 2 Clearly, (a2 + ab + b2 )/3 > (£? + c2 + c2 )/3 £?, where c is the smallest of the two numbers a, b. Therefore, (x2 y2 + z2 )/2 > £?, where c2 is the smallest of the two numbers (y - x ) 2 and (z - y) 2 • abc (iv) Suppose that the numbers a, b, c JR+ satisfy a + b c and a2 be. Show that a > v'3. By Vieta's relations (Chapter 1, Section 3) the numbers b and c are roots of the quadratic equation x2 - (a3 - a)x + a2 (i) Show that the positive numbers c are side lengths of some triangle if and only if they satisfy the inequality +
SOLUTION.
+
0.
= 0
=
=
if
+
0
E
+
=
SOLUTION.
=
X = -- =
+ ·
+
0
-------
E
SOLUTION.
+
=
+
0,
=
0.
=
0 -
+
+
=
o
+
E
=
SOLUTION.
= 0,
=
1 26
2. Algebraic Inequalities
3 2 . 2 2 -3) (a2 +1) , be = a2 b+c = abc-a a·a = a -a = -a 3 2 2 (a a 4a = (a -a) a2 3, a (v) Suppose that x1 is the smallest and Xn the largest among the n real numbers x 1 , x2 , . . . , Xn· Show that if x 1 + x2 + · + Xn = 0, t hen the sum x� + x� + · · · + x; + nx1xn is not positive. SOLUTION. The polynomial f(x) = x2 -(x1 + xn ) x+x l xn has X 1 and Xn as its zeros. Hence Theorem 5.1 implies that if x 1 < x < Xn , then 0 > f(x). Adding the inequalities 0 > f(xj ) for j = 1 , 2, . . . , n, we obtain (in view of the condition X 1 + X2 + · · · + Xn = 0) n n n n 0 > EJ(xj ) = Ex� - (x1 + Xn ) E xj + nx 1 Xn = Ex� + nx 1 Xn · D j=l j=l j=l j=l
It therefore has a non since and > 0. Since D negative discriminant: D = D we obtain > which implies the desired inequality > .J3. ·
·
(vi) The evaluation of the discriminant replaces several artificial tricks
that were necessary for solving problems by way of the method of squares in Section We illustrate this with the following example. Adding the > > 2 £, > v'ab inequalities ..JbC, > and E JR+ , (see 2 and (29)) we obtain for arbitrary
4. 4-v'ab, 2(a + b) b + c a + b + c + vfaC + 3(a+c) 6vfaC ( 7) a, b, c (36) 6a + 4b + 5c > 5v'ab + 7 vfaC + 3-Jbe. "Discovering'' this proof of (36) is certainly not easy; the reader should try to prove this inequality with the method of completing the square. However, the inequality (36) is quadratic in the variables x = yfa, y = Vii, and z = ...jC; with the help of the discriminant we will show that (37) 6x2 + 4y2 + 5z2 > 5xy + 7xz + 3yz for arbitrary x, y, z lR, that is, 6x2 - (5y + 7z)x + 4y2 + 5z2 - 3yz > 0. (Note. The authors know that this will work. However, sometimes such a generalization can be misleading: It could happen that the inequality (37) holds only for x, y, z JR+ , since x = VO,, etc. Then the discriminant of the last polynomial could be positive. ) It therefore suffices to show that the discriminant D (5y + 7z) 2 - 24(4y2 + 5z2 - 3yz) is not positive for any y, z JR. By an easy calculation we find that D = -71 (y - z ) 2 • This completes the proof of (37), and therefore also of (36). E
E
=
E
(vii) In contrast to the previous example we now deal with a situation
where the use of the discriminant is simply unsuitable. We show that the equation
F($) = (X - a) (X - C) + 2 (X - b)(X - d) = 0,
(38)
5 The Discriminant and Cauchy's Inequality
127
a < b < c < d,F(x) = 3x2 -(a+2b+c+2d)x+ac+2bd, if D = (a + 2b + c + 2d)2 - 12(ac + 2bd) a b < < c < d; x = b, F(b) = (b a)(b c) < a < xb2 0. The reader should attempt to derive this last inequality from the assumption this is not at all easy. Compare this with the following alternative solution: Substituting we obtain 0, since Therefore, the polynomial with the positive coefficient By 3 of attains a negative value at some point (namely when Theorem this means that has two distinct real zeros. (Note that we required only the inequalities in this proof.) We finally remark that the zeros and > of F( satisfy the inequalities 0 , and 0, F > 0. since > 0,
5. 3 Exercises >0 (i} Suppose that the numbers s E IR are such that > 0 hold for all E R. Show that >0 and for all E :JR. (ii) Let E R+ . Show that c are the side lengths of some triangle E 1R with + satisfy the inequality and only + > pq�. (iii) Show that are the side lengths of some triangle, then the inequality + + r? + be + ca ) holds. then + + z + z (iv) Show that > 8 z+ . * (v) Determine the values E IR for which x2 +
2 px x + q + 2x2 + prx + 2qs
p, q, xr,
x2x+rx + s a, b, c p, q a, b,p q = 1 if if all pa2 qb2 ifa2 a, bb,2 c 2(ab < ifx < y < < u, (x y u) 2 (x yu) p -2 a + a + . . + a l 2 n· an a1 a2 a3 SOLUTION. By (39) with Uk = ak / y'ak+ l and Vk y'ak+1 lln+ l = a 1 ) we have .
=
( 44)
(1 < k
R(n + 1) - R(n) for all n > 2.
The equality L(2) = R(2) therefore implies L(3) > R(3), and from this follows L( 4) > R( 4) , etc. Hence we have L( n) > R( n) for all n > 3. These examples illustrate the fact that in induction proofs of the in equality L( n) > R( n) we most often compare the absolute, resp. relative, growths of both sides L, R as the value of n is increased by 1. We express this in the following principle.
6. 2 The Induction Principle Theorem. Let L(no) > R(n0 ) and suppose that for each n > n0 one of the two conditions (i) and (ii) holds: (i) The numbers L(n) , R(n) , L(n + 1), and R(n + 1) are positive and satisfy L( n + 1 ) > R( n + 1) L(n) - R(n) . (ii) L(n + 1) - L(n) > R(n + 1) - R(n) . Then the inequality L( n) > R(n) holds for all n > no. If furthermore L(n 1 ) > R(n 1 ) for some n 1 > no, then L(n) > R(n) for all n > n 1 .
no N
Remark 1. In order to prove a finite number of inequalities L(n) > R(n),
no the inequality L(n) > R(n) holds, we derive the truth of the inequality L(n + 1) > R(n + 1). Thus, in Example 6.1.(i) we could have proceeded as follows: Multiplying the
137
6 The Induction Principle
>n >
> n
inequality 2n ( + 1) 2 by 2 , we obtain 2n + l 2 ( + 1) 2 . Hence it suffices to verify that 2(n + 1) 2 (n+2) 2 . In Example 6.1. (ii) , by adding 1/(n+ 1) 2 to both sides of (51) , we obtain 1 + 1 + 1 + + · · · 22 3 2 n2
(n
1 + 1) 2
> 127 - n +1 1 + (n 1 1) 2 = 127 - (n + 1)2 · +
n
Hence it suffices to verify the inequality
1 -127 - (n +n 1)2 > -127 - -n + 2·
6. 3 Exercises Show that the inequalities (i) -(viii) hold for all (i) (1 - a) n < 1 1 , 0 < a < 1). + a n (2n)! (ii) 4n 2) . ( < < +1 ( ) (iii) 1 + 1 + · · · + 1 1 1). 3 + (iv) 2 ! n(n- l ) 3) . (v) + + + ) +... + +X+ ) 1 1 1 (no 2). ( ) < 1+ + +...+ n 2 2 3 2 -1 < *(vii) (1 + a) n (1 + an+l )(1 + an+2 ) · · · (1 + a2n ) (no = 1, 0 < a < t2 ). 2) . (viii) ) [ + ··· (ix) Show that if 1 < k < then we have
n > n0 :
(no = l n 4 no = n n! 2 n+l n+2 n l > (no = n! (no = 1 > 2n y2n n > (xy) 2n- (xy 2n-� x(x,y xy y E lR,no = l . . n = n > 2!4! (2n ! > (n l)!]n (no = n, k k 1) k2 . 1 + - < ( 1 + - < 1 + -k + n n n2 n Remark. For k = n, (52) implies (compare with 3.8.(iv)) n ( 2 < 1 + :) < 3 for all n > l . (x) Show that for all n > 6 have ( n ) n > n! > ( n ) n 3 2 VI
we
·
(52)
(53)
138
2. Algebraic Inequalities
(xi) Show that if k > 2 and n > 2, then at least one of the numbers {in
and V'k is less than 2. (xii) Given the expression
1 3
2n
Q(n ) = 2 · 4 · · · 2 prove the estimates
Q(p)
�
2 we have An > Bn - l · With the induction method it is easier to show the stronger inequality An > 2Bn- l · By direct verification we find that A2 > 2B1 . If the inequality An > 2Bn - l holds for some n > 2, then 4"
(ii} Find the smallest number A for which one can prove
(1 + ..!.22 . ) (1 + ..!.32. ) · · · (1 + _!_n2 ) 0 can be verified by way of the principle 6.2 with the "division" condition (i) . first point out that the inequality (55) cannot be directly proven by induction, · since its left-hand side grows with increas ing n, while the right-hand side remains constant. We therefore consider the stronger inequality (56); condition (i) in 6.2 means that for each n > 2 we need 15/8. It is also clear that (56) with A = B = 15/8 holds for all n > 2; hence we may set A = 15/8 in (55) . Can we use this approach to determine whether (55) is true also for some A < 15/8? We note that when (55) holds for some integer n = p > 2, then it holds also for all preceding n = 2, 3, . . . , p - 1 . Therefore, it suffices to show that the stronger inequality (56) is true for all n > p, where p is a certain (arbitrarily large) number. This occurs if and only if A < B and
We can therefore set
( ..!.. ) (
) (
)
+1 1 · 1 ..!.. . · A=B= P + 2 1 + ..!.. 32 · · + p2 2 p We have thus shown that the estimate (55) holds with the constant A=P
; 1 ( 1 + ;2 ) (1 + ;2 ) (1 + �) , · · ·
(57)
where p > 2 is an arbitrary integer. If we denote the right-hand side of (57) by Ap, then it is easy to see that
[
.AzM-1 = 1 - (p
: 1)4 ] A,.
This means that the value of Ap decreases as p grows. Actual computations give , for instance, A2 = UV8 = 1.875, A 13 � 1.83828, and A3o � 1.83806.
2. Algebraic Inequalities
140
converges We finally remark that the sequence of numbers A2 , A3 , A4 , to a minimal value A that satisfies (55); this value is equal to the infinite •
product
(1 + _!_32 ) . . ( 1 + _!_n2 ) . . . (1 + _!_) 22 .
.
.
•
By means of complex analysis one can show that the value of this product is equal to the number (47r) - 1 ( e'��" - e -'��" ) � 1.83804 (compare this with D the value of A30 above) . ·
6. 5 Exercises (i) Eight fives and five eights are used to form the numbers
5
s·
8
5
Which one is larger? (ii) Show that if 0 < a < 1 and 0 < b < 1, then for each n > 1 we have (a + b - ab) n + ( 1 - an ) ( 1 - bn ) > 1 . (iii) Show that for each n > 1 we have 1 1 ... 1 < 25 . 1+ 1 + _2 1 + _ 2 n(n + 1) 1· ·3 To do this, replace the right-hand side by the expression B An+C with appropriate constants A, B , C. *(iv) Try to improve the upper bound in (54) in the form
(
< Q(n) -
)
) (
)(
Q(p)
(n > p), v'An + B where A and B are appropriate constants (depending on p). ( v) Show that for each n > 1 we have
(1 + D ( 1 + � ) · · ( 1 + � ) < 3. ·
2
* (vi) Let S(n) = 1 1 + 22 + · · · + nn Show that the inequality .
1 1 1 1 + + + · · > · nn S(n) S(n + 1) S(n + k) holds for arbitrary integers n > 2 and k > 0. (To do this, use the 2 k fact that the numbers ck = (1 + l) , where k = 1, , . . . , form an increasing sequence; see 3.4.(vi) and 3 .8.(iv) .)
141
6 The Induction Principle
6. 6 Induction on the Number of Variables We will now use the induction method to solve problems concerning > 0. As a rule one can de inequalities of the form > 0 from the preceding inequality rive an inequality > 0, where is some n-tuple, appropriately chosen and dependent on the given (n + 1)-tuple In the simplest case one can choose y 1 in more com plicated cases where the variables are dependent on each other through conditions that have to be satisfied, the chosen variables have to satisfy the same conditions. We will now consider five examples. (i) Suppose that the real numbers satisfy either the inequal > 0 (k = 1, 2, . . . , n) , or -1 < < 0 (k = 1 , 2, . . . , n) . Prove ities that (1 + )(1 + > 1+ + + + · (1 + (In the case a 1 we obtain Bernoulli 's inequality (17) .) = · · ·
, Q(x , x . , . . ) X 1 n 2 , , x . . . , Q(x ) X 1 n+l 2 Q(y�, Y2 , . . . , Yn ) y�, Y2 , . . . , Yn , , x�, x . . X . 2 = x�, x�, xY2 , =. . .x,2 , . . . , Yn = xn ; n+l· 2 Xn+l Yl , Y2 , . . . , Yn , , a�, a . . a . n 2 ak ak a 1 a2 ) . . an) a1 a2 . . an . = a2 = an ·
SOLUTION. Using the method of finite induction we show that for each
index k = 1 , 2, . . . , n we have
a2 )
a k ) > 1 + a 1 + a2 +
ak . (58) For k = 1, equality occurs in (58) . If (58) holds for some k < n, then upon multiplication of both sides by the nonnegative number 1 + ak+l we get (1 + a 1 ) (1 + a2 ) (1 + ak+l ) > (1 + a 1 + a2 + + a k )(1 + a k+l ) = 1 + a1 + a2 + + ak+l + bk , where bk a 1 a k+l + a2ak+l + + a k a k+l > 0, since by assumption the numbers aiak +l are nonnegative. Therefore, we have (1 + a 1 ) (1 + a2 ) (1 + ak+l ) > 1 + a 1 + a2 + + ak+ b (1 + a 1 )(1 +
· · ·
(1 +
· · ·
+
· · ·
· · ·
· · ·
=
all
· · ·
· · ·
· · ·
which completes the proof by induction . (ii) Suppose that
a�, a2 , a3 , . . . is a sequence of real numbers satisfying
0
(59)
for each n > 2. Show that for each n > 1 we have +···+ a1 + ------------� -> + +···+ n+1 n SOLUTION. We rewrite the inequality (60) in the equivalent form > (n + 1)(a2 + + · . . + n(a 1 + + · +
a3
a2n+l a2 a4
a3 . a2n+l ) .
a2n
a4
.
a2n) ·
(60)
1 42
2. Algebraic Inequalities
L(n) R(n) and prove it by induction. !or 2a n 1 L(n +L(1) a + a R(1), by (59) with n 2. The 1 3 2 1) - L(n) R(n + 1) - R(n), which is of the form We denote this inequality by = we have = > inequality
>
>
=
=
can be rewritten as
8(n) a1 -a2 + a3 - · · -a2n + a2n+l - (n + 2)a2n+2 + (n + 1)a2n+3 0 and will also be proved by induction. For n 0 we have S(O) a 1 - 2a2 + 0, by (59) with n 2. Since, again by (59) , a2n+5 2a2n+4 - a2n+3 , awe3 estimate the difference 8(n + 1) - S(n) as follows: 8(n + 1) - 8(n) (n + 2)a2n+5 - (n + 3)a2n+4 - na2n+3 + (n + 1)a2n+2 (n + 2)(2a2n+4 - a2n+3 ) - (n + 3)a2n+4 - na2n+3 + (n + 1)a2n+2 (n + 1)(a2n+4 - 2a2n+3 + a2n+2 ) · 0 By (59) the last term is nonnegative. Hence 8(n) 0 for all n 0 . 3, satisfy the (iii) Show that if the numbers a �, a2 , . . . , an E :JR+ , n inequality =
>
·
=
>
=
>
=
=
>
=
>
>
>
ai , ai , ak ( 1 < i < j < k < n) are side lengths of Since (61) for n 3 is actually identical with (35), the asser tion for n 3 follows from 5. 2.( i) . Now let n 3. In view of the symmetry it suffices to show that (61) is a consequence of the inequality (62) for arbitrary ab a2 , . . . , an+l E JR+ . If we set 8k af + a� + · · + a! for k 2 and k 4, then ( 62) can be rewritten as (82 +a;+ 1 ) 2 n(84 +a!+ 1 ), which is equivalent to ( ) 2 8 2 2 (n - 1) an+l - n 1 < n n 1 (82 ) 2 -n84 . (Here we have used the method of completing the squares.) If (62) holds, then the right-hand side of the last inequality is positive, and thus we have (82)2 (n - 1)84. But this, by the definition of 82 and 84 , is just the0 inequality (61). then any three numbers some triangle . SOLUTION.
=
>
=
=
=
=
_
>
_
>
·
143
6 The Induction Principle
and , , u U Ub n 2 V2 , . . . , Vn satisfy U1 > V1 , U1 + U2 > V1 + V2 , , U1 + U2 + · · · + Un > V1 + V2 + ' . . + Vn · Show that if X 1 > X2 > · · · > Xn > 0, then also
(iv) Suppose that the two n-tuples of real numbers
.
.
•
v� ,
.
•
•
SOLUTION. We show by finite induction that for k = 1, 2, . . . , n we have
u1 x 1
u1 1
> v1 x � , which follows from > v and For k = 1, (63) is of the form x > 0. We will now consider the question: Is it possible to derive (63) for a given k, 1 < k < n, from the inequality
1
U1Y1 + U2Y2 + . . . + Uk- 1Yk- 1 > V1Y1 + V2Y2 + . . + Vk- 1Yk- b (64) where Y1 > Y2 > · · · > Yk- 1 > 0 is an appropriate (k - 1)-tuple? To begin with, it is clear that we cannot choose Yi = Xj (1 < j < k - 1); this, in order to derive (63) from (64), would require UkX k > VkXk , which does not follow from the assumptions. However, we have ( u 1 + U2 + · · · + Uk ) X k > (v1 + V2 + · + Vk ) Xk · Now it suffices to note that we obtain (63) by adding this last inequality to (64) with Yi Xj -Xk (1 < j < k - 1) Since for these numbers we have > Y2 > > Yk- 1 > 0 (this follows from 1 > > · · · > Xk ) , the use Y1of (64) 0 justified, and the proof by induction is complete. (v) Show that if X I, . . , Xn E :JR+ and 1, then we have · · X n · · · Xn = 1 . X1 + X2 + · + Xn > n, with equality if and only if X1 SOLUTION. For n 1 the assertion is true. We assume that we have (n + 1) positive numbers , Xn+ 1 that satisfy X 1 X2 · Xn+ 1 1, and we assume further that the assertion holds for n . If we choose Y1 , X and Yn Xn Xn+ b then Y1 Y2 Yn = 1 , which X Y Y 1 -1 n n 2 2 ' means, by our assumption, that Y1 + Y2 + . . . + Yn = X1 + X2 + . . . + Xn- 1 + XnXn+1 > n. We have to show that 1 + + · · · + Xn+ 1 > n + 1. To this end, it clearly suffices to verify that Xn + Xn+ 1 > XnXn+ 1 + 1, that is, (xn - 1 ) (xn+ 1 - 1) < 0. This, in general, does not follow from the as sumption X 1 X2 · Xn+l = 1 alone; however, in view of symmetry we can order the numbers X 1 , . . . , Xn+l in advance such that Xn < Xi < Xn+l .
· ·
=
.
x
· · ·
is
x2 ,
·
x1x 2 •
.
x2
=
= X2 =
·
=
=
x� , x2 ,
•
.
.
·
·
=
=
x� ,
=
•
.
=
•
=
x
•
·
x2 ,
x2
•
•
.
144
2. Algebraic Inequalities
n+1 X1X2 · · · Xn+1 (xn+1 )n+1 , 1 , 2, . 1 -1 . (x ) n Xxn+1+, x + · · · + (x>n - +1)(x1; n+1 - 1) Xn X 1 1 n + 2 X 1 X · · · = X·n·-1· ==X XnXn=+1.1 = 1 (xn - 1)(xn+1 - 1) 1 n+ l Remark. 8. 8.1
< for i = < ,n Then we have < 0. This < < that is, which implies equality means that proves the inequality n 0, which = X2 = and 0 implies = X2 = The result of (v) follows immediately from the so-called AM Since that GM inequality, which will be studied in detail in Section inequality is homogeneous, we make use of the opposite direction: The will be based on the above result proof of the AM-GM inequality in of (v) . .
.
=
6. 7 Exercises (i) Show that for each n
> 2 we have (1 - ..!.22 . ) ( 1 - ..!..23 ) . . . (1 - _.!._2n ) > !2 . (ii) Suppose that the sequence of real numbers x�, x2 , X3 , . . . satisfies 1 1 + x2 + · + xn ) Xn+1 > -(x n for each n > 1. Show that for all n > 1 we have X1 + 2x2 + · · · + nxn > n +2 1 (x1 + X2 + · · · + Xn ) . (iii} Suppose that the sequence of real numbers x�, x2 , X3 , . . . is nonde creasing, that is, Xn+ 1 > Xn for each n > 1. Show that for all n > 1 we have the inequality n(x1 + 2x2 + 4x3 + · · + 2n- 1xn) > (2n - 1)(x1 + X2 + · · + Xn) . (iv) Suppose that the sequence of real numbers a�, a2 , a3 , . . . satisfies the inequality an+ 1 · an- 1 > a; for each n > 2. Show that if we set bb;n =holdsy'a1aas2well. · · an, then for each n > 2 the inequality bn+1 · bn- 1 > *(v) Suppose that the two n-tuples of positive numbers a 1 > a2 > · · · > aan +aand b1 > b2 >> · > bn satisfy a 1 > b�, a1 + a2 > b1 + b2 , . . . , that for each k E N we 1have af2 ++· a�· · +a+ ·n· · +ba!1 +b>2bf+ +· b�+b+n . Show b + · · · !. (vi) Use a method similar to 6.6.(v) to show that if the positive numbers x�, x2 , . . . , Xn satisfy X1 + X2 + · + Xn = n, then X1 X2 · · · Xn < 1. (vii} Show that if n > 4, then all x�, x2 , . . . , Xn E JRt satisfy (x1 + X2 + · · + Xn) 2 > 4(X1X2 + X2X3 + · · · + XnX1 ) · * (viii) Show that if a 1 , a2 , . . . , an E N are distinct, then (ai + · · · +a�) + (a� + · · · + a�) > 2(a� + · + a�) 2 . ·
·
·
·
·
·
·
·
·
·
·
·
·
·
7 Chebyshev's Inequality
7
145
Chebyshev's Inequality
7. 1
Estimating a Sum Using Permutations We will be concerned with the following problem: Given two n-tuples of real numbers x 1 , x2 , , and y� , y2 , , find the smallest and the largest value of the sum (65) is an arbitrary ordering of the numbers y� , y2 , , where z�, z2 , . . . A special case of this problem is the following situation: On a table there are four piles of coins, a pile each of one-crown, two-crown, five-crown, and ten-crown coins. From two piles you may take eight coins each, from the next one five, and from the last one three coins. How will you decide? Obviously, it will be convenient to first order both the given n-tuples and Let us therefore assume that x 1 < x2 < · · < and < Y2 < First we show that for each k = 1, 2, . . . , n the inequality · ·< (66) < + X2 Y2 + · · + X1 Z 1 + X2 Z2 + · · · + •
•
•
Xn
.
.
•
Yn ,
, Zn
.
. Yi · Yn ·
·
, Zk
Xn
Yl
.
•
Yn· Xj
· XkYk
XkZk Xl Yl
holds, where z� , z2 , . . . is an arbitrary ordering of the numbers also, equality occurs in (66) if and only if the condition y�, y2 , • . . (67) is . satisfied. The reader should carefully consider what the condition ( 67) means; it may be surprising that equality in (66) can also occur for a k tuple z�, z2 , different from the one for which z1 < z2 < · · < that is, = (1 < i < k ) . Note, however, that in the case where x = for some i =I= j , switching the numbers and does not change the value of the left-hand side of (66) . We prove the inequality (66) by finite induction. The case k = 1 is trivial. Let us now assume that the assertion holds for some k < n; we will show that for an arbitrary ordering z1 , z2 , . . of the numbers we have the inequality < X 1Y1 + X2 Y2 + · · + (68) X 1 Z1 + X2 Z2 + · · + is an ordering of the numbers then z� , z2 , . . . If = hence the inequality (66) follows, and upon adding y1 , y2 , • . • does not hold, then = we obtain (68) . If = < the ordering and = for some i < k. We denote by zf, z�, • . . by switching the terms and that is obtained from z� , z2 , • . . Obviously, - (x 1 z1 + · · + (x 1 z� + + = + + = > 0,
, Yk ;
, Z . . . k Zi Yi
· Zk , i Xj
Zi Zj
, Zk+l Yl , Y2 , , Yk+l · Xk+l Yk+l · Xk+l Zk+l , Zk Zk+l , ; Yk+b X Z Yk k k l +l + Z Z + Xk l Yk+l Z k+l Yk+l k Yk +l +l , z � Yk+l i +l Zi Zk+l· , Zk+I · Xk+l Zk+I ) · · · Xk+l Z�+l ) (Xi Zk+l Xk+l Yk+l ) - (XiYk+l Xk+l Zk+l ) (xk+l - Xi)(Yk+l - Zk+l ) .
•
.
•
•
146
2. Algebraic Inequalities
Xk+1 > xi and Yk+1 > Zk+1 ; to prove (68), it therefore suffices to
since verify that
(69)
Yk+1
z�, z2, . z/c . , . 1 + z/c+1 Yk+b
Xi Xk+1
Zk+1 Yk+1 ) ·
Y1, Y2 , · · · '
is the ordering of the numbers However, since for which the inequality (69) has already been proven. Through an easy examination of our approach we can verify the truth of criterion (67) for equality in (66); in the process it is determined that the left-hand side of (69) is larger than the left-hand side of (68) as long as (since This proves (66) for all k 1 , 2, . , n. < < To obtain a lower bound =
.
=
.
is some ordering of the numbers Z , z , . z . , . 1 n 2 an ordering of the 1 , - z2 , . . . ,by-Zthe n isprevious Y1 Y2 -y· n· · -YYnn 1 · · · -Y1-· zTherefore, discussion X1 ( - zl ) + x2 ( -z2 ) + · · · +xn ( - zn ) < X1 ( -yn ) +x2 ( -Yn- 1 ) + · · · +xn ( -y1),
we now use a simple trick; < < < if and only if < < < numbers we have
which is just (70); furthermore, equality occurs in (70)
if
and only if
(71) (1 < i < j < n) . We note that the conditions (67) and (71) are formulated in such a way
x1 , x2 , . . . , Xn
that for their verification in specific situations the n-tuple need not be previously ordered. To summarize everything concisely (Theorem 7 .3), we need the following definition. 7. 2
Orderings Definition. We say that the two n-tuples and of real numbers have the same ordering, respo the opposite ordering if
u�, u2 , . . . , Un v�, v2 , . . . , Vn
( 1 < i < n, 1 < j < n) , respectively
(1 < i < n, 1 < j < n) o
u�, u , . . . , and v�, v2 , . . . , Vn have U n 2 the same ordering does not necessarily mean that the n-tuples u 1 , u2 , . un have opposite orderings-consider, for instance, the and Vn , Vn- b v 1 quadruples 1, 1, 2, 3 and 2, 1, 3, 4. However, such a conclusion is true for instance, the numbers u�, u2 , , Un are distinct.) ( Caution. The fact that the n-tuples
0
0
0
0
.
,
,
0
0
0
if,
7 Chebyshev's Inequality
147
7. 3 Extremal Values of a Sum
be two n-tuples of real , . . . , Let and y�, y , . x�, x . , . Y X n n 2 2 numbers. , is an 1 where z�, z z Z The x + X sum S = x + + z n n 1 1 2 2 2 arbitrary , . . . , 1 is maximal, resp. minimal, , y ordering of the numbers y Y 1 n 2 if x1 , x2 , . . . , Xn and z�, z2 , . . . , Zn have the same, resp. opposite, orderings. Inordering, particular, if the n-tuples Xi, x2 , . . . , Xn and Yl , y2 , . . . , Yn have the same , have and at the same time x�, , . . . , and x . . , X b . Y Y Y n n n l 2 opposite orderings, then each of the n! sums S satisfies the inequalities X1Yn + X2Yn- l + · + XnYl < 8 < X1Y1 + X2Y2 + · · + XnYn · (72)
Theorem.
. . . , Zn
· · ·
· ·
·
The proof was already carried out in Section 7.1.
7.4 Examples The above theorem will be applied in the following two examples. (i) Show that for arbitrary numbers E the inequality
a�, a2 , . . . , an , p :JR+ aia2 + a�a3 + · + a:_ 1 an + a�a1 < af+1 + a�+l + + a�+l (73) holds, with equality if and only a 1 = a2 an SOLUTION. For p > 0 the n-tuples a�,a2 , . . . ,an and af,a�, . . . ,a� have the same ordering, by 1.10. Then by Theorem 7.3 the ordering a2 , a3 , . . . , an , ai satisfies ai · a2 + a� · a3 + · · · + a� · a1 < ai · a1 + a� · a2 + . . · + a� · an , and this is (73) . Here we obtain equality only in the case where the n tuples a�, a2 , . . . , an and a2 , a3 , . . . , an , a 1 have the same ordering. Let us assume that this is the case and that we do not at the same time have if we set M = max( a�, a2 , . . . , an ), then there exist aindices an ; that 1 = a2i =and· · ·j such < ai = M and ai+l < ai = M, where a i l athat an and an+l = a 1 . The inequality (ai- l - aj)(ai - ai+ d < 0 means o = the and a2 , a3 , . . . , an , a 1 do not have original n-tuples a�, a2 , 0 the same ordering, and this is a contradiction . (ii) Show that for arbitrary a, b, c E JR+ we have 2 2 2 t? t? + a2 b + b + a < a + b + c 2c + 2a + 2b SOLUTION. By 1 .9 and 1.10, the triples (a2 ,b2 , t? ) and (�, z, �) have opposite orderings, and thus Theorem 7.3 gives the inequalities a2 . .!a + b2 . .!b + c2 . .!c < a2 . .!b + b2 . .!c + c2 . .!a a2 . .!a + b2 . .!b + . .!c < a2 . .!c + b2 . .!a + c2 . !b . · ·
· · ·
if
=
. . . , an
C2
= · · · =
.
148
2 . Algebraic Inequalities
If we add these two inequalities and then divide by two, we obtain the D desired inequality. 7. 5 I?xercises (i) Show that if
a�, a2 , . .. , an , p E JR+ , then the inequality a1 + a2 + . . . + an- 1 + an a'f- 1 + a�- 1 + . . . a:- 1 (74) an a 1 a 2 a3 holds, with equality if and only if a 1 = a2 = · · · = an· (ii) Give a new prdof of the result of 6.6. (v): If x 1 , x2 , . . . , Xn E JR+ and = 1, then x1 + x2 + + Xn n, with the exception of x x X 1 n 2 the case x 1 = x2 1. (Hint: Use (74) with p = 1 and an = = X n appropriate n-tuple a b a2 , . . . , an .) (iii) Show that for arbitrary a, b, c E JR+ we have a2 + b2 + b2 + + + a2 < a3 + b3 + · 2a 2b be ac ab 2c if
p
p
p
p
+
>
· ··
·· ·
= ·· ·
c3
c2
c2
7. 6
>
Chebyshev 's Inequality
numbers real of n-tuples ordered two given e we that Suppose ar X1 < X2 < < Xn and Yl < Y2 < < Yn . Set Bmin = X1Yn + X2Yn- l + · · + XnYl , (75) Bmax = X1 Yl + X2Y2 + · · · + XnYn (this notation is consistent with Theorem 7.9). Then the inequalities nSmin < (xl + X2 + . + Xn)( Yl + Y2 + . . . + Yn) < nBmax (76) (76} occurring only simul of parts left and right the in equality with hold, taneously, and if and only if x1 = x2 = · = Xn or Y1 = Y2 = · = Yn .
Theorem.
···
· ··
·
. .
· ·
PROOF.
· ·
By Theorem 7.3 and definition (75) we have
Bmax = X1Y1 + X2Y2 + · · · + XnYn, Bmax X1Y2 + X2Y3 + · + XnYl , Bmax X!Y3 + X2 Y4 + + XnY2 , Bmax X1Yn + X2Y1 + + XnYn- 1 · >
· ·
>
· ··
>
···
Adding all the relations in (77) , we obtain
nBmax (xl + X2 + · + Xn )(Yl + Y2 + · + Yn ), >
·
·
·
·
(77)
7 Chebyshev's Inequality
149
and this is the right-hand part of (76) . The left part of (76) can be proved in a similar way. If equality holds, for example in the left part of (76) , then we have equality everywhere in (77). Let us assume that does not hold, that is, that < Then according to 7.1 the equalities in (77) mean that < < < which implies > > > However, by the theorem's hypothesis we have < < · · < thus 0 and the proof is complete.
x = = · X · 1 2 ·, x X 1 n , . , , 1 Y . . Y Yl Y Y Y1 n Y n 3 2 2 ·· · Yn · · Y Y1 2 Y1 = Y2 · · · = Yn ,
·
= Xn Y2Y , n
=
7. 7 Example We now consider a typical instance of the use of Chebyshev 's inequality: We will prove that arbitrary positive numbers b, c satisfy the inequalities
a,
(78)
and (79)
a = = c. ( ) n a a n n= = x 2 , 1 = b2 a-=1 . ( a)3 , = 1 = a X2 , X3 nY1= x1 =Y2a4 , x = YX3 == cl;c4, 3 2 1 1 xb x2 , x3 Y3 = a • Y1Y1 , =Y c, Y , Y2 = 2 3 ( (
with equality in both 78 and (79) if and only if b In view of the symmetry we may assume that 0 < < b < c. Then clearly < bn < d"' for 2, 3, 4, and c- < b- 1 < Hence 78 is the right part of (76) with 3, c? , l,S, and similarly, (79) is the left part of (76) with 3, b4 , In both cases the triples and b- 1 , and have the same ordering. This completes the proof. Further consequences of Chebyshev's inequality will be derived in the following exercises. Especially the result of v) deserves some attention; it is an interesting generalization of Cauchy's inequtlity 48). =
7. 8 Exercises ( i) Another proof of Chebyshev1s inequality: Show that the expression
n(X 1Y1 + x2Y2 + · · + Xn Yn) - (x 1 + X2 + · +xn )(Y1 + Y2 + · · · + Yn ) can be written as the sum 1 n Bij, 2 iE ,j=1 where = (xi - Xj )( Yi - Yi ) · (ii) Show that the inequality (46 ) is a consequence of Chebyshev ' s ·
Bij
inequality.
·
·
150
2. Algebraic Inequalities
(iii) How can inequalities (78) and (79) be generalized? (iv ) Show that for eacfl integer we have the inequality
n>2 1 2 + 3 + · · · +-n > n+ 1 ( -+ 1 1 + 1 + · · · + 11· ) . -+ n n- 1 n-2 1 2 n n-1 n-2 (v) Show that for a 1 , a2 , . . . , an E JRt and each integer k > 2 the strict inequality (a 1 + a2 + · · + an ) k < nk- 1 (ak1 + a2k + · · · + ank ) (80) holds, with the exception of the case a 1 = a2 = · · · = an · (vi) Show that for arbitrary a 1 , a2 , . . . , an E IR+ we have (a 1 + . . . + an ) (a 1 + . . . + an ) < (a51 . . . + an5 ) ( a1l + . . . + a1n ) . (vii) Show that the result of 6. 7.(iii) is a certain Chebyshev inequality. (viii) Suppose that the numbers x, y E satisfy xy + 1 > x + y . Show that for each integer > 1 we have the inequality (n - 1 )xy + 1 > (n - 1)2 xy + (nn- 1)(x + y) + 1 . (ix) Show that if 0 < ai1 < ai2 < · · · < ain (1 < j < k), then ·
+
3
3
n
Ill
(x) Show that the inequality
(x1 +x2 + · · · +xn )(Yl + Y2 + · · · + Yn ) < n (x 1Y1 +x2Y2 + · · · +xnYn) holds also in more general situations than in 7.6: The inequality Yi < holds for any indices i,j E {1, 2, . . . , n} for which Xi < A < Xj , Yiwhere A = (x 1 + X2 + · · · + Xn) fn.
8
Inequalities Between Means
mean value in many different situations; in arithmetic mean of the values x1 , x2 , . . . , Xn ,
We encounter the term most cases it refers to the determined by the relation
(8 1)
8 Inequalities Between Means
151
However, there are also other kinds of means; the best known among them is the of nonnegative numbers which is defined by
geometric mean
x1 , x2 , . . . , Xn , 9n (Xl , X2 , · · · , Xn ) = ytx1X2 · · · Xn .
(82)
Between the various kinds of means there are inequalities that can be successfully used to solve a variety of problems. Especially the inequal ity between the arithmetic and geometric mean, more concisely known as the AM-GM is among the most important results of the entire theory of inequalities, with numerous applications (see A gener alization of the AM-GM inequality to the case of will be discussed and applied in Sections the remaining sections are then devoted to the theory of with arbitrary real degrees.
inequality,
8 5) 8.1 . weighted means.
6 8. 8. 9 ; power means
8. 1
The AM-GM Inequality
For arbitmry numbers a1 , a2 , . . . , an E JRt the inequality a l + a2 + . . . + an .y a-(83) 1a2 · · · an n holds, with equality if and only if a1 a2 = · ·· = an . PROOF . If a k = 0 for some k = 1 , 2, . . . , n, then the statement is clear, since the right-hand side of (83) vanishes. Let therefore a k 0, 1 < k < n. Since the inequality (83) is homogeneous (see 2. 7) , the n-tuple a 1 , a2 , . . . , an can be normalized such that a 1 a2 · · ·an = 1 . But then the assertion (83) is identical with the result that was derived in Section 6.6.(v) . This completes Theorem.
-------
> �,,-
_
-
=
>
0 the proof of the AM-GM inequality. Other proofs of follow from the results of 3.4.(iv) and 6.7.(vi). The significance of the inequality has stimulated the interest of mathematicians in finding proofs by different approaches and different methods see, e.g. ,
Remark. (
(83)
(83)
[1], [3]).
8. 2 Applications of the AM-GM Inequality The proofs of numerous inequalities follow from comparing the sum and the product of appropriate n-tuples of numbers, such as in the AM-GM In some situations (examples (i) - (vii)) we require a inequality
(83). lower bound for the sum A1 + A2 + . . · + An n y'A1 A2 . . · An , (84) while in others (examples (viii) -(xi)) we need an upper bound for the product (85) >
·
152
2. Algebraic Inequalities
We emphasize that the inequalities (84) and (85) hold for an arbitrary n tuple of nonnegative numbers with equality if and only if the numbers are all equal. We have already seen the significance of (84) and (85) for n = 2 in Sections 4.3 and 4. 5. (i) Show that if 0, then 0, and that this inequality is sharp when 1=
, A , A , , A 1 n 2 A1 , A2 , . . . , An > all -3a5 + a4 + 1 > a a 1. SOLUTION. By (84) , the triple of numbers a ll , a4 , 1 satisfies all + a4 + 1 > 3 -?'a ll · a4 1 = 3a5 , and equality occurs and only if a ll = a4 = 1, that is, a = 1. This implies 0 the assertion. .
.
•
·
•
if
a > 0 and for each integer n > 1 we have a2 n 2n < 1 · 1 + a + a + · + a 2n + 1 SoLUTION. By (84) the numbers 1, a, a2 , , a2n satisfy 1 + a + a2 + · + a2n > (2n + 1) · 2n+�1 · a2 · a2n = (2n + 1) · 2n+\lan (2n+l) = (2n + 1)an , (ii) Show that for each
· ·
•
·
•
.
·
·
·
0
which is equivalent to the desired inequality.
2b2t? + p a ab + be+ pabc a, b, e SOLUTION. If we set a = b = e = 1, we obtain the condition p < 4. On the other hand, by (84) the four numbers a2 b2 t? , ab, be, satisfy a2b2 e2 + ab + be+ > 4 · �a2b2e2abbcca = 4abc. (iii) Determine the largest value E 1R for which the inequality holds for arbitrary numbers E IR+. ca > ca
ca
0
Hence the desired largest value of p is equal to 4.
b
(iv) Show that if 0 < < a, then we have the inequality
a + (a � b)b > 3'
SOLUTION. By (84) , applied to the number triple we have
b, a-b, and [(a-b)b] - 1 ,
a + (a � b)b = b + (a - b) + (a � b} b > 3 . 3 b(a - b) . (a � b} b = 3.
D
8 Inequalities Between Means
153
a > 0 satisfies the inequality a4 + 9 > 4 · lOa 5 SOLUTION. We apply (84) with an appropriate quadruple of numbers: a4 + 9 = a3 + 3 + 3 + 3 > 4 · a3 · 3 · 3 3 = 2 · lOa 10 lOa lOa lOa 10 lOa lOa lOa 5 {/27. 0 Now it suffices to note that {127 > 2, since 27 > 16 24 • (vi) Show that a , b E JR+ , then 3a3 + 7b3 > 9ab2 • SOLUTION. The term ab2 on the right is the geometric mean of the three numbers a3 , b3 , b3 • We therefore try to use the estimate (v)
Show that each
·
4
=
if
7 + PbP P P1 · 2 2 3 {/3P1P2 > 9, 3 P1P2 > 9. Pl P2 . (86) Pl P2 · (vii) Show that for any a, b, c E JRt we have the inequality 2(a + b + c)(a2 + b2 + �) > a3 + b3 + c3 + 15abc. SOLUTION. Upon expanding the left-hand side and subtracting the sum a3 + b3 + from both sides of the inequalitY we obtain a3 + b3 + c3 + 2(a2b + a2 c + b2a b2c + c2a + c2b) > 15abc. The left-hand side L of this last inequality is a sum of 3 + 2 · 6 = 15 terms; by (84) we have L > 15 · 1{!a3 b3 c3(a2 b) 2 (a 2 c) 2 (b2 a) 2 (b2 c) 2 (c2 a) 2 (c2 b) 2 = 15abc. 0 (viii) Prove the inequality 1 · 3 · 5 · (2n - 1) nn for all n > 1 . SOLUTION. Applying (85) to the numbers 1, 3, 5, . . . , 2n - 1, we obtain 1 . 3 . 5 . . . (2n - 1) [ 1 + 3 + 5 + � + (2n - 1) r = c:: r = n" ' since 1 + 3 + 5 + · · + (2n - 1) = n2 (see Chapter 1, Section 2). 0
< for appropriate positive numbers We require that and that is, We easily see that both conditions are satisfied by = = This solves the problem. It is left to the reader to D explain why the estimate is �'most valuable" in the case = ·
e
+
·
k ; 1 .
SoLUTION. We consider the system of the nk
k(k+l)
2
•
= � (k2 + k) numbers
(88)
1 1 1 1, ! ' ! ' ! ' ' ' 1 ' 1 ' 2 2' 3 3 3 ' " k k ' ' ' ' k' Since their sum is equal to k, by (85) we have 2 1 . _!_ . _!_ . _.!_k < 2 2 33 k - nk k+1 By writing down the opposite inequality for the reciprocals, we obtain the o inequality (88). . .
( �) n"' = ( ) n"'
·
8 Inequalities Between Means
155
8. 3 Exercises Using (84) , prove the inequalities (i)- (vii) , where a, b, c E :JR+ :
a4 + a3 - 8a2 + 4a + 4 > 0. a2 (b + c) + b2 (c + a) + c2 (a + b) a4 + 2a3 b + 2ab3 + b4 > 6a2 b2 . 2 vfci + 3 · {/b > 5 · {lab. a3 ; b6 > 2 (v) 3ab - 4.
(i} (ii) (iii) (iv}
;
>
6abc.
> 3. (vi) a + a .!!.±.! (vn•• } an - 1 > n a - a-2- , if a > 1 and n E N.
(
2
n-1)
Using (85) , prove the inequalities (viii)-(xv) , where a, b, c, d k, m E N: a + 2b � 3c + 4d ) (viii) ab2
(ix) (x) (xi) (xii) (xiii)
E
:JR+ and
10 .
(k2 + 3k) (k2 + 5k + 6) k+2 < (k2 + 5k + 4) k+3 • 27a2 b < 4(a + b) 3 • 432ab2c3 < (a + b + c) 6 • (1 - a) 5 (1 + a) (l + 2a) 2 < 1. (b - a) k (b + ka) < bk+ l , if a < b.
(xiv) (xv) (xvi) Show that for arbitrary a�, a2 , . . . , ak E :JRt we have
8.4 Further Applications of the AM- GM Inequality In order to solve the following problems, whic� are somewhat more difficult than those in 8.2, we will usually have to apply the AM-GM inequality several times over. For clarity we will use the symbols An and gn for the two means introduced in (81) and (82). It is important that the reader be well acquainted with these symbols, for instance through verifying the
2. Algebraic Inequalities
156
ak , bk , c E JRt) : a1 + a2 + · · · + an = · An (aba2 , . . . ,an ), a1 a2 · · ·an = Qn (a�,a�, . . . ,a�) , An (cab . . . ,can) = C· An (ab . . . ,an), Qn (cab . . . , can) = c · Qn (ab . . . , an ), Qn (ab . . . , an ) Qn (bb · · · , bn) = Qn (al bb · , anbn), 1 . . , _!_) = gn (_!_, a1 an gn (a1 , . . . ,an) (ak > 0, 1 < k
9abc.
SoLUTION. We prove the statement as follows: L=
9Aa (a, b, c) A3 (a2 , b2 , c2) > 9Q3 (a,3b, 3c) Q33 (a2 , b2 , c2 ) 9Q3 (a , b , c ) 9abc = R. =
=
9
A different proof follows from the AM-GM inequality applied to the numbers that are obtained by multiplying out the left-hand side of the 0 given inequality. (ii) Tlie follows:
harmonic mean
of the numbers
a1 , a2 , . . . , an E IR+ is defined (89)
as
Show that the harmonic and geometric means satisfy the inequality
'Hn (ab a2 , . . . , an ) < Qn (a l , a2 , . . . , an ), (90) with equality if and only if a 1 a2 = · · = an . SOLUTION. By definition (89) and by the AM-GM inequality we have ·
=
with equality if and only if ... . =
an
1/al I/a2 · · · 1/an , that is, a1 a2 =
=
=
=
=
8 Inequalities Between Means
1 57
(83)
and (90) we obtain an inequality between As a consequence of the ¥ithmetic and the harmonic means; this also follows from the inequality 46). 0
(
a, b, E JRt we have �+ �+ -Vca - 1 32 (a + b + )
(iii) Show that for arbitrary
c
2 and S = a 1 + a2 + · · · + an . (iv) Show that for arbitrary
SOLUTION. To obtain a lower bound for the left-hand side, we apply the AM-GM inequality twice: L
= n . ( S -S al , . . . , S -San ) n(in ( S -S al , . . . , S -San ) nS . gn ( S -1 a1 , . . . , S -1 an ) Qn (S - a1nS, . . . , S - an ) n2 S nS ----= ( S - a� , . . . , S - an ) ( S - a1 ) + · · + (S - an) >
An
=
=
> - An
n2 S n2 n2 S 0 nS - (a1 + · · · +an ) nS - S n - 1 ( ) Suppose that the numbers x�, x2 , . . , xn are all greater than -1 , and that their sum is equal to zero. Prove the inequality X l + X2 + · · + Xn < 0. --X l + 1 X2 + 1 · Xn + 1 v
·
.
158
2. Algebraic Inequalities
SOLUTION. After rewriting the right-hand side we apply the AM-GM inequality twice:
L (X 1X+1 +1) 1- 1 + . . . + (XnX+n +1)1- 1 = n ( X1 1+ 1 + . . . + Xn 1+ � ) = n - n · ( x1 � 1 ' · · Xn � 1 ) n - n . gn ( X1 � 1 ' · · . ' Xn � 1 ) n -------:n - Qn (x 1 + 1 ,n. . . ,Xn + 1) - n - -An,-(x1 + 1 , . . . ,Xn + 1) - n - (x1 + 1) + ·n·2· + (xn + 1) = n - (x1 + · · n+2 Xn ) + n 2 n n - O +n 0 (vi) Show that for each integer n > 2 we have the inequality 1 + !.2 + !.3 + · · + .!n > n( v'n + 1 - 1). SoLUTION . If we denote the left-hand side by Ln , then Ln +n = (1 + 1) + G + 1) + G + 1 ) + · · · + G + 1 ) = 21 + 32 + 34 + · · · + n+n 1 = n A ( 21 ' 32 ' . . . ' n+n 1 ) 2 3 n+1 (n + 1)! n · n + 1 , > nQn ( l ' 2 , . . . , n ) = n · n! 0 which, after subtracting the number n , gives the desired inequality. _
=
.4,
·
3 ( 1 + J3) abc. (v) Show that for arbitrary a 1 , a2 , . . . , an n > 3, we have a 1 - a3 + a2 - a4 + · · · + an-1 - a1 + an - a2 >- 0 . --a2 + a3 a3 + a4 an + a1 a1 + a2 (vi) Show that for each integer n > 2 we have the inequality 1 ) 1 + .!:.2 + !.3 + · · · + .!.n < 1 + n ( 1 - n y�n- . (vii) Show that for arbitrary numbers a, bb �, . . . , bn we have b1-b2 + ab2 -b3 + · · · + abn-bl > 2 n a--bn + b1 - 2(b1 + � + · · + bn) . b1 + � � + b3 (viii) Suppose that the product of the positive numbers a1 , a2 , . . . , ak is equal to 1. Show that for each n we have (n + a1 )(n + a2 ) · · · (n + ak ) > (n + 1 ) k . (ix) Suppose that the sum of the positive numbers a1 , a2 , . . . , an is equal to 1. Show that for each real number b > 1 we have the inequality a1 + a2 + . . . + an -> n b - an bn - 1 b - a 1 b - a2 (x) Suppose that the sum of the positive numbers a 1 , a2 , . . . , an is equal to 1. Prove the inequality E
ca
E IR+ ,
E IR+
·
EN
Prove the inequalities (xi) -(xii ) for arbitrary
ab a2 , , an •
•
•
E
JRt:
162
8. 6
2. Algebraic Inequalities
Weighted Means
1 , x2 , . . . , Xn , we weight coefficients Pi weighted arithmetic mean Aw n (X1 J X2 , · · · ,Xn) - P1 XP11 ++PP2X22++· ·· ·· ++PPnnXn · It is convenient to normalize the weight coefficients Pi in such a way that their sum is equal to 1. We achieve this by introducing new coefficients V . - Pl + P2 p+· + Pn (1 < j < n), since clearly v1 + v2 + · + Vn = 1 ; the relation for the mean then simplifies to (92) + VnXn · A: ( X2 , , Xn ) = V1 X 1 + V2 X2 + We also define the weighted geometric mean
Sometimes, in determining the mean value of numbers x associate with different numbers xi different levels of "importance," indi cated by E :JR+ . The is then defined by the expression ·
_
J
J -
· · ·
·
xb
·
•••
·
· ·
(93)
v1 = v2 · Vn =
then the question arises whether the AM-GM inequality also holds in this �, more general situation. (If in and (93) we set · we get formulas and for the original nonweighted means An and The following theorem gives an affirmative answer.
gn .)
(92 ) (81) (82)
=
·
=
8. 7 The AM- GM Inequality for Weighted Means
Let the sum of the positive numbers v be equal to 1. , v , , V 1 n 2 Then for arbitrary numbers a1 , we have the inequality
Theorem.
a2 , . . . , an E
:JR+
with equality if and only if a1 a2 = = an =
· · ·
.•.
(94) .
v1 , v2 , , Vn mi = kvi (1 < j
(cb )af(a+b) . (c-a ) bf(a+b) = 1 , a+b a+b which upon multiplying by a + b gives the desired inequality. 0 holds for arbitrary
use
=
=
2:
&
2. Algebraic Inequalities
1 64
(iii) Show that if 0 < x < 1, then
1 < XX < X2 - X + 1. 2-x
(97)
SOLUTION. If 0 < x < 1, we can use (94) with n = 2 and set v 1 a1 = x, v2 = 1 - x, and a2 = 1. Since a 1 < a2 , we obtain the strict inequality x . x + (1 - x) . 1 > xx . 1 1 - x , which is the right-hand part of (97). If we interchange the numbers a 1 and a2 , we get x · 1 + (1 - x) x > p= · x l - x , 0 that is, 2x - x2 > x 1 x , and this gives the left part of (97) . =
·
-
(iv) Show that if a, b, c are side lengths of a triangle, then a-b c c-a b b-c
(1 + )a . (1 + a
b
) . (1 +
c
)
< 1.
SOLUTION. We use the inequality (94) with n = 3 for the coefficients
v1 - a + ab + c ' V2 - a + bb + c' ---
_
_ _ _ _
and
c V3 = a + b + c .
and
a3 = 1 + a -c b
(98)
'.fhen for the (positive!) numbers
a1 = 1 + b -a e , a2 = 1 + c -b a ,
-
we obtain the inequality
c c c a 1 a2 a3 < v1 a 1 + v2 a2 + v3 a3 = a + b - + ab ++ b -+ ac + + a - b = 1 , 0 which upon raising to the power a + b + c gives the desired inequality. v1
v2
v3
(v} Show that for arbitrary numbers a, b, c E JR+ we have the inequality
SOLUTION. We use again (94) with However, this time we set a1 = 1/a, obtain since
= 3 and the coefficients (98) . a2 = 1/b, and a3 = 1/c; we thus n
1 V1 - V2 = V3 = -;; b c a + b + c ·
If we write the opposite inequality for the reciprocals and then raise both 0 sides to the power a + b + c, we obtain the desired inequality.
Inequalities Between Means
8
165
(vi) Prove the general Bernoulli inequality: If x E IR, p E JR+ , x > -1 , and
p 1= 1 , then
(1
+ x)P > 1 + px
(p > 1),
(99)
(O < p < 1),
(100)
respectively (1
+ x)P < 1 + px
(99) or (100) if and only if x = 0 (compare with 3.3.(iii)). SoLUTION. If 0 < p < 1 , we substitute in (94) n = 2, v 1 = p, v2 = 1 - p, a1 = 1 + x > 0, and a2 = 1; we thus obtain p(1 + X) + (1 - p ) 1 > ( 1 + X )P 1 l -p ,
with equality in
·
·
that is, inequality ( 100) , with equality if and only if a1 = a2 , which means X = 0. Now let p > 1. If 1 + px < 0, we obviously have a strict inequality in (99). If 1 + px > 0, then by the already proven inequality (100) we have 1
(1 + px) P
< 1 + -1 px = 1 + x. p ·
(Equality occurs only for x = 0.) From this we obtain both sides to the power p.
(99)
after raising
0
8. 9 Exercises (i) Show that the weighted means (92 ) and (93) of the positive num
bers x 1 , . . . , Xn have the fundamental mean property: Their value is at most (at least) equal to the largest (smallest) of the numbers X 1 , • . • , Xn · (ii) Show that if a, b,p E JR+ and p < 1 , then (a + b)Pa 1 -P < a + pb.
a, b, c, d E JR+ : adb- c + bdc- a + cda- b > a + b + c. acbd (c + d) c+d < ccdd (a + b) c+d . a + b + c a+b+ c > (b + c) a (c + a) b (a + b) c . 3 2a+b+ c Show that if the sum of the positive numbers a, b equals 1 , then the inequalities ab ba < 2ab and � < aa bb < a2 + b2 hold. When does
Show that (iii) -(v) hold for arbitrary
(iii) (iv) (v) (vi)
(
)
equality occur in these inequalities? (vii) Prove the inequality acbl - c + (1 - a) c (1 - b) 1 - c positive numbers a, b, c less than 1 .
1, then for arbitrary numbers x�, x2 , . . . , Xn , Yl , y2 , o . . , Yn E JR+ we have
=
Proceed as in 5.5o (iv) for the proof of the triangle inequality ( 45)0 (xiii) Let c, p� , p2 , . . . , Pn E JR+ and suppose that the positive variables x 1 , x2 , . . . , Xn satisfy the condition x 1 +x2 + · o +xn = c. Show that the expression af1 1 �2 . 0 xPn ·
2
0
n
attains its largest value if and only if
X1 P1
X2 P2
Xn Pn
- = - = ··· = -
8. 10 Power Means Up to the end of Section 8 we will assume that a 1 , a2 , . . . , an are arbitrary positive numbers. We will derive several useful inequalities between the sums a} + a2 + · · · + a;;_ for various r E JR. For this purpose it is convenient
8
Inequalities Between Means
to introduce the so-called power means. By a mean of degree numbers a� , a2 , . . . , an we understand the value
r
167
of the
(102) which is meaningful for each r E IR \ {0}. We note that for r = 1 ( resp. r = -1) this is the arithmetic ( resp. harmonic ) mean (compare with (81) and (89) ) . The following theorem deals with the comparison of the power means and the geometric mean.
8. 1 1
Power Means of Positive and Negative Degrees
Theorem.
If r < 0 < s, then
with equality anywhere in (103} if and only if a 1 = a2 = · · · = an . PROOF .
inequality
We obtain the right part of (103) if we raise the AM-GM
to the positive power 1 / s; the left part is obtained by raising the AM-GM inequality Yn (af, a�, . . . , a;;.) < An(af, a�, . . . , a:;_) to the negative power 0 1/r. This also implies the statement about equality in (103) .
Remark. Using mathematical analysis one can show that Therefore, the geometric mean is often called mean We now state the main result on power means.
of degree zero.
8. 12 Inequalities Between Power Means Theorem.
Let r, s E IR, r < s, and rs 1= 0 . Then (104)
with equality if and only if a1 = a2 = · · · = an . The inequality (104) for integers r = 1 and s
earlier inequality (80) in Exercise 7.8. (v). PROOF.
and
r
1
is equivalent to the
By Theorem 8. 1 1 it suffices to examine only two cases: 0 < s < 0. First let 0 < r < s. If we set p = s/r > 1 , then a�
r<s
=
�,
168
2. Algebraic Inequalities
where bk = a'k ( 1 < k < n ) . Therefore, upon raising to the power inequality ( 104) has the form
bl + � �
0
0
0
+ bn
1 holds for arbitrary positive numbers bk , with equality if and only if b1 = � = · · · = bn · Since we are dealing with a homogeneous inequality in the variables b 1 , b2 , . . . , bn (see 2. 7), we may assume that (106) This means that for the numbers
Xk = bk - 1 we have
Since X k > -1 and p > 1 , Bernoulli's inequality (99) implies 1
2 can be expressed as a primes. This expression is unique up to the order of the factors. is a prime, then by ''product" we mean just this prime).
product of (If n itself
First we prove by induction that every n > 2 can be decomposed into a product of primes. If n = 2, then the product is just the prime 2. We now assume that n > 2 and that we have already shown that any n', 2 < n' < n, can be written as a product of primes. If n is a prime, then the product consists of just this one prime. If n is not prime, then it has a divisor d, 1 < d < n. If we set c = n/d, then also 1 < c < n. The induction hypothesis implies that c and d can be expressed as products of primes, and therefore their product c · d = n can also be expressed in this way. PROOF.
3. Number Theory
186
To prove uniqueness, we assume that we have equality between the prod are primes such ucts ··· · · · where m s. We use that and 1 ··· induction on m to show that m s, would In the case m = 1 , is a prime. If s > 1 , then have a divisor such that 1 > 1 ) , which is (since ·· impossible. Hence s 1 and = We now assume that m > 2 and that the assertion holds for m - 1 . Since = divides the product the prime · ··· · ·· which by 2.2 is possible only when divides a qi for an appropriate i {1, 2, . . . , s } . But qi is a prime, so it follows that i (since > 1). In complete analogy it can be shown that for an appropriate j { 1, 2, . . . , m}. Hence
P1P1 P 5 there are at least two primes between n and 2n," which generalizes Chebyshev's theorem: "For any integer n > 3 there is at least one prime between n and 2n - 2." The proof uses only elementary methods, but is relatively long ( pp. 131-137) . Example 2.7. (ii) is generalized by Dirichlet's theorem on primes in arith metic progressions: "If a and m are relatively prime natural numbers, there exist infinitely many natural numbers k such that mk + a is prime." This is a deep theorem, and its proof requires methods that go well beyond elementary number theory.
2. 8 Exercises (i} Find ( a + b, ab) , where a, b are relatively prime integers. (ii) Given integers a and b, find ( a + b, [a, b]) . (iii) Show that if a > 4 is a composite integer, then a I ( a - 1 ) !. * (iv) Show that if p > 5 is a prime, then there is no m E N such that (p - 1 ) ! + 1 = pm . ( v) Show that there exist infinitely many primes of the form 4k + 3, where k E No. (vi) Show that a prime of the form 22n + 1 , where n E N, cannot be expressed as a difference of fifth powers of two natural numbers. *(vii) Suppose that a and n are natural numbers with the property that there exists an s E N such that (a - 1 ) 8 is divisible by n. Show that 1 + a + · · · + an- l is also divisible by n.
188
3. Number Theory
(viii) Show that if a, b E Z are such that (a, b) = 1 and 2 I ab, then (a + b, a2 + b2 ) = 1. 2. 9 Exponents in a Factorization To any prime p and any natural number n there is, according to The orem 2.5, a uniquely determined number of appearances of p in the decomposition of n into prime factors (if p does not divide n, we take this number to be zero) . We denote this number by the symbol Vp (n). For a negative integer n we set vp (n) = vp ( -n) . According to 2.6.(i) we can characterize this notation by saying that vp (n) is the exponent of the largest power of the p�ime p that divides n, or by n = pvp(n) · m, where m is an integer not divisible by p. From this it easily follows that any nonzero integers a, b satisfy
Vp (ab) = Vp (a) + Vp (b), vp (a) < vp (b) 1\ a + b 1= 0 ==? vp (a + b) > vp (a ) , Vp (a) < Vp (b) ==? Vp (a + b) = Vp (a), Vp (a) < Vp (b) ==? Vp ((a, b)) = Vp (a) 1\ Vp([a, b]) = Vp (b).
(8) (9) (10) (11)
2. 10 Example Show that any natural numbers a, b, c satisfy
([a, b], [a, c], [b, c]) = [(a, b), (a, c), (b, c)]. SOLUTION. By 2.5, we are done if we can show that vp (L) = vp (R) for any prime p, where L, resp. R, denotes the term on the left, resp. on the right. Now let p be any prime. In view of the symmetry on both sides, we may without loss of generality assume that vp (a) < vp (b) < vp (c). By (11) we have vp([a, b]) = vp (b), vp([a, c]) = vp([b, c]) = vp(c); vp ((a, b)) = vp ((a, c)) = vp (a), vp ((b, c)) = vp (b), and this implies vp (L) = vp( b) 0 vp (R), which was to be shown.
2. 1 1 Exercises (i} Give new proofs of 1.15.(i)-(v) , using the method of 2.10. (ii} Show that any natural numbers a, b, c, such that (a, b) = 1 , satisfy (iii)
(ab, c) = (a, c) · (b, c). Show that if for m, n E N the number yin is rational, then it is in
fact a positive integer. (iv) Show that if n, r, s E N, ( r, s) = 1 and \fiiT E N, then n = m8 for an appropriate m E N.
3 Congruences
189
(v ) Show that if for some positive rational numbers a and b the number
Va + v'b is rational, then ya and v'b are also rational. * (vi) Let a, b, c, d be integers such that ac, bd, be + ad are all divisible by the same natural number m. Show that then the numbers be and ad are also divisible by m. ·
3
Congruences
The concept of congruence was introduced by Gauss . Although it is a very simple notion, its importance and usefulness in number theory are enor mous. This is particularly obvious in that even complicated arguments can be presented in a concise and clear manner .
3. 1
Definition of Congruence Definition. If the two integers a, b have the same remainder r upon divi sion by the natural number m, where 0 < r < m, then a and b are called congruent modulo m, and we write
a = b ( mod m) . Otherwise, we say that a and b are not congruent, or incongruent, modulo m, a ¢. b ( mod m) .
3.2 Equivalent conditions
E
E
For any numbers a, b Z and m N the following conditions are equivalent:
Theorem. (i) (ii) (iii)
a = b ( mod m) , a = b + mt for an appropriate t E Z, m I a - b.
I. If a = q1 m + r and b = q2 m + r, then a - b = (q1 - q2 ) m, and therefore ( i) implies (iii ) . II. If m I a - b, then there exists a t Z such that m · t = a - b, that is, a = b + mt, and so ( iii) implies ( ii) . III. If a = b + mt, then it follows from b = mq + r that a = m(q + t) + r, so a and b have the same remainder r upon division by m, that is, a = b 0 ( mod m ) ; hence (ii) implies ( i) . PROOF .
E
190
3. Number Theory
3. 3 Basic Properties of Congruences It follows directly from Definition 3.1 that congruence modulo .m is reflexive (i.e., a = a (mod m) for any a E Z), symmetric (i.e., for each a, b E Z, a = b (mod m) implies b = a (mod m)) and transitive (i.e., for each a, b, c E Z, a = b (mod m) and b = c (mod m) imply a = c (mod m)). We now prove some further properties: (i) Congruences with the same modulus can be added . Any summand can be moved, with the opposite sign, fro:rp. one side of the congruence to the other. To any side of the congruence we can add an arbitrary multiple of the modulus. If a 1 = b1 (mod m) and a2 = b2 (mod m), then by 3.2 there exist h , t 2 E Z such that a 1 = b 1 + mh, a2 = b2 + mt2 . Then a 1 + a2 = b1 + b2 + m(t 1 + t2 ), and again by 3.2 we have a 1 + a2 = b1 + b2 (mod m). If we add a + b = c (mod m) and the obviously valid congruence -b = -b (mod m), we obtain a = c - b (mod m). If we add another obviously valid congruence, mk = 0 (mod m), to a = b (mod m), then we get a + mk = b D (mod m). PROOF.
(ii) Congruences with the same modulus can be multiplied together. Both sides of a congruence can be raised to the same positive integer power. Both sides of a congruence can be multiplied by the same integer. If a 1 = b1 (mod m) and a2 = b2 (mod m), then by 3.2 there exist t 1 7 t2 E Z such that a 1 = b1 + mt1 and a2 = b2 + mt 2 . Then PROOF .
a 1 a2 = (b1 + mt 1)(b2 + mt 2 ) = b 1 b2 + m(hb2 + b1 t 2 + mt 1 t 2 ), and from 3.2 we obtain a1a2 = b 1 b2 (mod m) . If a = b (mod m) , we prove by induction on n that an = bn (mod m) . For n = 1 there is nothing to show. If an = bn (mod m) for some fixed n, we multiply this congruence with a = b (mod m) and obtain an . a = bn . b (mod m), hence an+ l = bn+ l (mod m), and this is the assertion for n + 1 . The proof by induction is now complete. If we multiply the congruences a = b (mod m) and c = c (mod m), we o obtain ac = be (mod m).
(iii) Both sides of a congruence can be divided by a common divisor, as long as this divisor is relatively prime to the modulus. PROOF. We assume that a = b (mod m) , a = a1 d, b = b 1 . d, and ( m, d) = 1. By 3.2 the difference a - b = ( a1 - b 1 ) d is divisible by m. Since (m, d) = 1, by 1. 14.(ii) the number a1 - b1 is also divisible by m, and with 0 3.2 it follows that a1 = b 1 (mod m) . ·
·
(iv) Both sides of a congruence and its modulus can be simultaneously multiplied by the same natural number.
3 Congruences
191
If a = b (mod m), then by 3.2 there exists an integer t such that a = b + mt; hence for e E N we have ae = be + me · t, and using 3.2 again 0 we get ae = be (mod me) . PROOF.
(v) Both sides of a congruence and its modulus can be divided by a positive
common divisor.
We assume that a = b (mod m), a = a 1 · d, b = b 1 · d, m = m 1 · d, where d E N. By 3.2 there exists a t E Z such that a = b + mt, that is, a1 · d = b 1 · d + m 1 dt, which implies a1 = b1 + m 1 t; this means, by 3.2, that 0 a 1 = b 1 (mod m 1 ). PROOF.
(vi) If the congruence a = b holds with various moduli m 1 ,
it also holds modulo numbers.
[m 1 ,
•
•
•
, mk ] ,
•
•
.
, mk , then
the least common multiple of these
(mod m 1 ), a = b (mod m2 ), . . . , a = b (mod mk ), then by 3.2 the difference a - b is a common multiple of the mod uli m 1 , m2 , , mk , and is thus divisible by the least common multiple 0 [mb m2 , . . . , mk], which implies a = b (mod [m1, . . . , mk] ). PROOF.
If
.
a = b
•
•
(vii) If a congruence holds modulo m, then it also holds modulo d, where
d is any divisor of m.
If a = b (mod m) , then a - b is divisible by m, and therefore also D by the divisor d of m, which means that a = b (mod d). PROOF.
(viii) If one side of a congruence and its modulus are divisible by some
integer, then the other side will also be divisible by this integer.
We assume that a = b (mod m), with b = b1d, m = m 1d. Then by 3.2 there exists a t E Z such that a = b+ mt = b 1 d+m 1 dt = (b 1 +m 1 t)d, 0 and thus d I a . PROOF.
We remark that we have already used several properties of congruences without having taken note of this fact. For instance, 1.4. (i) can be rewritten as "if a = 1 (mod m) and b = 1 (mod m), then also ab = 1 (mod m) , '' which is a special case of the property in (ii) . This is, of course, not a coincidence. Any statement using congruences can easily be rewritten in terms of divisibilities. The usefulness of congruences lies therefore not in the fact that one could solve problems that could not be solved otherwise, but in the fact that they provide a very convenient notation. By making consistent use of this notation we can significantly simplify not only the exposition, but also certain arguments. This is a typical situation: Appropriate notation plays a very important role in mathematics.
192
3. Number Theory
3.4 Examples (i} Find the remainder when 5 20 is divided by 26. SOLUTION. Since 52 = 25 = -1 (mod 26) , we have by 3.3.(ii),
5 20 = ( -1) 10 = 1 ( mod 26), 0 so the desired remainder is 1. (ii) Show that for any n E N, 3�+ 2 + 16n+ 1 + 23n is divisible by 7. SOLUTION. Since 37 = 16 = 23 2 ( mod 7), by 3.3. (ii) and (i) we have 3�+2 + 16n+ 1 + 23n = 2n+2 + 2n+ 1 + 2n = 2n (4+2+1) = 2n 7 = 0 ( mod 7), =
·
0
which was to be shown.
(iii) Show that n = (8355 + 6) 18 - 1 is divisible by 112.
SOLUTION. We factor 1 12 = 7 16. Since (7, 16) = 1, it suffices to show that 7 1 n and 16 1 n. We have 835 = 2 ( mod 7), and thus by 3.3, ·
38 18 - 1 = 3 18 - 1 ( mod 7); = 276 - 1 = ( - 1) 6 - 1 = 0 hence 7 1 n. Similarly, 835 = 3 ( mod 16) , and thus n = (35 + 6) 18 - 1 = (3 81 + 6) 18 - 1 = (3 1 + 6) 1 8 - 1 1 = 9 8 - 1 = 819 - 1 = 1 9 - 1 = 0 ( mod 16); hence 16 I n . Together, 1 12 1 n , which was to be shown. n = (25 + 6) 18 - 1
=
·
·
0
(iv) Show that no number of the form 8k ± 3, k E N, can be expressed in the form x2 - 2y2 for any integers x, y. SOLUTION. Assume that x2 -2y2 = 8k±3 for some x, y E Z. Then x must be an odd number, and therefore x2 = 1 ( mod 8) by 1.5 . ( ii). If y is also odd, then x2 - 2y2 = 1 - 2 = -1 ( mod 8). If y is even, then 2y2 = 0 ( mod 8), and thus x2 - 2y2 - 1 ( mod 8). In neither case do we have x2 - 2y2 = ±3 0 (mod 8), and this is a contradiction.
(v} Show that for any prime p and any a, b E Z we have
aP + QP = (a + b) P (mod p). SOLUTION. By the binomial theorem we have
(a + b)P = aP +
(i) ap- 1 b + (;) aP- 2b2 + . . . + � P 1 ) ab"- 1 +
b" .
3 Congruences
For any k E { 1, . . . , p - 1 } we have implies the assertion.
(�) = 0
193
(mod p) by 2.3. (iv ) , and this D
(vi) Show that for any natural number m and any a, b E
(mod mn ), where n E N, we have am = bm (mod mn+ 1 ).
Z such that a = b
SOLUTION. By identity (5) in Chapter 1 we have
am - bm = (a - b)(am- 1 + am-2b + · · · + abm- 2 + bm - 1 ).
(12)
E N it follows that m I mn, and so by 3.3. (vii) we have a = b ( mod m). Hence all the summands in the right-hand expression in parentheses in ( 12) are congruent to am - 1 modulo m, and thus
From
n
am- 1 + am-2 b + + abm- 2 + bm - 1 = m · am- 1 = 0 ( mod m). Therefore, am- 1 + am- 2 + · · · + abm- 2 + bm- 1 is divisible by m. From a = b ( mod mn ) it follows that mn divides a - b, and thus mn+ 1 di vides the product, which by (12) leads to the conclusion that am = bm D (mod mn+ 1 ). ·
·
·
3. 5 Exercises Show that (i) the number 260 + 730 is divisible by 13; (ii) for any n E N, 722n+ 2 - 472n + 282n- 1 is divisible by 25; (iii} for any k, m, n E N, 55k+ 1 + 45m+2 + 35n is divisible by 11; (iv) for any integers a, b, the congruence a2 + b2 = 0 ( mod 3 ) implies a = b = 0 (mod 3) ; (v) for any integers a, b, the congruence a2 + b2 = 0 ( mod 7) implies a = b = 0 ( mod 7); (vi) there exist integers a, b such that a2 + b2 = 0 ( mod 5), while a = b = 0 ( mod 5) does not hold; (vii) for any integers a , b, c, the congruence a3 + b3 + c3 = 0 ( mod 9) implies abc = 0 ( mod 3) ; (viii) for any integers a�, a2 , . . . , a7, the congruence a� +a� + · · · +a� = 0 ( mod 9) implies a 1 a2 · · · a7 = 0 (mod 3) ; * (ix) no integer a satisfies a2 + 3a + 5 = 0 ( mod 121). (x) Find all natural numbers n such that n · 2n + 1 is divisible by 3. (xi) Determine whether there exists a natural number n such that the difference 4n - 2n- 1 is the third power of an integer.
194
3. Nwnber Theory
1, 2, . . . , 12 on the circum ference of a circle in such a way that for any three neighbors a, b, and c, with b between a and c, the number b2 - ac is divisible by 13. Show that 19 sn + 17 is composite for any natural number n. Show that (n - 1) 2 is a divisor of nn - n2 + n - 1 for any natural number n. Let a, ao , a h . . . , an be integers (n E No ). Determine whether it is
(xii) Show that one ca'il place the numbers
* (xiii) * (xiv) * (xv)
·
true that the integer
(a2 + 1) 0 a0 + (a2 + 1) 3 a 1 + (a2 + 1) 6 a2 + · · · + (a2 + 1) 3k ak is divisible by a2 + a + 1 (resp. a 2 - a + 1) if and only if ao - a1 + a2 · · · + ( -1) k - l ak- 1 + ( -1) k ak ·
-
is divisible by a2 + a + 1 (resp. a2 - a + 1).
3. 6 Euler's 1) are distinct primes and n�, . . . , nk E N, we define � (m) as follows: •
•
•
(13) If, in addition, we define �(1) = 1, we have a function � : N ---? N. We can easily convince ourselves that (13) can also be written as
( : J (1 - : J
� (m) = m · 1 -
···
and that for any relatively prime natural numbers m 1 , m2 we have � (m1 m2 ) = � (m1 ) � (m2 ). The reader should verify that the condition of relative primality of m 1 , m2 is indeed necessary. The clearest and most sug gestive representation of Euler's �function is given later by the assertion in 3.10.
3. 7 Examples (i) Show that for any n
E N we have � (4n + 2) = � (2n + 1).
SOLUTION. Since 2n + 1 is odd, the numbers 2 and 2n + 1 are relatively prime, and thus we have �(2 (2n + 1)) = � (2) · � (2n + 1). We are done if we realize that � (2) = L (More generally, for each prime p we have D �(p) p - 1.) ·
=
195
3 Congruences
E
m N for which cp( m) is odd. SOLUTION. We have cp(1) = cp(2) = 1. If m > 3, then m is divisible by an odd prime p, or m = 2n , n > 2 ( this follows from the decomposition of m into a product of primes) . In the first case, cp ( m) is divisible by the even number p - 1, and in the second case we have cp(m) = 2n- l , so cp(m) is always even. Therefore, the number cp(m) is odd only when m = 1 or D m = 2. (iii) Solve the equation cp(5 x ) = 100, where x is a natural number. SOLUTION. By (13) we have cp(5x ) = 4 · 5x- 1 ; we therefore have to solve the equation 4 · 5 x - 1 = 100, i.e., 5x - 1 = 5 2 , which means x - 1 = 2, or D x = 3. (iv) Show that for any m, n N and d = ( m, n ) we have (ii) Find all
E
cp(mn) = cp( m) · cp(n) · d . cp(d) SOLUTION. We first assume that m = pa and n = pb , where p is a prime and a, b No. H a = 0, then m = 1, d = 1, cp(d) = 1, and the assertion clearly holds. Similarly for b = 0. If a, b E N, then cp ( mn) = cp(pa+ b) = (p 1)pa+ b- 1 , and if we set c = min { a, b} ( i.e., c is the smallest of the numbers a, b), then d = pc and cp( m ) cp (n) . cp d) = (p - l)p" - 1 . (p - l)pb- 1 . P"- 1 - 1) = (p _ 1)pa+b- 1 ,
E
_
�
r;
E
and thus the assertion holds. Let now m, n N be arbitrary. We write m = pa1 1 · · · Pak�e , n = pb11 · · · Pbk�e , where a� , . . . , ak , b� , . . . , bk .11M"qo and Pb . . . , Pk are distinct primes. If we set 0 we have
3x - 5 = (-1)x - 1
(mod 4) .
The number ( -1)x - 1 is congruent to 0 modulo 4 if and only if x is even, i.e., x = 2k, where k E N0 • The solutions of the given equation are therefore all pairs X = 2k, where k
0
E No is arbitrary.
(ii) x(y + 1 ) 2 = 243y . SOLUTION.
We solve for x: X=
243y . (y + 1)2
In order for x to be an integer, (y + 1 ) 2 must divide 243y. Since y and y + 1 are relatively prime for any y E Z, then (y+ 1 ) 2 has to divide 243 = 35 • But this number has only three divisors that are squares of integers, namely 1 , 9, and 81, and this leads to the following possibilities: y+ 1 = ±1 , y+ 1 = ±3, or y + 1 = ±9. Hence we obtain the following six solutions of the given equation: y = 0, y = -2, y = 2, y = -4, y = 8, y = - 10, The equation has no other solutions.
(iii) .JX + v'Y = v'1988 .
X = 0, X = -2 243 = -486, X = 2 27 = 54, X = -4 27 = -108, X = 8 3 = 24, X = -10 3 = -30. ·
·
·
·
·
0
226
3. Nwnber Theory
SoLUTION.
If we subtract ..jY from both sides and then square, we obtain X=
1988 - 4vf7 71 ·
·
y + y.
If x, y are integers, then 4-/7 71y is also an integer, and thus J7 71y is a rational number. By 2.11.(iii) , -/7 · 71y = k is a nonnegative integer. Hence 7 · 71y = k2 , which means that k2 and thus also k are divisible by the primes and thus 7 and 71. Therefore, k = 7 · 71t for an appropriate t E ·
·
No ,
y=
k2
7 . 71
= 497t 2 .
In complete analogy we can find that there exists an
s
E
No such that
x = 497s2 • Substituting this into the original equation, we get
v'497s + v'491t = -/1988, which, after dividing, gives s + t = 2. Hence there are three possibilities: s = 0, t = 2, or s = t = 1, or s = 2, t = 0, and so the given Diophantine equation has the three solutions X = 0,
(iv) x2 + y2
=
1988; (x - y ) 3 . y=
X = y = 497;
X=
1988,
y
=
0.
0
SOLUTION. The equation is certainly satisfied by x = y = 0, so this is one solution. For what follows we assume that at least one of the numbers x, y is nonzero. We rewrite the equation as x2 - 2xy + y2 = ( x - y) 3 - 2xy, (X - y) 2 (X - y 1) = 2xy. -
We denote the greatest common divisor of x , y by k, and set x 1 = xfk, Yl = yfk. By 1 . 14.(iii) the integers x 1 1 y1 are relatively prime. By substituting into the last equation, we obtain
If x 1 - Yl = 0, then x = y and the original equation gives x2 + y2 = 0; this implies x = y = 0, which contradicts our assumption. Hence x 1 - y1 1= 0; moreover, since the original equation implies x > y, we have x 1 > y1 . If we divide the last equation by k2 (x 1 - y1 ) 2 , we obtain kx 1 - ky1 - 1
=
2X 1 Yl ( X l - Yl ) 2
·
5
Diophantine Equations
227
Hence (x 1 -y1 ) 2 is a divisor of 2x1 Yb but this can occur only if (x 1 -y1 ) 2 = 1. Indeed, if we assume that (x 1 - y1 ) 2 > 1, then by 2.5 there exists a prime p that divides ( x l -y1 ) 2 . aut then, by 2.2, p I X l -yb and thus from p I X l it. would follow that p I Yl , and vice vers�. Since (x 1 , y1 ) = 1, p divides neither x 1 nor Yl · However, in this case the condition p2 I (x1 - y1 ) 2 1 2x 1Y1 implies p2 I 2, which is a contradiction. Hence ( x 1 - y1 ) 2 = 1, and from x 1 > Yl it follows that x 1 - Yl = 1, i.e., x 1 = 1 + y1 . Substituting this into the last equation, we obtain and thus
x = kx 1 = (2(1 + Yl ) Yl + 1) (1 + Yl ) , y = ky1 = ( 2 ( 1 + Yl ) Yl + 1) Yl · If we set Yl = t, then all the solutions of the original equation are given by the pair x = 0, y = 0, and all pairs
X = ( 2t2 + 2t + 1) (t + 1), where t is an arbitrary integer.
0
5. 10 Exercises Solve the following Diophantine equations:
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 5. 1 1
5x + 3Y = 8z - 2. 2:1: = 3 + 13y. x2 - xy + 2x - 3y = 11. 2x3 - x2 y + x2 + 14x - 7y + 40 = 0. (x - 2) 4 = (y + 3) 5 • ..jX = 1 + v"Y· 2xy + 3y2 = 24. x2 y2 + x2 y + xy2 + xy = xyz - 1 . Solving Diophantine Equations Using Inequalities
This method is based on the fact that for any real numbers a, b there exist only finitely many integers x such that a < x < b. Therefore, in order to solve a given equation, we try to find numbers a, b such that the inequalities a < x < b, for some variables x, follow from this equation. The finitely many integers lying between a and b are then substituted for x in the given equation, which in this way becomes simpler. Let us consider the following examples.
228
3. Number Theory
5. 12 Examples Solve the Diophantine equations (i)-(iii) :
(i) 6x2 + 5y2 = 74. SOLUTION. Since all y equation must satisfy
E
Z satisfy 5y2 > 0, any solution x, y of the given 74
=
6x2 + 5y2 > 6x2 ,
which means x2 < 37/3, and thus -3 < x < 3, so x2 is among the numbers 74, 5y2 = 68, 0, 1 , 4, 9. Substituting these in succession, we obtain 5y2 5y2 = 50, 5y2 = 20. The first three contradict y E Z, while from the last one we get y2 = 4, i.e., y = ±2. Our equation therefore has the following four solutions: x = 3, y = 2; 0 X = 3, y = -2; X = -3, y = 2; X = -3, y = -2. =
(ii} x2 + xy + y2 = x2 y2 .
SOLUTION. Since this equation is symmetric in the variables x, y, we may assume that x2 < y2 , which implies xy < y2 , and thus
< y2 + y2 + y2 = 3y2 .
x2 y2 = x2 + xy + y2
We have therefore y = 0 or x2 < 3. Substituting these into the equation, we obtain x = 0 in the first case, and in the second case for x = 0 again y = 0, while for x = 1 we get y = -1 and for x = - 1 it is y = 1 . Our equation has therefore the following three solutions: X = 0,
y = 0;
X = 1,
y = -1;
x = -1 ,
y = l.
0
(iii) 2x = 1 + 3Y. SOLUTION . If y < 0, then 1 < 1 + 3Y < 2, which means that 0 < x < 1, but this is a contradiction. Hence y > 0 and therefore 2x = 1 + 3Y > 2, which means that x > 1 . We will now show that also x < 2. Indeed, if we had x > 3, then 1 + 3Y = 2x = 0 ( mod 8) ,
which would imply
3Y = - 1
(mod 8) ;
but this is impossible, since for even numbers y we have 3Y - 1 (mod 8) , and for y odd, 3Y = 3 (mod 8) . It therefore remains to consider the case 1 < x < 2. For x = 1 we get
and thus y = 0. The case x = 2 gives 3Y = 2 2 - 1
=
3·,
5 Diophantine Equations
1. The given equation has therefore the two solutions x y = 0 and x = 2, y = 1. (iv) Solve the equation x + y + z = xyz in natural numbers.
hence
y
=
229 =
1, o
SOLUTION. Since the variables in this equation are symmetric, we may assume that x < y < z. But then
xyz = x + y + z < z + z + z = 3z, which means that xy < 3 . Hence xy = 1, or xy 2, or xy = 3. If xy = 1, then x = 1 , y = 1, and by substituting into the given equation we get 2 + z z, which is impossible. If xy = 2 , then x = 1, y = 2 ( recall that we have x < y), which means that 3 + z 2z, and thus z = 3. If xy = 3, then x = 1, y = 3; hence 4 + z = 3z , and thus z = 2. But this is a contradiction to the assumption y < z. Our equation has therefore the unique solution x 1 , y = 2, z = 3 under the assumption x < y < z. The complete set of solutions in natural numbers is therefore given by all permutations of the numbers 1 , 2, 3 : (x, y, z) E {( 1 , 2, 3), (1, 3, 2), (2, 1, 3) , (2, 3, 1), (3, 1, 2) , (3, 2, 1) } . o =
=
=
=
5. 13 Exercises Solve the equations ( i) - ( iv ) in natural numbers, and the equations ( v ) and (vi) in integers: (i) (iii) (v)
6(x - y) = xy. + yY = 2X + y. 3Y = 1 + 2 x .
(ii)
XX
(iv) (vi)
l
!
X3+ y 3+
! Z
=
1.
x + y + z3 99. x2 - xy + y2 x + y. =
=
It is often advantageous to use contradictions to show that the set of values of the variable x is finite and bounded by the inequalities a < x < b; this is done by deriving a false statement from the assumption x < a ( resp. x > b) . In the following examples, such a false statement will be the pair of inequalities
en < an < (c + 1) n ,
where
5. 14
c,
d are integers and
n
is a natural number.
Exa�ples
(i) Solve the Diophantine equation SOLUTION.
Rewriting, we get
x(x + 1)(x + 7) (x + 8 ) = y2
y2 = ( x2 + 8x)(x2 + 8x + 7).
230
3. N"!liilber Theory
x2 + 8x = z, then our equation becomes y2 = z2 + 7z. We will show that z < 9 . We assume to the contrary that z > 9 . Then ( z + 3) 2 = z2 + 6z + 9 < z2 + 7z � < z2 + 8z + 16 = ( z + 4) 2 ,
If we set
=
which
is
a contradiction, since
inequalities would imply
z + 3, y, z + 4
are integers , and these
lz + 31 < I Y I < lz + 41. Hence z < 9, i.e. , x 2 + 8x < 9, which means that (x + 4) 2 = x2 + 8x + 16 < 25, and thus -5 < x + 4 < 5, or -9 < x < 1. Substituting these values into the original equation, we obtain all solutions: (x, y) E {( -9, 12) , ( -9, -12), (-8, 0), (-7, 0) , (-4, 12), (-4, -12), (-1, 0), (0, 0), (1, 12), (1, -12) } . 0 (ii) Solve the Diophantine equation (x + 2) 4 - x4 = y3 •
SOLUTION.
Rewriting, we obtain
4x
8x3 + 2 2 + 32x + 16 = 8(x3 + 3x2 + 4x + 2) , which means that y is even, and we can set y = 2z , z E Z. Then z3 = x3 + 3x2 + 4x + 2. If x > 0, we have (x + 1 ) 3 = x3 + 3x2 + 3x + 1 < x3 + 3x2 + + 2 3 = z3 < x3 + 6x2 + 12x + 8 = ( x + 2) , hence x + 1 < z < x + 2, which is impossible. The given equation therefore has no solutions x, y E Z such that x > 0. Let us now assume that it has a solution x 1 , Yl E Z such that x 1 < -2. Then (x1 + 2) 4 - xt = y�, and if we set x2 = -x 1 - 2 , Y2 = -y 1 , we obtain y3
=
4x
and so
x2 , Y2 is also a solution of the given equation. But x2 = -x1 - 2 > 0,
and from the above it follows that this case cannot occur. In summary, then,
-2 < x < 0, i. e. , x = -1. The original equation then gives y = 0; the pair 0 x = - 1, y = 0 is therefore the unique solution.
5 Diophantine Equations
231
5. 15 Exercises Solve the Diophantine equations (i)-(iv) :
(i) * (ii) * (iii) (iv) *(v)
x2 = 1 + y + y2 + y3 + y4 in natural numbers. x2 + x = y4 + y3 + y2 + y in integers. x6 + 3x3 + 1 = y4 in integers. x8 + 2x6 + 2x4 + 2x2 + 1 = y2 in integers. For any natural number n and all natural numbers d dividing 2n2 show that n2 + d is not the square of a natural number.
Some other, more complicated, approaches based on inequalities intro duced in Chapter
2
are sometimes also useful for solving Diophantine
equations. We will now consider several such examples.
Examples
5. 1 6
(i)
Find solutions in natural numbers of the equation
1 + 1 + 1 + 1 =1 2 2 2 . y X 2
Z U
is symmetric in the variables x, y, z, u, we may assume that x < y < z < u. It is clear that x > 1. If u > 3, we would have 1 +1 +1 +1 < 1 + 1 + 1 + 1 < 1· ' x2 y2 z2 u2 -4 -4 4 -9 SOLUTION .
Since the equation
-
2 < x < y < z < u < 2 . By substituting into the given equation we verify that x = y = z = u = 2 is indeed the D unique solution. hence
(ii)
u < 2,
and consequently
Given any
> 2,
s
show that the equation
1 + · +1 =1 1 +· · x2 x2 x2 1
has no solutions
in
2
natural numbers
s
x1 < x2 < · · · < X8 •
> 2, we have x1 > 1; hence from x1 < x2 < · · · < x8 and x b x 2 , . . • , x8 E N it follows that X k > k + 1 for all k = 1, 2, . . . , s . Since SOLUTION.
for any
Since s
k = 1, 2, . . , s , 1 < 1 < 1 = -1 - 1 X� - ( k + 1 ) 2 k ( k + 1 ) k k + 1 ' .
--
232
3. Number Theory
we obtain
-X1� + -X�1 + · · · + X-1� -212 + -312 + · · · + ( +1 1) 2 1( - 21 ) + ( 21 - !.3 ) . + ( ! - s+1 1 ) = 1 - s +1 1 < 1·'
z is 0
an
2, the equation
has no solutions in natural numbers. SOLUTION .
x,
to
Let us assume the contrary that for some natural numbers > > n, with n the given equation is satisfied , and set =
Then
2,
y x+ 1
2.
(29) and we get
0
= (y + l )n - yn - (y - 1)n - 1 - ( -l)n
(mod
y).
5 Diophantine Equations If n
is odd,
then 0
=2
which holds only for n we have (y (y which implies 0
(mod y) , and thus y
233
= 2 and
= 1. Hence n is even, and by the binomial theorem
(;) + (;)Y + 1 - 1t - (;) (;)Y + 1
+ 1} n =
2
(mod y3 ) ,
2 y -
3 (mod y ) ,
Y
3 (mod y ) ;
= (y + 1 ) n - yn - ( y - 1 ) n = 2ny
hence 0 = 2n (mod y2 ) by 3 . 3. (v) , and therefore 2n by yn , we get
(l + �r = 1 + (1 - �r
>
y2 . 1f we divide (29)
< 2.
On the other hand, Bernoulli's inequality (see 3 . 3. (iii) , Chapter 2) gives
( ) 1 + y-1
n
>
1+
n y
2n
=1+2y
>
-
y2
y > 2. 1+-=1+-2y
2
Putting everything together, it follows that the given equation has no solution
in
natural numbers for any integer n
>
0
2.
5. 1 7 Exercises Solve the following equations in natural numbers .
+ y3 + z3 = 3xyz. x u y z (ii) 2 = 7. + + + y (iii) (x + 2y + 3z ) 2 = 14(x2 + y2 + z 2 ) . . 1 1 + 1 + 1 + 1 = 1. v ) + 2 2 (• 2 2 2 y (i)
x3
( Z U X)
X
Z U V
5. 18 Solving Diophantine Equations by Decomposition This method is based on writing a given equation in the form (30) where
A 1 , . . . , An are expressions containing variables such that they take
on integer values when the variables are replaced by integers, and
B is
a
234
3. Number Theory
number ( possibly an expression ) whose prime decomposition we know. In
this case there exist only finitely many decompositions of the number
a� , . . . , an ·
B
If for each such decomposition we then into n integer factors consider the system of equations
..., we obtain all solutions of the equation illustrate this.
(30).
The following examples will
5. 1 9 Examples (i)
Solve the Diophantine equation
y3 - x3 = 91.
Factoring the left-hand side of the equation, we get
SOLUTION .
(y - x)(y2 + xy + x2 ) = 91.
2 :
Since
y2 + xy + x2 = (y + ; ) + x2 > 0, we will also have y - x > 0. The number 91 can be written as a product of two natural numbers in four different ways: 91 = 1 ·91 = 7 · 13 = 13·7 = 91 · 1. Therefore, we will separately solve the following four systems of equations: ( a) y - x 1, y2 + xy + x2 = 91. Substituting y = x· + 1 into the second equation, we obtain x2 + x - 30 = 0, which gives x 5 or x = -6. The corresponding values of the second variable are then y = 6, y = -5. ( b) y - x = 7, y2 + xy + x2 = 13. Then x2 + 7x + 12 = 0, so x = -3 and y = 4, or x = -4 and y = 3. ( c) y - x = 13, y2 + xy + x2 = 7. Here we have x2 + 13x + 54 = 0. But this equation has no real roots, and thus no integer roots . ( d) y - x = 91, y2 + xy + x2 = 1. In this case x2 + 91x + 2760 = 0, and =
this equation has no real roots either . The given equation has therefore the following four solutions:
{(5 , 6), (-6, -5), (-3, 4) , (-4, 3)}. (ii}
Solve the Diophantine equation
SOLUTION .
( x, y) E 0
x4 + 2x7 y - x 14 - y2 = 7.
We rewrite the left-hand side,
x4 + 2x1 y x 14 y2 = x4 (x7 y) 2 = (x2 x7 + y)(x2 + x7 y) , and note that the number 7 can be written in four different ways as a product of two integers: 7 = 1 · 7 = 7 1 = ( -1) . ( -7) = ( -7) . ( -1). We _
_
_
_
_
_
·
will therefore solve the following four systems of equations:
( a) x2 - x7 + y = 1,
x2 + x7 - y = 7. Adding these two equations, we obtain x2 = 4, thus x = 2 and y = 125 , or x = -2 and y = -131.
5 Diophantine Equations
235
x2 - x7 + y = 7, x2 + x7 - y = 1. Again x2 = 4, and thus x = 2, y = 131 or x = -2, y = -125. (c) x 2 - x 7 +y = -1 , x2 +x7 - y = -7. By adding we now get x2 = -4, which is a contradiction. (d) x2 - x 7 + y = -7, x2 + x7 - y = -1. Again x2 = -4, which is ( b)
impossible .
The equation therefore has the four solutions
(x, y) E {( -2, -1 31), ( -2, -1 25 ), (2, 125), (2, 131) }.
D
(iii) Solve the Diophantine equation
1 - + -y1 = p-1 .'
X
where
p is an arbitrary prime.
SOLUTION .
Multiplying by
xyp and rearranging., we get
xy - px - py = 0. Bringing this into the form of the equation,
so
(30) now requires a trick. Add p2 to both sides
that the left-hand side can be written as a product:
X
( - p) (y - p) = p2 .
p2 p2 = 1 . p2 = p . p = p2 . 1 = (-1 ) . (-p2 ) = ( -p) (-p) = ( -p2 ) · (-1). This leads to the following six systems of equations: (a) x - p = 1, y - p = p2 , hence x = p + 1 , y = p2 + p; ( b) x - p = p, y - p = p, hence x = 2p, y = 2p; (c) x - p = p2 , y - p = 1 , hence x = p2 + p, y = p + 1; (d) x - p = - 1 , y - p = -p2 , hence x = p - 1, y = p - p2 ; (e) x - p = -p, y - p = -p, hence x = y = 0, which is impossible; (f) x - p = -p2 , y - p = - 1 , hence x = p - p2 , y = p - 1. p
Since is a prime, can be written as a product of two integers in no more than the following six ways: ·
The given equation has therefore five solutions, as described in the cases
D
(a)-(d) and (f) .
(iv)
Solve the Diophantine equation
1 + x + x 2 + x 3 = 2Y . SoLUTION .
Again, we rewrite the left-hand side as a product:
3 . Number Theory
236
x
E Z, Since the left-hand side of the given equation is integer-valued for the right-hand side must be an integer as well, hence > 0. Now, the right hand side is a power of a power of
2
2,
so
each factor of the left-hand side must be < as well; hence there exists an integer m, 0 < such that
or X =
and
y
m
y,
2m - 1,
Squaring the first equation and comparing with the second one, we obtain
22m - 2m+ l + 1 = 2y- m - 1, or
(31) m = 0 , then 2Y = 1, which means that y 0 and x = 2m - 1 = 0. If m > 1, then the odd number 2 2m - l - 2m + 1 is, by (31), a power of 2 , which is possible only when 22m - l - 2m + 1 = 1 , which means that 2m - 1 = m, i. e. , m = 1, and thus x 1, y = 2. By substituting into the original equation, we convince ourselves that x = y = 0 and x = 1, y = 2
If
=
=
0
are indeed solutions.
( v)
Solve the equation
in natural numbers. SoLUTION .
We set
t = (x, y, z) , x 1 = xjt, y1
which upon division by
=
yjt, and z1 = zjt. Then
t2 =/= 0 gives (32)
where
(x�, Yb z1)
=
1
(see, e.g. ,
5.1).
pairwise relatively prime: Indeed,
The numbers
x1 , Yl , Z 1
are actually
if a prime p divides two of the numbers
x1, Yl , z1 , then by (32), it would also divide the third one, but because of (x 1 , Yl , z1 ) = 1 this is impossible. Thus at most one of the numbers x1 , y1 is
even. Let us assume that both are odd. Then it follows from the congruence
z� = x� + y� = 1 + 1
(mod
8)
that z� is an even number not divisible by 4, which is impossible. Therefore, exactly one of Xt ,
Yl
is even. Since the equation
(32)
is symmetric in
x1
5
Diophantine Equations
and Yl , we may as well assume that the even one it follows from
(32) that
is x 1 = 2r , r E N.
237 Then
4r2 = Z21 - Y21 ,
and thus
+ r2 - Z1 Yl Z1 Yl . 2 2 If we set u = (z1 + Yl)/2, v = (z 1 - yl )/2, then z 1 = u + v, Yl = u - v. Since Yl , Z1 are relatively prime, then so are u, v. Now it follows from the ·
---
equation
2.6.(iii) that there exist relatively prime integers a, b such that u = a2 , v = b2 , and furthermore from u > v it follows that a > b. Altogether and from
we therefore have
x = tx 1 = 2tr = 2tab, Y = ty1 = t(u - v ) = t(a2 - b2 ), z = tz1 = t(u + v) = t(a2 + b2 ), which indeed satisfies the given equation for any relatively prime
a, b E
N such that
a
be obtained by changing the order of problem we assumed that only
where again
>
b.
is
arbitrary
The remaining solutions could
x and y (in the course of solving this
x 1 is even) :
t, a, b E N are any numbers such that a > b, (a, b) = 1.
We remark that the Diophantine equation solved
t E N and for
called the
Pythagorean equation
x 2 + y 2 = z2
D
that we just
and describes all right-angled
triangles with integer sides.
5. 20 Exercises p is a fixed prime number: (ii) x(x + 1 ) = 4y(y + 1) . (iv) 2x 2 - x - 36 = p2 •
Solve the Diophantine equations (i)-(iv) , where
(i} (iii)
xy = x + y. x 2 - y2 = 105.
Solve the equations (v)-(viii)
in
natural numbers:
2 x2 +x -2 - 2x2 -4 = 992. 2x+ l - 1 yz + l . (ix) Show that no number of the form 2n , n E No , can be written as a
2x2 + 5xy - 12y2 = 28. *(vii) 2 x - l + 1 = yz+l . (v)
(vi) * (viii}
=
sum of two or more consecutive natural numbers .
3. Number Theory
238
* (x)
E
2n , n No ,
Show that any natural number that is not of the form can be written as a sum of at least two consecutive natural numbers.
Solve the systems of equations
xx + yyz+ z 18.14, + * (xiii} x2 + y2 z4 , (xi)
=
=
(xi) , (xii)
over the integers:
xxs++yy+3 +z z33, 3. x, y, z ( x, y) 7. xy (x, y) 1 . =
* (xii)
Show that for natural numbers =
the product
=
is always divisible by
this is generally not true without the condition
6
=
such that
=
1
and
Show that
Solvability of Diophantine Equations
In the previous section we saw that solving Diophantine equations is, in most cases, no easy matter. Although we have learned several methods,
there are many specific examples of Diophantine equations where none of these methods of solution can be successfully used. Nevertheless, it may be possible in these cases to obtain some information about the solutions . For
example, we may be able to find an infinite set of solutions, which would
show that the set of all solutions is infinite, even though we may not be able to determine them all. Or on the other hand, we may be able to show
that the set of solutions is empty equation) , or that it is finite.
6. 1
(which
would actually solve the given
The Nonexistence of Solutions
To prove that a Diophantine equation has no solutions at all, it is often pos sible to successfully use congruences . Indeed, if the Diophantine equation = has solutions (where are expressions that contain the variables,
L R
L, R
and take on integer values for integer values of the variables ) , then also the congruence = (mod m ) must have solutions for any m since, for instance, the solutions of the original equation are also solutions of all these congruences. This means that if we find a natural number m such
L
R
E N,
L = R ( mod m )
has no solution, then the original equation = cannot have a solution either. However, it is important to note that the converse is generally not true: If the congruence = that the congruence
L
R
( mod m)
L
R
has solutions for each natural number m, this does not yet imply that the Diophantine equation = has solutions (we will show this in 6 .2 . (v )) .
L
R
6. 2 Examples (i}
Solve the Diophantine equation
xi + x� +
· · ·
+
xi4
=
15999.
6 Solvability of Diophantine Equations
SoLUTION.
239
We show that the congruence xi + x� +
· ·
·
+ xi4 = 15999 (mod 16)
has no solution, which would mean that the given equation is also not solvable. Indeed, if an integer n is even, then n = 2k for k E Z, and thus n4 = 16k4 = 0 (mod 16). If n is odd, then n4 - 1
=
(n - 1)(n + 1)(n2 + 1) = 0 (mod 16) ,
since the numbers n - 1, n + 1 and n2 + 1 are even and one of the integers n - 1, n + 1 must even be divisible by 4. This means that n4 is congruent to 0 modulo 16 for even n, and congruent to 1 modulo 16 for odd n. Therefore, if exactly r of the numbers x1 , x2 , , x 14 are odd, then xi + x� + + xi4 = r (mod 16). •
•
•
· · ·
Now 15999 congruence
16000 - 1 = 15 (mod 16), and since 0 < r < 14, the Xf + X� +
· · ·
+ Xf4 = 15 (mod 16)
cannot have a solution, and thus neither can the given equation be D solvable. (ii) Find integer solutions of the system of equations x2 + 2y2 = z2 , 2x2 + y2 = u2 . It is easy to see that from x = y = 0 it follows that z = u = 0, and this is a solution of the given system. We will now show that there are no other solutions. We assume that x, y, z, u is a solution and that x 1= 0 or y 1= 0, and we denote by d = ( x, y) > 0 the greatest common divisor of x and y. From the first equation it follows that d I z, and from the second one that d I u. If we set x 1 = xjd, Yl = yjd, z1 = z/d, and u 1 = ujd, then it follows from 1.14.(iii) that (x 1 , Yl ) = 1, and upon dividing the two original equations by tP we obtain SOLUTION.
By adding, we get 3x� + 3y� = z� + u�, and thus 3 1 z� + u�. By 3.14. (vi) we have 3 I z1, 3 I u 1 and thus 9 I z� + u�. But then 9 I 3(x� + y�), and thus 3 I x� + y�. Again by 3.4.(vi) we have 3 I x 1 1 3 I y� , which is a contradiction to ( x� , Yl ) = 1. The system has therefore the unique solution = 0. D X=y=
Z=U
240
3. Number Theory
(iii) Find all positive integer solutions of the equation 1! + 2! + 3! +
· .
·
+ x! = y2 •
By direct computation we convince ourselves that for x < 5 the equation is satisfied only by x = y = 1 and x = y 3. We will now show that the equation has no solution for x > 5. Since n! is divisible by 5 for any n > 5, we have SoLUTION.
=
1! + 2! + 3! + . + x! = 1 ! + 2! + 3! + 4! = 33 = 3 (mod 5). ·
·
Now, the square of a natural number is always congruent to 0, 1, or 4 modulo 5. Hence the congruence 1 ! + 2! + + x! = y2 (mod 5) has no solution for x > 5, which means that the given equation also has no solution 0 for x > 5. · · ·
(iv) Find all positive integer solutions of the equation x2 - y3 = 7. We will show that the given equation has no solutions. Let us assume to the contrary that for appropriate x, y E Z we have x2 - y3 = 7. If y were even, we would have x2 = 7 (mod 8), which is impossible. Hence y is odd, y = 2k + 1 for k E Z. Then SOLUTION.
x2 + 1 = y3 + 23 = (y + 2) (y2 - 2y + 4) = (y + 2) ((y - 1) 2 + 3) = (2k + 3)(4k 2 + 3) . The number 4k 2 + 3 has to be divisible by some prime p = 3 (mod 4). Otherwise, since 4k2 + 3 is odd, its prime decomposition would contain only primes congruent to 1 modulo 4; the product of these primes, namely 4k2 + 3, would then also be congruent to 1 modulo 4, which is clearly not the case. Therefore, 4k2 + 3 is divisible by a prime p = 3 (mod 4), and thus x2 + 1 = 0 (mod p).
By 3.14.(vi) this implies x
=
1 - 0 (mod p), which is a contradiction.
0
(v} Show that the congruence 6x2 + 5x + 1 = 0 (mod m ) has solutions for every natural number m, while the Diophantine equation 6x2 + 5x + 1 = 0 has no solution.
Solvability of Diophantine Equations
6
241
SoLUTION. Since 6x2 +5x+1 = (3x+1) (2x+1), the equation 6x2 +5x+1 = 0 has no integer solution. Now let m be any natural number, m 2n k, where n E No and k is an odd number. Since (3, 2n) = (2, k) = 1, both congruences of the system =
·
3x = -1 (mod 2n) , 2x -1 (mod k) =
have solutions by Theorem 4.2, and since (2n, k) = 1, by Theorem 4.6 the whole system has solutions. Hence for any x satisfying this system, 3x + 1 is divisible by 2n and 2x + 1 is divisible by k, and therefore the product (3x + 1) (2x + 1) is divisible by 2n · k = m. Thus x solves the congruence 0
6.3 Exercises Solve the Diophantine equations (i)-(v) : (i) 3x2 - 4y2 = 13. (ii) 2x2 - 5y2 = 7. (iii) 3x2 + 8 = 2 . * (iv) 3x2 + 3x + 7 = (v) ( x + + (x + 2) + (x + 3) + (x + 4) = (x + 5)
1)3
y
3
3
3
y3 • 3 .
Solve the equations (vi)-(viii) in natural numbers:
19z .
(vi) 2x + 7Y = * (vii) 2x + 5Y = 19Z. *(viii) 1! + 2! + 3! + . . . + x! = (ix) Show that for any given odd numbers r, s , the Diophantine equation x 1 0 + rx7 + s = 0 has no solutions. (x) Given a prime p 3 (mod 4) , show that x2 + y2 = pz 2 has no solution in natural numbers.
yz+ l
=
Find all integer solutions of the systems of equations (xi) , (xii) , and of the equations (xiii), (xiv): (xii) x2 + 6y2 = z 2 , (xi) x2 + 5y2 = z2 , 6x2 + 2 = u2 . 5x2 + 2 = u2. *(xiii) x2 - + 1 = (4z + 2) . * (xiv) x2 + 5 = x - = z2 has no solution in natural *(xv) Show that the equation numbers.
y3 y
3
4xy- y
y 3 y•
6.4 Reductio ad Absurdum The method of reductio ad absurdum is a method for proving the nonexis tence of solutions of a Diophantine equation. To carry out this method of
242
3. Number Theory
proof, we characterize a supposed solution of the given Diophantine equa tion by some natural number (for instance, the greatest common divisor of the values of some variables, or the square of the value of some variable, or something like that) and show that if there exists a solution characterized by the natural number d, then there must exist another solution charac terized by the natural number d' < d. But then no solution can exist, of which we can easily convince ourselves by contradiction: If there were a so lution, we could choose one that was characterized by the smallest natural number d; then, of course, there would exist another solution characterized by a natural number d' < d, which, however, would be a contradiction to the choice of d. (We note that we could solve, for instance, Example 6.2.(ii) in this way if d denoted the greatest common divisor of the values of the variables x, y, z, u, or the exponent of 3 in the prime decomposition of this greatest common divisor.) 6. 5 Examples (i) Solve the Diophantine equation x3 + 2y3 + 4z3 - 6xyz = 0. SOLUTION. The equation is certainly satisfied by x = y = z = 0. We will show that there is no other solution. We set d = x2 + y2 + z2 and assume that for some solution x, y, z of the given equation we have d > 0. It follows from the original equation that x3 is an even number, and thus x = 2x1 for an appropriate x1 E Z. Substituting this into the equation, we get 8x� + 2y3 + 4z3 - 12x1yz = 0, and upon dividing by 2, 4x� + y3 + 2z3 - 6x1yz = 0, so y3 is also an even number, and we write y = 2y1 , with Y1 E Z. Substituting and dividing by 2, we get
an
appropriate
2x� + 4y� + z3 - 6X1Y1 Z = 0, which implies that z3 is also even, and thus z 2z1 for some z1 E Z. Substituting and dividing by 2 once more, we obtain =
X� + 2y� + 4z� - 6X1Y1Z1
=
0,
so x1 , Y1 , z 1 is a solution of the original Diophantine equation, and furthermore 2 x2 y 2 X21 + Y21 + Z 21 = + - + z = d < d. 4 4 4 4 According to the method described in 6.4, the given equation does not have any solution with the property d > 0, and thus x = y = z = 0 is its unique o solution. -
-
-
6
Solvability of Diophantine Equations
243
4z in natural numbers. SOLUTION. We use the method of 6.4 with d = z. Let us first assume that x, y, z is a solution of the given equation. Then certainly z 1= 1 , since for x y 1 we have x2 y2 2 < 4, and if at least one of the numbers x, y is greater than 1 , then x2 y2 > 4. Thus z > 1 , and we have x2 y2 4z 0 (mod 8). Since the square of an odd number is congruent to 1 modulo 8, and the square of an even number is congruent to 0 or 4 modulo 8, it follows (ii) Solve the equation x2 + y2 = =
+
=
+
=
+
from this congruence that both x and y are even, i.e., x appropriate x 1 , Yl E N. But then
X� + Y�
=
2 x2 4 4 +y =
=
=
=
2x1 , y = 2y1 for
4z- l '
and thus, if we set z1 = z - 1 E N, the numbers x� , y� , z1 satisfy the given 0 equation, with z1 < z. Therefore, the equation has no solution.
(iii) Solve the Diophantine equation x4 + y4 + z4 = 9u4 • SOLUTION. If u = 0, then necessarily x = y = z = 0, which is a solution of the given equation. We will show that there are no other solutions. Let us assume that the integers x, y, z, u satisfy the given equation and that u 1= 0; we set d u4 If the number u were hot divisible by 5, then Fermat's theorem gives u4 - 1 (mod 5) , and we would have =
•
x4 + y4 + z4 =
4
(mod 5).
This, however, is impossible since by Fermat's theorem the numbers x4 , y4 , z4 are congruent to 0 or 1 modulo 5. Thus, u is divisible by 5, i.e., u = Su 1 for an appropriate u 1 E Z, and we get which implies that x, y, z are divisible by 5, i.e., x = 5x�, y = 5y1 , z = 5z1 for appropriate x1 , y1 , z1 E Z. Substituting this into the original equation and dividing by 54 , we obtain
i
x + Yf + zi = 9ui, and thus x 1 , y1 , z1 , u 1 satisfy the given equation, and
u4 < u4 = d. u4l = 54 (iv) Solve the equation x2 - y 2 = 2xyz in natural numbers.
0
SOLUTION. We assume that some natural numbers x, y, z satisfy the given equation, and we set d = xy. If we let d = 1 , then x = y = 1 and the
244
3. Number Theory
equation would give z 0, which is impossible. Hence d > 1 . Let some prime dividing d. Since (x + y) (x - y) = x2 - y2 2xyz = 0 (mod p) , =
P
be
=
we have x = y (mod p) or x = -y (mod p) . In view of the fact that the prime p divides the product xy, either x or y is congruent to 0 modulo p, and together x y - 0 (mod p) . Hence x1 xjp and Yl = yjp are natural numbers, and =
=
from which, upon dividing by p2 , we see that x1 , Yl , z satisfy the given equation, and that d d. X y x 1 Y1 = < p p p2 0 By the principle of 6.4 the given equation has no solution. (v) Solve the Diophantine equation x2 + y2 + z2 = 2xyz. ·
-
=
-
SOLUTION. The equation is certainly satisfied by x y z = 0. We will show that it has no further solutions . In fact, we will prove the following stronger assertion: Given any u E N, the equation (33) x2 + y2 + z 2 2uxyz, =
=
=
where x, y, z E Z, has no solution other than x = y z = 0. Let us assume that x, y, z E Z and u E N satisfy (33) , and that d = x2 + y2 + z2 > 0. Since u > 1, 2u xyz is an even number, and thus x2 + y2 + z 2 is even. But this means that either exactly one of the numbers x, y, z or all three of them are even. However, in the first case we have x2 + y2 + z 2 = 1 + 1 + 0 = 2 (mod 4) , =
while
2uxyz = 0 (mod 4) , since u > 1 and one of the numbers x, y, z is even. Hence we consider the second case, where the numbers x 1 = x/2, y1 = y/2, z1 = z/2 are integers. We set u 1 = u + 1 and substitute everything into (33): 4x� + 4y� + 4z� = 2u1 - 1 • 2x1 • 2y1 2z 1 7 •
and upon dividing by 4, X� + � + Z�
=
2u1 X 1Y1Z b •
so that xl , Yl , zb ul satisfy (33) . In addition, 0 < x� + y� + z� = d/4 < d, since d > 0. By 6.4, the equation (33) can therefore only have a solution with the property d = 0, which, as we have already seen, leads to the solutions x = y = z = 0, with arbitrary u E N. In particular, the given o equation has the unique solution x = y z 0. =
=
6 Solvability of Diophantine Equations
245
6. 6 Exercises Solve the Diophantine equations ( i) -(viii) :
(i) x3 - 2y3 - 4z3 = 0. (ii) x4 + 4y4 2(z4 + 4u4 ). (iii) x2 + y2 + z2 = 7u2 . (iv) x3 + 3y3 + 9z3 = 9xyz. * (v) x2 + y2 + z2 = x2y2u2 . * (vi) x2 + y2 = 11 2z . * (vii) x2 + y2 + z2 + u2 = 2xyzu. (viii) 2x2 + y2 = 7z2 . We have now studied some of the most common approaches used for proving that a given Diophantine equation has no solution. We do not claim completeness of this list of methods. Indeed, even though there is no universal method for solving Diophantine equations, one can find in the mathematical literature a large number of methods; most of them, however, are useful only in very special cases. We conclude this overview of nonexistence proofs of solutions of Diophantine equations by presenting one of these special instances. =
6. 7 Example Given an arbitrary prime p, solve the Diophantine equation x4 + 4x = p. SOLUTION. For any integer x < 0, 4x is not an integer, and neither is x4 + 4x. Thus for any prime p, the given equation has no negative solution. For x = 0 we have 04 + 4° = 1 , which is not a prime, and for x = 1 we have 1 4 + 4 1 = 5, a prime. We will now show that for any integer x > 2, the number x4 + 4x is composite. If x = 2k is even, where k E N, then x4 + 4x = 24 k4 + 42k = 16(k4 + 42(k-l) ), which is a composite number. If x = 2k + 1 ( k E N) is odd, then x4 + 4x = x4 + 4 . 42k = [x4 + 4x2 (2k ) 2 + 4(2 k ) 4] 4x2 (2k ) 2 = [x2 + 2(2k ) 2 ] 2 - (2x . 2 k ) 2 [x2 + 2x 2k + 2(2k ) 2 ) [x2 - 2x 2k + 2(2k ) 2) = [(x + 2k ) 2 + 2 2k ] [(x 2 k ) 2 + 2 2k ] , which is again composite, since (x ± 2k ) 2 + 22k > 22k > 22 > 1. To sum marize, for p = 5 the given equation has the unique solution x = 1 , while D there are no solutions for any prime p 1= 5. _
=
·
·
_
6. 8 The Cardinality of the Set of Solutions In many cases when we are unable to find all solutions of a Diophantine equation, we may at least succeed in determining whether there are finitely
246
3. Nwnber Theory
or infinitely many solutions. Finiteness can, for instance, be established by showing that the values of the variables that yield solutions are in absolute value less than a certain number. If we can find this number and it
is
''rea
sonably small," we can then find all solutions with the method described in
5.11.
On the other hand, the infinitude of solutions of a given Diophantine equation can, for instance, be established as follows. For each variable we find an expression involving a parameter, in such a way that upon sub stituting, the equation becomes an equality, and that for infinitely many values of the parameter the variables have different values. Or we can find one solution of the equation and give a description of how to obtain a new solution from any known one. If we can guarantee that upon repeated appli
if
cation of this rule we always obtain different solutions (for instance, the solutions thus obtained get larger and larger ) , then we have again proved that the set of solutions is infinite. It
is
clear that with both approaches
there may exist more solutions than the ones detected.
6. 9 Examples (i) Show that the Diophantine equation 4xy-x -y = z 2 has infinitely many
solutions (it can be shown that this equation has no solution in positive integers; see
6.3.(xv)).
SOLUTION.
If we choose
simpler:
x
=
-1,
the equation becomes significantly
-5y = z2 - 1,
and this is an equation that we know how to solve by the method described in
5 . 5. Hence we first solve the congruence z2 - 1 = 0
(mod
which is satisfied by exactly those integers modulo
5.
Thus
z = ± 1 + 5t, with an -5y = z2 - 1
and therefore
=
5) ,
z that are congruent to 1 or - 1
arbitrary t E
Z,
and we get
1 ± lOt + 25t2 - 1 ,
y = =F2t - 5t2 .
By substituting we convince ourselves that the values
x = -1, y = 2t - 5t2 ,
z = -1 + 5t indeed satisfy the given equation . It is clear that different values of t give different values of z; hence the above solutions are indeed and
o
infinite in number.
(ii)
Show that the Diophantine equation (
x - 1 ) 2 + (x + 1 ) 2 = y2 + 1
6 Solvability of Diophantine Equations
247
has infinitely many solutions. SOLUTION. The equation can easily be rewritten as
Y2 - 2x 2 = 1 . We try to find some solution; it is easy see that the choice y = 3, x = 2 satisfies the given equation. Now we assume that we have an arbitrary solution x, y E Z, and we try to find another one. We have
(y + hx) (y - hx) = 1 . Substituting the values y = 3 and x = 2, we obtain the equality (3 + 2v'2) (3 - 2v'2) = 1 , and by multiplying we get ·
[(y + v'2x) (3 + 2v'2 )] · [ (y - v'2x) (3 - 2v'2 )]
= 1.
Manipulating both expressions in square brackets, we get
(y + v'2x) (3 + 2v'2 ) = 3y + 3v'2x + 2v'2y + 4x = (4x + 3y) + (3x + 2y)v'2, (y - v'2x) (3 - 2v'2 ) = 3y - 3v'2x - 2J2y + 4x = (4x + 3y) - (3x + 2y).J2. If we set u = 4x + 3y, v = 3x + 2y, then (u + hv) (u - v'2v) = 1 , which means that
u2 - 2v2 = 1 ,
and thus u, v E Z is a new solution of the given equation. If we set x 1 2, Y1 = 3 and Xn+ l 3xn + 2yn , Y�+ l 4xn + 3yn for any n E N, then for all n E N we have a solution Xn , Yn of the given equation. FUrthermore, since 0 < x 1 < x2 < · · · , 0 < Yl < Y2 < · · · , different indices n give different solutions Xn , Yn· The given equation has D therefore infinitely many solutions. =
=
=
(iii) Show that for any integer k the equation
k + x2 + y2 = z 2 has infinitely many solutions in natural numbers. SOLUTION. Factoring z 2 - y2 , we rewrite the given equation as
k + x2 = ( z - y) ( z + y).
248
3. Nwnber Theory
It is not necessary to find all solutions, and hence we may assume that z - y = 1, z + y = k + x2 . Any solution of this system will also be a solution of the given equation (the converse, however, is not true; the reader is encouraged to find, for some fixed k, an example of natural numbers x, y, z that satisfy the given equation but do not satisfy the above system of equations). If we solve the system for z and y, we get 1 z = (x2 + k + 1) , 2 y = !. (x2 + k - 1). 2 If we choose x = lkl + 1 + 2t, where t E N, then x is a natural number. Then x2 + k = k + 1 + 2t + k = 1 (mod 2), and thus z = ! ((lkl + 1 + 2t) 2 + k + 1) > 0, y = ! (( l kl + 1 + 2t) 2 + k - 1) > 0 are also natural numbers. Since for different t we obtain different x and O thus different solutions, the equation has infinitely many solutions. (iv) Show that for any natural number k the Diophantine equation 5x2 - Bxy + 5y2 - 4k2 = 0 has only finitely many solutions. We rewrite the given equation as (2x - y) 2 + (2y - x ) 2 = 4k2 , which implies (2x - y) 2 < (2k) 2 and (2y - x) 2 < (2k) 2 , and thus -2k < 2x - y < 2k and -2k < 2y-x < 2k. lf we add the first and twice the second equation, we get -2k < y < 2k, and in complete analogy -2k < x < 2k. Since x and y can take on only finitely many values for a fixed k, the given o equation has only finitely many solutions. SOLUTION .
6. 10 Exercises Show that the Diophantine equations (i)-(iv) have infinitely many solutions: (i) z3 + 5xz2 + 3y2 z + 2y3 = 9. * (ii) x2 + (x + 1 ) 2 = y2 . *(iii) (x + 1)3 - x3 = y2 . * (iv) x2 + y2 + z2 = 3xyz. (v) Show that for any integer n the Diophantine equation x2 + y2 - z2 = n has infinitely many solutions.
Solvability of Diophantine Equations
6
(vi)
Show that for any integer 2
x
n
249
the Diophantine equation
+ y2 + z2 + n2 = n6 - n (x + y + z)
has only finitely many solutions.
6. 1 1
Examples
We devote the remainder of this section to equations with an undetermined number of variables; as a rule, we will solve them by induction. (i) Show that for any natural number equation
s
and any rational number
w
the
1 + ··+1 +1 =w · 1 X
X8
X2
has a finite number of solutions in natural numbers. SOLUTION. We use mathematical induction. For s = 1 the equation has at most one solution, and thus the assertion is true for s = 1. We now assume that the assertion holds for s E N, and we will prove it for s + 1 . We assume that the natural numbers X 1 , . . . , X8 , Xs+l , with X 1 < X2 < · · · < X8 < Xs+l , satisfy the equation
1 +1 + -1 = u, 1 +...+1 Xs Xs+ X
X2
(34)
l
where u is a given rational number, which is obviously positive. But then
1 1 the equation
1 = 1 + 1 +···+ 1 2 x2 x2 x2
X
0
1
2
s
250
3. Number Theory
has infinitely many solutions in natural numbers such that xo < X 1 < . . . < Xs . SOLUTION. We convince ourselves that if x0 , x 1 , • . . , x8 is a solution of the given equation with the required condition, then txo, tx 1 , . . . , txs is also a solution for any natural number t , and we have txo < tx 1 < · · · < txs . Therefore, it is sufficient if we can show that for any natural number s > 1 the given equation has at least one solution xo, x 1 , . . . , X8 such that Xo < x 1 < · · · < X8 • To do this, we use induction on s. If s = 2, then x0 = 12, x 1 = 15, x2 = 20 is a solution, since it is easy to verify that 1 1 1 = + 122 152 202 ' We now assume that the assertion holds for some s > 2, i.e. , there exist natural numbers xo < x1 < · · · < X8 such that 1 1 1 - 1 + + · · + · x02 x21 x22 x2s . We set yo = 12xo, Yl = 15xo, and Yi = 20xi- 1 for i = 2, 3, . , s + 1. It is easy to see that Yo < Y1 < · · · < Ys + 1 . Furthermore, we have 1 1 1 1 1 1 1 l 1 1 .+ 1 . = . 2= = + + + . , y� X� 12 X� 152 202 152 X� 202 X� X� 1 1 1 1 = 2 + 2 + . . . + 2 + -2. Y1 Y2 Ys Ys+ l 0 This completes the proof by induction. .
(
)
(
.
)
6. 12 Exercises (i) Show that for any natural number s the equation Xo2 = x2l + . . . + X2s has infinitely many solutions in natural numbers. (ii) Show that for any rational number w and any natural number s the equation X8 X 1 + X2 . . . + ----,. . + =w 1 + X� 1 + X � 1 + X� has only finitely many solutions in natural numbers. * (iii) Show that for any integer s > 2 the equation -
1 1 . . 1 = x03 x31 + . + x3s has infinitely many solutions in natural numbers.
7
7
Integer Part and Fractional Part
251
Integer Part and Fractional Part
The integer part and the fractional part of a real number are encountered in many different situations in number theory. Therefore, we devote an entire section to these concepts; we will summarize some of their properties and uses, and solve several equations in which they occur. 1. 1
Integer and Fractional Parts Definition. The greatest integer not exceeding the real number x is called the integer part of x, and is denoted by [x] . This means that [x) is the (unique) integer satisfying the inequalities
[x] < x < [x] + 1.
(35)
The difference x- [x] is called the fractional part of x and is denoted by (x) . We note that both symbols could also be defined in a different way, for instance by the conditions
[x) + (x) = x,
[x] E Z, 0 < (x) < 1 ;
these, as the reader can easily verify, are equivalent to Definition 7 .1. It also follows from these conditions that for arbitrary real numbers x, y such that x < y, we have [x) < [y) , but not (x) < (y) . If, however, [x] = [y) , then (x) < (y) if and only if x < y. 7. 2 Examples Show that (i)-(v) hold for any x, y E lR, m E Z, and n E N: (i} [x + m] = [x] + m; (x + m) = (x) . SOLUTION.
From (35) it follows that
(x] + m < x + m < ( [x) + m) + 1.
Since [x] + m E Z, this means, by (35) , that
[x + m)
=
[x) + m,
which is the first of the desired equalities. Since
(x + m )
=
x + m - [x + m]
=
x + m - [x) - m = x - [x) = (x) ,
the second one follows as a consequence. (ii) m;-1 < r:J + 1 .
0
252
3. Number Theory
SOLUTION. From the inequality
: < [:] + 1 (see (35) , with x = m/n ) we obtain upon multiplying by n,
m n, then [! ] is the number of positive integers divisible by n and not exceeding x. SOLUTION. By (35) we have
[: ] < : < [ : ] + 1 , and multiplying by n we obtain
[:]
·
n<x
1 if and only if -i < nr - n, which occurs when -i { -(n - 1), -(n - 2) , . . , [nr - n)} . This set has exactly [nr - n) + (n - 1) + 1 = [nr - n) + n elements, and thus the sum within parentheses satisfies [r] + [r + �] + + [r + n : 1 ] = [nr - n] + n = [nr - n + n] = [nr], ···
·· ·
E
·
·
.
·
where we have used (i). Altogether, it follows that
[x] + [x + �] + . · . + [x + n : 1 ] = nk + [nr] = [nk + nr] = [n(k + r)] = [nx]
(where (i) has been used again) , and this proves the first identity. The second one follows as a consequence, since
(x) + (x + �) + + (x + n : 1 ) = x - [x] + (x + �) [x + �] + · · · + (x + n : 1 ) [x + n : 1 ] = nx + -n1 + . . + n n- 1 - [nx) by the first identity. Since 1 + 2 + ( n - 1) = n( n - 1) /2, we have (x) + (x + �) + . . . + (x + n : 1 ) = nx - [nx] n ; 1 = (nx) + n ; 1 ' ·
··
_
_
·
·· ·+
+
which completes the proof of the second identity.
0
3. Number Theory
254
7. 3 Exercises (i) How many integers m with 106 < m < 107 are divisible by 786? (ii) How many natural numbers less than 1000 are divisible neither by 5 nor by 7? (iii) How many natural numbers less than 100 are relatively prime to 36? (iv) For how many natural numbers m < 1000 is v5 (m) = 1 (see 2.9)? Prove (v)-(xxi) for any x, y E 1R and m, n E N: (v) (vii) *(viii} * ( ix) (x} (xi) (xii) (xiii) (xiv) (xv) *(xvi) (xvii) (xviii)
[ [:] ]
=
[:J .
(vi)
[ ; ] - [- ; ] = n.
the smallest integer not less than x is - [-x). [x + y] + [x) + [y) < [2x] + [2y). [x] · [y) < [xy) < [x] · [y) + [x) + [y] , if x > 0 and y > 0. If x > y, then [x) - [y) is the number of integers r such that y < r < x. [x) + [-x] is 0 for x E Z and - 1 for x E lR\Z. (x) + (-x) is 0 for x E Z and 1 for x E lR\Z. 0 < [2x) - 2 [x) < 1. -1 < {2x) - 2(x) < 0. 0. ( Vfx ] = [ vrxr] , if [vln + v'n + 1] = [v'4n + 2J . [:J + x 1 + . . . + x + - 1 = [x].
X�
[:]
[
]
:
( :) + ( x : 1 ) + . . . + ( x + : - 1 ) = (x) + n ; 1 .
[
l
* (xix) [mx] + [mx + :J + . . . + mx + (n - 1) m n (m l )n = [nx] + = + : J + . . . + = +
[
[
�
l
:) + . . . + (mx + (n -n1)m ) + m ; n (n -:n1 )n ) = (nx ) + ( nx + : ) + . . . + ( nx + . (
*(xx) (mx) + mx +
*(xxi) [r 2 n]
=
[r [rn) + 1), where r
=
(1 + ..;5)/2.
The following theorem illustrates one of the uses of the concept of integer part in number theory.
7
1.4
Integer Part and Fractional Part
255
Prime Powers in a Factorial
For any prime p and natural number n the exponent in the power ofp in the prime decomposition ofn! (see Section 2.9) is determined by Theorem.
vp
(n!J =
[;] [; ] +
+ ...+
where k is any natural number such that n
1 be an integer. The sequence ( )� 0 is determined by = = m and for n E N. Show that this sequence contains infinitely many odd and infinitely many even numbers. *(vi) Define f(n) = n + [Vn] for natural numbers n. Show that for any m E N the sequence m, f(m) , f(f(m)), f(f(f(m))), . . . contains at least one square of a natural number. *(vii) Suppose that several numbers are given, each of which is less than 2 000, and suppose that the least common multiple of any two of them is larger than 2 000. Show that the sum of the reciprocals of these numbers is less than 2. 7. 5 (i) *(ii) *( )
(�2n25-) l 2 .
(m�n).
111 ···
xo
Xn
n
Xn
Xn [�Xn- 1 ]
7. 6 Examples We devote the remainder of this section to several equations that contain the integer part or the fractional part of a real number. 4 (i) Solve the equation [ ) = 3 in real numbers.
x x
SOLUTION. Since
[x] E Z for any real number x, we have 3 x [x] m E Z, 4
=
=
256
3.
Number Theory
x = 4m/3. But then m = [x] = [4:] = [m + ; ] = m + [;] , which implies that [m/3) = 0. This condition, however, is satisfied by only three integers, 0, 1, 2, and so x can take on only the values 0, � , �- Fi and thus
nally, we verify that these three values are indeed solutions of the given 0 equation. (ii) Solve the equation
[x3] + [x2) + [x] = (x) - 1 in real numbers.
SoLUTION . We rewrite the equation as
(x) [x3] + [x2) + [x] + 1, and see that (x) E Z. Since 0 < (x) < 1, we have (x) = 0, i.e., x = [x) E Z. Then [x3] = x3, [x2) x2 , and it remains to solve the Diophantine equation x3 + x2 + x + 1 = 0. Factoring, (x + 1)(x2 + 1) = 0, and since for any integer x we have x 2 + 1 1= 0, we get x + 1 = 0, and thus x -1 . Finally, by substituting we verify that x -1 is indeed a 0 solution. =
=
=
=
(iii} Find all positive integer solutions of the equation
[ 1�] + [ 120 ] + · · · + [ n� 1 ] + [ � ] = 217.
SOLUTION. We first add the expression on the left. For i = 2, . . . , 9 we have If we set = then by the solution of 7.2.(iv) the number is the greatest multiple of not exceeding and therefore 2
1,
[i/10)10k= 0. k [n/10), 10 n, [ 110] + [ 10 ] + · · · + [n� 1 ] + [ �] ([��] + . . . + G�D + . . . + ( [ 10 we have we also have =
D
�'--v-'
•
'--v--'
·
a
8 Base Representations
267
This equality has to hold for all n, in particular for
t1 (9a - � ) = 2c - a - b, t2 (9a - c2 ) = 2c- a - b, and by subtracting we obtain (t 2 -t1)(9a- c2) = 0. Since t2 -t1 = 1 0 1= 0, this implies = 9a, 2 c = a+b. If, on the other hand, the digits a, b, c satisfy the conditions c2 = 9a, 2 c = a + b, then the identity ( 4 1 ) is satisfied for all n. It remains to determine the a, b, c that satisfy these two conditions. From c2 = 9 a it follows that c is divisible by 3. Since c 1= 0, we have c E { 3, 6, 9}. If c = 3, then a = 1, b = 5; if c = 6, then a = 4, b = 8, and finally, if c = 9 then a = 9, b = 9. Our problem has therefore three 0 solutions altogether. c2
8. 6 Exercises
(i} Show that for each natural number n, the difference between the number consisting of 2n ones and the number consisting of n twos is
(ii) (iii) (iv) (v)
the square of an integer. Determine this integer. Suppose that the number consists of n threes, and of n sixes, where n What are the digits of the product Show that for any n, the integer consisting of n ones is divisible by the integer consisting of ones if and only if I n. Given a natural number find the smallest number consisting only of ones that is divisible by the number consisting of threes. For which digits is there a natural number n > 3 such that
m1 kEN k k,
E N.
m m1 m2 ?2 k k
a
1 +2+ ···+n
(vi)
has no other digits than a in its decimal expansion? For > 1 we denote by M = 1 1 . . . 1 the number consisting of ones. Does there exist a natural number n divisible by M such that its digit sum is less than the digit sum of M?
m
m
m
To conclude this section, we describe a method that exemplifies the diver sity of methods that can be used for solving problems on digit expansions. We will take advantage of the concept, introduced in the previous section, of the integer part of a real number.
8. 7 Example
ab a2 , . . . , as ,
a1 f=
0, there exists a nat Show that for any digits with ural number n such that the decimal representation of n2 begins with
at, a2 , . . . , as .
268
3. Number Theory
t · 10] ++1.asThen · We m = a t · 1os-1ot +k-at2and1osset-2 +n =· ·[10+ kas·-Vm
SOLUTION. We set choose a k E N such that 2Vm < by (35) we have
·
·
which implies
102km < n22 1 be a natural number. Find all natural numbers that divide at least one of the integers an = an + an - l + . . . + a + 1, where n *(iii) Given a natural number n, show that from among 2n+l - 1 natural numbers one can choose 2n numbers such that their sum is divisible by 2n * (iv) Given a natural number n , show that from among 2 3n 1 natural numbers one can choose 3n numbers such that their sum is divisible by 3n .
E N.
·
9.4
-
Examples
(i) Suppose that the product of nine distinct natural numbers is divisible by exactly three primes. Show that from among these nine numbers we can choose two distinct ones whose product is the square of an integer.
SOLUTION. Let us denote the given primes by Pl , P2 , P3 and let A { a1 , a2 , . . . , ag } be the set of the nine given numbers. From the given con dition it follows that none of the numbers a1 , . . . , ag is divisible by any prime other than Pl , P2 , P3 . Each element of A can therefore be written in
9 Dirichlet's Principle
271
E
the form p�1 • p�2 p3'3 , where n�, n2 , n3 N0 . It is clear that the product p�1 P�2 Pa3 • pf1pr2 p33 is the square of an integer if and only if the sums n1 + m 1 , n2 + m2 , n3 + m3 are even, i.e., the exponents ni and mi have the same parity for each i = 1, 2, 3. We now consider all eight triples (c:�, c:2 , t:3 ) of zeros and ones, and for each such triple we consider the set of numbers p�1 • p�2 Pa3 from A such that all numbers ni + c:i, for i = 1, 2, 3, are even. Since in this way we distribute nine numbers over eight sets, one of these sets must contain at least two numbers p�1 p�2 • Pa3 and pf1 • pr2 • Pa3 • But then the exponents ni, mi have the same parity for i = 1 , 2, 3, and 0 thus the product of these numbers is the square of an integer. •
•
•
(ii) Suppose that four distinct natural numbers are divisible by exactly three primes. Show that for some k, 1 < < 4, one can choose k numbers
k from among these four such that their product is the square of an integer.
SOLUTION. Let us again denote the given primes by Pb P2 , P3 , and the set of given numbers by A = { a 1 , a2 , a3 , a4 }. The set A has 24 - 1 = 15 nonempty subsets. For each of these subsets we consider the product of its elements; each of these 15 products is of the form p�1p�2 Pa3 • Just as in (i) we group these products into eight groups according to the parity of their exponents n� , n2 , n3 . By Dirichlet's principle there exist two different subsets of A whose products are in the same group. The product of these two products is the square of an integer. Therefore, also the product of all numbers of the set A that belong to exactly one of the two subsets (i.e., the product of the elements of the symmetric difference of the two subsets) 0 is the square of an integer. (iii) Suppose that the product of 48 distinct natural numbers is divisible by exactly ten primes. Show that one can choose four of these 48 numbers such that their product is the square of an integer.
SOLUTION. We denote the given primes by Pl , . . . , Plo and form all two element subsets of the given set A of 48 numbers. The number of these subsets is 48 47 = = 1128.
(�)
�
We form the products of the elements of these subsets; they can be written as p�1p�2 p�/)0 • We consider all 10-tuples (c:�, . . . , c:10 ) of zeros and ones; 1 there are 2 0 = 1024 of them. We distribute the 1128 subsets over the 1024 groups determined by the 10-tuples (c:�, . . . , c: 10 ) such that the subset with product of elements p�1 • p�J0 belongs to the 10-tuple (c:1 , . . . , c:10 ) for which all the numbers n 1 + € 1 , . . . , n 10 + € 10 are even. By Dirichlet's principle there exists at least one group containing at least two such subsets A1 = { a, b} , A2 = {c, d} . The product abed is then the square of an integer. If A 1 , A2 are disjoint, then a, b, c, d are the desired four numbers. Otherwise, A�, A2 have exactly one common element, since A 1 =/:. A2 . Without loss of •
•
•
•
•
272
3. Nwnber Theory
generality we assume that a = c, b 1= d. Then the product bd is the square of an integer. Now we repeat the previous argument with the 46-element set B = A \ {b, d}. This time the number of two-element subsets is
(46) = 46 2. 45 = 1035 ' 2
which is again more than 1024. As a consequence there are two subsets B1 = { e, /}, B2 = {g, h} such that the product efgh is the square of an integer. If B1 , B2 are disjoint, then e , /, g, h are the desired numbers. If B1 and B2 have an element in common, say e = g, f 1= h, then the product D fh is a square, and the numbers b, d, J, h have the desired property. (iv) Suppose that the product of 55 distinct natural numbers has exactly three prime divisors. Show that the product of some three of these 55 numbers is the third power of an integer. SOLUTION. We denote the given primes by Pb P2 , P3 and consider all triples (c:�, c:2 , t:3 ), where c:�, c:2 , €3 { 0, 1 , 2}; their number is 33 = 27. We divide the given 55 numbers, which are of the form p�1p�2 Pa3 , into 27 groups by associating to each triple (c:�, c:2 , c:3 ) those numbers p�1p�2 Pa3 that satisfy n 1 = c: 1 (mod 3), n2 c:2 (mod 3) , n3 = c:3 (mod 3). By Dirichlet ' s principle, at least one of these groups contains at least three numbers, say a = pf1 pr2pr3 , b = p�1p�2p33 , c = p�1p�2p�3 • Then for all i = 1, 2, 3 we have mi = ni = ki (mod 3) , and thus mi + ni + ki = 3ki - 0 (mod 3) , i.e., mi + ni + ki = 3ri for appropriate ri N0 . This implies that
E
=
E
0
The product abc is therefore the third power of an integer .
(v} Given the natural numbers a� , a 2 , . . . , a 25 , whose product is divisible by exactly two primes, show that there exist i, j, 1 < i < j < 25, such that ai · ai+ l · · · ai is the fifth power of an integer. SOLUTION. We denote the given primes by p1 1 p2 and consider all pairs (c:� , c:2 ), where c:1 , c:2 No satisfy c: 1 < 5, c2 < 5. There are 5 2 = 25 such pairs. We divide the numbers j = 1 , 2, . . . , 25 into 25 groups such that j belongs to the group corresponding to (c: 1 , c:2 ) if and only if for a 1 · a2 · · · ai = P�1 P�2 we have n1 = c: 1 (mod 5) , n2 = c:2 (mod 5). If some number j belongs to the group associated with (0, 0) , it suffices to set i = 1; then these numbers i, j satisfy the given conditions, since a 1 · a2 . . . ai = P�1p�2 , where n1 = n2 0 (mod 5) , and thus for appropriate r1 1 r2 N0 we have n 1 = 5r 1 , n2 = 5r2 , which means that a 1 · a2 · . ai = (p�1p�2 ) 5 . Now we assume that none of the numbers j belongs to the group cor responding to the pair (0, 0) . Then the 25 numbers j are distributed over
E
=
.
E
9 Dirichlet's Principle
273
the remaining 24 groups, and by Dirichlet 's principle, at least one of these groups will contain at least two numbers j 1 , j2 , j1 < j2 . If we set i = i1 + 1, j = J·2 , then clearly 1 < i < j < 25, and we have where n 1 = m 1 (mod 5), n2 = m2 (mod 5). From the construction of the numbers n� , n 2 , mt , m2 it follows that n 1 < m 1 , n2 < m2 , and so there exist r1 , r2 No such that nt + 5r 1 = m t , n 2 + 5r2 = m2 . This implies that
E
a. . a· l t+
2- 2 - prl T2 )5 1 . . . aJ· - pml 1 -nlpm 2 n - ( 1 . P2 '
which means that the numbers i, j have the desired property.
0
9. 5 Exercises (i) Suppose that the product of five distinct natural numbers is divisible by exactly three primes. Show that from these five one can choose an even number of integers whose product is the square of an integer. (ii) Suppose that the product of 45 distinct natural numbers is divisible by exactly ten primes. Show that for some 1 < < 4, one can choose of those 45 numbers such that their product is the square of an integer. *(iii) Prove the following strengthening of Example 9.4. (iv): Suppose that the product of 29 distinct natural numbers has exactly three prime divisors. Show that the product of some three of these 29 numbers is the third power of an integer. *(iv) Suppose that the product of 27 distinct natural numbers is divisible by exactly three primes. Show that for some < < 3, one can choose of these 27 numbers such that their product is the third power of an integer. (v) Given the natural numbers s, n, with s > 1 , show that for any natural numbers a 1 , a2 , . . . , a8n. whose product is divisible by exactly n primes there exist i, j, 1 < i 2 the value F(k) is equal to the binomial coefficient (�) . This example was not arbitrarily chosen; indeed, in order to be able to easily write down polynomials with the desired property, it will be conve;nient to extend the concept of binomial coefficient. For any real number t and any natural number we define the generalized binomial coefficient (�) by
E
k
k
(kt ) = t(t - 1) . k!. (t - k 1) . We remark that if t is a natural number and t > k, then this new definition coincides with the original one. 1. For any integer t and any natural number k the generalized binomial coefficient (�) an integer. Given t E Z we find a rational number r such that r k! > k - t, and we set n t + r k! . By 10.1, the polynomial F(x) x(x - 1) (x - k + 1) satisfies F(t) F(n) ( mod k!), since t n ( mod k!). But then F (n) = G) · k! , .
+
Theorem
is
PROOF.
·
=
·
=
=
·
·
·
=
E n B) ).
where the binomial coefficient (�) is a natural number ( since n N and n > and so (:) is a coefficient in the binomial expansion of (A + Together, F (t) F(n) 0 (mod
k,
=
= (;)k!
=
k!) ,
10 Polynomials
and thus the integer F( t) is divisible by k!. Therefore, (�) = integer.
F��)
277
is an 0
Given any polynomial F(x) = bnxn + bn- 1 xn- 1 +· · ·+b1 x+bo , where bo , b�, . . . , bn are real numbers, there exist real numbers ao, a�, . . . , an such that x F(x ) = ao + a 1 + · · · + an- l n 1 + an PROOF. We use induction on the degree of the polynomial F(x). If F(x) is the zero polynomial or deg F (x ) = 0, then the assertion is clear. We will now assume that given any natural number n, the assertion holds for all polynomials of degree less than n. Our aim is to prove it for a polynomial F(x) = bnxn +· · · + b1 x+bo of degree n. We set an = bn · n! and consider the polynomial G(x) = F(x) - an (�) . Since the polynomials F(x) and an (�) have the same degree n and the same coefficient bn of xn , G(x) is either the zero polynomial or has smaller degree than F(x). It follows from the induction hypothesis that there exist real numbers a0 , , an- 1 such that G(x) = ao + a 1 (�) + · · · + an- 1 (n:_ 1) . But then x F(x) = G(x) + a, = ao + a 1 + · · · + an - 1 n 1 + a.. .
Theorem 2.
G)
( ) (:}
.
•
G)
(:)
which was to be shown.
•
•
( ) (:)
0
For any polynomial F(x) = ao + a 1 (�) + · · · + an (:) , where ao, . . . , an are real numbers, the following holds: a 0 , a�, . . . , an are integers if and only if F(t) is an integer for any integer t.
Theorem 3.
1,
a0 , a�, . . . , an are integers, then by Theorem F(t) must be Z. an integer for any t Let us now assume that F(t) is an integer for any integer t; we will prove by induction that a0 , a 1 , . . . , an are integers. Certainly, ao = F(O) is an integer. Next we assume that for some natural number k < n, the coefficients a0 , a�, . . . , ak - 1 are integers, and we aim to show that ak is an PROOF.
If
E
integer as well. We have
F(k) = ao + a 1 since (�) = 1 and
(�) + · · + ak- 1 (k k 1) + ak , ·
(!J = 0 for any integer m > k + 1. Hence
ak = F(k) - ao - a1
G) - · · · - ak- 1 (k k 1 }
and thus ak is an integer, which completes the proof.
0
3. Number Theory
278
We note that in the second part of the proof we used only the fact that F (O), F(1), . . . , F(n) are integers. Then it followed that ao, ab . . · , an are integers, and therefore F(t) has to be an integer for any integer t. This fact is worth being formulated as a separate result.
If a polynomial F(x) with real coefficients and degree n is such that F(O), F(1), . . . , .F (n) are integers, then F(t) is an integer for any integer t.
Consequence.
This follows from the above remark and the fact that by Theo rem 2 we can express the polynomial F(x) in the form ao + a 1 (�) + · · · +
PROOF.
D
an (:) .
1 0. 5 Examples (i} Find all real numbers a, b, c with the property that for each integer t the number at 2 + bt + c is an integer. SOLUTION.
as
Using Theorem 2 in 10.4, we express the trinomial ax2 +bx+c
ax2 + bx + c = 2a
(;) + ax + bx + c = 2a (�) + (a + b) G) + c.
Hence by Theorem 3 in 10.4, at2 + bt + c is an integer for all integers t if D and only if 2a, a + b , c are integers. (ii} Decide whether the polynomial F(x) = 2� (x9 - 2x5 + x) takes on integer values F (t) for all integers t. By the consequence in 10.4 it suffices to establish whether F(O), F(1), . . . , F(9) are integers. It is easy to check that F(O), F(1), . . . , F(4) are indeed integers. However, F(5) is not an integer, since 59 - 2 - 55 + 0 5 5 ¢ 0 ( mod 25). SOLUTION.
=
(iii} Decide whether there exists a polynomial F(x) with real coefficients and degree n E N such that F(O), F(1), . . . , F(n - 1) , and F (n + 1) are integers, while F(n) is not an integer. SOLUTION .
By Theorem 2 in 10.4 we can express F(x) as
F (x) = ao + a 1
G) + · · · + an (:) .
The numbers F(O), F ( 1 ), . . . , F(n- 1 ) must be integers, and thus a0 , a�, . . . , an- 1 Z, while F(n) cannot be an integer, which means that an ¢:. Z (see the proof of Theorem 3 in 10.4). Let us now evaluate F( n + 1):
E
F(n + l) = ao + a 1
e � 1 ) + · · · + an- l (: + D + an (n ; l }
10 Polynonti�
279
thus F ( n + 1) will be an integer if and only if an (n� 1) = (n + 1 )an is an integer. It is easy to see that if we choose, for instance, ao = a 1 = · · · = an- 1 = 0, an = 1/ ( n + 1), then all conditions are satisfied. The conditions of the problem are therefore satisfied by the polynomial
()
F(x) = 1 x = x(x - 1) · · · (x - n + 1) · n+ 1 n (n + 1)!
0
10. 6 Exercises (i) Find real numbers a, b, c, d with the property that at3 +bt2 + ct +d
all is an integer for all t E
z.
(ii) Decide whether the polynomial F (x) = 1� (x7 - x5 - x3 + x) takes
on an integer value F( t) for any integer t. *(iii) For each n decide whether there exists a polynomial F(x) with real coefficients and degree n such that F(O), F(1), . . . , F(n - 1), F(n + 1 ) , F(n + 2) are integers, while F(n) is not an integer. *(iv) Determine the n for which there exists a polynomial F(x) with real coefficients and degree n such that F(O), F(1), . . . , F(n - 1), F(n + 1), F(n + 2), F(n + 3) are integers, while F(n) is not an integer. (v) Let p be a prime and suppose that the polynomial F(x) of degree k < p and with integer coefficients has the following property: For each integer x the value F( x) is an integer multiple of p. Show that then coefficients of F(x) are integer multiples of p.
EN
EN
all
We will now deal with the decomposition of a polynomial with inte ger coefficient into a product of two polynomials with integer or rational coefficients.
10. 7 Example Show that there do not exist polynomials F(x), G(x) with integer coefficients and degrees at least one that satisfy the identity
F(x) . G(x) = x 10 + 2x9 + 2x8 + 2x7 + 2x6 + 2x5 + 2x4 + 2x3 + 2x2 + 2x + 2. We assume to the contrary that such polynomials F(x), G(x) exist, and we write F(x) = as x8 +· · · + a1x+ao , G(x) = btxt + · · ·+b1 x+bo , where a0 , ab . . . , a 8 , b0 , bb . . , bt are integers. Then a0 b0 = 2, so exactly one of the numbers ao, bo is even . We assume that it is ao (if it were b0 , we could simply interchange the polynomials F(x), G(x)). Hence a0 is even and bo is odd . Equating the coefficients of x, we obtain 2 = a0 b 1 + a 1 b0 ; thus a 1 b0 = 2 - aobb which is an even number. Since bo is odd, a 1 must SOLUTION .
.
280
3. Number Theory
be even. Comparing coefficients of x2 and simplifying, we find that a 2 bo 2 - a 1 b1 - aob2 is even, and thus a2 is even. If we continue in this way, we find after s steps that as is also an even number. But this is a contradiction to the fact that as · bt = 1 ; hence the polynomials F(x) , G(x) with the given 0 properties do not exist. =
With a similar argument one can obtain a proof of the following useful criterion. 10. 8 Eisenstein's Irreducibility Criterion Theorem. Let p be a prime and
a polynomial with integer coefficients such that ao, a 1 , . . . , an- 1 are divisible by p, the coefficient an is not divisible by p, and a0 is not divisible by p2 . Then there do not exist polynomials G(x), H(x) with mtional coefficients and with degrees at least one such that F (x) = G (x) H (x) . ·
A proof can be found, for instance, in [4] .
1 0. 9 Examples (i) Show that the polynomial 4x5 + 7x4 - 14x3+49x 2 -28 cannot be written as the product of two polynomials with rational coefficients and degrees at least one . SOLUTION. The assertion follows from 10.8 with p = 7.
0
(ii) Find a polynomial with rational coefficients and smallest degree that
has the irrational number a = 1980 as a zero.
SOLUTION. The number a is certainly a zero of the polynomial x 1 987 - 2. Let F(x) denote the desired polynomial, which has the smallest degree among all polynomials with rational coefficients and that have a as a zero. We will show that F(x) = c · (x 1987 2), where c Q \ {0}. To do this, we divide the polynomial x1 987 - 2 by F(x) with remainder: -
E
x1 987 - 2 = H(x) · F(x) + R(x) (see Theorem 3 . 9 in Chapter 1), where R(x) is either the zero polynomial or deg R(x) < deg F(x) . If we think about the division algorithm, we easily see that both H(x) and R(x) have rational coefficients . Also,
R (a) = (a 1 987 - 2) - H ( a) · F ( a) = 0
10 Polynomials
281
(since F ( a) = 0) , and thus the polynomial R(x) with rational coefficients has a as a zero. From the definition of F( x) it follows that the degree of R(x) (if it is defined) cannot be less than that of F(x). Hence R(x) is the zero polynomial, and therefore x 1987 - 2 = H(x) · F(x). By 10.8, however, the polynomial x 1987 -2 cannot be written as a product of two polynomials with rational coefficients and of degrees at least one. Since F(x) has a as a zero, we have deg F(x) > 1, and thus deg H(x) = 0, i.e., 0 H(x) = 1/c, where c =f= 0. Hence we have indeed F(x) = c· (x 1987 - 2).
10. 1 0 Exercises Show that the polynomials (i)-(iv) cannot be decomposed into a product of two polynomials (with rational coefficients and degrees at least one). (i} x5 - 5x4 + 25x - 5. (ii) x6 - 6x3 + 12. (iv) 5x5 +25x4 - 100x2 + 125x - 25. (iii) 3x7 - 21x4 - 63x2 - 21.
(v) Find a polynomial with rational coefficients and smallest degree that has 12if75 as a zero.
In the concluding example of this chapter we will present a result that simplifies the search for rational roots of a polynomial with integer coefficients. We have already illustrated its use in 3.27. (ii) of Chapter 1.
10. 1 1 Example Suppose that the polynomial F(x) = an xn + an - l Xn- l + · · · + a 1 x + ao
with integer coefficients has a zero a = r/ s, where r E Z, s E N, and r, s are relatively prime. Show that then for each integer k the number F( k) is divisible by r - ks . SOLUTION. We set G(x) = sn F (xjs) . Then G(x) is a polynomial with integer coefficients: G(x) = an Xn + an- 18Xn - l + . . . + a l sn- l x + ao sn . We have G(r) = sn F(a) = 0 and G( ks) = sn F( k ) . Let us set m = lr - ksl. If r- ks 0, then k = a, and so F( k ) = 0 is indeed divisible by r - ks = 0. We now assume that r - ks =I= 0, i.e., m E N. Since ks = r (mod m), we have by 10. 1 , G( ks) = G(r) = 0 (mod m), =
282
3. Number Theory
which implies m I G ( ks) , i.e., m 1 sn F ( k ) . Now , m and s are relatively prime integers (if there were a prime p dividing both m and s, it would also have to divide r, which contradicts (r, s) = 1), and thus by 1.14.(ii) 0 we have m 1 F ( k) , or r - ks I F ( k ) , which was to be shown.
4 Hints arid Answers
1
Hints and Answers to Chapter 1
1.4 (i) Add identities (3) and (4).
(ii) Subtract (4) from (3) . (iii) After multiplying by 2, use (1). (iv) The summand (��) ( v'2) ( v'3) (v) m = 3, n = 6.
22 28 .
1.7 Instead of B in (5) write -B and set n = 2m in (i) and n = 2m - 1
in
(ii).
2.3 (i) n 2 •
(ii) For n = 2k: S = - k; for n = 2k + 1: S = k + 1. (iii) 5n(n + 1)/2. (iv) 17. (v) 2 - 2. (VI) ( ) 3 ( 2 - 1) - n. ( ) 2n +
2n+(3.2-nl)n-l +l
'
n+ l
Vll n
•
n-l n -2 n2
Vlll " '
"
(x2nx_2nl).·((x2x2-n+l)2 +1) .
(n + 2) · 2 . (ii) 1. [Compare with the sum in 2.6 and use (4) .] (iii) n(n - 1) 2 (iv) n(n + 1 ) 2 - [Note that k = k(k - 1) + k and use (iii) .] (v) � [Consider (n + 1)8.] (vi) (:;;)�:t�) . [Consider (n + 1)(n + 2)8.] �+;n [Consider (n + 1)S; after manipulations as in 2.5 (ii) (vii) use 1.2.] (viii) 2 l /(2n)!. [Consider (2n)!S. ] 2.8 (i)
·
n l. 3n+ -3 . 2n -
.
•
2
284
4. Hints and Answers
2.10 (i) Consider the coefficient of
(1 + x) n (1 + x) m .
xP in the polynomial (1 +
x)n+m =
·
(ii) Follows from (i) by choosing m = n, p = n - r. [Use the method of 2 5 . ]
(iii) [�!��)l1k .
.
Sk (n) successively from the formulas B(3, n) = S3 (n) + 3 · S2 (n) + 2 · S1 (n), B(4, n) = S4 (n) + 6 · S3 (n) + 11 - S2 (n) + 6 · S1 (n), B(5, n) = Ss(n) + 10 · S4 (n) + 35 S3 (n) + 50 · S2 (n) + 24 · S1 (n) . (iii) n2 (n2 + 2n + 2). (ii) n4 . 2.22 (i) ( n3 + 2n)/3. (ii) k(k + 1) (k + 2) (3k + 2.25 (i) n(2n - 1) (2n + 1) (iii) n2 (2n2 - 1). (iv) For n = 2k: S = - k2 (4k + 3); for n = 2k + 1: S = (k + 1) 2 · (4k + 1). (v) n(4n2 + 15n + [First evaluate each term in the sum S .] 2.30 (ii) [n(n - 1)qn+ 1 - 2(n2 + n(n + 1) 1 (iii) For t = 1 : S(1) = n(n - 1)/2; for t 1= 1: S(t) (iv) For v = 1: S(1) = n(n + 1) (2n + 1)/6; for v 1= 1: S(v) = R2 (v, n)jv. (v) For X = 1: S( 1) = n2 ; for X 1= 1: S(x) = xn 0.] (iv) Verify by inducticn that (x2 + x + 1)Fk (x) = x21c+l - x21c + 1. 3.6 (i) S = F(1) = 1. (ii) After multiplication by a2 (x - p) (x - q) and simplification, find a quadratic polynomial with discriminant D = (p - q) 2 + 4a4 > 0 and with nonzero values for x = p and for x = q. (iii) Does not exist: For a = 0 it has at most one zero, or infinitely many zeros; for a > 0, A > -c + IJ2/(4a) it has no real zeros; for a < 0, A > -c it has one negative zero. (iv) a = ± 2{1220 - 1 , b = - � , p = -1, q = }. (Comparing coefficients of x20 , you obtain 220 - a20 = 1 . Substituting x = � leads to the equation (� + b) 20 + (� + � + q) 10 = 0, which implies b = - � . Then the original equation of polynomials can be rewritten as (x - �) 20 = (x2 + px + q) 10 , which implies x2 + px + q = (x - �) 2 .] 3.8 (i) F(x) = x2 + x. (ii) No such polynomial exists: F(x2 + 1) has even degree. (iii) F(x) = x2 - 3x. (Since deg F(x) = 2, set F(x) = ax2 + bx + c.] (iv) Set y = x - 1 and G(y) = F(y - 1), and compare with 3 . 7; apart from the zero polynomial, the solutions are the polynomials F(x) = (x + 1) n , where n No. (v) F(x) = xn . [From a comparison of the degrees of F(F(x)) and (F(x)) n it follows that deg F(x) = n; therefore, F(x) = an xn + an- 1 Xn - 1 + · · · + ao . Comparing the coefficients of the powers xn2 , you then obtain a: 1 = a:, that is, an = 1 . Assume that ak 1= 0 for some k < n and take the largest such k. The condition F(F(x)) = (F(x)) n can be rewritten as (iii)
E
+
ak (xn + ak xk +
· · ·
+ ao) k + ak - 1 (xn + ak xk + · · · + ao) k - 1 + · · · + ao = 0,
which is a contradiction, since the polynomial on the left has degree nk . ] 3.13 (i) (a) 6; (b) 6x. [Use (25) , and replace x by 1 and -1.) (ii) 0 is a zero only for even n, and it has multiplicity 2. [Use 1.2.) (iii) For even n. [Use (5) and determine when x + 1 divides xn+ 1 + 1 .] (iv) F(x) = 4x4 - 27x3 + 66x 2 - 65x + 24. [From the equality of the polynomials F(x) = (x - 1) 2 P(x) + 2x = (x - 2) 3 Q(x) + 3x it follows that you must find a polynomial of smallest degree such that (x - 1) 2 divides (3x - 4)Q(x) + x. From this you can see that deg Q(x) > 1.
286
4. Hints and Answers
1, that is, Q(x) = ax + b, you obtain the condition (3x - 4)(ax + b) + x R(x)(x - 1) 2 , which implies R(x) = 3a, and comparing coefficients, a = 4, b = -3.] (v) The desired remainder is F(1) 2 000. By Bezout's theorem there exists a polynomial H such that F(x) F(1) + (x - 1)H(x), which implies F (x5 ) = F(1 ) + (x5 - 1)H(x5 ) F(1) + (1 + x + x2 + x3 + x4 )(x - 1)H(x5 ) . 3.17 (i),(ii) Use the method of 3.16 (ii) ; verify the equality of the polynomi als by substituting 0, b, -b, c, -c for x (in (i), four of them suffice) . Special For deg Q(x)
=
=
=
=
=
care must be taken in the case where some of these quantities coincide. (iii) Set x = a, x = b, x = c. (iv) Set x = a, x = b, x = c. (v) Start with 2x3 + 3x2 + x = x(x + 1)(2x + 1), show that 0, -1, � are zeros of the given polynomial, and use 3.15 (ii). (vi) Show that F(x) has the zeros 0, 1, . . . , 25; therefore, F(x) = x(x 1 ) · · · (x-25 ) · G(x) for an appropriate polynomial G(x) . By substituting you see that G(x) G(x - 1), and therefore G(x) and the constant polynomial G(O) have the same values at all integers; thus G(x) = G(O) by 3.15 (i) . The given equation is therefore satisfied exactly by the polynomials a x · (x - 1) · · · (x - 25), where a is an arbitrary number. 3.20 (i) p = q 0; p = 1 and q = -2. [Use (28).] (ii) Proceed as in 3.19. (iii) Add the equations xf(ax� + bxi + c) = 0 for both roots Xi · 3.22 (i) If x 1 + x2 = 1, then by (29) you have X3 = - � , which implies -3 and x 1,2 (1 ± v'i3) /2. ..\ (ii) From (29) and from the hypothesis it follows that -q = X 1 X2X3 = x 1 + x2 = -x3 ; hence q X3 is a root, and therefore q3 + pq + q = 0. On the other hand, if the numbers p, q are such that at least one of them is nonzero and q3 + pq + q = 0, then one zero of the given polynomial is the sum of the reciprocals of the other two zeros. (iii) From (29) and from the hypothesis it follows that -p/2 is a zero of the polynomial, and therefore 8r = 4pq - p3 . On the other hand, if the numbers p, q, r satisfy 8r 4pq - p3 , then one zero of the given polynomial is the sum of the other two. (iv) G(x) = x3 - qx2 + prx - r2 ; H(x) x3 + 2px2 + (p2 + q)x + (pq - r). (v) r (9pq - 2p3 ) /27. [Show that one zero is -p/3.) (vi) Use (29) and substitute. (vii) x + y + z p + q + r - b - c. Consider the monic cubic polynomial -
=
·
=
=
=
=
=
=
=
=
F(t)
=
(1 - xt - t -Y b _ _
z
-
t - c )t(t - b)(t - c)
with three different zeros p, q, and r. Vieta's relation gives that -p - q - r equals the coefficient of t2 in F(t), which is -b - c - x - y - z, and the
1 Hints and Answers to Chapter 1 us
result follows. Let unique:
add that the numbers
(p - b) (q - b)(r - b) x = pqr y = ' be ' b( b - c) the polynomial
-
z=
3.24 to show that 4 6x - 40 in ( i) , resp. J5
3.25 Use the method of
x3
x, y,
z
287
always exist and are
(p - c)(q - c) (r - c) . c(c - b) is the unique real zero of is the unique real zero of
x3 - 3x - 2 · J5 in (ii). 3.28 (i) F(x) has a double zero at 1 and the two simple complex zeros -2 + i, -2 - i. (ii) F(x) has the zeros -3, ! , (-1 + J29)/2, (-1 - J29)/2. (iii) F(x) has the double zero -1 and simple zeros 1, -�, -2 + v'J, -2 -
JJ.
(iv) Let a denote the common irrational zero of F(x) and G(x).
a2 = -aa - b into the equation F( a) 0, you obtain 2 2 (a - b + p)a + (ab + q) = 0, which implies p = b - a , q -ab, and this leads to F(x) (x - a)G(x) . (v) The root in question is x = 1. 4.7 (i) x2 + (p3 - 3pq)x + ql. (ii) x2 + ( -p2 + 2q + p)x + ( -p3 + 3pq + q2 + q). (iii) b E {0, VJ5 - 1, -VJS - 1}. 4.9 (i) x2 - 6x + 8 or x2 + (13/3)x - 52/9. (ii) x2 + Bx + 15 or x2 - 8x + 15. (iii) x2 - 5x + 6 or x2 + (3/2)x - 19/12. ( iv) For the numbers p = a + b and q = ab, find two "independent" equations; from those you will realize by excluding the number p that q is a root of the given equation of degree 6. Since a3 (a + 1) = b3 (b + 1) = 1, you have 1 = a3 (a + 1) b3 (b + 1) = ql(p + q + 1). From the equation a4 + a3 = b4 + b3 , upon dividing by a - b, obtain the second equation for the numbers p and q: 0 = a3 + a2 b + ab2 + b3 + a2 + ab + b2 = (a3 + a2 ) + (b3 + b2 ) + q(p + 1) 1 1 = -a + b + q(p + 1) = p-q + q(p + 1) . Substituting
=
=
=
4.14 ( i) x3 - 2x2 - 47x - 144. (ii) Use the fact that a, b, c are three distinct zeros of a cubic polynomial
with vanishing quadratic term . (iii) Use the formula for s4 = xf + x� + x§ from 4.12. ( iv ) x3 + 2px 2 + (p2 + q)x + (pq - r).
288
4. Hints and Answers
Transform the supposed equation into (a + b) (a + c) ( b + c) = 0 and then discuss (for example) the case a + b = 0, in which an + b = 0 for any odd n.
n
(v )
(x l + X2 + X3 )(X 1 X2 + X 1 X3 + X2 X3 ). ( ii) (x 1 + X2 + X3 )(2x� + 2x� + 2x� - X 1 X2 - X 1 X3 - X2X3 ). ( iii) (x 1 X2 + X 1 X3 + X2 X3 ) 2 . ( iv ) 24x 1 X2 X3 . 5.3 ( i) { (x� , x2 , . . . , Xn ) }, where X k k - n/2 for k = 1, 2, . . . , n. ( ii) {(x1 , x2 , . . . , xn )}, where Xk = (n - 2k + 2)/2. ( iii) { (x 1 , x2 , . . . , xn )}, where x 1 = (n + 1)(2 - n)/2 and Xk = 1 (k > 1). 5.6 ( i) {((a 1 +a2 +a3 +a4 )j2, (a 1 +a2 -a3 -a4 )j2, (a 1 -a2 +a3 -a4 )j2, (a l a2 - a3 + a4 )/2)}. ( ii) For p = -3: no solutions; for p = 0: infinitely many solutions of the form (r, s, 1 - r - s ) , where r, s R; for p E 1R \ {0, -3}: the unique solution X1 = X2 = X3 1/(p + 3). ( iii) Necessary and sufficient condition for solvability: a 1 a2 + a3a4 a 1 a3 + a2 a4 a 1 a4 + a2 a3 , which holds if and only if at least three of the numbers a 1 , a2 , a3 , a4 are equal. If { 1, 2, 3, 4} = {p, q , r, s } and ap = aq = ar , there exists the unique solution Xp = Xq = Xr = !a�, Xs = ap (a8 - �ap) · (iv) {( aa �aa ' al �aa ' aa ; al )}, where a = a l + a2 + a3· (v) If a 1 = a2 = a3 , then there are infinitely many solutions of the form (r, s, 3 - r - s) , where r, s JR.; otherwise, there is the unique solution X1 X2 = X3 = 1. · 21c + l ) for k = 1 , 2, . . . , n. (v1. ) {(x 1 , x2 , . . , X )}, where Xk = a · 2""'+ 12-n+i(n_1 (vii) (x� , x2 , x3 , x4 ) }, where x 1 = ( 4a l + 3a2 + 2a3 + a4 ) /5, x2 = (3al + 6a2 + 4a3 + 2a4 ) /5, X3 = (2a 1 + 4a2 + 6a3 + 3a4 ) /5, X4 = ( a 1 + 2a2 + 3a3 + 4a4 )/5. (viii) Show that x 1 + x2 + · · + x 10o = 0; comparing this result with the sum of the 1st, 4th, . . . , 97th equation, you obtain x 100 = 0, and from the sum of the 2nd, 5th, . . . , 98th equation you get x 1 = 0. With this, the last equation immediately gives x2 = 0, the first one gives X3 = 0, etc. 5.9 ( i) (x 1 , x 2 ) {(-6, -2), (-4, -4)}. ( ii) (x� , x2 ) {(3, 2) , (2, 3)}. (1, 2), (i, �)} . [By subtracting you obtain a linear ( iii) (x1 , x2 ) equation. ] ( iv ) (x1 , x2 ) {(-3, 4), (4, -3)}. { (4, 3), (4, -3)}. [Solve the first equation for x 1 and ( v ) ( x 1 , x2 ) substitute into the second.] (vi) (x 1 , x2 ) {( -7, -3), ( 7, 3)}. [Add the two equations.] (vii) By adding five times the first and seven times the second equation, you obtain the homogeneous equation 5x� - 19x 1 x2 + 12x� = 0. Upon dividing by x� (x2 #= 0) you get a quadratic equation in x 1 jx2 , which leads to the solutions (x1 , x2 ) {( -4, -5), ( -3 J3, - J3) , (3 J3, J3) , (4, 5) } . 4.16 ( i)
=
E
=
=
=
E
=
.
{
n
·
EE E{ EE E
E
1 Hints and Answers to Chapter 1
289
E [Proceed as in (vii) .] (ix) (x t , X2 ) E { (1 , -1) , (3, -3), ( Jf57 - 13, ( Jf57 - 13)/2), ( -Jf57 13, ( -Jf57 - 13)/2) }. [The second equation is homogeneous; hence use it to determine x 1 /x2 .] (x) (x� , x2 ) E { (2, -�)} U {(c, 1) I c E R}. (Subtract twice the first equation from the second one.] (xi) (x t , X2 , X3 , X4 ) E {(1 + y'2, 1 - y'2, 2 - 3 y'2, 2 + 3y'2) , (1 - J2, 1 + J2, 2 + 2 We describe a procedure that applies to a more (viii)
(x�, x2 ) { ( - 1, -2) , (-v'2, -v'2) , (1, 2) , (J2, v'2) } .
3v'2, - 3v'2) }.
general system
x 1 + x2 = a� , X1 X3 + X2 X4 = a2 , X 1 X23 + X2 X42 = a3 , X1X� + X2 X� = a4 , with arbitrary parameters ai . Let x2 + px + q = 0 be the quadratic equation with zeros xa and x4 , i.e., x2 + px + q is the polynomial (x - xa)(x - x4 ). Since
X1 (x� + PX3 + q) + X2 (x� + PX4 + q) = a3 + pa2 + qa1 , x 1 xa (x� + px3 + q) + x2 x4 (x� + px4 + q) = a4 + paa + qa2 , the coefficients p and q satisfy the two equations aa + pa2 + qa1 = 0, a4 + pa3 + qa2 = 0, which in our case imply that p = -4 and q = -14 and hence xa ,4 = 2±3v'2. The computation of the unknowns x1 ,2 is now trivial. 5.12 (i) (x� , x2 ) {(2, 3) , (3, 2)}. (ii) (Xt , X2 ) {(3, 4), (4, 3)}. (iii) (xt, x2 ) {(( -16 + S .Ji0)/7, ( -16 - S .Ji0)/7) , (( -16 - S Ji0)/7, ( -16 + s.Ji0)/7), (3, 4), (4, 3) }. (iv) (x 1 , x2 ) { (3, -2) , (-2, 3), (1 + vf6, 1 - vf6) , (1 - vf6, 1 + vf6)}. (v) (x� , x2 ) { (6, 6), (( -3 + 3vf5)/2, ( -3 - 3 J5)/2), (( -3 - 3vf5)/2, (-3 + 3 vi5)/2) } . (vi) Eliminating 0'2 gives 0'1 (70"f - 430'� + 36) = 0. Since 0'1 1= 0, the
E EE EE
expression in parentheses is zero; solve it as a quadratic equation in O"�. The real solutions are (x� , x2 ) {(2, -1), (-1, 2), (-2, 1) , (1 , -2)}. In ad dition, there are the four complex solutions (x� , x2 ) { ((3 + i vf6)/ J7,
E
E
(3 - i vi6)/ V7) , ((3 - i vi6)/ V7, (3 + i J6)/ V7) , (( -3 + i vl6)/ v'7, ( -3 - i vi6)/ V7) , ((-3 - i vi6)/ V7, (-3 + i vi6)/ J7)}. (vii) First, deal with the cases x 1 = x2 , x 1 = -x2 . Then, in the case x� 1= x�, divide the first equation by x 1 - x2 , and the second one by x1 + x2 . The system has the following nine solutions: (x1 , x2 )
E
290
4.
Hints and Answers
{(0, 0), ( .;7, .;7) , ( - J7, - .;1), ( 09, - 09) , ( - 09, 09) , (2, 3), (3, 2) , ( -2, -3), ( -3, -2) }. (viii) (x 1 , x2 ) {(1, 4) , (-1, 4), (2, 1), (-2, 1)}. [Solve as a system of symmetric equations in x� and x2 .] (ix) (x1 , x2 ) {(a, O), (O, a)}, and also (x 1 , x2 ) {((1 + i v'7) a/2, (1 i .;1)a/2) , ((1 - i .;1)a/2, (1 + i 0)a/2)}. (x) (x� , x2 ) {(1, 2), (2, 1)}, and also (x 1 , x2 ) {((3 + i 09)/2, (3 i 09)/2), ((3 - i 09)/2, (3 + i 09)/2)}. (xi) Using u 1 = x 1 +x2 , u2 = x 1 x2 , set up a new system in the unknowns u1 , u2 , X3 . Use the first equation to eliminate X3 from the remaining two, and then eliminate u2 from the second equation of the new system. This system has the unique solution u1 = 19, 0'2 = 90, X3 = 12 , and thus the original system has the two solutions (x 1 , x 2 , x3 ) {(9, 10, 12), (10, 9 , 12)}. 5.14 (i) (x 1 , X2 , X3 ) = (3, 3, 3). (ii) (X 1 , X2 , X3 ) {(� , 1, 3) , (k, 3, 1), (3, k, 1) , (3, 1 , k) , (1 , k , 3) , (1, 3, k)}. (iii) ( X 1 , X2 , X3 ) {(0, -1, 3) , (0, 3, - 1) , (3, 0, -1), (3, -1, 0) , (-1, 3, 0) , (-1, 0, 3)}. (iv) ( X 1 , X2 , X3 ) {(2, 2, -1), (2, - 1 , 2) , (-1, 2, 2) }. (v) ( X 1 , X2 , X3 ) {(-1, -1, 2), (-1, 2, -1) , (2, -1, -1)}. (vi) ( x�, x2 , x3 ) {(O, O, a), (O, a, O) , (a, O, O) }. (vii) ( x�, x2 , x3 ) {(-1, 2, -2) , (-1 , -2, 2) , (2, -1, -2) , (2, -2, -1) , (-2, -1, 2) , (-2, 2, -1), (1, 2, -2) , ( 1 , -2, 2) , (2, 1 , -2) , (2, -2, 1), ( -2, 1, 2), ( -2, 2, 1) }. (viii) Substitute Y2 = 2x2 , Y3 = -x3 and solve for x� , y2 , Y3 · In real numbers, the system has the unique solution x 1 = 1 , x2 = ! , X3 = -1; in addition, it has the following six complex solutions: ( x�, x2 , x3 ) {(1, w/2, -w2 ) , (1, w2 /2, -w) , (w, 1/2, -w 2 ) , (w, w2 /2, -1), (w 2 , 1/2, -w) , (w2 , w/2, -1)}, where w = ( -1 + iv'a)/2. (ix) Solve the last equation for x 4 and substitute into the remaining equations. Solve the new system using the method of symmetric polyno mials, where the auxiliary polynomial has the solutions u1 = 2 , 0"2 = -3, 0"3 = 0 and 0"1 = -2, 0"2 = -3, 0"3 = 12. The first one of these gives six solutions of the original system: x4 = 2 and x 1 , x2 , X3 are all permu tations of the numbers -1, 0, 3. The second one gives six more solutions: X4 = -2 and x 1 , x2 , X3 are all permutations of the zeros of the polynomial x3 + 2x2 - 3x - 12; these, however, are not rational.
E E
E
E E
E
EE EE EE
E
(x) Although the polynomial on the left of the third equation is not symmetric, the method of symmetric polynomials can be used. Rewrite the third equation as x + x + x� = 5 + 2x� and express the polynomials on the left sides by way of elementary symmetric polynomials. Solution:
� �
(x� , x2 , x3)
E {(1, 2, 0), (2, 1 , 0)}.
(ii) -!. i· 6.5 (i) D = 0. (ii) L > 0, R = 0.
6.3 (i)
(i ii)
!(15 + 2 vfs).
(iv)
1.
1 Hints and Answers to Chapter 1
(iii)
291
For x > 0 you have Jx + 9 > 3, vfx > 0, so L > 3 > 2 = R. (iv) L > 0 > R. 6.8 (i) 5 . -4. (v) 3, �(iv) 4, (ii) 2¥'2, -2¥'2. (iii) 2. (vi) 4. (vii) �6.10 (i) 1, -1 · (ii) (iii) 2, -�. . (iv) -2, �- ( ) 2. £. 6.12 (i) �(ii) (iii) 2 (x) 0, - 1. (vi) 0, -5. (vii) 8. (ix) [2,oo). (viii) (1, 10]. (xi) 1, -2. (xii) 5. (xiii) 1. x 6.14 (i) -2. (Rewrite J2 < ; = 2 · /iiJ. ] (ii) 15. (iii) 1. (iv) -(5 + ¥'73)/14, �, g. [Square the given equation and use the substitution = 1/x, = 1/V1 - x2 .] (ii) 1, 3. (iii) � , - � . (iv) 0. (v) 2, 6. 6.16 (i) -15, 1. -2. (v) 1, 3. (ii) 1, 20. (iii) 1. (iv) 6.18 (i) 13, -15. (vi) 8, 8 ± 12J21/7. 6.21 (i) For a < 0 there is no root; for a = 0, every x E (0 , oo) is a solution; is the unique root x = 0. for(ii)a >For0 there a < -1 there is the unique root x = ((a + 1)/2) 2 ; no solution for a > -1. (iii) For a E [0, 1): the unique root x = a/ ( 1 -a2 ) ; otherwise, no solutions. (iv) For a < -1, one root x = for a = 0, every x E JRt a solution; for a E ( -1; 0) (0; oo ). no(v)solutions Denote x 1 = (1 - v4a + 1)/2; x2 = (1+v4a + 1)/2. Then for a < -i there are no solutions, for a = -i there is one root x = �, for a E (- i, 0] there are two roots x�, x2 , and for a > 0 there the unique root x2 . (Use the method of Example 6.20.] 6.23 (i) (x, y) E {(27, 1), (1,27)} . (ii)(iii) (x,=y)2,Ey{ (2,8), (8,2)}. = 1. (iv) (x, y) E { (10 + 6 ¥'3, 10 - 6 ¥'3), (10 - 6 ¥'3, 10 + 6¥'3) } . }. (v)( ) (x,(x,y)y EE { (8,(16,2),4),( -2,(5, -8) }. 15) ) { vi (vii) (x, y) E { (5,4), (-5,-4), (1 5,-12), (-1 5, 12)}. (In the first equa denominator; in the second one use the substitution t= tion, clear the y'x2 + xy + 4.] y= Condition for solvability.· a2 -> (viii)2 x a - 1. (Add and subtract the given equations, then multiply the new > equations together, thus obtaining x2 - y2 = a2 - b2 . Substituting this into the original equat�ons, you obtain two linear equations in x, y. Since the final formulas are obtained from the original system by interchanging the pairs (a, b) and (x, -y) , it is not necessary to check the result.] l.
v
l)
u
v
U
-1;
is
is
X
=
62
atbv'a2-b2 v'l-a2tb2 '
b+a v'a2-b2 v'l-a2tb2 .
292
4. Hints and Answers
w = ( -1 + i-/3)/2. (i) [2n + 2 cos (n7r/ 3)]/3 . [Rewrite (1 + 1) n + (1 + w) n + (1 + w 2 ) n with the binomial theorem and de Moivre 's theorem.] (ii) [2n +2 cos ((n - 2) 7r/3)]/3 . (Rewrite (1+l) n + w2 (1 +w) n +w( 1+w 2 ) n .] (iii) [2n +2 cos ((n- 4) 7r/3)]/ 3. [Rewrite (1+1) n +w(1+w) n +w 2 ( 1 +w 2 ) n .] (iv) 2n cos n37r . (Rewrite wn with the binomial theorem and de Moivre's theorem, and equate real parts.] (v) � sin n37r . [Rewrite wn and equate imaginary parts.] 7.5 (i) (6 tan a - 20 tan3 a + 6 tan5 a)/(1 - 15 tan2 a + 15 tan4 a - tan6 a). (Use 7.4. (i) for n = 6.] (ii) Express the real part of the sum of the complex roots of the equation 2n x + 1 = i . The result also follows from the formula for the sum 82 in 7.3 Set
7.4. (ii). (iii) For c = cos a + i sin a rewrite ( c + c - 1 ) n . (iv) 81 = 2n cosn (a/2) cos({3 + (na/2)) ; 82 = 2n cosn (a / 2) sin({3 + (na/2)). [Rewrite 81 + i 82 with the help of the binomial theorem and de Moivre ' s theorem, and use the identities cos o = 2 cos2 (a/2) - 1; sin a = 2 cos(a/2) sin(a/2) .] (v) Proceed as in 7.4.(ii) ; in order to make the denominator real, multiply numerator anq denominator of ((r · c) n+ 1 - 1)/(r · c - 1) by rc- 1 - 1 , where c = cos a + i sin a. (vi) Begin with the formula cos 2na =
t=O ( -1 )i (��)J cos2(n-i) a sin2i a
j
(see 7.4.(i)) , factor out cos2n a on the right, set a = ak = (2kin1 ) 7r , where k = 1 , 2, . . . , n, and thus determine that the desired numbers xk = tan2 ak are roots of the equation E; 0 ( -1) i (�;)xi = 0. Then by Vieta 's formula you have 'E�- 1 Xk = (2�� 2) = n (2n - 1). (vii) First verify that l1+ck l = 21 cos k: I for k = 1 , 2, . . . , n. Substitute this expression, for n = 2m + 1 , into 1 = 11 + cl · l1 + c2 1 · · · 11 + c2m l , which is a consequence of (65) for x = -1. Then you easily obtain (66) , if you use ( 1-k +k the relation cos (m 2m+)17r = cos m2+m+ 1 ) 7r £or k = 1 ' 2 ' · · · ' m · (viii) The polynomials Am (x) and Bm ( Y) are (up to a constant factor) determined as follows:
all
j
j
Am (x) = Bm ( Y) =
2 + 1 ) (1 - x) m-ia;J , - l) j ( � t( 2J + 1 =O
j
t( -l)j (�� : :) (1 - y)m-j yi . J J =O
1 Hints and Answers to Chapter 1
293
2m + 1, and the coefficient j (22J�++ 11 ) (-1) m-j = (-1)m . t (22J�++11 ) = (-1) m · 22m t(-1) j =O j =O by 1. 4 . (ii) . Therefore, X 1 X2 Xm = (2m + 1) /22m , which implies v'2mm+ 1 . 1r sm. 21r s1n. m1r = --s1n 2m+ 1 2m+ 1 2m+ 1 2 and this proves ( 6 3) for odd n. Similarly, from the coefficient ( -1) m 22m+ 1 of ym and the coefficient 2m + 2 of y0 in the polynomial Bm (y) it follows that 2 Y1Y2 Ym 2m+ 1 22m+ ' hence v'mm+ 1 21r · sm. m1r = --. 2m+1r 2 sm. 2m+ sm 2 2m+ 2 2 so (63) holds also for even n . Finally, from the equation
The constant coefficient of Am (x ) is obviously of xm is equal to
•••
·· ·
·
•• •
=
.
··
1 - Xk
in view of the fact that = cos2 2!: 1 , you obtain ( 66) . (ix) The equation �;- � = x6 = 0 has the six roots c = cos 2; i sin 2; , c2 , c4 , c5 , c7 , c8. Substituting y = x � , you obtain the cubic equation y3 = 0 with roots
- 3y + 1
+ x3 + 1
21r
+
+
Y3 = 2 cos -87r9 .
Y1 = 2 cos 9 ,
+ + 4( ) -1, + ( + B3 3 - 3�. (x) Set c = cos 2; + i sin 2; . Then c5 = 1 and c 1= 1, so c, c2 , c3 , c4 x4 + x3 + x2 + x + 1. Substituting y = x + ! you are roots of x:�l find that y1 c + : 2 cos 2; is the positive root of y2 + y - 1. Hence 2 Cos 25'��" - --1+12 iSV5. The other root is Y2 c2 + 1 2 cos 45'��" -- -2 cos '��"5 ' so -2 cos � 2 . 7.7 ( i) Proceed as in 7.6. ( i) , or use the fact that for an arbitrary root a of the polynomial x2 + x + 1 you have a2 +a+ 1 0, and thus (a+ 1 ) 2n+ 1 = 1 )n -i,-aand2 ausen . de Moivre's (a2 + 2ae+i and (a(ii)+ 1)Substitut theorem. To determine the sum � -?fii2 ?'fj4, use the same approach as in 7. . v . This time you get C = the pair of equations A3 = 3AB 3 and B3 = -3AB - 6, and w 3) 3 = 9, which implies A3 = 3� - 6 and =
=
=
=
=
� -
=
=
=
·
.
294
4. Hints and Answers
i,
(iii) Substitute -1 and -i. (iv) {O, ±l, ± J3, ± 1+2v's , ± 1 -2� }. Each = cos {3, where {3 (0, -�) . The polynomial roots cos {3 ± sin {3, and hence the polynomial different zeros,
a 2
aE
is of the form ag(x) E (-2,2) = x2 - ax + 1 has f(x) = x1 - ax7 + 1 has 14
E
i ±{3 + 2k1r + . . ±{3 + 2k1r , k E {0, 1 , . . . ,5, 6} . So the polynomial f (x) divides the polynomial h( x) = x 1 5 - ax77 + 1 if and only if each root of f (x) is also a root of h( x), i.e. , it is of the form ±{3 + 2l1r + . sm. ±{3 + 2l1r cos 7
COS
4
7
� Slll
4
77
with an integer
�
l. Especially, {3
±{3
7
77
+ 2l1r
77
l E Z, which gives {3 = ; or {3 = ; Hence {3 E { 1r6 ' 1r3 ' 21r ' 321r ' 657r ' 1r5 ' 521r ' 531r ' 547r } . For the values of cos ; see Exercise . 5 . ( x) . + 1) = F(w) · F(w2 ), 7.9 (i) From 3.14 derive (x� + x 1 + 1) · · · (x; + where w = ( -1 + iv'3)/2; show that F(w) = Cw2 + Bw + A , F (w2 ) = Cw + Bw2 + A. Multiply and simplify. (ii) From 3.14 derive (xf+l ) · · (x�+l) = G(r)G(r3)G(r5)G(r7) , where = (1 + i)v'2/2 ; show that G (r) · G( r5) = -q2 - ( + r ) 2 , G(r3) . G( r7) = -q2 + i( -pi + r) 2 , and multiply these two expressions together. z
for a suitable
z
z
.
7
,
Xn
·
T
2
i
pi
Hints and Answers to Chapter 2
72 00 5300 , since+ 27 >for53 •all n E N. (iii) The first number is larger. For all x E lR, x+ l + 1 lOx + 1 lo ( 10x + 1 )( 10xt 2 + 1 ) > (10x t l + 1 ) 2 . lOX±l + 1 > 10X±2 + 1 Expanding, you get 1ox +2 - 2 · lox+ l + lOx > 0, that is, 81 . l Ox > 0. (iv) L - R = (..Jb - a) · at1 l-2 v'b > 0 · +a ( v) Better than squaring: Multiply the inequality by ..ja + 1 + ...;a. 2.2 (i) \l'nf < ( ii) >
n+\f(n 1)!
{:::=}
2
295
Hints and Answers to Chapter 2
(vi) By squaring you obtain
(c + a) 2 > (c+b)2 + a2 + b2
1 + 2ac+ a2 > 1 + 2bc+ b2 b a > a2 + b2 + The last inequality can be rewritten as (a - b) t? > ab(a - b). (vii) This follows from 2. 1 . (2v) for l and2n k . 2 l· 1 . 2 2 . . �$� . (viii) Upon dividing by ( n! ) you get 1 < ; n;�1 n;�2 Each term on the right-hand side is greater than 1. 2.4 (i) Each term on the left-hand side is greater than 1/(a + b + c). (ii) Rewrite this as 2(x 1 + x2 + + x6 ) < 3(x7 + x8 + x9 + x 10 ). The left-hand side is at most 12x6 , and the right-hand side at least 12x7 . (iii) i, = 1!a + 1!b > 1+!+b + 1+!+b = 1 ==? L < R. (iv) Adding a + b > c and 2v'Cib > 0, you obtain ( ...fo + v'b) 2 > ( Jc) 2 , thus via+ v'b > Vc· The inequality (a + b) - 1 + (b + c) - 1 > (c + a) - 1 follows from a + b < 2(a + c) and b + c < 2(a + c). Changing the order of b,(v)c gives the remaining triangle inequalities. Distinguish between two cases: For x > 1 you have L = (x8 - x5 ) + > 1; for x < 1 it is nL = x8>+ (x2 - x2 5 ) + (1 -x)n- 1 > 0. + 1 (x2( -x) Therefore, +a ) vi) Upon rewriting, (n - 1)(a + 1) 2(a +a + k k n n it suffices to prove the inequalities a + 1 > a + a - and add them for k 1, 2, . . . , n- 1. (vii) Prove the inequalities c2
c2
c2
c2
c2 .
c2
s =
r =
.
···
and a,
· · ·
.
=
(m + 2j - � r > (m + 2j - l)(m+ 2j),
j 1, 2, a3 + b3 + 2abc > ab(a + b) + (a2 + b2 )c
multiply them for = . . . , k , and take the square root of the result. Show that then (viii) Without loss of generality you may take Adding and e these two, you obtain the desired inequality. (ix) The left-most term on the right-hand side is smallest. Hence
> b >> c. t?. a+ abc (a + b)
1 n+ ] (n 1 + 1)a + . 1 + (n+ 1)a > [ 1 + na (x) 31 11 < 1 714 , since 31 11 < 32 11 = 2 55 < 256 16 14 < 1 714 number is greater. Both can be rewritten as 29· 398 and (xi) The in view of the fact that 29 > 34 , it remains to show that 33984·2148,2 1and 8 . This last inequality is a product of 39 > 2 1 2 (which follows from 4 > 33 > 2 4 ) and 389 > 2 1 36 (which follows from 32 > 23 and 89/2 > 136/3 ). (xii) Rewrite the equation as 1 = F (x) where F(x) = ax1 + xa22 + . . . + xann . =
first
,
.
4. Hints and Answers
296
ai
It is easy to see that F(x) > F(y) if 0 < x < y and =/:. 0 for some i. a:tr J > 0. (xiii) L - R = xiv) Multiplying the first, resp. the last two inequalities together, and - cd) . simplifying, you obtain ) cd < < cd - resp . However, not both numbers cd cd can be positive. ( xv) By the inequality given in the hint you have
b)[ c (c a b) (c -a ( a2 + b2 ab, abab, (a2 + b2 ab( ab A + B + a + b > A + a and B + C + b + c > C + c B + C+ s C + s - b A+B+s A+s-b If you replace the denominators on the right-hand sides by the larger num ber A + C + s and subsequently add both inequalities, you obtain one of the desired inequalities; in view of symmetry, the remaining triangle inequalities are also true. ( i) By expanding both sides A(r)A(u), A(s)A(t) you will find that it suffices to show that ajaJ: + akaj > aJat + a�a} whenever ai =/:. ak Upon dividing by (ai ak )T you obtain the inequality from 2. 1.(v) , with the numbers r, s replaced by s -r, t r. 2.6 (i) Add J2 < Vk < Vii for k 2, 3, . . . , n and exclude the cases of equality. (ii) Multiply (k + 1)(n - k) > n for k 1, 2, . . . , n - 2. (iii) Add the inequalities 1 > n 1 ' resp. E 1 < n 1 · � kn+j (k + l)n k + l j= l kn + j - 1 kn k (iv) S(n) > 2 y'n + 1 - 2v'2 + 1. From this and from 2.5.(iv ) it follows that 1998 < 8(1 06 ) < 1999. (v) Use the estimate 2ki+ k < 2�2 for k 2, 3, . . , n. (vi) Add k13 < k3:_k for k 2, 3, . . . , n. (vii) Use the inequality k k2 1 (2k - 1) (2k + 1){2k + 3) (2k + 1){4k + 4k) 4(k + 1)(2k + 1) " (viii) Add 4k - 1 < (2k- l �2k+l) < 4k for k 1, 2, . . , n. (ix) Use the method of 2. 5 .(vii) with the inequalities :t;� < !z+� + , k > 1. (x) Estimate each term on the right of Q ( n) Q (p) . 2p + 1 . 2p + 3 . 2n - 1 2p + 2 2p + 4 2n as in Example 2 . 5.(vii). 2.8 (i) In view of the symmetry you may assume that 0 < a < b < c. Then c(c -a) > b(b-a), so that a(b-a)(c- a)+(c-b)[c(c-a) -b(b- a)] > 0. -
-----
-
�---=-
---
xv
·
-
=
=
n
n
=
=
=
=
.
=
>
=
.
.
=
L =
.
2 Hints and Answers to Chapter 2
297
< a < b c; then b = a+u and c = a+u+v, where u, v E JRt. < Substituting, you get L = u2 (a -v) + (u+v) 2 (a +v) +v2 (a+2u+v). The only term with negative coefficient, namely -u2 v, clearly disappears upon simplification, and so L > 0. (iii) The inequality t2 + 1 - t -> (t2 - 1) 2 (t2 + 3)(1 + 3t2 ) -2 8(t2 + 1)(t4 + 6t2 + 1) ( ii) Let 0
...;._-,---,-___;____;_.,..._..;. ..--,....;...., ---'-
can be transformed into
(n2 + n)(Lk - R) for= (1k =- 0,a){1 + a + + an- l - nan ) > 0, since , and a > an 1. 1, . . (1 (ii)- a)You> 0 obtain n . the inequality x - y < a - b, with x = f!an + c and = \Ibn + c, from the equation xn - yn = an - bn . To see this, expand both sides according to formula (14) and use the fact that the inequalities x > a and y > b imply the estimate 3.2 (i)
· · ·
y
(iii) Each inequality on the right-hand side of the implication is of the Under the assumption form a " -: k a k+ 1k__*a,1 k 1 , where k you can use formula with = and = k to change this But inequality into the equivalent form k( this is inequality (iv) It suffices to show that since
1 E N a > . (14), n 2k2k+l 1)n > a2 ++a22,+ + a2k . a + (6). 2(a2 -ab + b2 ) > a2 + b2 , s s b a + L = a + b = a2 . aa3 ++ bb2 . b3 > 2aa3 ++ b2b3 = 2(a2 - ab + b2 ) . (v) The equation a3 +b3 = a-b gives the expression a2 +ab+b2 = ::+:: ; from this obtain the stronger inequality a2 + ab + b2 < 1. - ( a- b) + (b -c) + ( c-a) + + ( - 0, Vl" ) Smce you have >
· · ·
.
a 3 - b3 a2 +a b+b2
b3 - c3 b2+bc+c2
c3 - a 3 c2+ca+a2
> 3 x, y E 3(x2 + xy + y2 ), (2(xx -+ y)y)2 > (16). (a + 1)n - 2n > n(a - 1) 2n- l , x +
x+
3 Y3 y + for all IR . Upon Therefore, it suffices to show that x2+x 2 + y y dividing by and multiplying by this inequality 0. changes to by the left part of 3.4 (i)
298
4. Hints and Answers
(1 a) n > n(1 -a)(1 + a) n - 1 = n(1 - a2 )(1 + a ) n - 2 > n(1 + )1(iii) Rewrite (1 - a)n - an > nan- 1 (1 - 2a). If 0 < a < � use (16) ' order, site with A = 1 - a and B = a; � < a < 1, substitute in the oppo namely A = a and B = 1 - a; if a = � , you obtain equality. (iv) Since An - An - 1 = (an - An - 1 )/ n, you have An > An - b unless a 1 = a2 = · · · = an . Then by (16), A: - A:_ 1 > n (An - An- 1 ) A�= � (an - An - 1 )A�= �. (v) Substitute n = k and = 1/k in (17) . k 2 ; inequality left the Rewrite (vi) [ (k 1) ] > 1 - k!2 ; this follows from (17) for n = k and (k;1)2 . The right inequality can be written in the form (17) for n = k + 2 and = k2!2k . in (17), obtain an + (n - 1)b- nr > na br( 1 -n) . 1 abr (vii) With Therefore, choose r = -1/n. The inequality (18) is strict if 1= 0, that a 1= v'b. (viii) Use ( 8) with a = 1 and n = k + 1, b = 1 + i, resp. n k + 2, b = 1 - k!1 · (ix) The inequality d'f < n- 1 d implies L < nn - 1 (d1 + · · · + dk )/k . Hence it suffices to show that a n - 1 < an + o:n Dn ; but this is {18) with b = O:nDn /(n - 1). 3.6 (i ) Use (n + 2k) ! > (n + 2) ! (n + 3 ) 2k - 2 for k > 2, and (19) with 1 2n a2 n - 2 . (ii)
as
if
=
x
as
x
=
1 -
-
x
x
=
is,
x
1
=
n
n
q -
i
·
- (n+3)2 · 2k- l ) satisfy (ii) The numbers a k = 1'3···{ < �3 . k·k 3 ! (iii) Let 28 a! be the largest of the numbers 2af, 22 a� , . . , 2n a: . For j = 2, . . . , n sum the inequalities ai < 2 s'ki a8 and then use with q =
a�.:a1c+ l
1,1/ {12.
.
(19) 3.8 (i) Since a < c and b < c, it suffices to show that a + b > c. Let us assume the opposite: a + b < c; then by ( 22) , an + bn < (a + b) n < d"', which a contradiction. (ii) In the expansion of (1 + a) n , use the inequality ak > an - 1 for k 1, 2, . . .( , n+1-) 2. a-1 (iii) a n�( )n 2 [ (�) � + (�)a\ + · · ·] . Estimate the term in square brackets from below by its first term, and from above by the sum : + ( : ) 3 + ( : ) 5 + · · · , and use { 19) with ( : )2• 1, B (iv) The expression {23) . follows fi:om (21) for A 1/ n, and exponent n k. Since 1 - t < 1 - m for j 1 , 2, . . . , k - 1 , you get the inequality Ck+ 1 > ck by comparing the two right-hand sides { 23) . The left inequality in (24) follows from ( 23 ) and from the estimates 1 - f < 1 for � > 0; the right inequality of (24) is obtained by summing ]1 2,� 1 for = 3, 4, . . . ' k. is
=
=
q =
=
J
4.2 (i)
=
=
=
0. (vi) Divide by then this is the sum of two inequalities (v) . (vii) �.
=
as
cJ3
+
+
_
ex
=
(xxxi) First estimate the right-hand side from above:
R < aP . b2 +2 c2 + . a2 +2 c2 + . a2 +2 b2 . bP
cP
4. Hints and Answers
300
Hence it suffices to prove the inequality
2(aP+2 + lJP+2 + e>+2) > aP (b2 + !?) + (a2 + !?) + (a2 b2 ) . Write this as the sum of three inequalities of the type aP+ 2 + lJP+ 2 > aPb2 + to prove; see Example 2.1.(v) . bPa2 , whichA >areBeasy , with equality only when a = b c = 1. Use the identity (xxxii) q,/Jc( abc+ 1 ) (A-B) = ab(b+ 1) (ca-l) 2 +be( c+ 1 ) (ab- 1 ) 2 +ca( a+ 1) (be-l )2 . (xxxiii) The desired largest p is �. For x = y z =I= 0 you obtain the necessary condition 6 > 9p; for p = � the difference L - R is equal to one-twelfth of the sum 5(x2 y2 ) 2 + 5(y2 z2 ) 2 +5(z2 - x2 ) 2 + 2(x2 -yz )2 + 2(y2 -xz) 2 + 2(z2 -xy)2 . (xxxiv) Answer: For p 1 all n; for p = � only n E {2, 3}; for p = t only n E {2, 3, 4}. In the case p 1 the given inequality is equivalent to X� + (xl - x2 )2 + (x2 -X3 ) 2 + + (xn- 1 - Xn ) 2 + x! > 0 , in the case p = � and n = 3 to (3x1 - 2x2 ) 2 + x� + (2x2 - 3x3 ) 2 > 0, and finally in the case p = ! and n 4 it is equivalent to (5x 1 - 3x2 ) 2 + x� + 2 + x� + (3x3 - 5x4) 2 > 0. (If n is less than mentioned in the x ) 15(x 3 2 last sentence, fill up the set of numbers x1, x 2 , . . . , Xn with an appropriate number of zeros.) lf p � and n > 4, consider the example x1 = X4 = 2, x2 = X3==9,3, and Xi= 15,0 for=i 16,> 4,andand if=p0 =for!i and5. n > 5, consider X1 = X5 X2 X4 X3 > Xi 4.4 (i) Follows from ( 2 9 ) with A a2 b and B = 1/b. (ii) By (29), L > 2yf6(c + 2)(2c + 3)ab. Prove the inequality J6(c + 2)(2c + 3) > 3(c + 2) by equivalent-v'ab transformations. and a3 + b > 2 Va3z>. One (iii) Multiply the inequalities a + b > 2 could also use the method of squares: L = (a2 - b) 2 + ab(a - 1) 2 . (iv) Use ( 29) to estimate from below each of the four expressions in parentheses on the left. (v) Use ( 2 9 ) to estimate from below each of the six expressions in parentheses under the square roots. (vi) By ( 2 9) 1b + 001 > 2 h b > 2 -2- -2- . (vii) Use ( 29) with A = a + b and B = 2 -v'ab. (viii) Since, by (2 9) , L > 4-v'ab(a + b) + 2 -v'ab and R = 4-v'ab( ..ja + Jb), it suffices to verify the inequality 4(a + b) + 2 > 4( J(i + vb) , i.e. , (2 J(i - 1)2 + (2 vb - 1) 2 > 0 . (i.x) The proof follows from adding the inequalit'ies (see also (vi)) 4 ' 1 + 1 >1 8 >aiaj (ai + aj )2 aiak ai ak (ai + ak )(aj + ak ) ' lJP .
·
If!>
cP
·
=
=
_
_
=
=
·· ·
=
=
=
=
=
�
' a
--
-
·
v
ab
·
--
v
cd -
--
·
a+ b
·
c+d
-:-------:�---
+
2 Hints and Answers to Chapter 2
301
A 1 2a and B = 1 + 2b in (31), or show that + 2. = (a b) L-R ( ii) y'b - (a + 2b-a2l ) (Vb- a){l++la -v'b) Use ( 31) with A a1 and 2 a+ a B = 1 - A. Alternative solution: Rewrite R = A+ B, where A 2af and B 2a�l , and use (29). (The condition a 2< v'b < a + 1 is not necessary.) (iii) By (31), with A = a + b and B a - ab+ a3 + b3 (a + b)(a2 ab + b2 ) < ( a + b + a:- ab + IJ2 y .
4.6 (i) Substitute
=
=
=
=
=
=
= Y'b D
·
-
�'
_
From this you obtain the result by taking the square root and setting
ab(iv)1.Use (31) to estimate from below the products ab, be, and on the left-hand side. (v) Using (31) for the products ab, cd, and (a + b)(c + d), you get R3 = ab(c+ d) +4 cd(a + b) -< ( � ) 2 (c + d) +4 ( � )2 (a + b) (a + b)(c + d) . a + b 1+6 c+ d 2 c d) [ (a + b) + ( + ] . a + b + c+ d = L3 . < =
ca
=
-
2
16
1 and Xk = 0 for the remaining indices k, obta� x x1 2 the necessary condition Pn < 4. On the other hand, show that for Pn 4 the given inequality holds everywhere: In view of its homogeneity it suffices to consider the case where the sum of all Xk is 1. (If Xk 0 for all k, then the inequality is clearly satisfied.) Then the inequality (with Pn = 4) can be rewritten as E�= l Xk Xk+ 1(1 - 4xk Xk+ l ) > 0. But this holds, since by (31) for i =I= j, (vi) Choosing
=
=
=
=
1 - 4xixi L < (a 1 +a + · · +a 1)(a +a + · ·+a ); · n 3 n 2 4 (31) ab a2 , . . . , ak- b ak + ak+ b ak+2 ,
> 0. so that (vii) If n is even, then apply to this product. For odd n, use the previous case for the even number . . . , an, where t he index k is of elements chosen such that the inequality ·
302
4. Hints and Answers
holds (one can usually take ao = an , an+ l = it suffices, for example, that ak+ l < ak+2 ·
a1
and
an+2 = a2 ) ·
For this
4.8 ( i) Follows from
(32) with A = yfa2 + 2 1= 1 . ( ii) Follows from (32) with A = yf4a2 + 1 ;/= 1. ( iii) L = (� + �) + (� + �) > 2 + 2 = 4. ( iv ) Upon dividing by 2ab rewrite as 2 (b + /;) + (2a + 2� ) > 6. (v) Divide by av'b and rewrite as ( � + �) + ( Ja,; + .Jab) > 4. 5.3 ( i) From p2 -4q < 0 and r2 -4s < 0 it follows that p2 < 4q and r 2 < 4s, which implies p2r2 < 16qs. This means that also the third polynomial has a nonpositive discriminant. ( ii) If you set q = 1 - p, you obtain the equivalent inequality
Cp2 + p( a2 - b2 - c) + b2 > 0. This holds for all p 1R if and only if D = (a2 - b2 - £?) 2 - 4l?b2 < 0. Use the factorization D = -(a + b + c)(a + b - c)(b + c - a)(c + a - b). ( iii) Let a > b > c. Set F(x) = x2 - 2x(b + c) + (b - c) 2 and show that F(a) < 0. Since b < a < b + c, by 5 . 1 it suffices to show that F (b) < 0 and F ( b + c) < 0. But this is easy. ( iv) L - R = x2 + 2(y-3z + u)x + (y + z +u) 2 - Byu. This polynomial has discriminant D = 4(y - 3z + u) 2 - 4(y + z + u) 2 + 32yu = 32(y - z)(u - z). Clearly, D < 0. This implies the assertion also under the following weaker assumption: The number z lies between y and u, and x is arbitrary. (v) Since x2 - 2x + 2 > 0 for all x lR, the inequalities are equivalent to
E
E
the pair
3x2 + 2(p - 2)x + 2 > 0
and
E
x2 - 2(p + 2)x + 6 > 0 .
This system is satisfied for all x 1R if and only if the discriminants of both polynomials are negative, that is, I P - 21 < v'6 and I P + 21 < VB. The common solution of the last two inequalities is 2 - VB < p < -2 + VB.
(39) with n = 2, u 1 = y'a, u2 = v'b, v 1 = -IC, and V2 = .Jd. ( ii) By (39), b2 = (a 1 x 1 + a2x2 + · · + anXn ) 2 < (a� + a� + · · · + a;)(x� + x� + · · · + x;), which implies x� + · · + x; > b2 /(a� + · · · + a;). Equality occurs for X k = (akb)/(a� + · · · + a;), 1 < k < n. ( iii) Use (39) with Uk = ..j7ik and Vk = Xk ..;ak for k = 1, 2, . , n. ( iv) Follows from ( 39) for Uk = ..j7ik and Vk = x k / ..j7ik, 1 < k < n. (v) By (39) with n = 2: (1 + p4 )(1 + q4 ) > (1 + p2 q2 ) 2 , ( 1 + r4 )(1 + 84 ) > ( 1 + r2 s2 ) 2 , (1 + p2 q2 )(1 + r2 s 2 ) > (1 + pqrs) 2 . ( vi) Use (v) with p = a314 , q = b31 4 , r = e/4 , and s = (abc) l /4 • ( vii) Apply (39) three times: for the pairs (uk , Vk ) equal to (pk qk , rk sk ), (p�, q�), and (r�, s�). 5.6 ( i) Follows from
·
·
.
.
2 Hints and Answers to Chapter 2 (viii)
303
Use ( 46) with n = 4, a 1 = a + b + c, a2 = a + b + d, a3 = a + c + d,
and a4 = b + c + d. (ix) In (vii) , substitute Pk = Xk and qk = rk = Sk = 1, or use ( 48) twice, with Uk = Xk , resp. Uk = X�. (x) Use inequality (48) with U 1 = X2 X3 · · · Xn , U2 = X 1 X3 · · Xn , · · · , ·
Un = X 1 X2 . ' · Xn- 1 ·
(xi) Suppose that, for instance, Xn < 0; then with the help of ( 48) reach a contradiction to the given inequality:
X 1 + X2 + · · ' + Xn < X 1 + X2 + · · · + Xn- 1 -- --- --- -----< J( n - 1) (x� + x� + · · · + x;_ 1 ) < V,_(n 1-) (x� +-x� + . . . + x� ). (xii) Add the inequalities ak bk ck < (a� + b� + �) /3, 1 < k < n, under the assumption a� + a� + · · · +a! = b� +b� + · · · + b! = � + � + · · · + c! = 1. (xiii) Use (xii) with bk = Ck = 1, 1 < k < n. (xiv) Use (xii) with bk = ak a Ck = 1 , 1 � k < n. (xv) By (39) with Uk = y'ak /(ak± l + ak+2 ) and Vk = y'ak ( ak± l + ak+2 ), where anti = a 1 , an+2 = a2 , and using the inequality aiai < (a� + aj)/2,
you get
(
)
( al + a2 + . . + an ) 2 < a a l + a a+2 a + . . . + a a+n a 1 2 2 + a3 3 4 x (a 1 a2 + a 1 a3 + a2 a3 + a2 a4 + · · · + an a 1 + an a2 ) a1 an < - a2 + a3 + . . . + a1 + a2 a21 + a22 + a21 + a32 + . . a2n + a2l + _an2;.+_ .;._ n� --=:o�: X · + 2 2 2 2 = 2( a� + a� + . . . + a;) a a+l a + a a+2 a + . . . + an . a + a 1 2 3 3 4 2 Upon dividing by 2(a� + a� + · · · + a;) you obtain the desired inequality. (xvi) Choosing Xk = 1 for all k, obtain for the sum S of all numbers ak the estimate S2 < S, or S < 1. Conversely, if S < 1, then n n 2 n (k=l L ak x� ) < L ak L akx� < L akx� , k=l k=l k=l where the left inequality is ( 39) with Uk = ylak and vk = ylak · x�. 6 3 L (n±l ) < R(ntl) - 1 - a < ltna . The 1ast 1nequaI1'ty can be ) ( 1 l t (nt l ) a R(n) L (n) • .
(
(
___
)
)
(
)
n
·
'
verified with an equivalent transformation. (ii) Write the given inequality symbolically Then
·
as
L(n) < P(n) < Q(n).
L(n + 1) < P(n + 1) < Q(n + 1) - 4( n + 1) < (2n + 1)(2n + 2) < 4. L(n) P(n) Q(n) n+2 (n + 1) 2
304
4. Hints and Answers
The inequalities on the right can be verified by equivalent transformations. (iii) L( n + 1) - L( n) 3n�2 + 3n�3 + 3n�4 - n�l = (3n+ 2) (3n�3) (3n+4) · + l) > R(Rn( +l) - n + 1 This last inequality can be verified (iv) L(Ln( n) n) by induction. (v) For n = 1 : x2 +y2 + 1 > xy+x+ y, which is inequality (27) . Next, find l that L(n + 1) - L(n) > R(n + 1) - R(n) � x2n+ + y2n+l + 1 - x2,.. - y 2,.. > (xy) 2,.. . The last inequality is again (27) . (vi) For the induction proof, use the estimates =
·
1 1 1 + 1 +...+ 1 < 2 2 n 2n + 1 2 n+ l - 1 < '
which can be verified using the approach in 2.5.(ii) . (vii) Li{�)l) > R�{�)l ) � (1+a2n+ l +a 2n+ 2 +a4n+3 ) < ( 1 +a)(1+an+l ). Since 1 +a > 1, the last inequality is satisfied as long as 1 +a 2n+ 1 + a2n+2 + a4n+3 < 1 + an+ l , which for a > 0 is equivalent to a1" + an+ l + a3n+2 < 1 . The left-hand side of the last inequality is largest for n = 1 and a = {2 , in which case it is, however, less than 0.992. (n+ l) � ( 2n + 2) 1 > (n + 1) 1 (n + 2) n+ l This is + l) > RR( (viii) L(Ln(n) n) equivalent to (n + 3)(n + 4) . . . (2n + 2) > (n + 2 ) n , which is the product of the inequalities n + k > n + 2, 3 < k < n + 2. (ix) Use finite induction on k. For k = 1, (52 ) holds. If (52) holds for some k < n, then ·
·
·
+l
(1 + : r = (1 + : r (1 + : ) > (1 + �) (1 + : ) k+1 = 1 + k +n 1 + � 1 + > n2 n
and
k+ l k 2 1 1 1+ n 1 + n1 < 1 + k + k 2 1 + 1 = 1+ n n n n 2 2 1 + k n+ 1 + ( k n+21) n( k +n1)3 - k 2 k 1 (k + 1) + < 1 + n + n2 ' since n ( k + 1) - k2 > 0. (This last inequality is obtained by multiplying n > k and k + 1 > k .) Remark. This approach shows that the left-hand part of (52 ) holds for all k > 0; this also follows from Bernoulli's inequality ( 17) or directly from
( )
=
( )( ) ( _
)( )
the binomial theorem. ( ) For n = 6: 729 > 720 > 64. Further, it suffices to show that x
n±! ) n+ l n+ l ( ( n±! ) 2 n -> n + 1 > ....:.3:::. ....:. n - , - (..- i )-=(�)
2 Hints and Answers to Chapter 2
305
n
which, upon dividing by ( + 1), can be rewritten as
1) ) + l ( n 2( n These last inequalities follow from (53). (xi) If, for instance, k < n, then V'k < y'n. Hence it suffices to show that y'1i < 2 for all n > 2. Rewrite this as n < 2n and use induction. (xii) For n = p you have equality in (54). Further, it suffices to verify that v'4n + 1 Q( n + 1) = 2n + 1 < v'3n + 1 < v'4n + 5 - Q(n) 2n + 2 - v'3n + 4 (use an equivalent transformation). In comparison with 2. 6 .(x), show that p 4p + 1 3p + 1 2p + 1 n < 4n + 1 < 3n + 1 < 2n + 1 (n > p). (xiii) Since L(1) 1 < 2 = R(1) and L(n + 1) = .Jn + 1 + L(n), it suffices to show that .Jn + 1 + ( fo + 1) < v'n + 1 + 1. This becomes clear upon squaring. n 1 >1> 1 + 3
1
l
""
·
=
(i) The power with the eight fives is greater. Denote by An , resp. Bn , the powers with fives, resp. eights, and show by induction that 6.5
n n An > 2Bn-3 , n > 4. (ii) Rewrite the inequality as (a+b-ab) n > an +bn -anbn . For n = 1 there is equality; show that (an +bn - an bn )(a + b-ab) > an+ l +bn +l _ an+ l bn+ l for all n > rewriting it as L - R = (bn - bn+ 1 )(a-an+ l ) + (an -an+ 1 )(b bn+l ) 0. >
1,
(iii) Rewrite the induction condition
1 A 1 + ( n ! c ) [ (n + 1) (n + 2) ] < A - n + :+ C the equivalent form A[n2 + (2C + 1)n + C2 + C] < B [n2 + 4n + C + 3] , which holds if A = B and 2C+ 1 4, that is, C = � (then C2 +C < C+3). Considering the inequality for n = 1, obtain then A = B = �. (iv) A more exact bound is of the form + 1) (n > p). Q(n) -< vf(Q4p(p+) 3(2p ) n+ p + 1 in
=
·
(v) Use induction to verify the more exact bound (vi) For a proof by induction (on k) of the stronger inequality
1 + 1 > nn
1 ... + 1) + +
S(n) S(n
S(n
1 + k) +
(n
1 + k + 1) n+k+ l
4.
306
Hints and Answers
all
it suffices to show that for n > 2 you have nin > sln) + (n+ i) n+i Show nl that this inequality is equivalent to 8fJ�_)I) < (n+�l + ; this last one can be proved by induction. If this inequality holds for some n (which is clearly the case for n = 2), then S(n + 1) = S(n) + (n + l) n+ I S (n - 1) + nn S (n)
·
(n + 1) n+ I
R(n + 1) - R(n) is obtained as follows: n+ 2 1 (n + 1)xn+ I > (xi + X2 + · + Xn ) + 2 Xn+ I , 2 ·
·
which is equivalent to Xn+ I > (xi + X2 + · · · + Xn )fn. (iii) The inequality L(n + 1) - L(n) > R(n + 1) - R(n) is obtained as follows: (n+ 1)2nxn+ I +xi + 2x2 + · · · + 2n- I xn > (2n+ I _ 1)xn+ I +2n (XI + X2 + · · + Xn ), which is equivalent to ·
(n2n - 2n + 1)xn+ 1 > ( 2n - 1)xi + (2n - 2)x2 + · · + (2n - 2n - 1 )xn . This is the sum of the n inequalities (2n - 2k - I )xn+ 1 > (2n - 2 k - I )xk for k = 1, 2, . . . , n. (iv) Show that the inequality bn+ I bn- I > b; is equivalent to (aia2 . . . an- I ) 2 (an+ I ) n2 -n > (an ) n2 +n - 2 ; ·
denote this by L(n) > R(n). Show that the sequence of inequalities L(n) > R(n) satisfies condition (i) of the principle 6.2. (v) The inequality for fixed k E N follows S(O) > S ( 1 ) > .J . from the-chain . . . k . k · · · > S(k), where S (j ) = ai -3b{ + a2 - � + · · + a! J b?t, 0 < j < k . The inequality S (j ) > S(j + 1 ) follows from the statement of Example 6.6 . (iv), . where ui = ai , vi = bi and xi = aik-3 - I Of. , 1 < i < n. (vi) For n = 1 the statement is clear. Assume that the statement holds for some n > 1 and consider an arbitrary set of n + 1 positive numbers X I , x2 , . . . , Xn+ 1 such that X I + X2 + · · · + Xn+ I = n + 1. Order them such that Xn < 1 < Xn+ 1 · Then for n positive numbers y1 = x 1, y2 = x2 , . . . , Yn-1 = Xn - I and Yn = Xn + xn+ 1 - 1 you have Y1 + Y2 + · · · + Yn = n, from ·
2 Hints and Answers to Chapter 2
307
which, by the hypothesis, X 1 X2 · · Xn - 1 (Xn + xn+ 1 - 1) < 1. It now suffices to add that Xn + Xn+ 1 - 1 > Xn Xn+ l · (vii) L(4) - R(4) = (x 1 - x2 + X3 - x4 ) 2 > 0, ·
L( n + 1) - L( n) > R( n + 1 ) - R(n) - 2Xn+ 1 (X2 + · · · + Xn - 1 ) + (xn+ 1 - 2x 1 )(xn+ 1 - 2xn ) > 0. The inequality on the right holds as long as Xn+ 1 < X1 and Xn+ l < Xn· By a cyclic permutation of the given (n + 1 ) -tuple x�, x2 , . . . , Xn+ b arrange Xn+ 1 < Xj (1 < j < n) ; doing this, the inequality L(n + 1) > R(n + 1) does not change. (viii) For n = 1 the inequality is equivalent to a� (a 1 - 1) 2 > 0. The induction step from n = k to n = k + 1 can be carried out with a "difference method," as long as
ai+ 1 + a�+ 1 > 2a�+ 1 + 4a�+ 1 (a� + a� + · · · + a� ) . You may clearly assume that a 1 < a2 < · · · < ak+ 1 · Then the expression in parentheses on the right-hand side of the last inequality is not greater than 1 3 + 2 3 + · · · + (ak+ 1 - 1) 3 . According to 2 . 15.(i) of Chapter 1, this sum is equal to (ak+ 1 - 1) 2 a�+ 1 /4. Hence 2a�+ 1 + 4a�+ 1 (a� + a� + · · · + a� ) < 2a�+1 + a�+ 1 (ak+ 1 - 1) 2 a�+ l = ai 1 + a tt- 1 · + 7.5 (i) Use the fact that the n-tuples af , a�, . . . , a� and a:t 1 , a2 1 , . . . , a� 1
have opposite orderings. (ii) Use ( 74) with p = 1, a 1 = X 1 X2 · · Xn = 1, a2 = X2 X3 · · · Xn , . . · , an 1 = Xn - 1 Xn , and an = Xn (iii) By Theorem 7.3, for two triples (a3 , b3 , e3 ) and ( bc :C bc) having the same ordering, the inequalities ·
-
a
_!_
_!_ _!_
' a
' a
> a3 . + b3 . _.!._ + e3 . _!_ a3 . _.!:._ + b3 . + e3 . _!_ be ab ae ab ae be 3 . _.!._ + e3 . _!_ > a3 . + b3 . _!_ + e3 . _.!._ + b a3 . _.!:._ be ae ab ab be ae hold. Add them and divide the result by 2 . 7.8 ( i) Since S ij = Xi Yi - Xi Yi + XjYi - YiXj, you will find that the sum s 1j + s2i + · · · + Snj for each j is equal to (X 1Y1 + . . . + Xn Yn ) - (x1 + . . . + Xn ) Yj + nXj Yj - ( Y1 + . . . + Yn )Xj · Hence the sum of all n2 numbers Sij is equal to 2n(X 1Y1 + + Xn Yn ) - 2(x 1 + · · · + Xn )( Y1 + · · · + Yn ) · · · ·
• ·
308
4. Hints and Answers
( ii) Assume that a 1 < a2 < · · < an , and use the left-hand part of (76) ·
Xk = ak and Yk = a�! 1 - k , 1 < k < n. ( iii) If a 1 , a2 , . . . , an IR+ , then for r > 0 and s > 0, (a! + a2 + · · · + a� )( a� + a� + · · · + a� ) < n(a!+s + a2+s + · · · + a�+ s ), while for r > 0 and s < 0 the opposite inequality holds. Both inequalities are strict, unless a1 = a2 = · · · = an . ( iv ) Use the right-hand part of (76) for the ordered n-tuples (1, 2, . . . , n ) and (!' n_: 1 , · · · ' f). k and the following consequence of Chebyshev)s for
E
(v) Use induction on
inequality:
(at + a� + · · · + a� )(a1 + a2 + · · · + an ) < n(a �+ 1 + a�+ 1 + · + a�+ 1 ). ·
·
(vi) Between the values of the left and the right sides there is the number
n(af + a� + · · · + a! ).
(vii) Use the right-hand part of (76) with Yk (viii) Rewrite the desired inequality as
n[(n - 1)xy + 1]
>
=
2k - 1 , 1 < k < n .
[(n - 1)x + 1] [(n - 1 ) y + 1] .
This is the right-hand part of (76) for the n-tuples x 1 = x2 = · · · = Xn - 1 = x, Xn = 1 and Y1 = Y2 = · · · = Yn - 1 = y, Yn 1. ( Since by assumption (x - 1)(y - 1) > 0 holds, these n-tuples have the same ordering. ) (ix) By Chebyshev's inequality, for each j = 1 , 2, . . , k - 1, =
.
n·
i:
(t t=1
alia2; · · aj; ·
) (tt=1 ai+ l,i) ;
hence the desired assertion can be proved by induction. (x) Choose ( if they exist) indices i and j such that Xi < A < Xj Check how both sides of the inequality in question change if in the n tuple x1 , . . . , Xn you replace the pair (xi, Xj) by the pair ( x� , xj ) , where x� min {A, Xi + Xj - A } and xj = max { A, Xi + Xj - A } . This change increases the number of those terms Xk that are equal to the ( fixed ) mean A of the whole n-tuple x 1 , . . . , Xn· If you repeat this change several times, you finally obtain the n-tuple x 1 , . . . , Xn consisting of n numbers A. =
8.3 The inequalities ( i) -(vii) follow from numbers: ( i) a4 , a3 , 4a, 4.
'L2 c2 a, c2 b. ( 11" ) a2 b, a2 c, b2 c, rra, ( iii ) a4 , a3b , a3 b , ab3 , ab3 , b4 ( iv) Ja, Ja, �' �' �(v) a3 , b6 , 8 . (vi) a/2, a/2, 4ja2 •
(84)
with the following sets of
2
Hints and Answers to Chapter 2
309
(Vll'') 1 , a, a2 , . . . , an- 1 . The inequalities (viii)-(xv) follow from (85) with the following sets of numbers: (viii) a, b , b , c , c, c , d, d, d, d. (ix) A 1 = k 2 + 3k, A2 = A3 = · · · = A k+3 = k 2 + 5k + 6. ( ) a/ 2, a/2, b. (xi) a, b/2, b/2, c/3, c/3, c/3 . (xii) A 1 = A2 = · · · = As = 1 - a, Aa = 1 + a, A7 = As 1 2a (for + a >(xiii1 trival). ) A 1 = A2 = · · · = Ak = b - a, Ak+l = b + ka . (xiv) 1 , 2, 2 , 3, 3, 3, . . . , k,k, . . . , k. (xv) A 1 = A 2 = · · = Am k, Am +l = Am+2 = · · · = Am+ k = (xvi) If you set S = a 1 + a2 + · · + ak , then by (85) , k k a + (1 (1 + ) k +a · · + ) a + (1 + ) Si S 6 · k l 2 L -< [ = 1 = + ] ( k ) LJ (J. ) ki ' k It suffices to add that (�) · f, < J1 (0 < j < k ) . 8.5 (i) L > n2 9n (a l , . . . , an ) 9n (a� - l , . . . , a: - l ) = n2 a 1 a2 · . . an . ( ii) (a 1 + · + an ) { a1 + · · · + a: ) > n2 9n ( ab . . . , an ) · 9n { a1 , , a: ) 1 1 = n(iii2 .) Multiply the two AM-GM inequalities x
=
·
m.
=
·
j =O
· ·
•
and take the square root of the result. (iv) For a lower boun
use
•
•
the three AM-GM
a2 + b2 + > 3 · {/a2b2c2 , ab + be + > 3 · {/a2b2c2 , a2b2 + b2c2 + Ca2 > 3 · {/a4b4c4. (a -a3 ) + (a2 + a3 ) + (a2 - a4 ) + (a3 + a4) + . . . (v) L + n = l +a a a + 3 3 4 a + a a + a ) (a a + (a + ) a . 1 1 3 n + a21 + a2 2 = a2 + a32 + a32 + a4 + . . + aan1 ++ aa21 . The sum of these n terms is at least n, since their product is 1. (vi) n + 1 - Ln = 1 + ! + i + · . . + n;; l > n · V1 · ! . . · n;; l = n {f. d'-
ca
�
4. Hints and Answers
3 10 (vii)
L > nQn
(�b:�' · · · ' ��: J = nQn ( b1 � b2 ' · · · ' bn � bJ
n n = R. > Qn (b1 + b2 , . . . , bn + b1 ) An. (b1 + b2 , · · · , bn + b1 ) (viii) Multiply the inequalities (j = 1, 2, . . . , k) =
n + ai
=
(n + 1) An+ 1 (aj , 1, 1, . . . , 1) > (n + 1) Qn+ 1 (aj , 1, 1, . . . , 1) .
(ix) an + ( b - a__..;_ n) L + n = a 1 +b -(b a- a l ) + . . . + ___:_;_ b - an 1 1 ,... 1 1 ,..., 1 > nbQn = nb . An b - a 1 ' b _ an b - an b - a1 nb n2 b . nb > = gn ( b - a� , . . . , b - an ) - An ( b - a� , . . . , b - an ) nb - 1 This implies L > nb�1 - n = R. (x) Multiply the inequalities (k = 1, 2, . . . , n ) a k 1 a k + 1 , , an a1 1 - 1 = "" ai L.., - > (n - 1) Qn - 1 -, . . . , - , · ··· ak a a a a a k k k k k i# k (xi) Use the AM-GM inequality for the following 2n+ 1 numbers: 2n times · a42 , , 2 tunes · ' the numb er 1 . an2n , and 2 trmes a21 , 2n- 1 t1mes k (xii) If G = yta 1 a2 · · · an , then G is the geometric mean of the set of (�) products ait ai2 ai�e , where 1 < i 1 < i2 < · · < ik < n, for each k = 1, 2, . . . , n - 1. Hence by the AM-GM inequality for these sets of ....:._ ._
_
)
(
(
.
)
(
)
•
•
•
•
•
·
numbers you get
n- 1
( 1 + a1 ) ( 1 + a2 ) · · · ( 1 + an) = 1 + L: {L: ait ai2 • • • ai�e ) + a 1 a2 . . · an k= 1 > 1+ Gk + an = (1 + G) n . k=1
E (�)
By taking roots you obtain the desired inequality. 8.9 (i) Under the assumption 0 < m < Xj < M , 1 < j < n, add, resp. multiply, the inequalities vi m < vi xi < viM, resp. mv3 < x;i < Mv3 . (ii) Use (94) with n = 2, v1 = 1 - p, a 1 = a, v2 = p, and a2 = a + b
( a1 :;f a2 ).
2 Hints and Answers to Chapter 2
(iii) Use (94) with n = 3 and with
Vj
according to (98) ,
311
a1 = db-c ,
a2 = de-a, a3 = da-b . (iv) By (94) with n = 2, � · % + t � > (%) etcl (�) cta . Raising this c d c d to the power c + d and simplifying, you obtain the desired inequality. ·
(v) Rewrite the inequality
where the
Vj
as
are
as
in (9,8 ) . By (94) the left-hand side is at most equal to
v1 ( b + c) + v2 ( c + a) + v3 ( a + b) = 2 · ab + bc + ca · a+b+c It therefore suffices to prove the inequality 3(ab + be + ca) < (a + b + c) 2 • (vi) By (94) with n = 2, v 1 = a, and v2 = b you have the following inequalities: 2ab = ab + ba > ba · ab , a2 + b2 = a · a + b · b > a" · bb , and finally
( ) a ( .!. ) b .!. 2 = a . .!. + b . .!. > = a
b- a
b
1 . aa bb ·
Equality occurs only when a = b = ! . (vii ) By ( 94) with n = 2, v1 = c, and v2
acbl -c < ca + (1 - c)b
and
= 1 - c you have the inequalities (1 - a) c (1 - b) l -c < c(1 - a) + (1 - c)(1 - b) ,
and by adding them you obtain the desired inequality. (viii) By raising this to the power 1/a, you obtain (99) with the exponent
p = (a + 1)/a > l.
(ix) By raising this to the power 1/b, you obtain {100) with p = a/b < 1 and x = 1/a. (x) The inequality (101) is homogeneous. Under the assumption xi + + x� = yr + + y� = 1 add n inequalities (96) for x = Xk and y = Yk · (xi) Use (101) with n = 2 and p = �, q = �' x1 = a2 , ·x2 = b2 , Yl = c3 , ·
·
·
·
·
·
Y2 = d3 •
(xii) Rewrite the sum of the two Holder inequalities
and
(xiii)
E Yk(Xk + Yk ) p- l < < E � ) � (E(xk + Yk ) P ) � . Use (94) with Vk = Pk /(p l + P2 + · + Pn ) and ak = Xk /Pk ,
1 < k < n.
·
·
312
3
4. Hints and Answers
Hints and Answers to Chapter 3
You have a · 0 = 0, and thus a I 0 for any a E Z. Similarly, a · 1 = a gives a[a for any a E Z. If a · x = b, b · y = c, then a · (xy) = c. If a · x = b, a . y = c, then a(x + y) = b + c, a(x - y) = b - c. For c 1= 0 it follows from ax = b that acx = be and conversely. If ax b, b > 0, then in the case a < 0 clearly a < b, and in the case a > 0 you have x > 0, hence x > 1, since x E Z, which implies b = ax > a. 1 .2 (i) If 3n + 4 I 7n + 1, then 3n + 4 J (.7 · (3n + 4) - 3(7n + 1)) , i.e., 3n + 4 E { ±1, ±5, ±25} , which means that n E { -3, -1, 7} . (ii) For n E {3, 14}. [Use 9(5n 2 + 2n + 4) - (15n - 4)(3n + 2) = 44.] (iii) Use the binomial theorem for (3 + 1) n , or prove the assertion by induction. (iv) Use the binomial theorem for ( 4 - 1) 2n+3 , or induction. (v) Use the binomial theorem for ( n + 1 ) n . (vi) Substitute 2n - 1 for n in (v). (vii) Use the identities 2a + 3b = 4(9a + 5b) - 17(2a + b) , 9a + 5b = 1.1
=
81(2a + 3b) - 17(9a + 14b). (viii) If 2n - 2 nt, then for m = 2n - 1 you have 2m - 2 = 2(2nt - 1) = 2m(2n(t- 1) + 2n(t- 2) + . . . + 2n + 1). 1.5 (i) Use the facts that a2 + (a + 1) 2 = 2a(a + 1) + 1 and that a( a + 1) is an even number for all a E Z. (ii) Take the square (2q + 1) 2 = 4q(q + 1) + 1 and proceed as in (i). =
(iii) Induction on n.
(i) 1. (ii) 17, if 17 I n + 8, and 1 otherwise. (iii) 3. (iv) 11, if 11 I n + 7, and 1 otherwise. (v) Proceed as in 1.12.(iii) , choose i k+ l = 6c + 1. (vi) For odd n choose 2 as one of the summands; for n = 4k or n = 4k + 2 (k E N) choose 2k - 1. 2 (vii) Use (22m - 1) = (22m- l + 1)(22m- + 1) · · · (22n + 1)(22n - 1) . (viii) d I a, d I b imply d I (a, b). Conversely, (a, b) I a, (a, b) I b, and thus
1 . 13
(a, b) I ka + lb = d. (ix) (2n - 1, 2m - 1) = 2d - 1, where d = (n, m). To prove this, use (5) of Chapter 1 to verify that 2d - 1 I 2n - 1, 2d - 1 I 2m - 1. For d = n or d = m the proof is easy. Otherwise, by 1.8 you have d = tn + sm for appropriate s, t E Z. Show that t > 0, s < 0 or t < 0, s > 0. In the first case set k = 2n(t- l ) + 2n (t- 2) + . . · + 2n + 1 , l = -2d (2m(- s - 1) + 2m( -s- 2) + . . . + 2 m + 1), show that (2n - 1)k + (2m - 1) l = 2d - 1, and use (viii). The second case
can be dealt with analogously. 1.15 (i) Use 1.8, or prove ((u, v), (x, y)) = (u, v, x, y ) and use 1. 14.(i) three times. (ii) Use a proof by contradiction, apply 1.8 or 1.15.(i) for d = a. (iii ) Show that ( ac, b) is a common divisor of ac, be, from which it follows
313
3 Hints and Answers to Chapter 3
1. 1 4.( i) that (ac, b) I c. Use this to show that the numbers (ac, b) and ( c, b) divide each other. (iv) Let m = (ab, ac, be), n = a;:: . Show that n E Z and n = [a, b, c] q, where q E N. Show that q is a common divisor of the numbers a b ac be and derive from 1.14.(i) that ( C:::, , � , �) = 1. (v) Substitute ab, ac, be for a, b, c in (iv) and divide both sides by abc. (vi) By 1.14.(iii), (x, y) = 30 implies the existence of u, v E N such that x = 30u, y = 30v, (u, v) = 1 and u + v = 5. Then x E {30, 60, 90, 120}, y = 150 - x. (vii) x = 495, y = 315. (viii ) X E {20, 60, 140, 420}, y � (ix) x = 2, y = 10, or x = 10, y 2. 2.4 (i) One of the primes has to be even, i.e. , 2. Then the second one is 6n + 3, which for n E N is not a prime. (ii) n = 3. [Of the numbers n, n + 1, n + 2, exactly one is divisible by 3; the same holds for n, n + 1 + 9, n + 2 + 12.) (iii) p = 3. [It follows from 1.3 that any prime p 1= 3 is of the form 3k ± 1, where k E N, while 2(3k ± 1) 2 + 1 is divisible by 3 .] (iv) p = 5. [Proceed as in (iii) , distinguish between the cases p 5k ± 1 and p 5k ± 2.) (v) n = 2. [For n = 1, 2n - 1 = 1; for n > 2, 2n is of the form 3k ± 1, where k E N, k > 1 . Then 2n =F 1 = 3k is composite.] (vi) If p = 3k ± 1 , n E N, then 8p2 + 1 is divisible by 3. If p = 3k, k E N, then p = 3 and 8p2 + 2p + 1 = 79 is a prime. (vii) n = 4. [Consider divisibility by 5.) (viii) Rewrite in the form 5 20 + 2 30 = (5 1 + 2 15 ) 2 - 2 . 5 10 . 2 15 ((5 1 + 2 1 5 ) + 5 5 . 2 )( (5 1 + 2 15 ) - 5 5 . 2 ). (ix) For n = 4k4 , where k E N, k > 2, you have m4 + 4k4 = ( m2 + 2k2 ) 2 - ( 2mk ) 2 = (( m + k) 2 + k2 ) (( m - k) 2 + k2 ). (x) Use the fact that p - 1 is even, and add the first and the last by
·
m '
=
m '
m
84 0 .
=
=
=
0
8 0
0
8
summands, the second and the second-to-last, etc. Then rewrite it as ';: = (p�1}! , which means that m(p - 1)! pqn, and note that p does not divide (p - 1)!. (xi) The condition is satisfied, for instance, by the numbers ai = i n! + 1, where i = 1, . . . , n. (xii) If x = (a, d) and y = (b, c), the natural numbers a 1 = i, d1 = � , b1 = ky , c1 = �y satisfy (a�, d1 ) = (bb c1 ) = 1, by 1 . 14.( iii). From ab = cd it follows that a 1 b1 = c1 db and by 1 14 . ( ii) you obtain a 1 I c 1 and c1 I a 1 . Hence a1 = CI , b 1 = d1 and finally an + bn + en + � = ( ar + br )(xn + yn ). =
·
.
2.6 (i) For any divisor d of a decompose d and a according to 2.5; consider the primes that can occur, and how the exponents of a particular prime in the decompositions of d, �, a are connected to each other. ( ii) This follows directly from ( i).
314
4. Hints and Answers
(iii) Write a, b, c according to 2.5 and note that in the decomposition of a, b the same prime cannot occur. 2.8 (i) If ( a+b, ab) > 1, there exists a prime p dividing a+b and ab. By 2.2., p divides one of the numbers a, b, and from p I a + b it follows that it also divides the other one; hence p I (a, b), which is a contradiction . Therefore,
(a + b, ab) = 1. (ii) (a, b). [Apply (i) to X = (a�b) , Y = (a�b) . ] ( iii) If p is the smallest integer bigger than 1 dividing a, then in the case a _g_ 1= p the numbers _g_ , p are factors of (a - 1)!. In the case _ = p you have p p P a = p2 , and from a > 4 it follows that p > 2, and thus p, 2p are factors of (a - 1)!. (iv) Write (p - 1)! = pm - 1 = (p - 1)(pm - 1 + pm - 2 + . . · + p+ 1), which implies m > 1 and (p - 2)! = pm - 1 + · · · + p + 1. By (iii) , p - 1 divides (p - 2)!; hence it also divides pm- 1 + · · · + p + 1 = (pm- 1 - 1) + (pm- 2 1) + · · · + (p - 1) + m . This means that p - 1 1 m, and so m > p - 1, thus (p- 1)! pm - 1 > pP- 1 - 1. However, by multiplying the inequalities p > 2, p > 3, . . . , p > p - 1 together, you obtain pP-2 > (p - 1)!, a contradiction. (v) Proceed as in 2 . 7. (ii) . Assume that P 1 = 3, P2 = 7, P3 = 11, . . . , Pn are all the primes of the given form, and consider the number N = 4p2 · · · Pn + 3. (vi) If 2 2n + 1 = m5 - k5 is a prime, then it follows from the factorization m5 - k5 = (m - k) (m4 + m3 k + m2 k2 + mk3 + k4 ) that m = k + 1, and thus 2 2n = (k + 1)5 - k5 - 1 = 5k(k3 + 2k 2 + 2k + 1 ) , a contradiction. (vii) Show that for any prime p and r E N such that p7 I n, the number m = 1 + a + · · · + an- 1 is also divisible by p'�"" . Use the expression =
r m = ( 1 + aPr + . . . + an-pr ) . II ( 1 + aPi- 1 + a2pi - 1 + . . . + a(p- 1 )pi - 1 ) ' i= 1 which is clear for a = 1, and which for a 1= 1 can be sho:wn by �ultiplying both sides by a - 1. Then any one of the r factors 1 + aP· - 1 + a2P• - 1 + . . . + aCP- l )pi-l = p + (aPi- 1 - 1) + (a2Pi - t - 1) + · · · + (a CP- 1 )Pi- l - 1) is divisible by p, since from pr I (a - 1) 8 it follows that p I a - 1. (viii) If the prime p divides a + b and a2 + b2 , then it also divides 2a2 = a2 + b2 + (a + b) (a - b) and 2b2 • For p 1= 2 you then have p I a, p I b, a contradiction. For p = 2 it follows from 2 I ab, 2 I a + b that 2 I a, 2 I b, a contradiction .
0, and thus vp (ab) = vp (b). 2.11 (ii) If vp(a) < vp (b), then vp (a) Analogously in the case vp (a) > vp (b) . (iii) Assume that yin = .;, where s > 1 and (r, s) = 1. Then rm = sm ·n, from which for any prime p dividing s it follows that p I rm , hence p I r, which contradicts (r, s ) = 1. (iv) For any prime p you have vp ( y'nT) = .;vp (n) E N and thus s I rvp (n), =
3 Hints and Answers to Chapter 3
315
which implies with 1 .14.(ii) that s I vp(n). If therefore n = Pi1 p�lc , where Pb , Pk are distinct primes, then m = p�1/8 p�lc I 8 E and m8 = n. (v) Set c = ..jO. + v'b. Then the number d = a�b = y'a - Vb is also rational, and therefore .JO, = cid and y'b = c;d are rational numbers as well. (vi) The assertion is clearly true if one of the numbers a, b, c, d is zero. In the case abed =I= 0 it is enough if for any prime p you show that vp (ac) > k, vp (bd) > k, vp (bc + ad) > k, or be + ad = 0 implies vp(bc) > k, vp (ad) > k . In the case vp (bc) = vp(ad) use (8); otherwise, use (10). 3.5 (i) Use the facts that 24 = 3 (mod 13) and 72 = -3 (mod 13) . (ii) Use 72 2n+ 2 _ 472n +28 2n- 1 = ( -3) 2n+ 2 _ ( -3) 2n +32n- 1 = 32n- 1 (33 3 + 1) (mod 25). (iii) This follows from 55 = 45 = 35 = 1 (mod 11). (iv) By 1 . 3, any a· E Z is congruent to 0, 1, or 2 modulo 3. If a = 0 (mod 3), then b2 = 0 (mod 3) , and by 2.2 you have b = 0 (mod 3). If a = 1 (mod 3) or a = 2 (mod 3), then you always have a2 = 1 (mod 3); this implies b2 = 2 (mod 3), which is impossible. (v) Proceed as in (iv): If a ¢ 0 (mod 7), then a2 is congruent to 1, 2, or 4 modulo 7, which implies that b2 is congruent to 6, 5, or 3 modulo 7, but this is impossible. (vi) It suffices to choose a = 1 , b = 2. (vii) Verify that for abc ¢. 0 (mod 3) each of the numbers a3 , b3 , c3 is congruent to 1 or - 1 modulo 9, and thus a3 + b3 + c3 ¢. 0 (mod 9). (viii) Proceed as in (vii) . (ix) Since 121 = 11 2 and a2 + 3a + 5 = (a - 4) 2 (mod 11), it follows from a2 + 3a + 5 = 0 (mod 121) that 11 I (a - 4) 2 , and thus 11 I a - 4, i.e., there exists a t E Z such that a = 11t + 4. But then 0 = a2 + 3a + 5 = 121t2 + 121t + 33 = 33 (mod 121) , a contradiction. (x) The solution is given by all n congruent to 1 or 2 modulo 6. [Consider six possibilities, i.e., n congruent to 0, 1 , 2, 3, 4, 5 modulo 6.] (xi) Such a number does not exist. For, if 2n- 1 (2n+ 1 - 1) were a third power of a natural number, then by 2.6 . (iii) , both factors would be third powers of natural numbers. First of all, this means that n - 1 is divisible by 3, i.e., n = 3k + 1 for k E Z, k > 0. Then 2n+ 1 - 1 = 4 gk - 1 = 3 (mod 7); however, as one can easily see, a third power of a natural number can only have remainders 0, 1, or 6 upon division by 7. (xii) Write down consecutively the remainders of 2 1 , 22 , 23 , . . . , 2 12 upon division by 13. (xiii) For n = 2 k (k E show that 3 I 19 sn + 17, for n = 4k + 1 show that (k E show that 13 1 19 · sn + 17, and for n = 4k + 3 (k E 5 1 19 . sn + 17. (xiv) For n = 1 and n = 2 this is clearly true . For n > 2 use equation (5) from Chapter 1: nn - n2 + n - 1 = (nn- 2 - 1)n2 + (n - 1) = (n - 1) (nn- 1 + nn- 2 + . . . + n3 + n2 + 1) . Derive from n = 1 (mod n - 1) that ·
·
•
·
•
•
N
•
•
all
·
No)
No)
·
No)
•
316
4.
Hints and Answers
nn- 1 + . · · + n2 + 1 = 0 (mod n - 1). (xv) For e E { -1, 1 } set be = a2 + ae + 1. Use the fact that (a2 + 1) 3 = (be - ae) 3 = -a3 e -ae(be - ae - 1) a2 + ae = -1 (mod be) · Both assertions are therefore true. =
=
(i) 400, 504, 48, 96, 720. (ii) For 5 I n you have �(5n) = 5�(n); otherwise, �(5n) = �(5) · �(n) = 4�(n). (iii) p = 2, x > 1. [px- 1 E N implies x > 1, and thus �(px ) = (p - 1)px - 1 .] (iv) x = 2, y = 3. [From 3 I 600 it follows that x > 1 , and similarly y > 1; hence �(3x 5Y) = 2 · 3x- 1 · 4 5y- l .] (v) p = 3, x > 1. [From 2 · 3px-2 E Z it follows that x > 1, and thus �(px ) = (p 1)px - 1 = 2 . 3 . px-2 .] (vi) (p, q) E {(3, 61), (5, 31) , (11, 13) , (13, 11), (31, 5), (61, 3) } . [Distin guish between p = q, where p(p- 1) = 120, and p 1= q, where (p- 1) (q - 1) = 120, and consider decompositions of the number 120 into two factors.] (vii) p = 2, m E N arbitrary odd. [If p I m, then �(pm) = p� (m); otherwise, �(pm) = (p - 1)�(m) .] (viii) p any prime, m E N any number divisible by p. (ix) p = q, m E N arbitrary, or p = 2, q = 3, m E N even and not divisible by 3, or p = 3, q = 2, m E N even and not divisible by 3. [See the hint to (vii) .] (x) No solution. [From �(m) = 14, and since neither 8 nor 15 is prime, it follows that 72 I m, hence 6 7 1 �(m) , a contradiction.) (xi) m E { 17, 32, 34, 40, 48, 60 } . [�(m) = 16 implies m = 2a 3b · 5c · 17d, where a, b, c, d E N0 , b < 1, c < 1, d < 1 . So �(m) = 2b . 4c · 16d if a = 0 and �(m) = 2a - 1 · 2b 4c 16d if a > 0. Hence either a = 0 and b + 2c + 4d = 4, or a > 0 and (a - 1) + b + 2c + 4d = 4.] (xii) m = 2a, where a E N. [ m E Z implies m = 2a · t, where a E N and t E N is odd. Then �(m) = 2a-¥�(t ) ; hence = �(t), i.e., t = 1.] (xiii) m = 2a · 3b, where a, b E N. [ � E Z implies m = 3b t, where b E N, t E N, � f/_ N. Then �(m) = 2 3b- 1 · � ( t ) , so � ( t ) = � - Use (xii) .] (xiv) No solution. [� E Z implies m = 2a · t, where a E N, a > 2 and t E N is odd. Then �(m) = 2a-1�( t ) , so = � - Use (xii) .] (xv) Proceed as in 3.7.(iv) , or substitute (m, n) and [m, n] for m and n in 3.7.(iv) and use 1.9. (xvi) pn . [With the exception of the first summand, factor out (p - 1) and use (5) of Chapter 1.] (xvii) Write m = p� 1 • • • p�lc and note that by 2.6.(i) the positive divisors of m are of the form pf1 • • • p�lc , where 0 < m 1 < n t , . . . , 0 < m k < nk , and that you have �(pf1 • • • p�lc ) = �(pf1 ) • • • � (p�lc ). Consider the product 3.8
·
_
all
·
·
·
·
t
·
·
� (t)
( �(p� )+ �(pl )+ �(p�)+ · . · +�(p�l )) . . . ( � (p�)+ �(pl)+ � (p�)+· . · + � (p� k )) , which by (xvi) is equal to p�1 • • p� lc = m. By multiplying out you obtain exactly the sum of the products � (pf 1 ) • • • �(p�1c ), where 0 < m 1 < n 1 , . . . , 0 < m k < nk · •
3 Hints and Answers to Chapter 3
317
(i) 0 for integers divisible by 5, and 1 for all others. [Use Euler's theorem for m = 53 .] (ii) Distinguish between the cases where is or is not divisible by the prime and use Fermat's theorem. (iii ) Decompose 730 = 3 5 7 13 and use (ii) and 3.3.(vi). (iv) Decompose 341 = 31 11. Using (ii) with 341 = ( 11 - 1 ) +1 (mod 11), and from 3 1 = (31 1) 11 + 1 + 10 it you obtain 1 (mod 31) . Similarly, (mod 31) . Use follows that = ¢. 3 (mod 31) , since = 3 = ( - 5) 2 -6 (mod 31) . (mod 10). = (v) Consider divisibility by 11, and use (mod and (vi) Consider divisibility by (vii) Use for divisibility by (mod (mod 36) for divisibility by 37. (viii) Use (ix) First convince yourself that the given number is an integer. Consider (mod 6) if is odd, divisibility by 7, using the fact that for E N, (mod 6) if is even. and and that the given numbers (mod (x) Show that + are divisible by E N0 ) the given numbers are divisible by 7, for (xi) For n = + n = + 1 (k E No) they are divisible by 13, and so on. (xii) For n = 1 this is clearly true. For n > 1 write n = where < P2 < are primes and E N. Note that from < 1< 1< 1< 1 < n - 1 it follows that 1) < 1) 1 ) I (n - 1)!. Also, I n, and together � (n) I n!, which by Euler's theorem implies that 1 (mod n). E N) , and for > consider (xiii) For = consider n = n = 1 ) - 1 ) E N) . 4.4. (i) x = 6 + 9t ( t E Z). (ii) x = 7 + 10t (t E Z) . (iv) x = 1 + 5t (t E Z) . (iii) x = -5 + 3lt ( t E Z) . (v) x = 1 + 3t (t E Z) . (vi) No solution. (vii) Use induction on n. For n = 1 it suffices to choose = 1. If +b are pairwise are chosen such that +b, < < < relatively prime, denote by m their product and show that m) = 1. Let denote some solution of the congruence ax 1 - b (mod m) > (existence is guaranteed by Theorem then + b 1 (mod m) and solve the problem. the numbers 4.8. (i) x = -3 (mod 55) . (ii) No solution. (iii) x 20 (mod (iv) x 1 3 (mod 36). [The first congruence follows from the second.] (mod 105) . (vi) x (v) x 11 (mod 30). (vii) For even no solution; for odd a the solution is x 4a-3 (mod (viii) No solution when a is not divisible by when is divisible by (mod 10) is the solution . [The second congruence gives x = then x + lOt, where t E Z, and substituting this into the first one, 3.15
a
p,
2 2· · 34 · 1 233441 = 2 10 4 · 1 0 2 2 2·2 = 2 32 3341 · 310 310 243242n+=1 2 · 2=4n = 2 n + 2 = 4 23. 28). 26 29, 12 0n+ 1 = 2 22) 26n+2 = 4 k k 2 = 2 k k2 = 4 k n+ 229. · 36 S 4 = 2 28) 4 (k 6k 12k �1 p:1c , p · · · 7 a a , P1 Pk k 1 · · (p (p · P1 1 P Pk 2 2 1 1 1 �c· · · (pk p� · · 2· p�n! = 2 p (k 2k 2 p (kp - (p (k ·
·
·
=
•
•
k2 · · · kn- 1 kn- 1 k1 , . . . , kn 42). == = a 2 = -2 k1 kn
•
•
•
ak1 . .. , akn- 1 (a, = 4.2); akn =
=4
•
4;
=a
k1
24).4, -4 + 20t = a
4.
318
Hints and Answers
(mod 4). For a not divisible by 4, this has no solution , while for 4 I a it is satisfied by any t E Z.) (ix) If the integer c satisfies the system (1 7) , then this system is equivalent to the system obtained from ( 1 7) by substituting all the right sides by c. Use 3.3.(vi). (x) Use induction on k with the help of Theorem 4.6. (xi) Choose k distinct primes Pl , P2 , . . . , Pk · The numbers p'f , p�, . . . , p: are pairwise relatively prime, and by (x) there exists an m E Z such that m = -i (mod pf) for i = 1, 2, . . . , k. The numbers m + 1, m + 2, . . . , m + k have the required property. (xii) To prove the more difficult part of the problem, set mi = ni · ' t'met prrmes d'1pni Ps , z = 1 , . . . , k , where P b . . . , Ps are all the dIS 1 • viding m = [m 1 7 , mk] , ni,j E No for i = 1, . . . , k; j = 1, . . . , s . Note that the system (17) is equivalent to the the system (i 1, . . . , k; j = 1, . . . , s) . x = Ci (mod p;i ,; ) 1
•
•
s
•
•
'
•
•
•
=
For each j = 1, . . . , s choose i(j ) E { 1 , . . . , k} such that for all i E { 1, . . . , k } you have ni,i < ni (j) ,j . If the condition of the problem is satisfied, this last system is equivalent to the system n (;) ,; - --i j ) (j = 1 , . . . , s ) , X = ( (mod p3- i ) which, according to (x) , always has solutions. 4.13 (i) x = 1 (mod 7) . (ii) x = 1 (mod 5) or x = 3 (mod 5). [It is convenient to write the given congruence in the form 2x2 + 2x+ 1 = 0 (mod 5) , respectively x2 + x + 3 = 0 (mod 5).] (iii) No solution. (iv) x = 1 (mod 5). (v) x = 1 (mod 6) or x = 3 (mod 6). [It is convenient to first solve the congruence x2 - 4x + 3 = 0 (mod 2) .] (vi) x 0 (mod 5). [By 3.11 you have x5 = x (mod 5); rewrite the given congruence as -x2 = 0 (mod 5).] (vii) x = 1 (mod 11). [Using 3.11, rewrite as x - 1 = 0 (mod 11).] (viii) x = 1 (mod 5) or x = 2 (mod 5). [Using 3.11, rewrite as x2 - 3x + 2 = 0 (mod 5); the trinomial on the left can be factored: x2 - 3x + 2 = (x - 2) (x - 1).] (ix) x = 3 (mod 35) . (x) x = 24 (mod 45). [Solve separately for the moduli 5 and 9. For modulus 5 rewrite as 3(x + 1) 2 = 0 (mod 5). In the case of modulus 9 it is convenient to first solve modulo 3, which gives x = 3s + 1 or x = 3s; the first case gives a contradiction, while the second one leads to x = 6 + 9t.] (xi) No solution. [First solve modulo 11 and substitute the solution x = 11t - 5.) (xii) All x = a (mod 35) are solutions, where a E { 1 , 4, 6, 9, 1 1 , 16, 19, 24, 26, 29, 31, 34}. r,
=
3 Hints and Answers
to
Chapter 3
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m m) �(m) m
(xiii) This condition is satisfied if and only if is a power of a prime. [For = pn , where p is a prime, use Euler' s Theorem 3. 13 to show that the choice F (x) = xcp(m) gives F(a) = 1 (mod for any integer a that 1 1 is not divisible by p, while the inequality = pn - (p - 1) > pn - > n shows that F(a) = 0 (mod for any integer a divisible by p. On the other hand, suppose that an integer is divisible by two different primes p and q and that a polynomial F(x) with integer coefficients satis fies the given condition. Then there are integers a, b such that F ( a) 0 (mod and F (b) = 1 (mod Due to Theorem 4.6 there is an in teger c such that c = a (mod p) and c = b (mod q), so Theorem 4.10 gives F(c) = F (a) = 0 (mod p) and F(c) F (b) 1 (mod q). But in this case you have neither F(c) = 0 (mod nor F(c) = 1 (mod a contradiction.] 4.15 (i) If a is a solution of the congruence x2 = t (mod p), then a2 = t (mod p), so (a,p) = 1 and by 3 . 1 1 you have t{p- 1) /2 = aP- 1 = 1 (mod p). (ii) Evaluate ( ±tk+ 1 ) 2 t2k+ 2 = t2 k+ 1 t = t (mod p). 5.4 (i) x = 3t + 2, y = 2t + 1 , t E Z arbitrary. [Other forms of the solution are also possible.] (ii) x = 1 + 15t, y = -1 - 17t, t E Z arbitrary. (iii ) No solution, since (16, 48) = 16, which does not divide 40. (iv) x = 2t, y = 4 - 5t, t E Z arbitrary. (v) x = -3 + 13t, y = 4 - 17t, t E Z arbitrary. (vi) x = 9 + 77t, y = 7 + 60t, t E Z arbitrary. (vii) x = -7 - 5s - 5t, y = 7 + 3s + 6t, z = 1 + 2s, s, t are arbitrary integers. [This expression is not unique; the solution can, for instance, be written as x = 3 + 5s, y = -5 - 6s - 3t, z = 1 + 2t, s, t E Z, and in many other ways. This is similar in (viii) and (x) -(xii) below.] (viii) x = -3 + 5s + lOt, y = 4 - 12s - 9t, z = 7s, s, t E Z arbitrary. (ix) No solution, since (105, 119, 161) = 7, and 83 is not divisible by 7. (x) x = -3 + s + 2r - llt, y = 4 - 2s - 3r + lOt, z = s, u = r, s, r, t E Z arbitrary. (xi) x = 4 - 5s - 4r - 10t, y = 4s - r + t, z = -l + r + 3t, u = r, s, r, t E Z arbitrary. (xii ) x = 2t, y = 2s - t, z = 2r - s, u = 2 - r, r, s, t E Z arbitrary. Or more easily, simplifying to the form x = -2y-4z - 8u+ 16, we immediately see the solution x -2r - 4s - 8t + 16, y = r, z = s, u = t. 5.8 (i) x = 2 + llt, y = 4t + llt 2 or x = 9 + llt, y = 7 + 18t + llt2 , where t E Z is arbitrary. (ii) No solution. (iii) x = 2 + 2lt, y = 44lt3 + 168t2 + 20t + 1, where t E Z is arbitrary. (iv) x = 125t - 12, y = 15625t3 - 4500t2 + 434t - 14, where t E Z is arbitrary. (First solve the congruence x3 + 2x + 2 = 0 (mod 5), obtaining the solutions x = 1 + 5r, x = 3 + 5r, r E Z; substitute them back into x3 + 2x + 2 and solve the same congruence modulo 25. The case x = 1 + 5r
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leads to the contradiction 5 = 0 (mod 25), while the second case gives x = -12 + 25s, s E Z. Substitute this again into x3 + 2x + 2 and solve the congruence (25s - 12) 3 + 2(25s - 12) + 2 = 0 (mod 125 ), which leads to s = 5t , t E Z, i.e., X = -12 + 125t. J (v) x = 11t - 4, y = 121t3 - 253t2 + 136t - 22, t E Z arbitrary. (vi) No solution. (vii) x = 3 + 30t, y = -1 + 6t + 30t2 or x = -3 + 30t, y = -1 - 6t + 30t2 , where t E Z is arbitrary. [The congruence x2 -39 = 0 (mod 30) is easiest to solve using a system of three congruences modulo 2, 3, 5 (see the paragraph before 4.12). ] (viii) x = 2t, y = 2s + 1, z = - !(t + s + l)(t + s) or x = 2t + 1, y = 2s, z = -!(t + s + 1)(t + s), where t, s are arbitrary integers. (i.x) x = t, y = t + 3s, z = t2 + 3ts + 3s2 - 2t - 3s, where t, s E Z are arbitrary. [Show that x = y (mod 3).] (x) No solution. [Use the fact that for arbitrary a E N you have a2 = 0 (mod 8 ) or a2 = 1 (mod 8) or a2 = 4 (mod 8). Hence x2 + (2y) 2 + (3z) 2 = 7 (mod 8) has no solution.] a ( xi ) x = -600 - 2t + t2 , y = 300 + t ( ;t) , t E Z arbitrary. [Choose the substitution x + 2y = z.] (xii) x = -1 - 5s - 7s 2 , y = 2 + 7s - (1 + 5s + 7s 2 ) 2 or x = -3 - 9s - 7s2 , y = 4 + 7s - (3 + 9s + 7s2 ) 2 . [Choose the substitution t = x2 + y and solve the equation t2 + t + 1 = -7x.] 5.10 (i) x 2k + 1, y = 2s, z = �(5 · 25 k + 9 8 + 2 ), where k, s E No are arbitrary. [Note that 52k = 1 = 328 (mod 8), 52 k+ l = 5 (mod 8), 328+ 1 = 3 (mod 8). ] (ii) x = 4 + 12k, y = 1� (24+ 1 2k - 3) , where k E No is arbitrary. [Use the fact that the smallest natural number n for which 2n = 1 (mod 13) is n = 12, and therefore by 3.14.(v) for any x E No you have 2x = 24 (mod 13) if and only if x = 4 (mod 12).] (iii) (x, y) E {(- 2, -11), (-4, 3), (-1, -6), (-5, -2) , (1 , -2), (-7, -6), (5, 3) , ( -11, -11)}. [Express y = x - 1 - x!a and consider the cases where x + 3 divides 8.] (iv) x = 2 , y = 8 or x = -2 , y = 0. [Express y = 2x + 1 + xl!7 and consider the cases where x2 + 7 divides 33 .] (v) x = t 5 + 2, y = t4 - 3, t E Z arbitrary. (Express y = -3 + {/(x - 2) 4 , which, by 2.11.(iv), implies x - 2 = t5 for appropriate t E Z. (Consider the cases x < 2, x = 2, x > 2; 2.11.(iv) is stated only for natural numbers) .] (vi) x = (t + 1) 2 , y = t2 , t E N0 arbitrary. [Use 2..jY = x - 1 - y and 2.11.(iii) to obtain y = t2 , where t E No.] (vii) (x, y) E { (3, 2), ( -3, -2), ( -3, 4), (3, -4), ( -7, 6), (7, -6), ( -17, 12), ( 17, -12) } . [It follows directly from the problem that y is even . If you solve the equation for x, you find that y has to be a divisor of 12; hence y can only be among the numbers ±2, ±4, ±6, ±12. ] (viii) (x, y, z) E {( 1, 1, 5), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)}. [By solv=
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ing for z, you find that xy 1 1, which is possible only when lxl = I Y I = 1.] 5.13 (i) (x, y) E { (3, 2) , (6 , 3), (12, 4) , (30, 5)}. [Rewrite as (x + 6)(y - 6) = -36, which implies y < 6.] (ii) (x, y, z) E { (2, 4, 4) , (4, 2., 4) , (4, 4, 2) , (2, 3, 6), (2, 6, 3) , (3, 2, 6) , (3, 6, 2) , (6, 2, 3) , (6, 3, 2), (3, 3, 3) }. [Assume that x < y < z and show that 1 < x < 4 ( if 4 < x < y < z, you would have � + � + ± < ¥ ). Distinguish between the cases x = 2, from which (y - 2) (z - 2) = 4, and x = 3, which gives (2y - 3) (2z - 3) = 9.] (iii) x = 2, y = 1. [If y = 1 , then x(xx- 1 - 2) = 0, which implies x = 2. If y > 2, then y(yY- l - 1) > 2, which gives x(xx - 1 - 2) < -2. Of course, for x > 2 you have x(xx- 1 - 2) > 0 and for x = 1, xx - 2x = -1.] (iv) (x, y, z) E { (2, 3, 4) , (2, 4, 3) , (3, 2, 4) , (3, 4, 2), ( 4, 2, 3) , ( 4, 3, 2)}. [Suppose x < y < z and show that z > 5 or z < 3 leads to a contradiction.] (v) x = y = 1 or x = 3, y = 2. [As in 5.12.( iii) , show that x > 0, y > 1. If y is odd , then 2x = 3Y - 1 = 2 (mod 4) , which means x < 1. For y = 2k you have 2x = 32k - 1 = (3k - 1) (3k + 1); hence 3k - 1 and 3k + 1 are powers of 2. Therefore, from 3k - 1 < 3k + 1 it follows that 2(3k - 1) < 3k + 1, which implies 3k < 3, i.e ., k < 1 .] (vi) (x, y) E { (0, 0) , (1 , 0) , (0, 1), (2, 1), (1, 2) , (2, 2)}. [Rewrite as (x-1) 2 + (y - 1) 2 + (x - y) 2 = 2, which implies 0 < x < 2, 0 < y < 2.] 5.15 (i) x 11, y = 3. [If x, y is a solution of the given equation, then 4x2 = 4 + 4y + 4y2 + 4y3 + 4y4 , and thus (2y2 + y) 2 = 4y4 + 4y3 + y2 < ( 2x ) 2 < 4y4 + 4y3 + 9y2 + 4y + 4 = (2y2 + y + 2) 2 , which implies 4x2 ( 2x) 2 = (2y2 + y + 1) 2 = 4y4 + 4y3 + 5y2 + 2y + 1. Equating this with the given equation, you get y2 - 2y - 3 = 0, i.e., y = 3 (y = -1 ft. N) , and thus X = 11.) (ii) (x, y) E { (0, -1) , ( -1, -1), (0, 0), ( -1 , 0), (5, 2), ( -6, 2)}. [A solution x, y of the given equation satisfies (2x + 1 ) 2 = 4y4 + 4y3 + 4y2 + 4y + 1 = (2y2 +y) 2 +3y2 +4y+ 1 (2y2 +y+ 1) 2 - (y2 -2y). Show that for any y E Z, y 1= -1, you have 3y2 +4y+ 1 = (3y+ 1)(y + 1 ) > 0, and that for any y E Z, y 1= 0, y 1= 1, y 1= 2, you have y2 - 2y (y - 2)y > 0. Hence for arbitrary y E Z, y ft. { -1, 0, 1 , 2 }, you get (2y2 + y) 2 < (2x + 1) 2 < (2y2 + y + 1) 2 , which is a contradiction. It remains to consider y E { -1, 0, 1, 2 } .] (iii) x = 0, y = 1 or x = 0, y = -1. [Show that a solution x, y of the given equation satisfies the following: If x > 0, then (x3 + 1 ) 2 = x6 + 2x3 + 1 < x6 + 3x3 + 1 = y4 < x6 + 4x3 + 4 = (x3 + 2) 2 ; if x < -2, then x3 + 3 < 0; hence (x3 +2) 2 = x6 +4x3 +4 < x6 +3x3 + 1 = y4 < x6 +2x3 + 1 = (x3 +1) 2 , and in both cases you get contradictions. Therefore, x = -1 or x = 0; the case x = -1 leads to y4 = -1, a contradiction. ] (iv) x = 0, y = 1 or x = 0, y = -1. [For x2 > 0 you have (x4 + x2 ) 2 = xB + 2x6 + x4 < y2 = x8 + 2x6 + 2x4 + 2x2 + 1 < x8 + 2x6 + 3x4 + 2x2 + 1 = (x4 + x2 + 1 ) 2 , a contradiction.] (v) For an appropriate k E N you have 2n2 = kd. Suppose there exists an x E N such that x2 = n2 + d. Then k2 x2 = k2n2 + k2 d = n2 (k2 + 2k), =
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which implies k2 + 2k > 0 and v'k2 + 2k = � . By 2 .11.(iii) you have v'k2 + 2k = m E N. But then k2 < k2 + 2k = m2 < k2 + 2k + 1 = ( k + 1 ) 2 , a contradiction. 5.17 (i) x = y = z, where z E N is arbitrary. (Use the AM-GM inequality for x3 , y3 , z3 . ] (ii) No solutions. (iii) y = 2x, z = 3x, where x E N is arbitrary. [Use Cauchy's inequality for the triples x, y, z and 1 , 2, 3; see Chapter 2, Section 5.4.] (iv) No solutions. [By symmetry you may assume that the natural num bers x < y < z < u < v satisfy the equation. Then clearly x > 2, but at the same time u < 3, since for u > 3 you would obtain 1 1 1 1 1 1 1 -1 -1 -1 = 35 1. < + + + + -2 + -2 + -2 + -2 + -2 < x y z u v - 4 4 4 9 9 36 Then x = y = z = u = 2 , from which upon substitution you get 1/v2 = 0, a contradiction.] 5.20 (i) x = y = 0 or x = y = 2. [Rewrite as (x - 1) (y - 1) = 1 .] (ii) (x, y) E { (0, 0) , (0, -1), ( -1, 0), ( -1, -1)}. [Rewrite as (2(2y + 1)) 2 (2x + 1) 2 = 3 and use the identity A2 - B 2 = (A + B )(A - B ) . The number 3 can be written as a product of two integers in only four ways.] (iii) (x, y) E { (11, 4), (11, -4) , ( -11, 4), ( -11, -4), (13, 8) , (13, -8) , ( -13, 8), ( -13, -8) , (19, 16), (19, -16) , ( -19, 16) , ( -19, -16) , (53, 52), (53, -52) , (-53, 52) , (-53, -52)}. [Note that if (xo, Yo) is a solution, then (xo, -yo) , ( -xo, Yo) , ( -xo, -yo) are also solutions, and therefore it suffices to solve the equation in N0. If you rewrite the equation as (x + y) (x - y) = 105, then the conditions x > 0, y > 0 imply x + y > 0, and therefore also x - y > 0, and since x + y > x > x - y, it suffices to solve only the four systems of equations.] (iv) For p = 3 the equation has the solution x = 5, for p = 17 it has the solution x = 13, and for all remaining primes it has no solution. [Decompose 2x2 - x - 36 = (2x - 9) (x + 4) and consider the six possible decompositions of p2 as a product of two integers. For each of these decompositions solve a system of two equations in the two unknowns x, p.] (v) x = 8, y = 5. [Decompose 2x2 + 5xy - 12y2 = (2x - 3y)(x + 4y) and use the fact that x + 4y > 5 holds for x, y E N, and solve the three systems of equations corresponding to the decompositions 28 = 4 · 7 = 2 · 14 = 1 · 28.] (vi) x = 3. [Rewrite the equation as 2x2 -4 (2x+2 - 1) = 25 31. Since for x E N the number 2x+2 - 1 E N is odd, it follows that 2x+2 1 = 31, i.e., x = 3. Convince yourself by substituting that x = 3 satisfies the equation.] (vii) X = 4, y = 3, z = 1 . [Rewrite as 2x- 1 = (y- 1)(yz +yz - 1 +· . ·+y+1). Then y - 1 is a power of 2, which means that y = 2 or y > 3 is odd. From y = 2 it would follow that 2x- 1 is odd; thus x = 1 and z = 0 fj. N. If y > 3 is odd, then z is odd, since yz + yz - 1 + · · · + y + 1 > y + 1 > 4 is a power •
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of 2. Hence z = 2v - 1 for an appropriate v E N, and 2x- 1 = y211 - 1 = ( Y11 - 1) (y11 + 1). This implies y11 - 1 = 2 8 , y'lJ + 1 = 2t , where 8, t E No satisfy 8 + t = x - 1 , and thus 2 = (y11 + 1) - (y11 - 1) = 2t - 28 • Therefore, t > 8 and 2 = 28 (2t - 8 - 1), i.e., 8 1 , t = 2. Then x = 8 + t + 1 = 4 and y'lJ = 3, so y = 3, v = 1, and finally z = 2v - 1 = 1.] (viii) No solution. [Rewrite as 2x+ 1 = yz+ 1 + 1, which implies that y is odd. If z is odd or y = 1, then 2x+ 1 = yz+l + 1 = 1 + 1 (mod 4) , which would imply x = 0 ft. N. Therefore, z is even and y > 1. Then 2x+ 1 = (y + 1)(yz yz - 1 + . . y + 1). Now yz yz 1 + . . . y + 1 = (y - 1)(yz 1 + yz - 3 + + y) + 1 > 1 is an odd divisor of 2x+ 1 , which is a contradiction.] (i.x) Assume to the contrary that for appropriate k, r E N you have k + (k + 1) + . . + (k + r) = 2n , which implies (2k + r)(r + 1) = 2n+ 1 . Since r+1 > 1 and 2k+r > 1 , both numbers are even. Then also (2k+r)-(r+1) = 2k - 1 has to be even, which is a contradiction. (x) Any natural number N that is not of the form 2n , n E N0 , can be expressed as N = 2n (2m + 1), where n E N0 , m E N. You have to show that there exist k, r E N such that k + (k + 1) + + (k + r) = N, i.e., (2k+r )(r+ 1) = 2n+ 1 . (2m+ 1). In the case 2n > m set r = 2m, k = 2n -m, and in the case 2n < m set r = 2n+ 1 - 1, k = m + 1 - 2n . (xi) (x, y, z) E { (6, 6, 2) , (6, 2, 6) , ( 18, -4, 0) , (18, 0, -4)}. [Subtracting the two equations from each other, rewrite them as (y - 1) ( z - 1) = 5 and solve four systems of equations in terms of y, z.] (xii) (x, y, z) E { (1 , 1, 1), ( -5, 4, 4) , ( 4, -5, 4) , ( 4, 4, -5) }. [From Table 4.12 in Chapter 1 (for x 1 = x, x2 = y, X3 = z) you get with 0'1 = 3, 8 3 = 3, (3 - x)(3 - y)(3 - z) = 27 - 90'1 + 30'2 - 0'3 0'1 0'2 - 0'3 31 ( 0'31 - 83 ) = 8, (3 - x) + (3 - y) + (3 - z) = 9 - 0'1 = 6. =
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Therefore, either of the numbers 3 - x, 3 - y, 3 - z are even, or exactly one is even. In the first case 3 - x = 3 - y = 3 - z = 2, and in the second case two of the numbers 3-x, 3-y, 3 - z are equal to - 1 and the remaining one is 8.) (xiii) Use the result of Example 5 . 1 9 .(v) to show that there exist a, b E N, a > b such that xy = 2ab (a2 -b2 ) , z2 = a2 +b2 • H 7 does not divide xy, then it divides neither a nor b. It is easy to verify that the square of a number not divisible by 7 is congruent to 1, 2 or 4 modulo 7. Since none of the sums 1 + 2, 1 + 4, 2 + 4 is divisible by 7, nor congruent to 1, 2, 4 modulo 7, z is not divisible by 7 and a 2 = b2 (mod 7) must hold. Then 7 I a2 - b2 and thus 7 I xy, which is a contradiction. The condition ( x, y) 1 is necessary, since, for example, 152 + 202 = 54 , while 7 does not divide 15 20. 6.3 (i) No solution. [Solve as a congruence modulo 3 and use the fact that the congruence -y2 = 1 (mod 3) does not hold for any y E Z.] =
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(ii) No solution. [The left-hand side is congruent to 3 or 5 modulo 8, since from the equation it follows that y cannot be even.] (iii) No solution. [y2 = 2 (mod 3) holds for no y E Z.) (iv) No solution. [For x = 1 (mod 3) you have 3x2 + 3x + 7 = 4 (mod 9), and for x ¢ 1 (mod 3), 3x2 + 3x + 7 = 7 (mod 9) holds. On the other hand, y3 can be congruent only to 0, -1, or 1 modulo 9.] (v) No solution. (By Fermat's Theorem 3.1 1 you have (x + 1) 3 + (x +2) 3 + (x+3) 3 + (x+ 4) 3 - (x +5) 3 = x + 1 +x +2 +x+3+x +4-x - 5 = 3x +5 = 2 (mod 3).] (vi) No solution. [You have 19z = 1 (mod 3), while 2x + 7Y = ( -1)x + 1 (mod 3).] (vii) No solution. [When x is even, then 2x +5Y = 1 + ( -1)Y ¢. 1 (mod 3), while 19z = 1 (mod 3). For odd x = 2n + 1 , n E No, you have 2x + 5Y = 2 · 4n = 2 · ( -1) n (mod 5), while 19z ( -1)z (mod 5).] (viii) x = y = 1 , z E N arbitrary, or x = y = 3, z = 1. [The case z = 1 was dealt with in 6.2.(iii). For z > 1, first deal with x < 7: 1!+21+ · · · + 7! = 5913 is divisible by 73, but not by 732 , and for x = 2, 3, . . . , 6, 1! + · · + x! is divisible by 3 but not by 27. Since for n > 9, n! = 0 (mod 27) holds, for x > 8 you have 1! + 21 + · · · + x! = 1! + 2! + · · + 8! = 46233 = 9 (mod 27), and thus 1! + 2! + · · + xl is divisible by 9 but not by 27.] (ix) For arbitrary x E Z you have x10 + rx7 + s = x + x + 1 = 1 (mod 2), and thus x 10 + rx7 + s 1= 0. (x) Suppose that x, y, z is a solution, and set d = (x, y) , x 1 = �' Y1 = � Use 2.5. to conclude from d2 I pz2 that d I z holds and set Zl = a · Then x� + y� = pz� = 0 (mod p), and from 3.14.(vi) it follows that x 1 = y1 = 0 (mod p), which is a contradiction to (x1 , Yl ) = 1. (xi) x = y = z = u = 0. [Proceed as in 6.2.(ii).] (xii) x = y = z = u = 0. [Proceed as in 6.2.(ii) and use 3.14. (vi) for p = 7.) (xiii) No solution. [Assume that x, y, z E Z is a solution of the given equation. Then y is odd, since otherwise x2 = y3 +( 4z+2) 3 - 1 = 7 (mod 8) , which is impossible. Rewriting, x2 + 1 = ( 4z + 2) 3 + y3 = (( 4z + 2) + y) . A, where A = (4z + 2) 2 - y(4z + 2) + y2 = (2z + 1 - y) 2 + 3(2z + 1) 2 • Hence A > 3 and A = 3 (mod 4). As in 6.2.(iv) show that there exists a prime p dividing A such that p = 3 (mod 4) . For such a p you have x2 + 1 = 0 (mod p) , and by 3.14.{vi) it follows that x = 1 = 0 (mod p), which is a contradiction.] (xiv) No solution. [Assume that x, y E Z is a solution of the given equa tion. Then x is even, since otherwise y3 = x2 + 5 = 6 (mod 8) , which is impossible. Hence y3 = x2 + 5 = 1 (mod 4) , and therefore y is odd, y = y · y2 = y3 = 1 (mod 4) . So for appropriate m, n E Z you have x = 2n, y = 4m+ l. Rewriting, n2 +1 = ! (x2 +4) = t (y3 -1) = ! (y- 1)(y2 +y+ 1) = m · A , where A = y2 + y + 1 . You have A = (y2 + y + 1) = 3 (mod 4) , 4A = (2y + 1) 2 + 3 > 0, and thus A > 3. As in 6 . 2 . (iv) prove the existence =
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of a prime p dividing a such that p = 3 (mod 4). Then n2 + 1 = m . A = 0 (mod p) , and by 3.14. (vi), n = 1 = 0 (mod p), which is a contradiction.} (xv) Assume that x, y, z E N satisfy the given equation. Then (4x 1)(4y-1) = 4(4xy-x -y) +l = 4z2 +1, while 4x-1 > 3, 4x-1 = 3 (mod 4). As in the preceding exercises show that there exists a prime p dividing 4x-1 such that p = 3 (mod 4) Then (2z) 2 + 1 = (4x - 1)(4y - 1) = 0 (mod p) , and by 3.14. (vi) you have 2z = 1 = 0 (mod p), a contradiction. 6.6 (i) x = y = z = 0. [Show that if x, y, z E Z is a solution, then � ' � ' � E Z is also a solution.) (ii) x = y = z = u = 0. [Show that if x, y, z, u E Z is a solution, then � ' � ' � ' � E Z is also a solution.) (iii) x = y = z = u = 0. [Proceed as in 6.5.(iii) , but modulo 8 .) (iv) x y = z = 0 . [Show that if x, y, z E Z is a solution, then � ' f , � E Z is also a solution. ) (v) x = y = z = 0, u E N arbitrary. [Show that if x, y, z, u E Z is a solution, then xy is even (otherwise, 1 + 1 + z2 would be congruent to 1 or 0 modulo 4, which is impossible) , and moreover, both x, y must be even (if exactly one of the numbers x, y were even, then 1 + z2 = 0 (mod 4) , which is impossible). Then, of course, z must also be even. Set x 1 = � , y1 = � , z1 = � ' u 1 = 2u, d = x2 + y2 + z2 and show that x� , y� , z1 , u 1 satisfy the given equation and x� + y� + z� < d if d > 0.) (vi) x = 0, y = 11 t, z = t or x = 11 t , y = 0, z = t, where t E N0 is arbitrary. [First show that from x, y E Z it follows that z > 0. Set d = x2y2 z and for an arbitrary solution x, y, z with the property d > 0 show with the help of 3.14.(vi) that x = y = 0 (mod 11). Hence z > 1 and x 1 = {1 E Z, Yl fr E Z, Z1 z - 1 E N, where =
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x2 y2 z x2 y2 z1 < x2 y2 z = d. < 4 11 By the principle 6.4 the equation has no solution satisfying d > 0 . ) (vii) x y = z = u 0. [Proceed as in 6.5.(v) and solve the auxiliary equation x2 +y2 + z2 +u2 = 22v- l xyzu, where x, y, z, u E Z, v E N. Assume that x, y, z, u, v is a solution of this equation such that d = x2 +y2 + z2 +u2 > 0. Then x 2 + y2 + z2 + u2 is even. For all numbers x, y, z, u to be odd, you would need x2 + y2 + z2 + u2 = 1 + 1 + 1 + 1 = 4 (mod 8), while 22v - 1 xyzu is congruent to 2 or 6 modulo 8 for v = 1, and is congruent to 0 modulo 8 when v > 2. For exactly two of the numbers x, y, z, u to be even, you would need x2 + y2 + z2 + u2 = 0 + 0 + 1 + 1 = 2 (mod 4), while 22v - l xyzu would be divisible by 8. Hence all the numbers x, y, z, u are even, and thus x 1 = � , y1 = � , z1 = � and u1 = � are integers. Furthermore, x 1 , y1 , z1 , u1 , v1 , where v1 = v + 1 , is a solution of the auxiliary equation, and 0 < x� + y� + z� + u� < d.] (viii) x = y = z = 0. [Show that if x, y, z E Z is a solution such that d = x2 + y2 + z2 > 0, then yz must be even (otherwise, 2x2 = 7z2 - y2 = 0 < x�y�z1
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7 - 1 = 6 (mod 8), which would imply x2 = 3 (mod 4), but this cannot occur for any x E Z). Hence one of the numbers y, z is even, and from the equation it follows that the other one is also even. But then 2x2 = 7z2 - y2 is divisible by 4, and thus x is even. Setting x 1 = � , Yl = � , Z1 � , you obtain an integer solution satisfying 0 < x� + y� + z� < d.] 6.10 (i) For instance, x = -50t3 - 75t 2 - 36t - 4, y = 5t + 2, z = 1, where t E Z is arbitrary. [Choose z = 1 and use the method of 5.5.) (ii) Rewrite as -1 = ( 2x + 1 ) 2 - 2y2 = ( ( 2x + 1) + y'?.y) ( ( 2x + 1) - y'?.y). The equation is clearly satisfied by x = 3, y = 5. As in 6.9. (ii) use the fact that the equation u2 - 2v2 = 1 has as a solution u = 3, v 2, and multiply out as ((2x + 1) + y'2y)(3 + 2v'2) = (6x + 3 + 4y) + v'2( 4x + 2 + 3y) . If you set X l = 3, Yl = 5 and Xn+ l = 3Xn + 2yn + 1 , Yn+ l = 4xn + 2 + 3yn for any n E N, then Xn , Yn for all n E N are solutions of the given equation, and 0 < X l < X2 < . . . ' 0 < Yl < Y2 < . . . . (iii ) Rewrite as 1 = (2y) 2 -3(2x+1) 2 (2y+ v'3(2x+1))(2y-v'3(2x+1)). The equation is satisfied by x = 7, y 13. Use the fact that the equation u2 - 3v2 = 1 has the solution u = 2, v = 1. Multiplying (2y + v'3(2x + 1)) by 2 + J3 will not lead to a convenient form (the term not belonging to J3 is odd), hence you have to multiply by (2 + v'3) 2 7 + 4v'3. You get (2y + J3(2x + 1))(7 + 4v'3) = (14y + 24x + 12) + v'3( 14x + 7 + By). If you set X l = 7, Yl = 13 and Xn+ l = 7xn + 4yn + 3, Yn+l 12Xn + 7yn + 6 for any n E N, then Xn , Yn are solutions of the given equation for all n E N, and 0 < x1 < x2 < · · · , 0 < Yl < · · (iv) For fixed y, z the equation x 2 + y2 + z2 = 3xyz is quadratic in x, and its roots x� , x2 satisfy, by Vieta's relations, x 1 + x2 3yz. Hence, if x, y, z are a solution of the given equation, then so are y, z, 3yz - x. The equation is certainly satisfied by x 1 , y 1 , z 1. Set u 1 = u2 = ua = 1 and Un+3 = 3un+ 1 Un+2 -Un for n E N. Prove by induction that u1 = u2 = ua < u4 < us < · · · (for an arbitrary n E N assume that 1 < Un < Un+ l < Un+2 and show Un+2 < Un+3 ) and that for all n E N, Un , Un+ l , Un+ 2 is a solution of the given equation (assume, for an arbitrary n, that Un , Un+ b Un+2 is a solution and show that Un+ b Un+2 , Un+3 is also a solution). (v) For instance, x = n - 1 + 2t, y = 2t( n - 1) + 2t 2 - 1 + �(n - 1)(n - 2) , z = 2t(n - 1) + 2t 2 + !(n - 1)(n - 2). [Choose z = y+ 1 and use the method of 5.5.) (vi) Rewrite the equation as (2x + n) 2 + (2y + n) 2 + (2z + n) 2 + n2 = 4n6 and show that an arbitrary number triple x, y, z solving this equation is such that none of these numbers exceeds In 3 1 - � , and none is less than =
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(i) Proceed as in 6 . 1l.(ii). [If xo , x� , . . . , X8 is a solution for s E N, set Yo = 5xo, Yi = 3xi for i = 1 , 2, . . . , s, Ys + l = 4xo and show that Yo, y� , . . . , Ys+ l is a solution for s + 1 .) (ii) Proceed as in 6. 11.(i) . [Use the fact that for s = 1 the equation can be written as a quadratic, and has therefore no more than two solutions. 6.12
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Further, use the fact that natural numbers a, b with a > b satisfy a + !a > a > b + 1 > b + t, which implies lta2 < l;b2 . From u < ( s + 1)xl /(1 + x�) it again follows that x1 < x 1 + ( 1 f x 1 ) < ( s + 1)/u.] (iii) Proceed as in 6.11.(ii). [Use the fact that (1/12) 3 + (1/15) 3 + (1/20) 3 = (1/10) 3 and (1/(5 7 · 13)) 3 + (1/(5 . 12 . 13)) 3 + (1/(7 . 12 . 13))3 + (1/(5 7 · 12 13)) 3 = (1/(5 · 7 12)) 3 ; these identities can be ob tained from the identities 33 + 43 + 53 63 , 1 3 + 53 + 73 + 123 = 133 . If xo, x1, . . . , Xs E N are such that ( 1/xo ) 3 = (1/x1) 3 + + (1/x8 ) 3 , set Yo = 10xo , Yi = 12xi for i = 1, 2, . . . , s , Ys+l = 15xo , Ys+ 2 = 20xo and show that (1/yo ) 3 = (1/y1) 3 + + (1/Ys+ 2 ) 3 .] 7.3 (i) By 7 . 2.(iv) the number of such m is exactly [(107 - 1)/786) [106 /786) = 1 1450. (ii) By 7.2.(iv) the number of such m is exactly 999 - [999/5] - [999/7) + [999/35] 686 (the numbers divisible by 35 were subtracted once as divis ible by five, and a second time as divisible by 7; hence they must be added in again). (iii) In a similar fashion as in (ii) you see that there are exactly 99 [99 /2) - [99 /3) + [99 /6) = 33 such numbers. (iv) By 7.2.(iv) there are exactly [1000/5] - [1000/25] = 160 such numbers. (v) Use 7.2.(ii) and show that ·
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0 and 81 = 7, 82 = 38, S3 = 1 1 1 , 84 244, the equation has the unique solution x = 5 .] (ix) For all k E N it follows from (35) that k [xJ < kx < k [x) + k, which means that k [x] < [kx) < k [x] + k - 1 . Therefore, S = [x] + [2x] + [4x] + [Bx] + [16x) + [32x) < 63[x] + (1 + 3 + 7 + 15 + 31) = 63[x] +57 and S > 63[x] . Hence for some m = [x] E Z you have 63m < S < 63m + 57, and therefore S can be congruent only to 0, 1 , 2, . . . , 57 modulo 63. On the other side, 12345 = 60 (mod 63), and thus the equation cannot have solutions. Remark. Show that the sum S , upon division by 63 with different values of x, gives only 32 different remainders. To do this, use the expression .
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as a (x) = a 1 + a2 + a3 + 4 + + r, 2 4 8 16 32 where ak E {0, 1} for k = 1 , 2, . . . , 5 and 0 < r < � . 3 (x) n2 - n + 1 solutions. [If n = 1 , then x = 1 is a solution. Now assume that n > 1 . By rewriting you obtain ( [x] + (x) ) 2 - [[x] 2 + 2[x] (x ) + (x) 2 ] {x) 2 , and thus, in view of [x] 2 E Z, you get 2[x] (x) = [2[x] (x) + (x) 2 ]. Since 0 < (x) 2 < 1 , this last equation is equivalent to the condition 2[x] (x) E Z, i.e., (x) = 2�1 , where k E {0, 1, . . . , 2[x] - 1 }. For each of the values [x] = 1 , 2, . . . , n - 1 the term (x) must take on exactly 2 [x] different values, and for [x] = n you have (x) = 0 (since x < n) , and thus the number of all solutions is 2 + 4 + . . + 2(n - 1) + 1 = n(n - 1) + 1 = n2 - n + 1 .) (xi) Answer: Exactly the squares of natural numbers cannot be expressed in this form. Denote f(n) = [n + v'n+ !J . The difference f(n + 1) - f(n) = 1 + [y'n + 1 + �] - [y'n + �] > 1 is greater than 1 if and only if there exists a natural number m satisfying y'n + � < m < y'n + 1 + � , which is equivalent to n < m2 - m + � � n + 1 , i.e. , n = m2 - m . On the other hand it follows from y'1i + 1 > v'n + 1 that f(n + 1) - f(n) < 2. Hence 2 if n = m2 - m for some m E N, m > 1 , f(n) = f(n + 1) 1 otherwise. =
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89. [Find a number lOa + b satisfying the equation lOa + b = a + b2 , which means 9a = b( b - 1). ·Use the fact that the ·numbers b and b - 1 cannot be both divisible by 3.] (ii) 8281. [Find a number lOOOa + lOO(b + 1) + lOa + b = x2 • Rewrite it as 101 (10a + b) = (x + lO) (x - 10) and use the fact that 101 is a prime.) (iii) 552 = 3025. (Find -m x E N such that (10x+5) 2 = 3005 + 100a+ 10b. Then b = 2, x2 + x = 30 + a, which means that 30 < x2 + x < 40, and thus X = 5.) ( iv) 192 = 361 . [Show that n is odd and is not divisible by 5 and check all such n, 1 1 < n < 31, among which only 19 satisfies the condition of the 8.2 (i)
problem.] (v) 432. [Find a number x = lOOa + lOb + c for which 2x = 3a!b!c!. Since 100 < x < 999, you have 67 < a!b!c! < 666. This implies that none of the numbers a, b, c exceeds 5, since 6! = 720 > 666. Furthermore, if all the numbers a, b, c are less than 4, then two of them must be 3, and the remaining one must be 2 . Then a!b!c! is always divisible by 8, which means that x is divisible by 4. Now, x is clearly divisible by 3 . So altogether
1 < a < 5, 0 < b < 5, 0 < c < 5, 3 I a + b + c, 4 I 2b + c, 67 < a!b!c! < 666. If c = 0, then b = 4, a = 5; if c = 2, then b = 3, a = 4 or b = 5, a = 2; if c = 4, then b = 0, a = 5 or b = 2, a = 3 or b = 4, a = 1 . Of the numbers 540, 432, 252, 504, 324, 144, only 432 satisfies the given problem.] (vi) x = 112, y = 896. [Set x = 100a+ 10b + c. Since y = 800a+80b+8c < 1000 and x > 100, you have a = 1. Then y - x = 700 + 70b + 7c = n2 for an appropriate n E N, which means that 700 < n2 < 1000, or 26 < n < 32. Since 7 I n, you have n = 28, which means that x = n2 /7 = 1 12, y = 896. In this case 112896 = 336 2 .] (vii) The number n is divisible by 5 and is odd, so n = lOx + 5, where x E N. Then n2 = (lOx + 5) 2 = lOOx(x + 1) + 25 . Since x(x + 1) is an even number, its final digit is even, and this is the hundreds digit of n2 • (viii) n = 1 or n = 9. [Assume that n has k digits, i.e. , 1 ok - 1 < n < lOk . Then n2 < 102k , and thus the digit sum of n2 is at most 9 2k, so I ok- 1 < n < 18k. For k > 3 it is easy to show by induction that I ok- 1 > 18k, hence k < 2. Therefore, n2 has at most four digits, i.e. , n2 = 1000a+100b+ 10c+d, where at least one of the digits a, b, c, d is nonzero, and n = a+b+c+d. This implies n (n - 1) = 9(1 11a+ llb+ c). Since n, n - 1 cannot both be divisible by 3, either n or n - 1 is divisible by 9. Since 1 < n = a + b + c + d < 36, n can only be among the numbers 1 , 9, 10, 18, 19, 27, 28, 36, among which only 1 and 9 work out.] (ix) 153846. [Write the desired number x in 1the form x = IOn + 6 and assume that it has k digits. Then 4x = 6 · I ok - + n, which implies 13n = 2(1 ok - 1 - 4). The smallest k such that I ok - 1 = 4 (mod 13) is k = 6, which means that x = 153846.) (x) n can be an arbitrary number divisible by 10, or any of the numbers 1 1, 22, 33, 44, 55, 66, 77 88, 99, 12, 24, 36, 48, 13, 26, 39, 14, 28, 15, 16, 17, 18, 19. [It is easy to see that upon removing the final digit, the number n ·
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will be decreased by at least a factor 10, and that the factor 10 is achieved exactly for all multiples of 10. Write n in the form 10m + a, where m E N, 0 < a < 9, and assume that for some x > 1 you have 10m + a = m( lO + x), which means that a = mx, so x < 9 . For x 1 you have 1 < m = a < 9; for x = 2 it is 2 < 2m = a < 8; for x = 3, 3 < 3m = a < 9; for x = 4, 4 < 4m = a < 8; for 5 < x < 9, m 1 , a = x.] (xi) Show by induction that for each m E N there exists an n E N in the decimal representation of which there occur only ones and zeros and such that s(n) = m and s(n2 ) = m 2 • For m = 1 one can choose n = 1 . If n E N satisfies s(n) = m and s(n2 ) = m2 for some m E N, and if r E N is determined by the condition 10r > n > l or - 1 , then set n' l Orn + 1 . It is easy to see that the decimal representation of n' again consists only of zeros and ones, and that s(n') = s(n) + 1 m + 1 . Since 102r > 2n · 10r , you also have s((n') 2 ) = s(n2 ) + s(2n) + 1 = (m + 1) 2 • (xii) It is clear that for any m, n E N you have s(m + n) < s(m) + s(n). If ak , . . . , a� , a0 are the digits of n, i.e., if n = I:� 0 lOi ai , then s(mn) s( L:� 0 lOi mai) < E� 0 s(mai) < I:� 0 ais(m) = s(n)s(m) . Hence s(n) = s(lOOOn) < s(125)s(8n) = 8s(8n). The example n = 125 shows that s(n) < 8s(8n) does not generally hold. (xiii) The case m < 2 is clear, so assume m > 2. Set Bn = n + s(n). If n does not end in a 9, then Bn+ l = Bn + 2. If, on the other hand, n does end in a 9, then Bn+1 < Sn. Let k be the largest natural number satisfying Sk < m . Then Sk+1 > Sk ; hence m < Sk + 1 = Sk + 2 < m + 2. =
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8.4 (i) Use the fact that 2 4 = 1 (mod 5) , and thus 2 2n = 1 (mod 5) for every integer n > 2. (ii) 07. [Use the congruences 74 = 1 (mod 100) and 99 = 1 (mod 4) .]
(iii) 6667. [Since �(24 ) = 8 and �(54 ) = 500 divide 1 000, you have 3 1000 = 1 (mod 104 ) , and thus for x = 3999 you have 3x = 1 = 1 - 104 (mod 104 ) , which implies x = -3333 (mod 104 ) .] 14 (iv) 36. [Note that 14 14 = 7 14 14 · 2 14 14 , where 7 14 14 = 1 (mod 100) , because 74 = 1 (mod 100) and 4 I 14 14 . Since 14 14 = 1 (mod 5) and 1414 = 0 (mod 4) , you have 14 14 = 16 (mod 20) , and since �(25) 20, it follows from Euler 's Theorem 3.13 (or from 3. 14. (v)) that 2 14 1 4 = 2 16 = 164 = 94 = 81 2 = 6 2 = 36 (mod 25). Since also 2 1414 = 0 = 36 (mod 4) , you get 2 14 14 = 36 ( mod 100) , and altogether 14 141 4 = 36 (mod 100) .] (v) Since n = k (mod 4) and 74 = 1 (mod 100) , it follows from 3.14. (v) that � = 7k (mod 100) . Since � \53 ) = 100 and �(23 ) = 4 divide 100, it follows from 3.14. (v) that 7-rn = 77 (mod 103 ). In complete analogy: Since k �(5 4 ) and �(2 4 ) divide 103 , you get 777n = 777 (mod 104 ) , and similarly 77n 771c 77 n 77 k 7 7 7 7 7 5 7 (mod 10 ) and 7 7 = 7 =7 (mod 106 ) .] (vi) For n = 100 both numbers end in 001 . [Find the smallest n E n N for which 19 = 313n (mod 1000) . Since �(5 3 ) = 100 and �(2 3 ) = 4 =
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divide the number 100, by Euler ' s theorem, 191 00 = 1 313 100 (mod 125) , 1 9 100 = 1 = 313 1 00 (mod 8) , so that 19 100 = 1 = 313 100 (mod 1000). Show that therefore the desired n has to divide 100 (divide the number 100 by n with remainder : 100 = nq + r , where 0 < r < n, then (19n ) q · 19r = (313n )q · 313r (mod 1000) , which implies 19r = 313r (mod 1000) , and thus r = 0). If n 1= 100, then n I 50 or n I 20 must hold. Now neither 1950 = 31350 (mod 1000) nor 1920 = 31320 (mod 1000) will hold (you have 192 = 1 (mod 5), 3132 = -1 (mod 5) , which with 3.4.(vi) implies 1950 = 1 (mod 125) , 31350 = -1 (mod 125) ; similarly, from 194 21 (mod 25) , 3134 = 11 (mod 25) it follows that 1920 = 21 5 101 (mod 125) , 31320 = 11 5 = 51 (mod 125)).] 8.6 (i) Note that �(102n - 1) - �(10n - 1) = �(102n - 2 · 10n + 1) = (�(10n - 1)) 2 , where (Ion - 1)/3 E N. Furthermore, (Ion - 1)/3 = 33 . . . 3. � n (ii) Note that m 1 m 2 = � ( 102n -2· 10n +l) = �(102n - 1) - �(10n - 1) , and so m 1 m2 is the difference between the number consisting of 2n twos and the number consisting of n fours. Therefore, m 1 m2 = 222 . . . 21777 . . . 78, with n - � twos and as many sevens. (iii) Divide n by k with remainder: n = qk + r , where 0 < r < k. Then 10n - 1 = lOqk+r - 1 = 10r(10qk - 1) + 10r - 1 = 10r(lO (q- 1) k + 10 (q -2) k + · · ·+ 10k+1)(10k -1)+(10r - 1). Since 0 < 10r - 1 < lOk - 1 , the remainder in the division of 1 on - 1 by lOk - 1 is 1 or - 1. Hence lOk - 1 I Ion -1 if and only if k I n. Now of course, lOk - 1 1 10n - 1 if and only if � (lOk - 1) I � (10n - 1) , i.e., � 111 . . . 1 I � 1 1 1 . . . 1. k n (iv) Assume that the desired number is written as m ones. The condition of the problem means that m > k, and furthermore (iii) implies k I m. Since for m = 2k you have �(102k - 1) = � (lOk - 1) � (lOk + 1) and the number lOk + 1 is not divisible by 3, then (lOk + 1)/3 is not an integer, and therefore the number consisting of 2k ones is not divisible by the number consisting of k threes. Hence m > 3k. However, for m = 3k you already have �(103k - 1) = � (lOk - 1) · �(102k + lOk + 1), where � (102k + lOk + 1) is an integer. The number consisting of 3k ones is therefore the desired smallest number. (v) a = 5 (for n = 10) or a = 6 (for n = 1 1 or n = 36). [Clearly, a 1= 0. The given sum is �n(n+l). For a = 1 it follows from ! ( lOk - 1) = �n(n+ l) that (3n + 1) (3n + 2) = 2 lOk = 2 k+ 1 · 5k , which is a contradiction, since (3n + 1 , 3n + 2) = 1 and 2 k+ 1 < 22k = 4k < 5k. Furthermore, 8( !n( n + 1)) + 1 = (2n + 1 ) 2 , and so the digit a is not among the digits 2, 4, 7, 9, since otherwise (2n + 1) 2 would be congruent to 3 or 7 modulo 10, which is impossible. If the digit a were equal to 3 or 8, then (2n + 1) 2 would be congruent to 65 or 5 modulo 100, which is impossible. Hence a = 5 or a=6] (vi) Prove by contradiction that no such n exists: Assume to the con trary that such numbers exist, and denote by no the smallest one among =
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them. Let r E N be determined by the condition 1 0r > no > Ior- l . Since M, 2M, . . . , 9M do not have a digit sum less than m , it follows from M I n0 that no > 10M and thus r > m. But then the number n1 = no - lor- 1 + lor-m- l = no - 9M lor-m- 1 is again divisible by M and its digit sum does not exceed the digit sum of no . Since n1 < no , you have obtained a contradiction to the definition of no . 8.8 Proceed as in 8.7, but choose a k such that 3 {'ffl2 < lOk- 2 and n = (lOk rmJ + 1. Show that then 103km < n3 < 103k m + (103k - 1). 9.3 (i) Set ai = � (lOi - 1) for i = 1, 2, . . . , n + 1 and proceed as in 9.2.(iii) . The difference ai -a; for i > j has the decimal representation of the desired form. (ii) The natural number m divides an for an appropriate n E N if and only if (a, m ) = 1. [If m I an , then (a, m ) must divide (a, an ) , but of course (a, an ) = (a, 1 + a · an- 1 ) = 1. If, on the other hand, m is an arbitrary natural number relatively prime to a, use the same argument as in 9.2.(iii) to show that m divides ai - a; for appropriate 1 < j < i < m + 1. Then ai - ai = ai · ai-i ' and since ( m, a) = 1 , then m divides ai-i· l (iii) Prove this by induction on n, using the arguments of 9.2. (iv). (iv) Use induction on n, as in (iii) . To do this, use the fact that from among any five natural numbers divisible by 3t one can choose a triple of numbers such that their sum is divisible by 3t+ 1 . [Distribute these five numbers over three sets according to the remainder upon division by 3t+ 1 . If it is possible, choose one number from each set. If one of the sets is empty, then by Dirichlet ' s principle one of the remaining sets must contain at least three numbers, which you then choose.] 9.5 (i) Proceed as in 9.4.(ii), but consider only nonempty subsets with even numbers of elements. Use the fact that the symmetric difference of sets with even numbers of elements has an even number of elements. (ii) Proceed as in 9.4. (iii) , but consider only nonempty subsets with one or two elements, the number of which is ( 415) + (4;) = 45 + 45;_44 = 1035. (iii) If the given numbers are ab a2 , . . . , a29, set bi = ai a� for i = 1 , 2, . . . , 29 and continue to work with the numbers b� , b2 , . . . , b29• (Show that the product akaman is the third power of a natural number if and only if bk bmbn is a third power.) Introduce the same triple (c: 1 , c:2 , c:3 ) as in 9.4.(iv) , but divide the numbers bb . . . , b29 into 14 groups: Keep the group that in 9.4. (iv) corresponds to the triple (0, 0, 0) , while of the remaining 26 you combine in one group always those groups corresponding to (c: 1 7 c:2 , c:3 ) and (c:]. , c:� , �), wpere € 1 + c:]. , €2 + c:� and €3 + c:3 are divisible by 3; thus you obtain 13 groups. If in the group corresponding to (0, 0, 0) there are at least three numbers, choose three numbers from this group. If, on the other hand, there are no more than two numbers in this group, then the remaining 27 are distributed over 13 groups, and by Dirichlet 's principle at least one of them contains at least three numbers. If you divide this group into the two original groups from 9.4.(iv) , then either one of them contains ·
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Hints and Answers to Chapter 3
335
at least three numbers, in which case choose these three numbers; or both contain at least one number, in which case choose one number from each group, and take b1 as the third number. (iv) Proceed as in (iii) and divide the numbers into the same 14 groups. H the group corresponding to (0, 0, 0) contains some number, then choose it and set k = 1; otherwise, there are again 27 numbers distributed over 13 groups; hence at least one of them contains at least three numbers. Divide them again into two groups and either choose one from each (k = 2) , or choose three numbers from one group (k = 3). (v) Proceed as in 9.4.(v) . Consider all n-tuples (c:� , c:2 , . . . , en ) , where 0 < < s for i = 1 , 2, . . . , n, of which there are sn . The rest is almost the same. 9.7 Note that if x is an arbitrary integer, then x2 is congruent to one of the numbers 0, 1, 22 , 32 , . . . , ( P2 1 ) 2 , of which there are P!1 < p. 10.3 (i) Since 19 = 62 (mod 43), a polynomial F(x) with integer coefficients must satisfy F(19) = F(62) (mod 43) . (ii) Proceed as in 10.2.(iii). (iii) Assume that for some n E Z you have F(n) = 0 and show that for each i = 1, . . . , 5 you have ai = n ± 1 or ai = n ± 73, and thus the numbers ab a2 , . . . , as are not all different. (iv) Divide the five numbers a 1 , . . . , a5 into two groups according to which value of the polynomial F( x) they take on. H none of these groups is empty, then by Dirichlet ' s principle one of them must contain at least three numbers, and the other one at least one, which by (ii) is a contradiction. (v) Find r, s such that F(r) = 0 and F(s) = 1. Then by (ii) you have s = r + 1 or s = r - 1. In the case s = r + 1 prove by induction that F(r + m) = m holds for each m E No . (This is true for m = 0 and m = 1. For integers m > 2 assume that F(r + m - 2) = m - 2, F(r + m - 1) = m - 1. Given an m, there exists a k E Z such that F(k) = m, and by (ii) you have k = r + m - 2 or k = r + m. The first case does not occur, since F(r + m - 2) = m - 2 =f= m = F(k). Hence k = r + m, and so F(r + m) = F(k) = m.) By 3. 15 in Chapter 1 you have F(x) = x - r. Similarly, in the case s = r - 1 show that F(x) = -x + r. The solution of the problem is therefore given exactly by all linear polynomials a 1 x + ao , where a 1 = 1 or a 1 = - 1 , and ao E .Z is arbitrary. 10.6 (i) Exactly when 6a , 2b, a + b+c, d E Z. [Rewrite ax3 + bx2 +cx+d = 6a (;) + 2(b + 3a) (�) + (a + b + c) (�) + d.] (ii) Yes, since F(O), F(1) , . . . , F(7) are integers. (iii) For n > 2 it exists, fot n = 1 it does not exist. [Proceed as in 10.5.(iii) , obtain the conditions ao, . . . , an- 1 E Z, an fj. Z, (n + 1)an E Z, ! (n + 2)(n + 1)an E .Z. For n > 2 choose an = n!l ; for n = 1 it follows from the conditions 2a 1 E Z, 3a l E Z that also a 1 E Z; for n = 1 it is therefore not possible to satisfy all conditions.
C:i
336
4. Hints and Answers
( i ) Such a polynomial exists for n = 3, n = 4, and n > 6, but does not exist for n < 2 and n = 5. [Proceed as in 10.5.(iii) , and see whether the conditions an ft. Z, (n + 1)an E Z, � (n + 2)(n + 1)an E Z, �( n + 3) (n + 2)(n + 1)an E Z are satisfied. For n = 3, n = 4, or n > 6, an = n! l satisfies them. For n = 1 it follows from 2a 1 E Z and 3a1 E Z that a 1 E Z . For n = 2 it follows from 3a2 E Z and 10a2 E Z that a2 = 10a2 - 3 3a2 E Z. For n = 5 it follows from 6a5 E Z, 21as E Z and 56as E Z that also a5 = -57 · 6a5 + 19 21a5 - 56a5 E Z (this last construction can be carried out by use of Bezout 's equality; it follows from (6, 2 1 , 56) = 1).] ( ) Let F(x) = bk x k + · · · + b1 x + b0 • By Theorems 2 and 3 in 10.4 there exist integers ao, . . . , ak such that F(x) , = ao + a 1 + · · · + ak k 1 P v
·
o
v
X( )
X( )
which implies that the polynomial k!F(x)/p has integer coefficients, so plbi · k! for all i = 0, 1, . . . , ko Since (p, k!) = 1 , this means that plbi. 10.10 ( i ) ( i ) Use 1008 in (i) for p = 5, in (ii) for p = 3 (it does not work for p = 2), in (iii ) for p = 7 (p = 3 not possible). In ( i ) , first divide by 5, then use 1008 for p = 5, and finally show that the original polynomial cannot be decomposed either ( ) x125 - 75. [Proceed as in 10.9.(ii) , use 10.8 for p = 3.] -
v
v
0
v
Bibliograp hy
A. Books cited in text
[1] E. Beckenbach and R. Belhnan, Inequalities, Springer-Verlag, Berlin Gottingen-Heidelberg, 1961. [2] A. I. Borevich and I. R. Shafarevich, Number Theory, Academic Press , New York-London, 1966. [3] P. S. Bullen, D. S. Mitrinovic, and P. M. Vasic, Means and Their Inequalities, Reidel, Dordrecht, 1987. [4] L. Childs, A Concrete Introduction to Higher Algebra, second edition, Springer-Verlag, New York, 1995. [5] P. Franklin, A Treatise on Advanced Calculus, Dover Publications, New York, 1964. [6] H. W. Gould, Combinatorial Identities; a Standardized Set of Tables Listing 500 Binomial Coefficient Summations, revised edition, Morgantown, W.Va., 1972. [7] S. L. Greitzer, International Mathematical Olympiads 1959-1977, Mathe matical Association of America, Washington, D.C., 1978. [8} S. Mac Lane and G. Birkhoff, Algebra, third edition, Chelsea Publishing Co., New York, 1988. [9] J. lliordan, Combinatorial Identities, Wiley, New York, 1968. [10] G. E. Shilov, Linear Algebra, Dover Publications, New York, 1977. [11] W. Sierpi:Dski, Elementary Theory of Numbers, Pa.D.stwowe Wydawnictwo Naukowe, Wa.rszawa, 1964. [12] I. M. Vinogradov, Elements of Number Theory, Dover Publications, New York, 1954. [1 3] B. L. van der Wa.erden, Algebra, Vol. I, Springer-Verlag, New York, 1991 .
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[14} E. J. Barbeau, M. S. Klamkin, and W.
0. J. Moser,
Five Hundred Mathe
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1995. [15} E. Beckenbach and R. Bellman, An Introduction to Inequalities, Mathemat ical Association of America, Washington, DC, 1961. [16] G . Birkhoff and S . Mac Lane, A Survey of Modern Algebra , third edition, Collier-Macmillan, London, 1965. [17} M. Doob {ed.} , The Canadian Mathematical Olympiad, 1969-1999 , Cana dian Mathematical Society, Ottawa, 1993. [18] H. Dorrie, 100 Great Problems of Elementary Mathematics , Dover Publica tions, 1965. Reprinted 1989. [19) M. J. Erickson and J. Flowers, Principles of Mathematical Problem Solving, Prentice-Hall, 1999. [20] A. Gardiner, The Mathematical Olympiad Handbook. An Introduction to Problem Solving Based on the First 32 British Mathematical Olympiads 1965-1996, Oxford University Press , 1997. [21] G. T. Gilbert, M. I. Krusemeyer, and L. C . Larson, The Wohascum County Problem Book, Mathematical Association of America, Washington, DC,
1993. [22] G . H. Hardy, J. E. Littlewood, and G. P6lya, Inequalities , second etition, Cambridge University Press, Cambridge, 1952. [23] K. Hardy and K. S. Williams, The Red Book of Mathematical Problems, Dover Publications, Mineola, NY 1996. [24] K. Hardy and K. S . Williams , The Green Book of Mathematical Problems, Dover Publications, Mineola, NY 1997. [25] R. Honsberger, Prom Erdos to Kiev. Problems of Olympiad Caliber, Mathematical Association of America, Washington, DC, 1996. [26] M. S. Klamkin, USA Mathematical Olympiads 1972-1986, Mathematical Association of America, Washington, DC, 1988. [27] P. P. Korovkin, Inequalities , Little Mathematics Library, Mir, Moscow; distributed by Imported Publications, Chicago, ill . , 1986. [28] J. Kiirscha.k, Hungarian Problem Book I, Based on the Eotvos Competitions, 1894-1905, Mathematical Association of America, Washington, DC, 1963. [29] J. Kiirscha.k, Hungarian Problem Book II, Based on the Eotvos Competitions, 1906-1928, Mathematical Association of America, Washington, DC, 1963. [30] L. C. Larson, Problem-Solving Through Problems, Springer-Verlag, New York, 1983. [31] D. J. Newman, A Problem Seminar, Springer-Verlag, New York, 1982. [32] G. P6lya, Mathematical Discovery. On Understanding, Learning, and Teach ing Problem Solving, reprint in one volume, John Wiley & Sons, Inc . , New York, 1981. [33] G . P6lya, How to Solve It. A New Aspect of Mathematical Method , Princeton University Press , Princeton, NJ, 1988. [34] J. Roberts, Elementary Number Theory. A Problem Oriented Approach, MIT Press , Cambridge, Mass.-London, 1977. ,
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[36] [37]
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[41] I. L. Babinskaya, Zadachi matematicheskikh olimpiad, Nauka, Moskva, 1975. [42] V. I. Bernik, I. K. Zhuk, and 0. V. Melnikov, Sbornik olimpiadnykh zadach po matematike, Narodnaya osveta, Minsk, 1980. [43] J. Fecenko, 0 nerovnostiach, ktore suvisia s aritmetickym a geometrick'!/m priemerom, Rozhledy mat.-fyz. 66 (1987/88}, no. 2, 46-49. [44] M. Fiedler and J. Zemanek, Vybrane Ulohy matematicke olympiady {kate gorie A + MMO) , SPN, Praha, 1976. [45] G . A. Galperin and A. K. Tolpygo, Moskovskiye matematicheskiye olimpiady, Prosveshcheniye, Moskva, 1986. [46] V. M. Govorov, Sbornik konkursnykh zadach po matematike, Nauka, Moskva, 1986. [47] J. Herman and J. Sim.Sa, 0 jednom uiit{ Dirichletova principu v teorii clsel, Rozhledy mat.-fyz. 66 (1987/88}, no. 9, 353-356. [48] K. Horak et al. , Ulohy mezinarodn{ch matematickych olympiad , SPN, Praha, 1986. [49] S. V. Kopyagin et al. , Zarubezhnyye matematicheskiye olimpiady, Nauka, Moskva, 1987. , [50] I. Korec, Ulohy o velk?}ch clslach, Mlada fronta, edice SMM, Praha, 1988. [51] V. A. Krechmar, Zadachnik po algebre, Nauka, Moskva, 1972. [52] G . A. Kudrevatov, Sbornik zadach po teorii chisel, Prosveshcheniye, Moskva, 1970. [53] V. B. Lidskiy et al. , Zadachi po elementarnoy matematike, Gos. izd. fiz.-mat. lit. , Moskva, 1960. [54] P. S. Modenov, Sbornik zadach po spetsialnom kurse elementarnoy matem atiky, Vysshaya shkola, Moskva, 1960. N-tY rocnz'k matematicke olympiady, SPN, Praha, (from 1953 to 1993}. [55] [56] 0 . Odvarko et al., Metody fesen{ matematickych uloh I, ucebni text MFF UK, Praha, 1977. [57] I. Kh. Sivashinskiy, Neravenstva v zadachakh, Nauka, Moskva, 1967. [58] I. Kh. Sivashinskiy, Teoremy i zadachi po algebre i elementarnym funktsiyam, Nauka, Moskva, 1971 . [59] S. Straszewicz, Zadania z olimpiad matematicznych, I, II, Warszawa, 1956, 1961. v
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Index
Abel, Nie1s Henrik 25 AM-GM inequality 109,
Cauchy's inequality
144, 151, 155, 160, 162, 163, 167, 232 for weighted means 162 algebra 23 algorithm division 28 Euclidean 1 78, 181 alternating sums 15 arithmetic mean 33, 109, arithmetic progression 5
150
base of the natural logarithm 113 base representati� 258 Bernoulli nwnbers 13 Bernoulli 's formula 12 Bernoulli ' s inequality 109, 141, 166,
168, 233 general 165 binomial coefficients 2, generalized 276 binomial theorem 2, 8, Bezout's equality 179 Bezout's theorem 28
75 175, 263
127, 131 , 133,
149, 166 Chebyshev's inequality
145, 148,
149, 150, 159 Chebyshev's theorem 187 Chlnese remainder theorem combinatorial identities 6 combinatorial numbers 1 common divisor 178, 180 common multiple 178, 180 completing the square 1 13 complex number 75 composite number 183 congruences 189 in one variable 201 linear 202 of higher degree 211 quadratic 216 conjugate 65 conjecture Goldbach 173 Shimura-Taniyama 173 twin prime 173 coprime 180 Cramer's rule 46 cubic polynomial 24
211
342
Index
cubic ( continued) de Moivre's theorem 75, 77, 78, 84 decimal expansion 261 degree of a polynomial 24, 38 of a power mean 167 digit patterns 265 digit sum 261 , 267 digits 258 final 261 Diophantine equations 217 cardinality of solution set 245 linear 217 linear in some variable 220 Diophantus of Alexandria 217 Dirichlet's principle 268 Dirichlet's theorem 187 discriminant 124 divisibility of numbers 174 division algorithm 28 division theorem 176
Gauss, Carl Friedrich 189 Gaussian elimination 46 generating functions 9 geometric mean 151 , 167 geometric progression 6 geometric series 1 10 geometric-power mean inequalities
167 Goldbach conjecture 173 greatest common divisor 178, 180 harmonic mean 156 homogeneous inequality 105 homogeneous polynomial 38 Holder's inequality 166 implication method 61 incomplete quotient 176 induction principle 135, 136 finite 134 inequality 89 AM-GM 109, 144, 151, 155, 160,
162, 163, 167, 232 Eisenstein's irreducibility criterion
280 elementary symmetric polynomials
32, 39 in three variables 43 in two variables 39 elimination method 54 equivalent transformations 94 estimation method 101 Euclidean algorithm 178, 181 Euler's cp-function 194 Euler's criterion 216 Euler's theorem 198, 262, 264, 266 factorial ! prime decomposition of 255 false roots 61 Fermat's last theorem 173 Fermat's theorem 197 finite induction 136 finite sums 5 fractional part 251 fundamental mean property 165 fundamental theorem of algebra 29
AM-GM, for weighted means 162 Bernoulli's 109, 141, 166, 168,
233 Bernoulli's, general 165 between power means 167 Cauchy's 127, 131 , 133, 149, 166 Chebyshev's 145, 148, 149, 1 50,
159 geometric-power mean 167 homogeneous 105 Holder's 166 Jensen's 105, 169 Minkowski's 130, 166 symmetric 104 triangle 129 weak 91 Young's 163 infinite product 140 infinitude of primes in arithmetic progressions 187 of the form 3k + 2 187 of the form 4k + 3 187 integer part 251 , 267 integer-valued polynomials 276 interpolation 23
Index irrational equations 60 with a parameter 71 irrational numbers 169 irreversible transformations 96 Jensen's inequality 105, 169 least common multiple 178, 180 length of a vector 130 linear congruences 202 linear Diophantine equations 217 linear polynomial 24 lower bound 101 mathematical induction 5, 135 mean of degree zero 167 mean value 150 method of squares 1 13 substitution 66 symmetric polynomials 55 undetermined coefficients 19 Minkowski's inequality 130, 166 monomial 38 multiple 174 multiple root 28 multiple zero 28 multiplicity of a zero 28 '
nth roots 60 number theory 173 ordering of an n-tuple 146 pairwise relatively prime 180 partial fraction decomposition 21 partial summation 16 pigeonhole principle 268 polynomials 23, 84, 274 cubic 24 degree of 24, 38 elementary symmetric 32, 39, 43 homogeneous 38 in several variables 38 integer-valued 276 irreducible 280 linear 24
343
Taylor 23 with integer coefficients 274 polynomial division 27 power mean 151, 166, 167 prime number 183 Pythagorean equation 237 quadratic congruences 216 quadratic polynomial 24 quotient 28, 176 rational zeros 35 regular n-gon 80 relatively prime 180 remainder 28, 176 repunit 265 root of a polynomial equation 25 scalar product 128 Shimura-Taniyama conjecture 173 Sierpitiski, W. 187 simple zero 28 strict inequalities 91 swn of digits 261 , 267 sums of powers 1 1 , 39 symmetric inequality 104 symmetric polynomial 38 systems of equations 46 equations of hlgher degree 54 irrational equations 73 linear congruences in one variable 206
linear equations 46 linear equations with parameters 48
Taylor polynomials 23 theorem binomial 2, 8, 175, 263 Bezout's 28 Chebyshev's 187 Chinese remainder 21 1 de Moivre's 75, 77, 78, 84 Dirichlet's 187 division 176 Euler's 198, 262, 264, 266
344
Index
theorem ( continued) Fermat's 197 Fermat's last 173 Wilson's 204 transitivity of inequalities 91 triangle inequality 1 29 trichotomy law 91 trigonometric functions 78 twin prime conjecture 1 73 unique factorization 185 upper bound 101 Vieta's relations 32, 38, 41 , 56, 78,
80, 82, 83, 125, 132
Vinogradov, I. �- 216 weak inequalities 91 weighted arithmetic mean 162 weighted geometric mean 162 weighted means 151, 162 Wiles, Andrew 173 Wilson's theorem 204 Young's inequality 163 zero of a polynomial 23, 25, 275 multiple 28 simple 28 zero polynomial 24, 38
lhis book
presents methods of solv1ng problems 1n three areas of dass1cal elementary
mathematics: equations and systems of equat1ons of vanous kmds. algebraiC mequahtles
and elementary number theory, 10 particular, d1v1sibihty and d1ophant1ne equatiOns In each top1c. brief theoretical discussions are immediately followed by carefully worked-out examples of mcreasmg degrees of difficulty, and by exercises that range from routine to rather challengmg problems While th1s book emphames some methods that are not usually covered 1n begmnmg um versity courses. 1t nevertheless teaches techniques and sk1lls that are useful not only 1n the speofic top1cs covered here. There are approximately 330 examples and 760 exercises.
l"
1@
Springer
ISBN
0 387 98942 0
www.spnnger-ny.com WWW.l.lllS. Ill&lh.ca
9
.I �
780J87 989426 )