[3] khinchin a.y.-three pearls of number theory

8/13/2019 [3] Khinchin a.Y.-three Pearls of Number Theory

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THREE PEARLS OFNUMBERTHEORY

byA Y Khinchin

Translated byF Bagemihl H Komm

and W Seidel

DOVER PUBLICATIONS INC.Mineola New York


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CopyrightCopyright 1952 by Graylock PressAll rights reserved under Pan American and International

Copyright Conventions.

Bibliographical NoteThis Dover edition, first published in 1998, is an unabridged

and unaltered republication of the edition published by TheGraylock Press, Baltimore, Maryland in 1952. t was translatedfrom the second, revised Russian edition published in 1948.

ibrary of Congress Cataloging in Publication Data...

Khinchin, Aleksandr IAkovlevich, 1894-1959.[Tri zhemchuzhiny teorii chisel. English]Three pearls of number theory by A.Y. Khinchin ; trans

lated by F Bagemihl, H. Komm, and W Seidel.p. em.An unabridged and unaltered republication of the edition

first published by the Graylock Press, Baltimore, Maryland in1952. t was translated from the second, revised Russianedition published in 1948 -T.p. verso.

Includes bibliographical references (p. ) and index.ISBN 0-486-40026-3l. Number theory. I TitleQA24l.K513 1998

512.'7-dc21 97-48530CIP

Manufactured in the United States of AmericaDover Publications, Inc., 31 East 2nd Street, Mineola, N. Y 11501


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FOREWORD

This little book is devoted to three theorems in arithmetic,which, in spite of their apparent simplicity, have been the objectsof the efforts of many important mathematical scholars. The proofswhich are presented here make use of completely elementary means,although they are not very simple).

The book can be understood by beginning college students,nd is intended for wide circles of lovers of mathematics.


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ONTENTS

A LETTER TO THE FRONT (IN L IEU OF A PREFACE) 9CHAPTER I VAN DER WAERDEN S THEOREM ON

ARITHMETIC PROGRESSIONS

CHAPTER I I THE L A N D A U ~ S C H N I R E L M A N N HYPOTHESISAND MANN S THEOREM 8

CHAPTER I I I AN ELEMENTARY SOLUTION OF WARING SPROBLEM 7


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A L T T R TO THE F R O N T(IN LIEU OF A PREFACE)

March 24 1945Dear Seryozha

Your letter sent from the hospital gave me threefold pleasure.First of all, your request that I send you "some little mathematical pearls" showed that you are really getting well, and are notmerely trying, as a brave fighter, to relieve your friends' anxiety.That was my first pleasure.

Furthermore, you gave. me occasion to reflect on how it isthat in this war such young fighters as you happen to pursue theirfavorite occupation-the occupation which they cherished alreadybefore the war and from which the war has torn them-so passionately during every little respite. There was nothing like thisduring the last World War. In those days a young man who had arrived at the ffront almost invariably felt that his life had been disrupted, that what had been the substance of his life before hadbecome for him an unrealizable legend. Now however, there aresome who write dissertations in the intervals between battles, anddefend them on their return during a brief furlough Is it not because you feel with your whole being, that your accomplishmentsin war and in your favorite occupations-science, art, practical activity-are two links of one and the same great cause? And if so, isnot this feeling, perhaps, one of the mainsprings of your victorieswhich we, he.re at home are so enthusiastic about? This thoughtgratified m very much, and that was my second pleasure.

And so I began to think about what to send you. I do not knowyou very well-you attended my lectures for only one year. Nevertheless I retained a firm conviction of your profound and seriousattitude toward science, and therefore I did not want to send youmerely some trinkets which were showy but of little substance scientifically. On the other hand I knew that your preparation was notvery great-you spent only one year in the university classroom,and during three years of uninterrupted service at the Front youwill hardly have had time to study. After several days' delibera-


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10tion I have made a choice. You must judge for yourself whether itis a happy one or not. Personally I coq.sider the three theorems ofarithmetic which I m sending you, to be genuine pearls of ourscience.

From time to time, remarkable, curious problems tum up in arithmetic, this oldest, but forever youthful, branch of mathematics.In content they are so elementary that any schoolboy can understand them. They are usually concerned with the proof of somevery simple law governing the world of numbers, a law which turnsout to be correct in all tested special cases. The problem now is toprove that it is in fact always correct. And yet, in spite of the apparent simplicity of the problem, its solution resists, for years andsometimes centuries, the efforts of the most important scholars ofthe age. You must admit that this is extraordinarily tempting.

I have selected three such problems for you. They have allbeen solved quite recently,. and there are two remarkable commonfeatures in their history. First, all three problems have been solvedby the most elementary arithmetical methods do not, however, con-fuse elementary with simple; as you will see, the solutions of allthree problems are not very simple, and it will require not a littleeffort on your part to understand them well and assimilate them).Secondly, all three problems have been solved by very young, beginning mathematicians, youths of hardly your age, after a series ofunsuccessful attacks on the part of "venerable" scholars. Isn'tthis a spur full of promise for future scholars like you? What anencouraging call to scientific daring

The work of expounding these theorems compelled me to penetrate more deeply into the structure of their magnificent proofs, andgave me great pleasure.

That was my third pleasure.I wish you the best of success-in combat and in science.

Yours,A Khinchin


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C H P T E R IVAN DER WAERDEN S THEOREM ON

ARITHMETIC PROGRESSIONS

lIn the summer of 1928 I spent several weeks in Gottingen. Asusual, many foreign scholars had arrived there for the summer se

mester. I got to know many of them, and actually made friends withsome. At the time of my arrival, the topic of the day was the brilliant result of a young Hollander, van der Waerdent who at that timewas still a youthful beginner, but is now a well-known scholar.This result had just been obtained here in Gottingen, and in fact,only a few days before my arrival. Nearly all mathematicians whomI met told me about it with enthusiasm.

The problem had the following history. One of the mathematicians there I forget his name**) had come upon the following problem in the course of his scientific work: Imagine the set of all natural numbers to be divided in any manner whatsoever into two partse. g., into even and odd numbers, or into prime and composite num-

bers, or in any other way). Can one then assert that arithmetic progressions of arbitrary length can be found in at least one of theseparts? By the length of an arithmetic progression I mean here, andin what follows, simply the number of i ts terms.) All to whom thisquestion was put regarded the problem at first sight as quite simple;i ts solution in the affirmative appeared to be almost self-evident.The first attempts to solve it, however, led to nought. And as themathematicians of Gottingen and their foreign guests were by tradition in constant association with one another, this problem, provoking in i ts resistance, soon became the object of general mathematical interest. Everyone took it up, from the venerable scholar to theyoung student. After several weeks of strenuous exertions, theproblem finally yielded to the attack of a young man who had cometo Gottingen to study, the Hollander, van der Waerden. I made hisacquaintance, and learned the solution of the problem from him personally. It was elementary, but not simple by any means. The problem turned out to be deep, the appearance of simplicity was deceptive.

B. L. van der Waerden, Beweis einer Baudetschen Vennutung Nieuw Arch.Wiskunde 15 212-216 1927). Trans.)Most probably Baudet; cf. the preceding footnote. Trans.)


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12Quite recently, M A. Lukomskaya of Minsk) discovered andsent me a considerably simpler and more transparent proof of van

der Waerden s theorem, which, with her kind permission, I am goingto show you in what follows.

Actually, van der Waerden proved somewhat more than what wasrequired. In the first place, he assumes that the natural numbers aredivided, not into two, but into arbitrarily many, say k, classes(sets). In the second place, it turns out that it is not necessaryto decompose the entire sequence of natural numbers in order to

~ r n t e e the existence of an arithmetic progression of prescribed(arbitrarily large) length l in at least one of these classes; a cer-tain segment of it suffices for this purpose. The length, n k, l),of this segment is a function of the numbers k and l Of course i tdoesn t matter where we take this segment, so long as there aren k, l successive natural numbers.

Accordingly, van der Waerden s theorem can be formulated asfollows:Let k and l be two arbitrary natural numbers. Then there existsa natural number n k, l) such that,, i an arbitrary segment, o lengthn k, l), of the sequence of natural numbers is divided in any mannerinto k classes some of which may be empty), then an arithmeticprogression of length l appears in at least one of these classes

This theorem is true trivially for l =2. To see this, it sufficesto set n k, 2) =k l; for if k l numbers are divided into k classes,then certainly at least one of these classes contains more than onenumber, and an arbitrary pair of numbers forms an arithmetic pro-gression of length 2, which proves the theorem. We shall prove thetheorem by induction on l Consequently, we shall assume through-out the following that the theorem has already been verified forsome number l 2 and for arbitrary values of k, and shall show thatit retains its validity for the number l+ l (and naturally also for allvalues of k).

3According to our assumption, then, for every natural number k

there is a natural number n k l) such that, if an arbitrary segment,of length n k, l , of the natural numbers, is divided in any mannerinto k classes, there exists in at least one of these classes anarithmetic progression of length l. We must then prove that, forevery natural number k an n k, l+ l also exists. We solve thisproblem by actually constructing the number n k, l l). To this endwe set


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13q 0 = 1, n 0 =n(k,l)

and then define the numbers q 1, q2 , , n 1, n 2, successively asfollows: I f q5 _1 and n 5 _ 1 have already been defined for some s >0,we put1) q =2n q 1 n 5 =n(k \ l ) (s=1,2 , ... ).s s-1 s -

The numbers n 5 , q5 are obviously defined hereby for an arbitrarys60. We now assert that for n(k, l+1) we may take the number qk.We have to show then that if a segment, of length qk, of the sequence of natural numbers is divided in any manner into k classes,then there is an arithmetic progression of length l+ 1 in at least oneof these classes. The remainder of the chapter is devoted to thisproof.

In the sequel we set l +1 = l for brevity.4

Suppose then that the segment 11 of length qk, of the sequenceof natural numbers is divided in an arbitrary way into k classes. Wesay that two numbers a and b of 11 are of the same type, i a and bbelong to the same class, and we then write a ' b. Two equally longsubsegments of 11 o '(a,a+1, ... ,a+r) and S =(a :a +1 , ... ,a '+r) ,are said to be of the same type, if a 'a: a+ 1 ' a'+ 1, ... a+ r ' a +r,and we then write o 'o . The number of different possible types forthe numbers of the segment 11 is obviously equal to k. For s ~ n t sof the form a, a+ 1) (i.e. , for segments of length 2) the number ofpossible types is k 2 ; and in general, for segments of length m it iskm. Of course not all these types need actually appear in the s ~ment 11.

Since qk=2nk_ 1qk_ 1 (see (l)), the segment 11 can he regarded asa sequence of 2nk_ 1 subsegments of length qk_ 1 Such subsegments,as we have just seen, can have kq k 1 different types. The left halfof the segment 11 now contains nk_ 1 such subsegments, wherenk_ 1=n(kqk- 1, l according to 1). Because of the meaning of thenumber n(kqk- 1 [ , we can assert* that the left half of the segment11 contains an arithmetic progression of l of these subsegments of

*Work with the initial numbers of the nk-l subsegments. (Trans.)


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14the same type,

111 /12, , 1lof length qkl; here we say for brevity that equally long segments1i form an arithmetic progression, i f their initial numbers form

such a progression. We call the difference between the initial numbers of two neighboring segments of the progression /11, 2, , 1lthe difference d1 of this progression. Naturally the difference between the second or third, fourth, etc.) numbers of two such neighboring segments is likewise equal to d1.

To this progression of segments we now add the succeeding,(l+ 1)-st, term 1l' we recall that l = l + 1) which may already project beyond the boundary of the left half of the segment 11 butwhich in any case still belongs entirely to the segment 11. The segments 1 , 2, , 1l, 1l' then form an arithmetic progression, oflength l '=l+ 1 and difference d 1, of segments of length qk-1' where/11o /12, , 1l are of the same type. We know nothing about the typeof the last segment 1l

This completes the first step of our construction. It would bewell i f you thought i t through once more before we continued.

5We now proceed to the second step. We take an arbitrary one of

the first l terms of the progression of segments just constructed.Let this term be 11 so that ~ i < ; ; l ; 1. is a segment of length1 - - 1qk-1 We treat it the same way as we treated the segment 11. Sinceqkl =2nk_ 2qk_ 2 , the left half of the segment 1 1i1 can be regardedas a sequence of nk. 2 suhsegments of length qk_2 For subsegmentsof this length there are kqk- 2 types possible, and on the other handnk_ 2 =n(k'lk2 , l because of l) . Therefore the left half of 1 1i 1 mustcontain a progression of l of these subsegments of the same type,1i 1i 2 l i ~ l , of length qk _2 Let d2 be the difference of this progression (i.e., the distance between the initial numbers of two

neighboring segments). To this progression of segments we add the(l+ l)-st term 1i 1l about whose type, of course, we know nothing.The segment 1i 1l does not have to belong to the left half of the


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15segment fli 1 any more, but must obviously belong to the segmentfl . '1 We now carry over our construction, which we have executedup to now in only one of the segments fli 1, congruently to all theother segments fli 1 (l i 1 0 . We thus obtain a set of segments~ ( l ~ i 1 ~ l , 1 ~ i 2 ~ with two indices. t is clear that two arbitrary segments of this set with indices not exceeding l are of thesame type:

You no doubt see now that this process can be continued. Wecarry it out k times. The results of our construction after the firststep were segments of length qk l after the second step, segmentsof length qk_ 2, etc. After the k-th step, therefore, the results of theconstruction are segments of length q 0 = 1, i .e. , simply numbers ofour original segment fl. Nevertheless we denote them as before by

fl,. (l i 1 , i2t ... , i k ~ l ).1 2 kFor 1 ~ s ~ k a n d 1 ~ i 1 ..., i 8 , i1, i ; ~ l w e have

. _2) fl. . . =fl . ' , . ,'1'2's '1'2' 8We now make two remarks which are important for what follows.l ) In (2), if s


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164) /1. . . , . .1 s-1 s s+1 k /1. . . . . = d .1 s -1 s s+1 k 5

6Now we are near our goal. We consider the following k + l num-

bers of the segment /1:

ao = 11z z z ... la1 = 111z z ... z

(5) a 2 = / 1 1 1 z . . z........ak = /11 1 1 1

Since the segment has been divided into k classes, and we havek + l numbers in 5), there are two of these numbers which belongto the same class. Let these he the numbers a, and a 5 (r


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7

s -rc,. - c .= k (c. - c . 1 .t m=l t ,m t ,m-

Because of (4) we havec. -c . =L l. . --L l. . . -=d .t ,m r,m-1 l . . . t t t t l l . .. t t t t r m____ - - ~ ~ . - - ~ - . . t - ---...- ~ . . -

m s-r-m k-s m-1 s-r-m 1 k-sThus the difference

and is indeed independent of i which completes the proof of ourassertion.

You see how complicated a completely elementary construction can sometimes be. And yet this is not an extreme case: in thenext chapter you will encounter just as elementary a constructionwhich is considerably more complicated. Besides, it is not out ofthe question that van der Waerden s theorem admits of an evensimpler proof, and all research in this direction can only be welcomed.


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CHAPTER IITHE LANDAU-SCHNIRELMANN HYPOTHESIS

AND MANN S THEOREM

lYou have perhaps heard of the remarkable theorem of

Lagrange, that every natural number is the sum of at mostfour squares In other words, every natural number is eitheritself the square of another number, or else the sum of two,or else of three, or else of four such squares. For the purposeat hand i t is desirable to understand the content of this theoremin a somewhat different form. Let us write down the sequence ofall perfect squares, beginning with zero:S) 0, l , 4, 9, 16, 25, . . . .

This is a certain sequence of whole numbers. We denote it by S,and imagine four completely identical copies of it, S1, S2 , S3, S4 ,to be written down. Now we choose an arbitrary number ~ f rom Span arbitrary number ~ from s2 an arbitrary number ~ from s3 andan arbitrary number ~ from S4 , and add these numbers together.The resulting sum(*)can be

l) zero(ifa 1 =a 2 =a3 =a4 =0);2) the square of a natural numb (if in some representation (*)

of the number n three of the numbers a 1, a2 a3 , a4 are zero andthe fourth is not zero);3) the sum of two squares of natural numbers (if in some rep-

resentation (*) of the number n two of the numbers a 1, a2 a 3, a 4are zero and the other two are not zero);4) the sum of three squares of natural numbers if in some rep-

resentation (*) of the number n one of the numbers a l a 2 a 3, a4Is equal to zero and the remaining three are not zero);18


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195) the sum of four squares of natural numbers if in some rep-

resentation of the number n all four numbers are different from zero).Thus the resulting number n is either zero or else a natural

number which can be represented as the sum of at most foursquares, and i t is clear that conversely every natural number can beobtained by the process which we have described.

Now let us arrange all natural numbers n which can be obtainedby means of our process i.e., by the addition of four numbers takenrespectively from the sequences S1 , S 2 , S 3, S4 ), in order of mag-nitude, in the sequenceA)where 0


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2

The content of Lagrange's theorem is that the sum S S S S con-tains the entire sequence of natural numbers.

Perhaps you have heard of the famous theorem of Fermat, thatthe sum S S contains all prime numbers which leave a remainderof 1 when divided by 4 (i .e., the numbers 5, 13, 17, 29, ... . Per-haps you also know that the famous Soviet scholar Ivan Matveye-vitch Vinogradov proved the following remarkable theorem, on whichmany of the greatest mathematicians of the preceding two centurieshad worked without success:

f we denote by P the sequenceP) 0, 2, 3, 5, 7, 11, 13, 17,consisting of zero and all prime numbers, then the sum P P Pcontains all sufficiently large odd numbers.

I have cited all these examples here for only one very modestpurpose: to familiarize you with the concept of the sum of se-quences of numbers and to show how some classical theorems ofnumber theory can be formulated simply and conveniently with theaid of this concept.

2As you have undoubtedly observed, in all the examples mention-

ed we are concerned with showing that the sum of a certain numberof sequences represents a sequence which contains either com-pletely or almost completely this or that class of numbers (e. g.,all the natural numbers, all sufficiently large odd numbers, andothers of the same sort). In all other similar problems the purposeof the investigation is to prove that the sum of the given se-quences of numbers represents a set of numbers which is in somesense dense in the sequence of natural numbers. t is often thecase that this set contains the entire sequence of natural numbers(as we saw in our first example). The. theorem of Lagrange says thatthe sum of the four sequences S contains the whole sequence of nat-ural numbers. Now i t is customary to call the sequence A a basis(of the sequence of natural numbers) of order k if the sum of k iden-tical sequences A contains all the natural numbers. The theorem ofLagrange then states that the sequence S of perfect squares is a


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2

basis of order four. t was shown later that the sequence of perfectcubes forms a basis of order nine. A little reflection shows thatevery basis of order k is also a basis of order k lIn all these and in many other examples the "density" of thesum which is to be established is determined by particular prop-erties of the sequences that are added together, i .e. , by the spe-cial arithmetical nature of the numbers which go to make up thesesequences (these numbers being either perfect squares, or primes,or others of a similar nature). Sixteen years ago the distinguishedSoviet scholar Lev Genrichovitch Schnirelmann first raised theqnestion: To what extent is the density of the sum of several se-ql ences determined solely by the density of the summands, irre-spective of their arithmetical nature. This problem turned out to benot only deep and interesting, but also useful for the treatment ofsome cla ; sical problems. During the intervening fifteen years itreceived the attention of many outstanding scholars, and it hasgiven rise to a rich literature.Before we can state problems in this field precisely and writethe word "density" without quotation marks, it is evident that wemust first agree on what number (or on what numbers) to use tomeasure the "densi ty" of our sequences with (just as in physicsthe words "warm" and "cold" do not acquire a precise scientificmeaning until we have learned to measure temperature).

A very convenient measure of the "density" of a sequence ofnumbers, which is now used for all scientific problems of the kindwe are considering, was proposed by L. G. Schnirelmann. LetA)

be a sequence of numbers, where, as usual, all the n are naturalnumbers and an


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density of the sequence A) in the segment from l to n of the sequence of natural numbers. Following the suggestion of Schnirelmann, the greatest lower bound of all values of this fraction iscalled the density of the sequence A) in the entire sequence ofnatural numbers). We shall denote this density by d A).

In order to become familiar at once with the elementary prop-erties of this concept, I recommend that you convince yourself ofthe validity of the following theorems:

l If a 1 > l (i.e. , the sequence A) does not contain unity), thend A) =0.2. I f an=l+r{n-1) (i .e., the sequence A), beginning with a 1,is an arithmetic progression with initial term l and difference r),

then d A) = l /r .3. The density of every geometric progression is equal to zero.4. The density of the sequence of perfect squares is equal to

zero.5. For the sequence A) to contain the entire sequence of nat

ural numbers an =n, n = l, 2, ... ), it is necessary and sufficient thatd A) = l6. I f t(A) =0 and A contains the number l and if E >O is arbi

trary, then there exists a sufficiently large number m such thatA(m)


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3

the sequence of natural numbers contains A(n) numbers of the sequence A, each of which also appears in the sequence C. Let akand ak+l he two consecutive numbers of this group. Between themthere are ak l -ak - l=l numbers which do not belong to A. Theseare the numbers

ak+l, ak+2, .. , ak+l=ak+ 1 l.Some of them appear in C, e. g., all numbers of the form ak+ rwhere r occurs in B (which we abbreviate as follows: rEB). Thereare as many numbers of this last kind, however, as there are num-bers of B in the segment (l , l), that is, B l) of them. Consequentlyevery segment of length l included between two consecutive num-bers of the sequence A contains at least B l) numbers which belongto C. t follows that the number, C n), of numbers of the segment(l, n) appearing in C is at least

where the summation is extended over all segments which are freeof the numbers appearing in A. According to the definition of density, however, B l)?J3l, so that

C n) A n ) +f3 .l =A(n)+ {31n -A n)l,because Il is the sum of the lengths of all the segments which arefree of the numbers appearing in A, which is simply the numbern -A n} of numbers of the segment (l , n) which do not occur in A.But A(n) a n , and hence

C n) (n){l-{3)+ {3n ~ a n { l - { 3 } + {3nwhich yields

C n)/n ~ a+f3-a{3.Since this inequality holds for an arbitrary natural number n, wehave

y=d(C) a + { -a{3, Q.E.D.Schnirelmann s inequality (l) can be written m the equivalent


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fonnd A + B ) ~ ~ 1-d(A)I 1-d(B)I,

and in this form can easily be generalized to the case of an arbi-trary number of summands:

k1-d A 1 +A 2 + +Ak) l l -d (A) l .It is proved by a simple induction; you should have no trouble incarrying it out yourself. If we write the last inequality in the fonnk2) d A +A 2 + .. A k ) ~ 1- l , }1-d(Ai ) l 'it again enables one to estimate the density of a sum from the den-sities of the summands. L. G Schnirelmann derived a series of veryremarkable results from his elementary inequality, and obtainedabove all the following important theorem:

Every sequence of positive density is a basis of the sequenceof natural numbers.

In other words, i f a=d A)>O, then the sum of a sufficientlylarge number of sequences A contains the entire sequence of nat-ural numbers. The proof of this theorem is so simple that I shouldlike to tell you about it, even though this will divert us a bit fromour immediate problem.

Let us denote for brevity by Ak the sum of k sequences, eachof which coincides with A. Then by virtue of inequality 2),

d A k ) ~ 1 - 1 - a ) k .Since a >0, we have, for sufficiently large k3)

Now one can easily show that the sequence A 2 k contains thewhole sequence of natural numbers. This is a simple consequenceof the following general proposition.

LEMMA. If A n)+B n)>n-1, then n occurs in A +B.Indeed, i f n appears in A or in B everything is proved. e may


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therefore assume that n occurs in neither A nor B. Then A(n) =A n -1) and B n}= B n -1), and consequently

A n-1) +B n-1) >n-1.Now let al a 2 a and bl b2 b8 be the numbers of thesegment 1, n-1) which appear in A and B respectively, so that

r=A n-1), s=B n-1). Then all the numbersa 1, a 2, ,a,,

n-b 1, n-b 2 , n-bsbelong to the segment 1,n-1). There are r+s=A n-1)+B n-1) ofthese numbers, which is more than n-1. Hence one of the numbersin the upper row equals one of the numbers in the lower row. Letai=n-bk. Then n=ai+bk, i .e. , n appears in A+B.

Returning now to our objective, we have, on the basis of 3),for an arbitrary n:

and therefore

According to the lemma just proved, it follows that n appears inAk +Ak =A 2k. But n is an arbitrary natural number, and hence ourtheorem is proved.This simple theorem led to a series of important applicationsin the papers of L. G. Schnirelmann. For example, he was the firstto prove that the sequence P consisting of unity and all the primenumbers is a, basis of the sequence of natural numbers. The sequence P it is true, has density zero, as Euler had already shown,so that the theorem which we just proved is not directly applicableto it. But Schnirelmann was able to prove that P P has positivedensity. Hence P P forms a basis, and therefore P indeed also.From this it is easy to infer that an arbitrary natural number, withthe except ion of 1, can, for sufficiently large k be represented asthe sum of at most k primes. For that time 1930) this result wasfundamental and evoked the greatest interest in the scientificworld. At present, thanks to the remarkable work of I. 1\1 Vinograd-


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26ov, we know considerably more in this direction, as I already relat-ed to you at the beginning of this chapter.

3In the preceding it was my purpose to introduce you in the

shortest way possible to the problems of this singular and fasci-nating branch of number theory, whose study began with L. G.Schnirelmann s remarkable work. The immediate goal of the presentchapter, however, is a specific problem in this field, and I now pro-ceed to its formulation.

In the fall of 1931, upon his return from a foreign tour, L. G.Schnirelmann reported to us his conversations with Landau inGottingen, and related among other things that in the course ofthese conversations they had discovered the following interestingfact: In all the concrete examples that they were able to devise, itwas possible to replace the inequality

d A +B) ?,d(A )+ d B) -d A) d (B),which we derived in 2, by the sharper (and simpler) inequality4) d A +B) ?,d(A)+ d B).

That is, the density of the sum always turned out to be at least aslarge as the sum of the densities of the summands (under the as-sumption, of course, that d A ) + d B ) ~ l ) . They therefore naturallyassumed that inequality 4.) was the expression of a universal law,but the first attempts to prove this conjecture were unsuccessful.t soon became evident that if their conjecture was correct, the road

to its proof would be quite difficult. We wish to note at this pointthat i f the hypothetical inequality 4) does represent a universallaw, then this law can be generalized immediately by induction tothe case of an arbitrary number of summands; i .e . under the as-sumption that

we have5) d I Ai) ?, I d Ai).

i= l i=l


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27

This problem could not help hut attract the attention of scholars,because of the simplicity and elegance of the general hypotheticallaw 4) on the one hand, and on the other because of the sharp con-trast between the elementary character of the problem and the dif-ficulty of its solution which became apparent already after the firstattacks. I myself was fascinated by it at the time, and neglectedall my other researches on its account. Early in 1932, after severalmonths of hard work, I succeeded in proving inequality 4) for themost important special case, d A) = d B) (this case must be consid-ered as the most important because in the majority of concreteproblems all the summands are the same). At the same time I alsoproved the general inequality 5) under the assumption that d A 1 =d A 2 )= .. =d Ak) (it is easy to see that this result cannot bederiv-ed from the preceding one simply by induction, hut requires a spe-cial proof). The method which I used was completely elementary,hut very complicated. I was later able to simplify the proof some-what.

Be that as it may, it was but a special case. For a long timeit seemed to me that a none too subtle improvement of my methodshould lead to a full solution of the problem, but all my efforts inthis direction proved fruitless.

In the meantime the publication of my work had attracted theattention of a wide circle of scholars in all countries to the Landau-Schnirelmann hypothesis. Many insignificant results were obtained,and a whole literature sprang up. Some authors carried over theproblem from the domain of natural numbers to other fields. In short,the problem became fashionable . Learned societies offeredprizes for its solution. y friends in England wrote me in 1935that a good half of the English mathematicians had postponedtheir usual work in order to try to solve this problem. Landau, inhis tract devoted to the latest advances in additive number theory,wrote that he should like to urge this problem on the reader . Butit proved to be obstinate, and withstood the efforts of the most ablescholars for a whole series of years. It was not until 1942 that theyoung American mathematician Mann finally disposed of it: he founda complete proof of inequality 4) (and hence also of inequality5)). His method is wholly elementary and is related to my work in


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28form, al thongh it is based on an entirely different idea. The proof islong and very complicated, and I could not bring myself to presenti t to you here. A year later, however, in 1943, Artin and Scherk pnb-lished a new proof of the same theorem, which rests on an altogeth-er different idea. It is considerably shorter and more transparent,though still quite elementary. This is the proof that I should like totell you about; I have written this chapter on its account, and itforms the content of all the sncceeding sections.

4Suppose then that A and B are two sequences. We set A B =C.

Let A n), d A), etc. have their usual meaning. We recall that all oursequences begin with zero, but that only the natural numbers ap-pearing in these sequences are considered when calculating A n),B n), C n). We have to prove that(6) d C) d A) +d B)provided that d A) +d B) l For brevity we set d A) =a, d B) = { inwhat follows.

FUNDAMENTAL LEMMA If n is an arbitrary natural number,there exists an integer m l ~ m ~ n ) such that

C(n) - C n - m f; (a+ {3 m.In other words, there exists a remainder n-m+ l , n) of the

segment (l, n), in which the average density of the sequence C is atleast a f3.We are now faced with two problems: first, to prove the funda-

mental lemma, and second, to show that ineqnality 6) follows fromthe fundamental lemma. The second of these problems is incompa-rably easier than the first, and we shall therefore begin with thesecond problem.

Suppose then that the fundamental lemma has already beenproved. This means that in a certain remainder n -m + l , n) ofthe segment (l , n) the average density of the seqnence C is at leasta f3. By the fundamental lemma, however, the segment l ,n -m)again has a certain remainder n -m -m + l , n - m in which theaverage density of Cis at least a f3. It is clear that by continuing


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29this process, the segment l , n is eventually divided into a finitenumber of subsegments, n each of which the average density of Cis at least a f3. Therefore the average density of C is also atleast a f3 in the whole segment l, n). Since n was arbitrary, how-ever, we h : : ~

d C ?, a {3 Q.E.D.Thus the problem is now reduced to proving the fundamental

lemma. We now turn to this proof, which is long and complicated.5NORMAL SEQUENCES

In all that follows we shall regard the number n as fixed, andall sequences which we investigate will consist of the number 0and certain numbers of the segment l , n). We agree to call such asequence N rwrmal i f it possesses the following property: f thearbitrary numbers f and f of the segment l, n do not appear in Nthen neither does the number {+ f n appear in N where the casef=f is not excluded).

f the number n belongs to the sequence C, thenC n)- C n - l = l?, a+ {3 1,

so that the fundamental lemma is trivially correct m =l). Consequently we shall assume in the sequel-1 beg you to keep this mmind-that n does not occur in C.

To begin with, the fundamental lemma is easy to prove in casethe sequence C is normal. Indeed, let us denote by m the smallestpositive number which does not appear in C m ~ n because n, by assumption, does not occur in C). Let s be an arbitrary integer lyingbetween n m and n; n-m


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30On the other hand, by the lemma on p. 24, since m does not oc-

cur in C =A+ B we have A m) + B m) ~ m - l . Consequently7) C n) -C(n -m A m ) + B m) ~ a + {3)m,

which again proves the validity of the fundamental lemma.6

CANONICAL EXTENSIONSWe now turn our attention to the case where the sequence

C=A+ B is not normal. In this case we shall add to the set B, ac-cording to a very definite scheme, numbers which it does not contain, and thereby pass from B to an extended set B1 The setA+ B1 = C1 evidently will then be a certain extension of the set C.As I said before, this extension of the sets B and C (the set A re-mains unaltered) will be defined precisely and unambiguously; it ispossible if and only if the set C is not normal. We shall call thisextension a canonical extension of the sets B and C. Some important properties of canonical extensions will be derived, with whoseaid the proof of the fundamental lemma will be completed.

I now come to the definition of the canonical extension of thesets and C. I f C is not normal, there exist two numbers c and cm the segment 0, n), such that

cC, c C, c +c 1- n t:C.Since C=A+ B, it follows that8) c+ c 1- n=a+b at:A, bt:B).

Let 0 be the smallest number of the set B which can play the roleof the number b in equation 8). In other words, f 0 is the smallestinteger be B which satisfies equation (8)for suitably chosen numberscC, c C, at:A of the segment O,n). This number 0 will be call--ed the basis of our extension.

Thus the equation9) c+c 1-n=a+{3 0

necessarily has solutions in the numbers c, c a satisfying the conditions


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31c C, c 1 C, a e A

where all three numbers belong to the segment (0, n). We write allnumbers c and c which satisfy equation 9) and the enumerated con-ditions, to form a set C*. Clearly the sets C and C* do not have asingle element in common. We call their union* (i .e. , the totalityof all numbers which occur either in C or in C*)

the canonical extension of the set C.Let U s now examine the expression f 0 + n- c f c here is al-lowed to run through all the numbers of the set C* just constructed,

the values of this expression form a certain set B*. According toequation (9), every such number 0 +n-c (ct:C*) can be written inthe form c -a, where c t:C*, a eA.

Let b* be an arbitrary number occurring in B*. Since it is ofthe form { 0 +n-c, it is ?,{3 0 ?,0; and since i t is also of the formc -a (c eC*, aeA), it is ;;;;c ;;;;n. Hence all numbers of the set Bbelong to the segment (O,n). Moreover, if b* t:B , then b* B,because otherwise it would follow from b* =c -a that c =a+ b* eA +B=C, which is false.

Accordingly, the set B is embedded in the segment (0, n) andhas no elements in common with the set B. We put

B UB =B 1and call the set B 1 a canonical extension of the set B.Let us show that

A +B 1=C 1First, let a eA, b1 t:B 1. We shall prove that a+ b1 t:C 1. Fromb1 t:B 1 it follows that either b1 t:B or b1 t:B . f b1 t:B, then a+ b1

eA +B =CCC 1 f b 1 t:B , however, then a+b 1 either occurs inC,and hence also in C1 , or a+ b1 C. But in this case (since bl> as anelement of the set B*, is of the form f 0 +n-c: c C) we obtain

c=a+b1 =a+f3o + n - c C.Therefore

*Here and in the sequel we use the symbol U o denote the union of sets, since weare using the symbol+ in another sense.


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32c+c -n=a+f3oE A+ B =C,

where c C and c C. But then according to the definition of theset C*,

Q.E.D.Thus we have shown that A +B 1 cC 1

To prove the inverse relation, let us assume that c EC 1 whichmeans that either cEC or cEC*. I f cEC, then c=a+b, aEA, bEBCB 1 If however, c EC*, then, for a certain a EA, the number b =c-a as we know, occurs in B . e have c=a+b* EA+B*CA+B 1 Therefore C1CA+B 1. e also proved above that A+B 1CC 1. Consequently C1 A B 1

Now recall that according to our assumption, n C. It is easy tosee-and this is important-that the number n does not appear in theextension c1. For if we had n EC*, we could, by the definition ofC*, put c =n in equation 9), which would yield c=a+f3 0 EA +B =C,whereas c C according to 9).f the extended sequence C 1 is not yet normal, then, because ofA+B 1=C 1 and nCh the sets A B 1, and C1 form a triple with allthe properties of the triple A B, C that are necessary for a newcanonical extension. e take a new basis : 1 of this extension, define the complementary sets B i_, C i_ as before, put

and are able to assert once more that A+ B 2 = C2 and n C 2 It isevident that this process can be continued until one of the extensions Ch proves to be normal. Obviously this case must certainlytake place, because in every extension we add new numbers to thesets B and C without overstepping the hounds of the segment(0, n).

In this way we obtain the finite sequences of setsB = B 0 c B1 c ... c Bh ,C=C 0 cC 1 c ... cch,

where every BJL l (respectively C11+l) contains numbers which donot appear in B C11 ) and which go to make up the set Bt C ~ ) , so


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that81l+ 1 =81l u 8 ; , ell+1 =ell u e ~ < o ~ l l ~ h - 1 .

e denote by 31l the basis of the extension which carries (811 ell)into (81l+ 1, ell+1 . e have

A+81l=ell ne l l O ~ I l ~ h ) .Finally the set Ch is normal whereas the sets ell (0 ll h -1 arenot.

7PROPERTIES OF THE CANONICAL EXTENSIONS

e shall now formulate and prove in the form of three lemmasthose properties of the canonical extensions which are needed lat-er. Only Lemma 3 will have further application; Lemmas I and 2 arerequired solely for the proof of Lemma 3.

LEMMA I . (31l>(31l_ 1 l ~ l l ~ h - 1 ) ; i .e . , the bases o f successivecanonical extensions form a monotonically increasing_ sequence.In fact since 31l E81l =81l_1U8t- l either 31l E 8 ~ _ or 31l E

8 ll_1 I f 31l E8t- l then 31l is of the form(31l=(31l-1 +n-c ,

where c E e ~ - 1 cel l and therefore c(31l_l and LemmaI is proved. I f 31l E81l-l however then by the definition of the num-ber 31l there exist integers aEA, cel l c e l l such that

c+c -n=a+f31l E ell .But for 31l E 8 ll_1 we have(10) c+c -n=a+{31lEA+81l_ 1=ell_ 1 ,

where c e l_1, c e l_1 Hence because of the minimal propertyof 3 l_ 1 (31l?,(31l_1 I f f31l=f31l-l it would follow from (10) and thedefinition of the set t 1 that

c E e t - 1 c e c E e ~ _ 1 c ell Both are false however and therefore 31l > 31l_ 1

In the sequel we shall denote by m the smallest positive integer


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which does not appear in chLEMM 2. I f c E C ~ ( O ~ ~ t ~ h - 1 ) and n-m< cn-m+

f3w That is, all numbers c of the set C ~ which lie in the intervaln-m< c


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LEMMA 3. We haveC* n) -C* n -m) = B*(m -1)

p p pThat is, the number of integers c E c ~ in the segment n-m< c


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where the sets appearing in any one of these two unions are mutually exclusive, so that h 1Ch n)- C h n -m = C n)- C n -m) l l ~ n ) - C ~ n -m) l,

h-1 0Bh m)=Bh m-1)=B m-1)+ l B ~ m - 1 ) ;

1=0we have of course put C ~ = C*, B ~ =B*. On account of 12) it followsthat h-1C n)-C n-m)+ ~ n ) - C ~ n - m ) l

1=0h 1

~ A m ) + B m - 1 ) + l B ~ m - 1 ) .1=0

By Lemma 3, however,C ~ n ) - C ~ n -m B ~ m -1 0 ~ / l ~ h - 1 ) ,

so that the preceding inequality becomesC n)- C n -m) A m ) B m -1 =A m) B m) a +{3) m,

which proves the fundamental lemma.As we saw in 4, this also completes the proof of Mann s theo

rem which solves the fundamental metric problem of additive numher theory.

Doesn t Artin and Scherk s construction have the stamp of amagnificent masterpiece? I find the outstanding combination ofstructural finesse and the extremely elementary form of the methodespecially attractive.


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CHAPTER IIIAN ELEMENTARY SOLUTION OF

WARING S PROBLEM

You will recall the theorem of Lagrange, which was discussed

at the beginning of the preceding chapter. t says that every naturalnumber can he expressed as the sum of at most four squares. I alsoshowed you that this theorem could he stated in entire y differentterms: f four sequences, each identical with

are added together, the resulting sequence contains all the naturalnumbers. Or even more briefly, the sequence A 2) is a basis (of thesequence of natural numbers) of order four. I also mentioned that, ashad been shown later, the sequence of cubes

was a basis of order nine. All these facts lead in a natural mannerto the hypothesis that, for an arbitrary natural number n the se-quence

is a basis (whose order of course depends on n). This conjecturewas also actually propounded y Waring as early as the eighteenthcentury. The problem proved to he very difficult, however, and itwas not until the beginning of the present century that the universalvalidity of Waring's hypothesis was demonstrated, y Hilbert (1909).Hilbert 's proof is not only ponderous in its formal aspect and basedon complicated analytical theories (multiple integrals), hut alsolacks transparency in conceptual respects. The eminent Frenchmathematician Poincare wrote in his survey of Hilbert 's creative

37


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38scientific work, that once the basic motivations behind this proofwere understood, arithmetical results of great importance wouldprobably flow forth as from a cornucopia. In a certain sense he wasright. Ten to fifteen years later, new proofs of Hilbert's theoremwere furnished by Hardy and Littlewood in England and by I. MVinogradov in the USSR. These proofs were again analytic and for-mally unwieldy, but differed favorably from Hilbert's proof in theirclarity of method and their conceptual simplicity, which left nothingto be desired. In fact, because of this, both methods became mightyscources of new arithmetical theorems.

But when our science is concerned with such a completely elementary problem as Waring's problem, it invariably attempts to finda solution which requires no concepts or methods transcending thethe limits of elementary arithmetic. The search for such an elementary proof of Waring's hypothesis is the third problem which I shouldlike to tell you about. Such a fully elementary proof of Hilbert'stheorem was first obtained in 1942, by the young Soviet scholar Y.V. Linnik.You are already accustomed to the fact that elementary doesnot mean simple . The elementary solution of Waring's problemdiscovered by Linnik is, as you will see, not very simple either, andit will take considerable effort on your part to understand and digestit. I shall endeavor to make this task as easy as possible for youthrough my exposition. But you must remember that in mathematics(as probably in any other science) the assimilation of anything really valuable and significant involves trying labor.

The ideas of Schnirelmann which I described to you in the begin-ning of the second chapter play an essential role in Linnik's proof.You will recall I mentioned it at that time) how Schnirelmannproved his famous theorem that the sequence consisting of zero,unity, and all the primes, is a basis of the sequence of natural num-bers: He showed that the sequence P has a positive density.This immediately yields the assertion, however, because, accordingto the general theorem of Schnirelmann which we proved on pp.24-25, every sequence of positive density is a basis of the sequence ofnatural numbers. The same method also lies at the basis of the proofof Hilbert's theorem discovered by Linnik. t all boils down to the


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proof that the sum of a sufficiently large number of sequences An)is a sequence of positive density. As soon as this is accomplished,we can, by virtue of the same general theorem of Schnirelmann, re-gard Hilbert's theorem as proved.

2THE FUNDAMENTAL LEMMA

f we add together k sequences, identical with An, according tothe rule in Chapter II, we evidently obtain a sequence A > whichcontains zero and all those natural numbers which can be expressedas a sum of at most summands of the form xm, where x is an arbi-trary natural number. In other words, the number m belongs to thesequence A ~ k > if the equation(l) n n nX1+X2+ +Xk= mis solvable m nonnegative integers x l ~ i ~ k ) . As we saw in lthe problem is to show that, for sufficiently large k the sequenceA ~ k has a positive density.

For preassigned and m equation (l) in general can be solvedin several different ways. In the sequel we shall denote by rk(m) thenumber of these ways, i .e., the number of systems of nonnegativeintegers x 1 x 7 xk which satisfy equation (l). It is clear that thenumber m occurs in A ~ k) if and only if rk(m)>O.

In the following, we shall assume the number n to be given andfixed and shall therefore call all numbers which depend only uponn, constants. Such constants will be denoted by the letter c or c n),where such a constant c may have different values in different partsof our discussion, provided merely that these values are constants.Perhaps you are rather unused to such freedom of notation, butyou will soon become familiar with it. It has proved to be very con-venient, and appears more and more frequently in modern research.FUNDAMENTAL LEMMA. There exist a natural number k=k n),depending only on n, and a constant c, such that, for an arbitrarynatural number N,2) ) N k/n)-1rk m


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problems: first, to prove the fundamental lemma and second to drawfrom the fundamental lemma the conclusion that we need viz. thatthe sequence A ~ k has a positive density. This time again the sec-ond problem is considerably easier than the first, and we shall there-fore begin with the second problem.

t follows immediately from the definition of the number r,. m),that the sum

represents the number of systems x 1, x 2 , , xk) of k nonnegativeinregers for which3)

Every group of numbers for whichO ~ x . ~ N i k ) 1 n- ( l ~ i ~ k )

obviously satisfies this condition. To satisfy these inequalitiesevery xi can evidently he chosen in more than Nik)lln differentways xi= 0 l ... , [ NI k l n ] . After an arbitrary choice of thissort the numbers x 1, x 2, , x k may he combined and so we havemore than NI k)kl n different possibilities for choosing the completesystem of integers xi l;;;;i;;;;k) so as to satisfy condition 3). Thisshows that

4)We assume that the fundamental lemma has been shown to be

correct an.d that inequality 2) is satisfied for an arbitrary N Wenow have to verify that inequality 2) is consistent with inequality4) which we proved only if the sequence A ~ k has a positive den-

sity. The idea behind the following deduction is very simple: In thesum R,. N), only those summands rk m) are different from zero forwhich m occurs in A ~ k ) . f A ~ k ) had density zero then for largeNthe number of such summands would he relatively small; because of2), however every summand cannot he very large. Their sum R,. N),

therefore would also he relatively small whereas according to 4)it must he rather large .

[a] denotes the largest integer;;;; a.


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It remains to carry out the calculations. Suppose that d A ~ k l ==O.Then, for an arbitrary small E>O and a suitably chosen N,

Here the number may be assumed to be arbitrarily large, because~ k (for an arbitrary k contains the integer (bear in mind Problem

6 on p. 22, which you solved). Applying the estimate (2) we getn NR N)= r m)=r 0)+ l r m)


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5)let a 1, az m be integers with ia 2 i la 11 A , and let a 1 and a 2 berel tively prime. Then the number of solutions of equation 5) s t-isfying the inequ lities lz 11 ;;;A lz 2 1 A , does not exceed 3A/Ia11.

Proof: We may assume that a 1>0, because otherwise we havemerely to replace z1 by - z 1 in every solution.

Let lz1o z 2 l and lz:l, ~ l he two different solutions of equation5). Then from

a 1z 1 + a 2 z 2 =ma1zi + a z ~ =mwe get

by subtraction. Accordingly the left-hand side of this equation mustbe divisible by a 1 But* a 1, a 2 = l , and consequently z ~ - z mustbe divisible by a 1 Now z ~ , J , z 2 , and therefore l z ~ - z as a multi-ple of a 1, is not smaller than a1. Thus, for two distinct solutionslz 1, z 2 } and {z{, z ~ l of equation 5), we must have l z ~ - z l ~ a

In every soluti on lz 1, z 2 of equation 5), let us agree to call z 1the first member and z 2 the second. It is obvious that the number ofsolutions of equation 5) which satisfy the conditions Iz11 A ,

I ~ A , is not more than the number t of second members which oc-cur in the interval . Since we have proved that two such second members are at least the distance a 1 apart, the difference be-tween the largest and smallest second members occurring in the interval is at least a 1 t- l) . On the other hand, this differencedoes not exceed 2A so that

a t - l ) ~ 2 A ,t - l } 2 A / a 1 ,

t ~ 2 A / a + l ~ 3 A / a 1b e c a u s e , b y a s s u m p t i o n , a ~ A , and therefore l ~ A / a This proves

Lemma l .LEMMA 2. In the equation

6) a 1, a 2 ) denotes the grea test common divisor of the in tegers a1 and a o


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let the ai and m be integers satisfying the conditions*

Then the number of solutions of equation 6) satisfying the inequalities l z i i ~ A l ~ i ~ l ) , does not exceed

c(l)Al1 /H,where H is the largest of the numbers ia 1, ia2, . , iazl, and c l) isa constant depending only on l.

Proof: f l=2, Lemma 2 obviously becomes Lemma 1 with c(2)=3). Accordingly Lemma 2 is already verified for l=2. We shall therefore assume that l ~ 3 and that the truth of Lemma 2 has already beenestablished for the case of l 1 unknowns. Since the numbering isunimportant, we may assume that iall is the largest of the numbersa1i a2 , ...,Iall i .e. , H I all

There are two cases to consider.1) a 1 =a 2 = =al-l =0. Since (a 1, a 21 , al)= 1, we have Iall =H =

1, so that the given equation is of the form Zl=m. In this equationeach of the unknowns z 1, z 2 , , zl_ 1 can obviously assume an arbitrary integral value in the interval and hence at most2A 1 3A values all told. As for z however, it can assume at mostone value. Consequently the number of solutions of the given equation satisfying the inequalities lzil ~ A l ~ i ~ l ) , does not exceed

3A )l- 1 =c l) A 1 =c l) A 1/Hwhich proves Lemma 2 for this case.

2) f at least one of the numbers a1> a 2, , al_ 1 is different fromzero, then

exists. Let us denote by H the largest of the numbers1 ~ i ~ l - 1 ) .

Suppose now that the numbers z1> z 21 , zl satisfy the given e-quation 6) and the inequalities lzil ~ A 1 ~ i ~ l ) . We set7) a1/o)z1 a ~ o)z 2+ ... + al-1/o) zl-1 = m ',

*(a1, a 2 , , al) denotes the greatest common divisor of the integers in parentheses.


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and hence

Then obviously( ~and

l-1j 8 m j ~ i i l a i j j z i j ~ l 8 H ' A ,

which implies that Im l Z H ' A .Thus, if the numbers z1 , z 2, , z satisfy equation 6) and the inequalities jzij A ( 1 ~ i ~ l ) , then the integer m exists, which, withthese numbers, satisfies equations 7) and 8), where m 1 Z H A.But in equation 8) clearly 8 ~ jalj and 8, al) = 1 (otherwise we shouldhave (a 1, a 2 , , al-1 al)> 1). Hence, by Lemma 1, the number of solutions of equation 8) in the unknowns m , zl), for which m I ~ZH A, j z l j ~ A < l H ' A , does not exceed 3lHit/jalj. For the same m ,equation 7), according to Lemma 2 for equations in l 1 unknowns,has. at most c l)Al 2/H solutions in integral zi with J z ; I ~ A .

It is evident, from what has been said, that the number of solutions z 17 z 2, , zll of equation 6) which satisfy the inequalitiesjzil A ( 1 ~ i ~ l ) , does not exceed(3ZH 11/j ali) c(l)A l- 2/H = c(l)A l- l iati

=c l)Al- 1/H,which completes the proof of Lemma 2.*

We shall now investigate the totality of equations of the form(9)where jail A ( 1 ~ i ~ l ) and, as always, all ai are integers. Let B be apositive number whose relation to the number A is described by the in-equalities 1 A ~ B c(l)A l- 1, and let l> 2. We now want to estimate

You have probably noticed that in the last chain of equations the symbol c(l) occurred in different places with different meanings. On p. 39 I prepared you for such a ust.of this symbol.


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5

the swn of the nwnbers of solutions zi, \ z i \ ~ B l ~ i ~ l ) of all thee-quations 9) of this family.

l? First let us make a separate examination of equation 9) fora 1 =a 2 = ... =a1=0 it is a member of our family) and estimate thenwnber of its solutions which satisfy the inequalities \zi\ ~ B l ~

i ~ l ) . Our equation is obviously satisfied by an arbitrary system ofnwnbers z 1 , z 2 , , z 1 and we have merely to calculate how manysuch systems exist which satisfy the inequalities \ ~ B , \ z \ ~ B ,... \z1\ B . Since the interval contains at most 2B + in-tegers, each zi can assume at most 2B + different values. Conse-quently the nwnber of systems lz 1, z2, , z1l of the type in whichwe are interested does not exceed 28 + 1)1 ~ 38)1= c(l)B 1 By our hy-pothesis, however, B ~ c l ) A l-1 so that c(l)B 1=c(l)Bl-1B c l ) A B ) l - , 1Hence, for the case where a 1 =a 2= ... =a1=0, equation 9) has atmost c(l)(AB)11 solutions of the type we are interested in.

2? Even i only one of the coefficients ai is different from zero,the greatest common divisor of these coefficients, a 1, a 2, , a1)=o,exists. Suppose first that O= l and let H be the largest of the nwn-bers \ai\ i = l 2, ... , l). Clearly H is one of the integers in the inter-val Hence, H is either between A and A/2, or between A/2and AI4, or between A/4 and A/8, etc. t is therefore possible tofind an integer m 0 such that10)

According to Lemma 2, for an equation of the form 9) in whicho= and H satisfies the inequalities 1 0), the number of solutionszi \ z i \ ~ B , does not exceed

c(l)Bl- 1/H c(l)Bl- 1/ A/2m+1) =c(l)B 1-12m /A.On the other hand, it follows from 10) that

11)Consequently the nwnber of equations of type 9) for which the ine-qualities 10) are satisfied is at most equal to the number of equa-tions of the same type which satisfy the conditions 11), i. e.,atmost


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Thus the sum of the numbers of solutions lzil ~ B of all such equations of type 9) for which 8=1 and A 2 m + l ) < H ~ A 2 m , doesn texceed

Summing this estimate over all m ~ O , we reach the following conclusion: The sum of the numbers of solutions izil ~ B of all equations 9) for which iaii::;;A U ~ i ~ l ) and8=1 is at most

c l) AB)l1.3 ? It remains for us to figure out the numbers of solutions of the

required type for equations with > 1. In this case equation 9) isevidently synonymous with the equation

where onlya,/8, ar/8, , az/8) = 1

and the number A has to be replaced by the number A/8. As we sawin 2, the sum of the numbers of solutions I il ~ B of all such equations, for a given, fixed 8 does not exceed*

c(l) (A81.B)l1 = c l) AB)l18Clearly now we have merely to sum this expression over all the possible values of 1 ~ 8 ~ A ) .

Thus we find that the sum of the numbers of required solutionsof all equations of the form 9), where iaii ~ A l ~ i ~ l ) and not allai are equal to zero, does not exceed the value

Ac l) AB)l-1 I8 0 1 > c l) (AB)l1.H=c(l)(AB)l1.8=1 l-2[To obtain the first relation we employ the inequality

AI U nq+l )< q+ 1 /q,n=1

*Since instead of A we now have to talte the smaller number AI8, i t is conceivable that the assumed condition B ~ c l A l - is violated. You can verify, however,without any trouble, that we made no UBe of this usumption in Case 2, and that theresult in fl therefore does not depend on it.


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47

which is valid for an arbitrary natural number q and for an arbitraryA ~ we denote by q the number l-2 which is positive because weassumed that l 2). Here is a simple proof: For n ~ 1 we have

nq- (n+ l)-q =l n+ 1 q -nq l / nq(n+ 1 q=(nq + qnq-1 + ... + 1-nq / nq(n+ 1 q~ q n q - / n q n + 1 q >q/ n+ 1 q+1,

and hence

By substituting successively n= 1, 2 . . . _A 1 in this inequality andadding all the resulting inequalities together we find that

AIn q+l)


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48words than in signs. This is of course a difficulty of presentation,however, and not of the subject itself, and I shall take pains to out-line all questions that arise, and their solution, as concretely aspossible.

We shall denote by A a finite complex (i.e., collection) of numbers, not all of which are necessarily distinct. f the number a oc-curs A times in the complex A we shall say that its multiplicity isA. Let a 1a 2 a he the distinct numbers which appear in A, andlet A1 , A2, , A, he their respective multiplicities (because the corn-rplex A contains all together I\ numbers). Let B be another corn-

t lplex of the same type, which consists of the distinct numbers b1 ,b D ... , bs with the respective multiplicities /11 /12 , 5 Let us investigate the equation12) x+y=c,

where c is a given number and x au.d y are unknowns. We are inter-ested in such solutions lx, yl of this equation in which x is one ofthe numbers of the col lplex A (abbreviated x EA and y is one of thenumbers of the complex B (yEB).If the numbers x=a; andy=bk sat-isfy equation (12), this yields Aip.k solutions of the required kind,because any one of t h e \ specimens of the number ai, which oc-cur in the cornpl ex A can he combined with an arbitrary one of theIlk specimens of the number bk appearing in the complex B. But \\oehave* A;llk ~ ~ A r + P . D Therefore the number of such solutions ofquation (12), where x=a;, y=bk, is not greater than ~ A r + p . ~ ) . I tfollows that the number of all solutions x EA y EB of equation (12)is not more than the sum I ~ A t + p.z). Here the summation is over allpairs of indices li, kl for which a;+ bk =c. Our sum is enlarged i f wesum Ar over all i and 1 1 ~ over all k (because every bk can he combined with at most one ai. It finally follows, therefore, that the number of solutions x EA y EB of equation (12) does not exceed thenumber

* The geometric mean is not greater than the arithmetic mean . Here is the sim-plest proof:

O ~ \ - p . k ) = A t p . z - 2 A i l 1 k and hence 2 \ p . k ~ t + P i .


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49

On the other hand let us consider the equation13) x-y=O

and calculate the number of its solutions x EA yEA. Clearly everysuch solution is of the form x=y=a; l ~ i ; : : ; ; r ) . For a given i we ob-tain A.t solutions because the numbers x and y can coincide inde-pendently of one another with any one of the A specimens of thenumber a i appearing in A Accordingly the total number of solutionsx EA yEA of equation 13) is equal to ~ A . ? . In exactly the samet = lway we find, of course that the number of solutions x EB y EB ofsthe same equation is equal t o k ~ f ~ If we compare these results withthe one found above we reach the following conclusion:

LEMM 4. The number of solutions of the equationx y=c, xEA, yEB

does not exceed half the sum of the numbers of solutions of the e-quations

x-y=O, x EA yEAand

x-y=O, x EB y EB.For the special case in which the complexes and B coincidewe obtain the followingCOROLLARY. The number of solutions of the equauon

x y=c, xEA,yEAdoes not exceed the number of solutions of the equation

x-y=O, x EA yEA.Now let k and s be two arbitrary natural numbers. e put k2 5 =l

and investigate the equation

Let A 1 A 2 , , A 1 be finite complexes of numbers. Suppose thatthe complex A; l ~ i ~ l ) consists of the distinct numbers a i l aiZ


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50

with the respective multiplicities \ 1 Ai 2, We are interested inthe number of solutions of the equation14)

f we setX1+X2++Xl/2=X, X(l/2)+1+ Xl=y

(l/2 is of course an integer), then the given equation can he writtenin the form

x+y=c,and Lemma 4, which we have just proved, can he applied to it. Wehave only to find out to which complexes the numbers x and y he-long. Since xi EAi l;:;;i;:;;l), x can he an arbitrary number of the formZ1+z2+ +zll2 where zi EAi l;:;;i;:;;l/2). Similarly r can he an ar-bitrary nwnher of the same form, where, however, zi EA(l/ +i1;:;; i; ;; l/2).

Hence, by Lemma 4, the number of solutions of equation 14)does not exceed half the sum of the numbers of solutions of theequation15) x - y=O

under the following two hypotheses:1) X =z 1 + z 2 + + z l/2r = z 1 z ~ +zil2

where16)

2) X and Y have the same form, hut17)

In both cases equation 15) may he rewritten in the form18)We conclude therefore that the number of solutions of equation 14)does not exceed half the sum of the numbers of solutions of equation


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51

(18) under the hypotheses (16) and (17), i. e., it does not exceedhalf the sum of the numbers of solutions of the equations

l /2(l8a) I z .-z. ')=O, zi EA . z(EAi l ~ i ~ l / 2 )i=1 ' 'and

l /2(l8b) I z . - z ~ ) = O , zi EA(l/ 2 )+i ' z[EA(l/ 2 )+i l ~ i ~ l / 2 ) .i=l ' ' .E'quation (18) has l /2 summands ~ n the left-hand side, i. e., half

as many as the original equation (14).We set

l/4.I z . - zn=x,i=1 ' 'l/2I (z.-z. ')=y,i=(l /4 +1 ' '

and thereby bring equation (18) into the formx+y=O.

To this we can apply Lemma 4 anew. t is evident that, just as wearrived at equation (18) from equation (14), we now get from equation(18) to the equation

l/4.(19) I u. u.'-u. -u : ) =0,i=1 ' ' ' 'where we have to consider the sum of the numbers of solutions ofthis equation under the following (now four) hypotheses:

4) , , , , , ,u,, u,, ui ' u, E (3l /4 )+i.Since l =k28 , we can repeat this process s times. We evidently end~ p then with the equation

k(20) I y ~ 1) + r 2 >+ ... +yf2s 1> y ~ 2 s - 1 + 1 >- ... -yf2s>l =0,i=1

where we have to consider the sum of the numbers of solutions ofthis equation under 28 different hypotheses, viz.:


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521 > A 1 > A < > AI) Yl..1 E 1 y i E 2Yk 1 E k

) 1 > 1 > A < > A2 r1 EAk+l n / E k+2 Yk E 2k2s) rV> EAk2sk+1 ... ,yv> EAk2s.I f we put

yU> = yfil +y2


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53

5PROOF OF THE FUNDAMENTAL LEMMA

We are going to prove the fundamental lemma by the method ofinduction on n. It is often the case in inductive proofs, that astrengthening of the proposttion to be proved, considerably facilitates i ts proof by the given method and sometimes is actually whatmakes the proof feasible in the first place). The reason for this iseasy to understand. In inductive proofs, the proposition is assumedto be correct for the number n -1 and is proved for the number n.Hence, the stronger the proposition, the more that is given to us bythe case n -1; of course, so much the more has to be proved for thenumber n, but in many problems the first consideration turns out tobe more important than the second.

And so it is, in fact, in the present case. Of immediate interestfor us is the number of solutions of the equation x1 x ~ + ... xk =m

1 ~ m ~ N ) where, according to the very meaning of the problem,O ~ x i ~ m l l n ~ N 1 n . But xn is the simplest special case of an n-thdegree polynomial

n ) n n-11,x =a0x a 1x + ... +an_ 1x an,and it will be to our advantage to replace the given equation l) bythe more general equation22)

where the unknowns are subjected to the weaker conditions lx l ~Nlln l ~ i ~ k ) . The proof of our proposition for equation 22) willgive us more than we really need; but, as you will see, it is justthis strengthening of our proposition which creates the possibilityo f an induction. And so, for m ~ N , let us denote by rk m) the numberof solutions of equation 22) which satisfy the conditions lx I N 1 In

l i ~ k . Of course we are still free to dispose arbitrarily of the coefficients of the polynomial / x) in the interest of the induction to beperformed provided only that the imposed conditions are satisfiedin the case f x)=xn). We are going to prove the following proposition:

Let the coefficients o the polynomial f x) satisfy the inequalities23)


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54Then, for a suitably chosen k =k(n),

rk(m) l , and suppose that our assertion has already beenverified for the exponent n l . Put k(n - l)=k and choose

k =k(n) =2n2 [4 log 2 k ] ,where the exponent means the greatest integer not exceeding 4log 2k.'In the sequel we shall set [4log 2 k 1 - l = s , for brevity, so that24)

To estimate the number, rk(m), of solutions of equation (22), wefirst apply Lemma 4 to it, setting

Xk kX = ; - ~ _ / ( x i ) J = ~ f(x.). i ~ k l

The complex A (and the complex B which coincides with it inthis case) consists of all sums of the form

ki' if(xi), where lx l :;;Nlln ( l : ; ; i : ; ; ~ k ) .By the Corol ary of Lemma 4, rk m) does not exceed the number ofsolutions of the equation x-y=O, where x EA, yEA, i .e. ,


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k /2 k/2X= I,f(x.), y= I,f(y.),i l i l

In other words, rk m) does not exceed the number of solutions of theequation

k 2(25) . I . I f x)- f(y,)l =0,=1

where i x , i ~ N i r ; i ~ N 1 n ( l ~ i ~ k / 2 ) . e now set x,-y,=h,l ~ i ~ k / 2 ) and replace the system of unknowns lx,,y,l by the sys

tem ly,, h;l; here we allow Y and h, ( l ~ i ~ k / 2 ) to assume all possible integral values in the interval , which canonly increase the number of solutions of our equation. This meansthat every summand f(x,)-f(y,) in equation (25) is replaced by theexpression

n 1f(y.+h.)-{(y.)= I.a l y . + h . ) n - v _ y ~ v l' ' v=O v ' 'n 1 n-v= I a I. (n-v) ~ y ' v - .v=O v t=1 t

I f we change the variable t of summation by puttingv+t=u,

so thatn - v - t=n-u , t=u-v,

we obtainn 1 nf(y.+h.)-f(y.)=h. I a I (n-v)h':'v1y' u v=O v u=v+1 u-v n u-1=h. I y' U I a (n-V) hlfV-1

u=1 v=O v u-vn=h. I a. y ~ u = h . c p . { y . ,u=l u ' ' '

wheren .(y)= I a. yn-u' u=l 1.,u


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56is a polynomial of degree n 1 with coefficients

u-1a. = I a (n-v)h f-v- 1 l ~ i ~ k / 2 )r. u v=O v u v

which depend on the numbers hi.Thus, in the new variables ly i hi I equation 25) assumes the

form26)

In this equation the numbers hi and y i may take on arbitrary integral values in the interval where we must bearin mind that the coefficients of the polynomials i y) of degree n-l)depend on the numbers h

Mark well that we have proved the following so far: The numberrk m) which we are estimating, does not exceed the sum of the num-bers of solutions in integers Yi lr;l 2 N / n l ~ i ~ k / 2 ) , o f all theequations 26) which can be obtained from all possible values o f thenumbers hi, lh;l 2 N 1 n l ~ i ~ k / 2 ) .

6CONTINUATION

e are now going to examine one of the equations 26), i .e. , weshall regard the numbers hi l ~ i ~ k / 2 ) for a while as fixed. Let usapply Lemma 5 to this equation; the numbers h;;(r;} play the roleof the unknowns X; the number 7:ik=2n25 plays the role of the num-ber l, and we set 2n =ko for brevity. Recall once more that the num-bers h; appear in equation 26) not only explicitly but also throughthe coefficients of the polynomials ; y). The complex A. to whichthe numbers X; =h;;(y) must belong consists, in the p_resent case,of all numbers of the form h;/y , where the numbers hi have given,fixed values and the numbers yi run through the interval .

According to Lemma 5, the number of solutions of equation 26)satisfying the requirements just described, does not exceed the sumof the numbers of solutions of the equation(21) yO ) + y 2 ) + +y 2s -1 )_y 2s -1 +1)_ - y 2 s )= Ounder the following 2 5 hypotheses which correspond to the values


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of the parameter w =0, 1, ... 25 -1:r< : =r1 + rP + .. + r ~ ~ . }r l leAwko+i ( 1 ~ _ i ~ k o )

57

where, remember, A, ( 1 ; ; ; _ r ~ 2 is the complex of numbers of theform h,if>,(r,.)with prescribed h and arbitrary y,, l r , I ~ 2 N / n .

For the case w O (which we choose merely as an example),equation (21) in expanded form looks as follows:

l r ~ > >+ + r ~ ; l+ l r P + r ~ P + ... r ~ ; l+ .

I 2s-1) (2s-1) (2s-1lj+ Y +Y2 + . +Yk0I 2s-1+l) (2s-1+ 1) (2s-1+1lj- Y +Y2 + +Yk0

or, rearranging the summands,

++ l (l) + (2)+ + (2s-1)- y 1 v1< 1l) if> 1 ( v P ~ + . +if>l(vf2s- 1l) -if>l(vP s- 1+1> - i f > v P+ h2lif>2) + ... -if>2 vPs l)j + .. + h , . 0 1 k 0 ( v k ~ l)+ ... ( v k ~ ' ~ ) 1 =0.

By putting, for brevity,


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58

this equation c n be written quite shortly as follows:(27)All together we have 25 equations of this sort, and their totality canbe written down in the compact form

o~ h ko+ z ko+ =0

i=1 w wFor the present, however, we shall confine our investigation to e-quation (27), which may of course be regarded as typical. To esti-mate the number of solutions of this equation which interest us, wemust first see within what limits the quantity ;(vi


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But every zi is the sum of 28 =c(n) summands of the form cpi(vli>),and therefore

(with another c(n), naturally). This means that in equation 27) everyzi can assume only the values lying in the interval ) + + . v.)'1-', ' '1-', ' f

This is possible because for k =k(n-1)> l (and we have seen thatalready k(l) =2) we have

2s -1= 2[4log2k ]-2>k .On detail: k ~ 2 , l o g 2 k ~ l , 3 l o g k ~ 3 , l o g k ] - 2 > 4 l o g k - 3 ~1 k , 2s - 1 _ 2[4log2k ]-2 k )og2 , -

I f we denote the right-hand side of the last equation by m weget(30) ,/,. (1)) ,/,. k >) - ,fl v + + f v m Let us choose some particular values for the numbers

vp ( k + l ~ j ~ 2in the interval < N 11 , +2N 11 >, naturally); then m also acquires


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60a definite value. To equation 30) we now apply the theorem to beproved, since /y is a polynomial of degree n l . We have to verify that all the necessary hypotheses are fulfilled. We have

ncp. y)= I a. ynu,u l 1. u

where, according to 28),u-1 n-1 u-131) Ia I< c n)N_n_=c n) N_n_fii=T,,,u

and, as is easily seen,n-1lm I< c n)N_n_

(because mand all ;


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61

isfy the equation) does not exceed the number 34).This result makes it possible to reduce the whole problem to an

estimation of the numbers of solutions of linear equations. For atthe end of 5 we reduced the estimation of rk(m) to the estimationof the numbers of solutions of equations of the form 26). But as weproved by an application of Lemma 5, the number of solutions ofequation 26), for which lril 2 N 1 1 is at most equal to the sum ofthe numbers of sol iitions of 2 5 equations of type 27), i. e., alreadylinear equations. In this connection we obtained limits within whichthe unknowns zi are allowed to vary. A certain new difficulty theprice we have had to pay for the transition to linear equations) isthat the new unknowns z have to be considered with certain multiplicities for which we have also determined limits).

Finally we must not forget that all these calculations are madeunder the assumption that the numbers hi are chosen and fixed.Therefore we still have to multiply the result obtained, by the num-ber of all such possible choices.The final result of this section, which we have to keep in mind,reads: Our estimated number rk(m) does not exceed the sum of thenumbers of solutions in integers zi, I il ~ c(n)N


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62ih l ~ . l N l ~ i ~ k / 2 ) , and by Uw K) the number of solutions ofequation (35) for this fixed combination K and for a certain prescribed w, where we are concerned with those solutions z whichsatisfy the inequalities lz I ~ c n)N


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63

sible combination of the values of all hi ( l i ~ k/2); the numberU(j(K), however, is completely determined by the values of the firstk0 = ln of these values 1 i 2n), because they alone appear in equation (36). Of course when w choose a certain fixed combinationK, we thereby also uniquely define a certain combination K of thevalues h1o h2, .. , h2n. But if, conversely, a certain combination K'of the numbers h1o h2.. . , h2n is selected, there corresponds to itnot the single combination K, but rather as many as there are waysof choosing the remaining supplements hi ( 2 n < i ~ k / 2 ) . Sinceevery hi must belong to the interval , it is evident that to a combination K there correspond at most

c(n) (N1/n)(k/2)-2n = c(n)N


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64weeks work with pencil and paper to understand and digest it com-pletely. It is by conquering difficulties of iust this sort, that themathematician grows and develops.

[3] khinchin a.y.-three pearls of number theory

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