Three Pearls of Number Theory

OTHER

CIlAYWCK

PUBLICATIONS

PONTRYAGIN:

Foundation., of COllbinator i a I To polo g y

NO\OZHILOV:

Founda t ions of the \,." I ineor Th'ory of Elastici Iy

ALEKSA'lIOROV:

Combtnatorial Va I. 1

Topology,

TlIH

E:E

PEAHLS

N U \1 U E H

o F

T /I E 0 H Y

BY

A. Y. KIIINCIIIN

G RAY L 0 C K

PRE S S

ROCHESTEH., N. Y.

1952

OTHER

GR4YWCK

PUBLI CA nONS

PONTRYAGIN:

Foundattons of Co.~ina­ tor i a I To polo g y

NOV07.IIILu\':

F 0 u n d a t ion s 0 f t I,· v"" I t n e ar Theory of Elasticity

ALEKSANDROV:

Combtnatorial Va 1. 1

Topology,

T H R E E

N U

~I

PEARLS

TilE 0

BElt

o

n y

BY

A. ). KIIINCHIN

GRAYLOCK

PRE S S

ROCHESTEH, N. Y. 1952

F

TRANSLATED FROM

Tin:

SECOND (1948), REVISED RUSSIAN EDITION

BY F. [HGEMIRL

H.

KLlM~1

W. SEIDEL

Copyright, 1952, by

GHA Y LOCK PHESS Rochester, N. Y.

Second Print in g, 1956 All rights reserved. This book. or parts thereof, may nol be reproduced in any form, or IrBnslet .. d, without permission in "riting from th e pu blish ers.

Manufactured in the United States of America

FOREWOHD

Thi~ little hook is devoted to three lheorems in arithmetic, y,laich, in spite of their apparent simplicity, have been the ob.iects of the efforts of many important mathematical scholars. The proofs \~hicl, ilre presented here make use of completely elementary means, (although lhey are not very simple).

The book can be unden,lood by beginning college students, and is intended for wide circle.,; of lovers of mathematics.

CON T

A

LETTJo:II

CIIAPT[':R

TO I.

TilE: VAN

FnO~T

!)ER

CHAPTER Ill.

NT S

(IN

LIEU

I\.\ERDE:-f'S

A"ITHMt~TIC

CII.\PT.:R II.

j;\2' ... , \/,!\/' then form an arithmetic progJ.'ession, of length l'=l+] and difference dlo of sewnents of length qk-l' where A 1 ':\2, ... ,I\l are of the same type. \\'e know nothing about the type of the last segment A l " This completes the first step of our construction. [t would be well if you thought it through once more before we continued. ~5

We now proceed to the second step. We tuke an arbitrary one of the first l terms of the progression of segments just constructed. Let this term be \1' so that l::;;i1::;;l; "i1 is a segment of length qk-l' We treat it the same way as we treated the segment _1. Since qk-l =2n k _2Qk_2' the left half of the segment :~i1 can be regarded as a sequence of n k _2 subsegments of length Qk-2' Forsubsegments of this length there are k qk-2 types possible, and on the other hand nk_2=n(k Qk - 2 ,l) because of (]). Therefore the left half of -\1 must contain a progression of l of these subsegments of the same type, Ai1i2 (I::;;i2~l), of length qk.2' Let d 2 be the difference of this progression 0. c., the distance between the initial numbers of two neighboring segments). To this progression of segments we add the (l+ 1)-st term l1i 1 l ', about whose type, of course, we know nothing. The segment l1i 1 l ' does not have to belong to the left halI of the

segment

tJ.. •

"1

15 any more, but must obviously belong to the segment

'1

\\Ie no .... carry over our construction, which we have executed up to now in only one of the segments .'\ l' congruently to all the other segments ,1\'i l O;;;il;;;l1. \\e thus obtain a set of segments \li2 O;;;il~l~ l;;;i 2 ;;;lj with two indices. It is clear that lwo arbitrary segments of this set with indices not exceeding l arc of the same type:

You no doubt see now that this process can be cont inued. We carry it out k times. The ('esults of our construction after the first slep were segments of length qk-l' after the second step, segments of length qk .. '1.' elc. After the k-th step, therefore, the results of the construction are segments of length go= I, i. e., simply numbers of our original segment Ll. Nevertheless we denote them as before by

1\.. " '1 2'" k

,~..

(2)

(I ;;;il' i 2 ,

. '" /'t;. . '. ,

- '1'2""s

....

ik;;;l').

.,

'1'2""s

We now make two remarks which are important for what follows. l) In (2), if s 1 (i. e., the sequence (A) does nnL contmn unity), then dCA) =0. 2. If an=l+r(n-I) (i.e .. the sequence ('0, beginning with ai' is an arithmetic progression with initial term I and difference r), then d(A)=l/r. 3. The densi ty of every geometric progression is equul to zero • .t.. The density of the sequence of perfect squares is equal to zero. 5. For the seqnenct' (/1) to contain th" {'nt ire "equellec of natural numbers (an =n, Tl = 1. 2 .... ), it is necessary and sufficient that d(:!)=l. 6. J[ aI::I) =0 and ;1 contains the number I, and if (>0 is arbitrary, then there exists a sulficiently large number m such that A(m)<wl. If you have proved all this, you are familiar I'nough with the cflflcepL of density to be abl", to use it. ]\0\\ r want to acquaint you with the pruof of the following remarkable, albeit very simple, lemma of Schnirelrnann:

d( 1+ n) ~ d("!) + d( B) -d(.,f) dun. The meanmg of this inp.qualily is clear: the den~ity of the sum of 1\\0 arbitf'dfY sequences of numbers is not smaller than the slim of their densi ties diminished by the product of these densities. This "Schnirclmann inequality" represents the first tool, still crude to be SlIre, for estimating the density of a sum from tllc densities of the summands. lIere is its proof. We nennte by ,l(TI) the !lumber of natural numbers ",hich arrear in the sequence ,I dnd do not exceed n, and by B(n) the analogous number for the seljuence B. For brevity we set dCI)=a, d(B)=/3. A+B=C, d(C)=y. The segment n,n) of

23 the sequence of natural numbers contains A(n) numbers of the sequence A, each of \\hich also appears in the sequence C. I.et a k and a k + 1 be two consecutive numbers of this group. Ilet\\et"n them there are a k + 1 -ak -] =l numbers which do not belong to A. These are the numbers

Some of them appear in C, e. g., all numbers of the form a,,+r, where r occurs in B (which \\1' abbreviate as follo\\s: r eR). There arc as many numbers of this last kind, however, as there arc numbers of B in th(' segment (I, 0, that is, B(l) of them. Consequently every segment of length I included between two consecutive numbers of the sequence -1 contains at least B(l) numbers which belong to C. It follows that the number, C(n), of numbers of the segment appearing in C is at leust

n, n}

A(T!) +I n(l)

where the summation IS extended over all segments \\hich are free of the numbers appearing in A. According to the definition of density, however, B(l)~{3I, so that C(71) ~ ·f(n) + PIl == A(n) + f3111 -A (71)1,

because "il is the sum of the lengths of all the segments which are free of the nllmbers appearing in 4, which is simply the number n-A(n) of numbt"rs of the segment (I, n) which do not occur in 1. But A(n) ~an, and hence

C(n) ~A (n)(I -(3) + (3n ~an(l-fJ) + {3n, which yields

C(n)jn ~ a+{3-a{3. Since this inequality holds for an arbitrary natural Dumber n, \\e have

y=d( (;) ~a +{3 -a{3. Schnirelmann's inequality (1) can be written

Q. E. D.

In

the equivalent

24

fonn I-d(A +Rhll-d(A)l1

-d(m!,

j

and in this form can easily be generalized to the case of an arbitrary number of summdnds: k

l-d(A 1 +A 2 + ... +A..):>; ft

-

n fl-d(.1.)I. i==l '

It is proved by a simple induction; you should have no trouble in carrying it out yourself. If we write the last inequality in the fonn k

d(A 1 +A 2 +···+A k );;;1-

(2)

}J 1Il-d(" i )l,

it again enables one to estimate the density of a sum from the densities of the summands. G. Schnirelmann derived a series of very remarkable results frolll his elementary inequality, and obtained above all the following important theorem:

r..

Every sequence of positive density is a of natural numbers.

baSIS

of the sequence

In other words, if a==d(A»O, then the sum of a sufficiently large number of sequences A contains the entire sequence of natural numbers. The proof of this theorem is so simple that I should like to tell you about it, even though this will divert us a bit from our immediate problem. Let us denote for brevity by A k the sum of k sequences, each of v.hich coincides with A. Then by virtue of inequality (2), d(·1 k);;; l-O-a)k. Since a>O, we have, for sufficiently large k, (3)

Now one can easily show that the sequence A2k contains the ",hole sequence of natural numbers. This IS a simple consequence of the following general proposition. LEM!'1:\.

If A(n)+B(rt»n-l.

then

n occurs

in A+B.

Indeed, if n appears in A or in 8, everything is proved. We may

25 therefore assume that n occurs in neither :f nor [;. Then A(n) = f( n -0 and B(n)= B(n·- 1), and consequently

l1(n- 0+8(n-]»,,-1.

°

Now let a p a 2 , •.• , a r and 1 ,1>2' ... , Os be the numbers tlf tLe segment (l,n-l) which appear in ·1 and H, respectivdy, so that r~ 1(n-l). s=Rlrl-l). Then all the numbers a l , a z'

... , ar'

n-b l , n-b 2 ,

... ,

n-b s

1(71 - 1) + R(n -1) of these IIl1mlJf'I"S, which is more than n-1. IIpncc one of the numbers III Ihl' lJl'pl'r ro\\ equals one of thc numbers in the lower ro\\. r .el '1,. -1/ -lJ k • Thl'n n=Cli+b k • i. e., rt appears in tI +B. HClllrning now to our objccti~e, w(' have, all the basis of (3), ror an arhitrary n:

(wifing t(l the l>ef,'ffient (1, n -1), There arc r+ s

=

ilnd tllf'refore

,\ccording to the lemma just proved, it follows Ihal n appears in

I k +A k =:l 2k • Hut n is an arbitr.u·y natural number. and hence our theorem is proved. This simple theorem led to a series of important applicdtions in the papers of [ .. C. Schnirellllann. For example, he was Ihe first to prove that the sequence P consisting of unity and all the prime numbers is a basis of the sequence of natural numbers. The sequence P, it is true, has densi ty zero, as Euler had already shown, so that the theorem which we just proved is not directly applicable to it. Uut Schnirelrnann \\fJS able to prove that P +1' has positive density. lIence p+p forms a basis, and therefore I' indeed also. From this it is easy to infer that an arhitrary natllrdl numher, with the except ion of J, can, for sufriciently large k, be represented as the sum of fit most Ir primcs. For thaI lime (1.930) this result was fundamental and evoked the greatrst intC'rest in the scientific \\orld, At present, thanks to the remarkable \Iork of I. \1. \"inograd-

26 ov, we know considerably more in this (Iireclilln. us I already related to you at the beginning of this chapt er.

In the preceding it was my purpose to introdllce )011 in the shortest way possible to the problems of this singular ana fascin v.ith l.andau III COILingen, ano related among other things that in the course of these conversations they had discoHre(1 the following interesting fact: In all the concrete pXHlllple!o> that they were ahle to devise, it was possible to replace the inequality d(A +R)~d(A)+d(H)-d( t)d(/n,

which we derived in *2, by the sharper (.rnd simplPr) int·quality

That is, thl' density of thc sum alvyays turueo flul to be at least us large as the sum of the densities of the sllmrnanrls {under Ih£> assllmpt ion, of course, that d(,.1) +d(B) ~ I). Thcy thereforp naturally assumed that inequality ('1) was the expression or a universal law, bllt the first atlempt& to pruvc this conjecturc \\pre Ilnsllcces~ful. It soon became t'vid"nt that if their c()nJecture WiIS correct, the road to its proof would bc quitc dilTiculL. \\e "ish to note at this point that if the hypothetical inequality (4) does represPllt " univprsal law, then this law can be gencralizeo immcdiately by induction to thc cuse of an arbitrary numbcr of summands: i. ('., IIn(ler the assumption that k

lJ(n~l

i=1

,-

Voe have k

(5)

k

d( ~ Ai)~ 2 d (-Ii). i=1

i=1

27 This problcm coule! not help hut attract the attention of scholars, bN'ause of the simplicity and elegance of the p;eneral hypothetical IlIw 0) on the one hand, and nn the other becallse of the sharp conIrllst bet¥.een the elementary character of the problem and the diflieulty of its sollltion which became apparent already after the first i1l1acks, I myself was fascinated by it at the time, and neglected 1111 my other researehcs on its uccount. ~:arly in 19:12, aILer several IIIllnths of hard ¥'ork, l succeeded in proving inequality (4) for the luost important special case, d(A) = dW) (this case must be considNetl as the m()st important becalls!' in th". majority of concrete proLlpllIs all the summands are the same). At the same time l also I'rnvcJ the ~cneral inequality (5) under the assumption that d(A 1)= d(:l z )= ... =d(A,J (it is eus)' to see that this result cannot bederivI'c! from the preceding onc simply by induction, but requires a spel'ial proof). The method which I used was completcly elementary, but v('ry complicatNl. I "US late-r able to simplify the proof somewhat. Be that as it may. it was bllt a special case. For a long time it scemed to me that a 1I0n(' too subtle improvemcnt of Illy method should Ipac! tll a full solution of the problem, but all my efforts in this direction prov('d fruitless. ! In the meantime the publication of my work bad all~'arted the utt~lItion of a wide circle of scholars in all countries to Ihi! I andallSchnirclmann hypothesis. \Iany insignificant results were obtained, and a whole litl.'rature spran~ lip. Some authors carried over the problem from the domain of natural numbers to other fields. In short, the problem became "fashionable". r,earned societies olTered prizes for its solution. My friends in Englanl"! wrote me in 1935 that a ~ood half of the English mathematicians had postponed their usual work in order to try to solve this problem. Landau, in his tract devoted to the latest .ldvance as an element of the set B*, is of the form f3o+n-c: c'¢e) WI' ohtain

Therefore ·Here and in the sequel we use the symbol are using the sytl1bol + in another sense.

Uto

denote the union of seLs, since ....e

32 c+c'-n=a+{3oEA + n = L. where c set

¢C

and c' ¢

r:.

Hut thpn ,Icrordi II).! tu the dcfini tion of the

C"'. 0.1' .. II .

Thus wp ha,,!' sh,mn that 1 -t 8

1 ( ( : 1.

Tu prove the imerse relation. let liS assllme thut r E('l, "llIdl ntc,m!-l thut pithPr c EC or c EC*. If c EC, tiwn c=u+b, a 1"

eli l ' If.

1. bEN

ho\~e\'er, c EC1, then, for a certain a E,I, the nllTliLer

G-a. "'" \\e knnw, "l:cur!-l in U*. \\c have c=a+b* E I +/{* - t Th('refurc C 1 C.-I + U l' \\ e al so pruved above that A + H 1 :-.e'jlJently (;1 = ., + 8 1 , No\\ r("call th'lt according to 01lr .Jssumrtion,

7/

¢c. It

r: l' i:-.

b*=

,-n 1 • l. "n-

I'H0 and a suitahly chosen ,V,

Here the number ,\' may be assumed Lo be arbitrarily large, because A~k) (for an arbitrary k) contains the integer 1 (bear in mind Problem 6 on p.22, which you solved). Applying the estimate (2) we get n

Rk(\)=

~ rk(m)=rk(O)+ m=O

N

~ rk(m)0. n But, as \\e alrf'ady kno\\. this proves lIilbert's theorem. 'ou see how simply it all cornes t~ut. But we still have to prove the fundamental lemma, and to du this we shall have to travel a long and din'jeult road, as in the preceding chapter.

§3 LEMMAS

CONCERNDIG

LlJliEAR

EQI,ATIO""S

\\e shall have to go far back. It will therefore be well for you to forget completely for a while the problem which has been posf'd. r shall call your attention to it when we return Lo It later. !light now, ho\',ever, \',e have to find some estimatt"s fo/' the number of solutions of systems of linear equations, The lemmas of this paragraph, moreover, are perhaps also of intrin"ic interest. independent of the problem for" hose sol ntion they are required here.

LE \1 \J:\ 1. In the C'7ua1i on

42

(5) let ai' a2, m be integers with la21 ~lall ~A, and let al and a2 be relatil'ely prime. Then the number of solutions of eqlwtioTl (5) satisfying the inequalities IZ11~A, Iz:;>I~A, does not exceed 3.t/la11. Proof: We may assume that a i >O, because otherwise we huve merely to replace Zl by - Z l in every solution. Let IZ1> z 2 1 ane! 1z{, z;1 be two different solutions of equation (5). Then from alz1

+a2z2=m,

atd -l-a:,zb =m we get

by subtraction. I\ccordill/?:ly the left-hand side of this equation must be divisible by ai' But* (ai' a2)= 1, and consequently Z;'-Z2 mllst be divisible by a1' 'Jow Z~';'Z2' and therefore IZ;-Z21, as a multiple of al, is not smaller than a1. Thus, for two distinct solutions Iz 1, z) and lz~, z;1 of equation (5), v.e must have IZ;'-z21~a1' In every solution lz l' zd of equation (5), let us agree 10 call Z1 the first member tim! Z2 the sec(lnd. [t is obvious thai the llumber of solutions [)C equation (S) which satisfy the conditions IZ11 ~A, IZ21 ~A. is not more than the numLer t of second members which occur in the interval . Since we have proved that two such second members are at least the distance al apart, the difference between the largest and smal\t'st second members occurring in the interval is at least a 1 (t- 1). On the other hand, this difference does not exceed 2.1, so that a1(t-l):?;2·/'

(t-l) ~2A/a1' t~(24/ai)+ 1~~4/a1

(hecause, by assumption, a 1~A. and therefore 1 ~A/(t1)' This proves Lemma 1.

LE MM A 2. In the equation (6) • (al. a:;»

denotes the greatest Common divisor of the integers a1 Bnd a:;>.

43

let the a, and m be integers satisfying tlte conditions'"

Then thp number of solutions of equation (6) satisf""ing the i"pqualities IZil~A (l~l~l). does not exceed

ell) 11- 1 /11, u,here II is the largest of the numbers la11.la21 • •.. , a constant depending only on l.

lall,

and c(l) is

Proof: If l=2, I.emma 2 obviously becomes Lemma 1 (with c(2)= 3). Accordingly Lemllla.2 is alrcully vt>rifierl for l=2_ We shull therefore assume that 1~3 and that the truth of Lemma 2 has already been established for the case of l-I unknowns. Since the numbering is nnimportant, we may assume that lall is the largest of the numbers la1i, IU21 • ... , la,l, i. e., fI = lall. There arE' 1\"'0 cases to com,i.\cr. I) Ul =a?=... =a[.l=O.Since (a1>(l;?1 ... ,tl/)=I, ""t' hdve la l l=lI= 1, sn that the ~iv('n eqnation is of the form ±;:l=m. In this equation euch of the unknowns ;:l';:, .... ,z/_l can obviollsly assume an arbitrru'Y illte~ral valuc ill the interval . and hencc at 1II0st 2:J +] ~3A values all toln. As for zl' ho""ever, it can assume at 1II0st one value. Conseqllentl) the nwnLer of solutions of the given equation satisfying the ine'lllUlities IZil~4 (1~i~I), rlOI'S nnl exceed

which proves Lemma 2 for this case. 2) If at least one of the numbers zero, then

a1, a 2 , ... , a l _ l

is dilTerent from

(U\.CL 2 , .. ··a l _1 )=o

exists. Let us denote hy II' the lar~est of the number~

Suppose now that the numhers ;:1, Z2> ... , zl salisfy the p;iven equation (6) and the inequalities IZil~A (1~i~l). \\'e set *(a1'~'

•••• al) denotes tht'

grpatest

common 2). Here is a simple proof: For n~ 1 we have n- q - {n + l)-q =I(n+ l)q -nql/ nq(n+ l)q =(nq+qnq-l+ ••• +l-nq)!nq(n+ 1)q ~qnq-l/nq(n+ l)q > q/(n+ l)q+l, and hence

By substituting successively n=1,2, ..• ,-1-1 in this inequality and adding all the resulting inequalities together we find that A

I n-(q+l) < q-l(l-A-q) < 1/ q.

n=2 which implies that A

I n-(q+l)< 1 +O/q)={q+ l)/q,

Q.E.O.]

n=l

Comparing this with the result in 1°, where we ohtained an esti· mate for the case a1=a:;= ... =a l ",O, we reach the following conclusion:

LE MMA 3. Let l> 2 (OLd l;;;;A;;;; B;;;; c(l)A l-l. Then the sum of the numbers of solutions IZ i I; ; B (l~i;;;;l) of all equations of the form (9)

where lail;;;;A (l;;;;i~l), does not exceed c(l){AB)l-l.

§4 TWO MORE LEMMAS Before proceeding to prove the fundamental lemma, we have to derive two more lemmas of a special type. They are both very simple, in idea as well as in form, and yet their assimilation might cause you some difficulty because they are concerned with the enumeration of all possible combinations, whose construction is rather invol ved. The difficulty with such an abstract combinatorial problem is that it is hard to put it in mathematical symbols: one has to express more in

48

words than in signs. This is of course a difficulty of presentation, however, and not of the subject itself, I shall take pains to outline all questions that arise, and their solution, as concretely as possible. \\ e shall denote by .4 a finite complex (i. e., collection) of numbers, not all of which are necessarily distinct. [f the number a occurs A times in lhe complex A, we shall say that its multiplicity is A. Let a 1, a?, ... , aT be the distinct n umbers which appear in A, and let '\l,A" ..... Ar be their respective multiplicities (because the com-

ann

T

plex A contains all together .~\ numbers). I.et B be another com-

,=J

pie" of the same type, which consists of the distinct numbprs b 1 b',2, ... , b s with the respective mlJltiplicities fl1,fl", ... , fl s ' Let us investigate the equation (12)

,

x+y=c,

where c is a given number and x and y are unknowns. We are interested in such solutions lx, yl of this equation in which x is one of the numbers of the complex A (abbreviated x 10 A) and y is one of the numbers of the complex B (yEB). If the numbers x=a i and y=b k sati sfy equat ion (12), this yields \IL k sol lit ions of the required kind, because anyone of lhe >-; "specimens" of the number ai' which occur in the complex A, can be combined with an arbilrary one of the flk specimens of the number bk appearing in the complex B. Hut we have* Aiflk ~~(At+flV, Therefore the number of such solutions of quation (12), where x = ai' Y = bk' is not greater than ~~(,\} + flV, ) t follows that the number of all solutions xEA, yEll of equation (12) is nol more than the sum ~~(A7 + flV, J1l're the summation is over all pairs of indices Ii, kl for whi ch a i + bk = c. Our sum is enlarged if we over all i and fl~ over all k (because every bk can be comsum bined with at most one a i .) It finally follows, therefore, that the num-

'\t

ber of solutions x 10/1, y EB of equation (12) does not exceed the number r

1'( /2

S

~\2 .... fl..

2)• + " ~ iJ-, i=l' k=l'

"'''The geometric mean is not greater than the arithmetic

mean'~.

plest proof:

O;i,(\-/1k)2=A;+fl~-2\ILk'

and hence

2'\h~At+IL~.

Here is the sim-

49

On the other hanel, let us consider the equation (13)

x-y=O

and calculate the !lumber of its soilltions x fA, Y fA. Clearly every such solution is uf the form x=y=a i (] ~i.

Let m b(' nne of these numbers. The equation Z i =m can Le salin general not only in one hut in several way!'!, hecause the definition of the number zi (p.58) is such that one and the SdIne value of zi can very ..... ell result from different choices of the nllmhers vii> (1 ~j~2$). \\e now have to estimate the number of solutions of the relation zi=m, i. e., of the equation i~lied

"'.( v.(l » + •.• + ¢.( v( 2 5 -I» _'"

(29)

'PI.

,

r.

.(v.(2 5-1+ 1)

'PI.

I.

r.

_ ... _¢( v(2 5 » t

t

=

m.

For this purpose .... e shall finally have to apply the long-promised induction. \\e proceeo as follows. First we rewrite equation (29) in the form

=

m-'"'-Pt,.(v.r.( k '+1)

_ ••• +"'.( v.{, 2 'f.lt

5-1

This is possihle because f,)f Ie' = already k(J) =2) we have

($)

+1) + ••. + "'.( '1-'1. v.I. 2

ken -1) > I

2 s - 1 -_2[t !og2k']-2 >,1:"

) •

(und we have seen that

•

(In detail: k'"?,2, log2!c'"?,1, :nog'2k'"?,3, [1Iog 2 k']-2>11og 2 k'-3"?, /") k ' 2s_I_2l41og2k']-2 I Og2, "?, I: • If we denote the right-hand side of the last equation by m', ,,'e

get (30)

I,et

liS

choose some particulur values for the numbers

t,P)

(k'+ 1 ~j~2s)

(in the interval . naturally); then ",'also acquires

60 a definite value. To equation (30) we now apply the theorem to he proved, since ¢j(Y) is a polynomial of degree n-1. \\e have to verify that all the necessary hypotheses are fulfilled. \\e have n

¢.(y)= ,

la. yn-u, u=l I. u.

\\ here, aecordi n/1; to (28), u-1

(31)

n-1

u-1

la.'0 u I < c(n)l't-n-=dn)(I\-n-)Ii"=T,

and, as is easily seen, n-1

1m ' 1< c(n)!''' --;r(because m and all chj(yF» satisfy this inequality). In virtue of the last inequality, the role of lv' can be assumed by the number c(n) N(n-1 lin; then the conditions (30, whi cli the coefficients of the polynomial ¢/y) satisfy, are precisely the conditions (2.1) wi tb n rep lacE"d by n - 1. Thus all the hypotheses are indeed fulfilled, and we can assert that the number of solutions of equation (30). for ¥.-hich lui{j)1 ~Z,\'1/n=2(N(n-1lIn)1/(n-l~ does not exceed the number

l!.:!

(32)

L _1

c(n.}{ JV n )n -1

= c(n} tI,'

I.. '.n+l

n

•

.

This estimate IS obtained for the fixed values u.uc;l\lt (lbtollllt,c1. b} Ihe

1111111-

uer nf all 'illl h I'o". ... ibl.· dllll ce,... The' final rt'!-olllt of Ihlt' ,.,e( lIon, "hich \\e ha\ c 1(. kpep in rllind. re where U~(I\. ") is the number of solutions ill Integers zi' Izil'$.d1£)' N(n-llln n~i~2n) of equation (36) for the given combination 1\ ' of the numbers hi' Ihil~2 \lln (l~i~2n), and the summation is to bl:' extended over all such combinations. From (:37) \~e therefore obtain* (38) r. (m)~c(n)N2(2s-n+l)i\(k/2n)-2 ~ li8'