Ernie Croot 1. Department of Mathematics, Georgia Institute of Technology, Atlanta, GA Abstract

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1 SUMS OF THE FORM 1/x k /x k n MODULO A PRIME Ernie Croot 1 Deartment of Mathematics, Georgia Institute of Technology, Atlanta, GA ecroot@math.gatech.edu Abstract Using a sum-roduct result due to Bourgain, Katz, and Tao, we show that for every 0 < ɛ 1, and every integer k 1, there exists an integer N = N(ɛ, k), such that for every rime and every residue class a (mod ), there exist ositive integers x 1,..., x N ɛ satisfying a (mod ). x k 1 x k N This extends a result of I. Sharlinski [5]. I. Introduction In the monograh [2], among the many questions asked by P. Erdős and R. L. Graham was the following: Is it true that for every 0 < ɛ 1 there exists a number N such that for every rime number, every residue class a (mod ) can be exressed as a 1/x /x N (mod ), where the x i s are ositive integers ɛ? This question was answered in the affirmative by I. Sharlinski [5] using a result due to A. A. Karatsuba [4] (actually, a simlified version of Karatsuba s result, due to J. Friedlander and H. Iwaniec [3]). A natural question that one can ask, and which Sharlinski recently osed to me, was whether this result can be extended to recirocal owers. Unfortunately, in this case, the methods of Karatsuba do not give a bound on N (at least not using an obvious modification of his argument). Fortunately, there is a owerful result due to J. Bourgain, N. Katz, and T. Tao [1] which can be used to bound certain exonential sums, and which can be used to solve our roblem: Theorem 1 (Bourgain,Katz,Tao) Let A be a subset of a finite field Z/Z. If δ < A < 1 δ for some δ > 0, then A + A + A A c A 1+θ, where θ = θ(δ) > 0 and c = c(δ) > 0. 1 Partially Suorted by an NSF grant.

2 Using this result, we rove the following theorem, which is just a restatement of the roblem osed by Sharlinski: Theorem 2 For every 0 < ɛ 1, and every integer k 1, there exists an integer N = N(ɛ, k) such that for every rime 2, and every integer 0 a 1, there exist integers x 1,..., x n such that 1 x i ɛ, and a 1 x k x k N (mod ). Comments. A more general theorem can erhas be roved here, as was suggested to me by I. Sharlinski in an . Basically, suose S = S() {1,..., 1} is an infinite sequence of sets, indexed by rimes satisfying the following two conditions 1. The sets S() are multilicative in the sense that 1 S, and if s, t S satisfy st 1, then st S; and, 2. There exists an absolute constant 0 < θ 1 so that for every 0 < ɛ 1, and sufficiently large, the set S() contains at least ɛθ elements ɛ. Then, there exists an integer J = J(ɛ) 1 such that for every sufficiently large rime, and for every residue class r (mod ), there exist integers x 1,..., x J S(), all of size at most ɛ, such that r (mod ), x 1 x J II. Proof of Theorem 2 First, we note that it suffices to rove the result only for sufficiently large rimes, as we may enlarge N = N(ɛ, k) as needed so that the theorem holds for all rime < o, for some 0. We also may assme 0 < ɛ < ɛ 0 (k), for any function ɛ 0 (k) that we might haen to need, since if the conclusion of the theorem holds for these smaller values of ɛ, then it holds for any larger value of ɛ. Let 0 < β < 1/5k be some arameter, to be chosen later, and let u be the largest integer less than β 1 /(2k), and consider the set { 1 S = } (mod ) : 2 k 1 < < u β, i rime, u k 1 2

3 which will be non-emty for sufficiently large. We claim that S > 1/(2k) β u! log u (1) for sufficiently large, which would follow from the rime number theorem if we had that all the sums in S were distinct modulo. To see that they are, suose that we had 1 k k u 1 q k q k u (mod ), where the left and right side of the congruence are elements of S, where the i s and q i are in increasing order. Multilying through by ( 1 u q 1 q u ) k on both sides and moving terms to one side of the congruence, we get that u (q 1 q u ) k j=1 k i ( 1 u ) k q k i 0 (mod ). (2) Since all the terms in the sum are smaller than (2u 1)kβ < /u (for sufficiently large), we deduce that if (2) holds, then u (q 1 q u ) k k i ( 1 u ) k = 0; j=1 and so, = k 1 k u q1 k qu k It is obvious then that the i = q i, and (1) now follows. q k i Let S 0 = S, and consider the sequence of subsets of Z/Z, which we denote by S 1, S 2,..., where S i+1 = { Si + S i, if S i + S i > S i S i ; and S i S i, if S i + S i S i S i. We continue constructing this sequence until we reach the set S n satisfying S n > 2/3. Using Theorem 1 we can roduce a non-trivial uer bound on the size of n for β < 1/5k: Let δ = 1/4k, and let c = c(δ), θ = θ(δ) be as in Theorem 1. Then, for sufficiently large, we will have δ < S 0 = S < 1 δ, and the same inequality will hold for S 1, S 2,..., S n. Now, alying Theorem 1, we deduce that S i+1 > c S i 1+θ ; 3

4 and so, S 1 > c S 0 1+θ, and for j = 2,..., n, S j > c 1+(1+θ)j 1 S 0 (1+θ)j. From this inequality, we deduce that ( ) 1 2k β (1 + θ) n + o(1) > 2 3 ; so, since β < 1/5k, we need take n no larger than log(3k)/ log(1 + θ) + o(1) in order for S n > 2/3. Now, every element of S 0 is a sum at at most u terms; each element of S 1 is a sum of at most u 2 terms; and, by an induction argument, each element of S n is a sum of at most u n+1 terms. Also, each element of S n is a sum of terms of the form 1/q1 k qk 2n, where q 1 q n+1 2nβ. Now, let β = ɛ/2 n+1. If ɛ < 1/5k, then this value of β < 1/5k (recall we said that ɛ is allowed to be bounded from above by a function of k). Let h be the smallest integer satifying h > log(3k) log(1 + θ), and define { 1 T = } (mod ) : 2 q q1 k qh k 1,..., q h ɛ/2 Here, the q i s are not restricted to being rime numbers. Since T S n, we have that T > 2/3 for sufficiently large. Now we use the following lemma: Lemma 1 Suose that T Z/Z satisfies T > 1/2+β. Then, every residue class modulo contains an integer of the form x x J, where the x i s are all of the form t 1 t 2, where t 1, t 2 T, and where J = 2(1 + 2β)/β + 1. Proof of the Lemma. First, we consider the exonential sums h(a) = ( ) at e, t T and f(a) = t 1,t 2 T e ( ) at1 t 2. 4

5 We have from Parseval s identity and the Cauchy-Schwarz inequality that for a 0 (mod ), f(a) ( ) at1 t 2 e t 1 T t 2 T ( ) 1/2 ( ) 1/2 1 h(at 2 ) 2 t 1 T t 2 T = 1/2 T f(0) (1+β)/(1+2β). Now, if we let J be the least integer greater than ( β ) 1 = 1 + 2β 2(1 + 2β), β then we have that for a 0 (mod ), f(a) J < f(0) J(1+β)/(1+2β) < f(0) J f(0) Jβ/(1+2β) Thus, given an integer r, the number f(0) J 2 f(0) J. #(x 1,..., x J : x i = t 1 t 2 ; t 1, t 2 T ; and x x J r (mod )) = 1 1 a=0 f(0) This roves the lemma. f(a) J e( ar/) f(0) ( 1)f(0) 2 > a 1 f(a) J From this lemma, we deduce that for every residue class r modulo, there exist integers t 1,..., t 16, such that r t 1 t 2 + t 3 t t 15 t 16 (mod ), where t 1,..., t 16 T. This sum can be exressed as a sum of at most 16h 2 terms of the form 1/(qq ) k, where q, q < ɛ/2. This then roves the theorem, since h deends only on k and ɛ. Acknowledgements I would like to thank Igor Sharlinski for the interesting roblem, which is the one solved by Theorem 2. 5

6 References [1] J. Bourgain, N. Katz, and T. Tao, A Sum-Product Estimate in Finite Fields, and Alications, Prerint on the Arxives. [2] P. Erdős and R. L. Graham, Old and New Problems and Results in Combinatorial Number Theory, Univ. Genève, Geneva, [3] J. Friedlander and H. Iwaniec, Analytic Number Theory (Kyoto, 1996), Cambridge University Press, Cambridge, [4] A. A. Karatsuba, Fractional Parts of Functions of a Secial Form, Izv. Ross. Akad. Nauk Ser. Mat. 59 (1995), [5] I. Sharlinski, On a Question of Erdős and Graham, Arch. Math. (Basel) 78 (2002),