Uniform bounds for the number of powers in arithmetic progressions

We give sharp, in some sense uniform bounds for the number of ℓ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\ell $$\end{document}-th powers and arbitrary powers among the first N terms of an arithmetic progression, for N large enough.


Introduction
The problem of giving (sharp) upper bounds for the number of powers among N consecutive terms of an arithmetic progression is a classical one with many deep results and open problems and conjectures. Here we only give a brief introduction; for a more precise account of the topic the interested reader may consult e.g. the papers [1] and [6].
Let a, b, be integers with a > 0 and let ≥ 2. Write P a,b;N ( ) for the number of -th powers among the first N terms of the arithmetic progression ax + b (x ≥ 0). Denote by P N ( ) the maximum of these values taken over all arithmetic progressions ax + b. (Note that this maximum obviously exists.) The case of squares (i.e, = 2) has been studied by many authors. Erdős [3] conjectured and Szemerédi [10] proved that P N (2) = o(N ). Later, by deep tools (such as e.g. elliptic and higher genus curves, Falting's theorem, the distribution of primes etc.) Bombieri, Granville and Pintz [1] proved P N (2) < O(N 2/3+o(1) ), which subsequently was improved to P N (2) < O(N 3/5+o(1) ) by Bombieri and Zannier [2]. See also Granville [5] for related results and remarks. A strong conjecture of Rudin (see [9], end of paragraph 4.6) predicts that P N (2) = O( √ N ), or in an even more precise form, that should hold. In case ≥ 3 there is hardly anything known. The authors of [1] noted (without proof) that their methods probably make it possible to prove P N (3) N 3/5+ε and P N ( ) N 1/2+ε ( ≥ 4). Hajdu and Tengely [6] showed that (up to equivalence) for any ≥ 2 there is a unique arithmetic progression ax + b which contains the most -th powers asymptotically, that is, which maximizes the expression (In fact, for = 4 there are two such progressions.) They could describe these arithmetic progressions a x + b explicitly. Based upon their results, they extended Rudin's conjecture (1) for any ≥ 2 (by replacing 24x + 1 by a x + b and changing the right hand side accordingly), and proved that for = 3, 4 for certain small values of N . Note that this asymptotic ('global') version of the problem is simpler than the original 'local' one, namely when we concentrate on a finite part of the progressions. The reason is that the asymptotic approach brings in an 'averaging' effect, which roughly speaking makes it possible to concentrate on a complete (finite) period of a progression ax + b modulo a.
In this note we prove that for any positive ε there is an 0 depending only on ε such that for > 0 the number of -th powers among the first N terms of any integral arithmetic progression is below (1 + ε) √ N , provided that N is large enough in terms of ε, and the parameters of the progression. The important feature of 0 is that it is uniform in the sense that it depends only on ε, it is independent of the progression. This result is sharp in the sense that for infinitely many , one can find a constant c 1 = c 1 ( ) > 1 and an arithmetic progression having more than c 1 √ N -th powers among its first N terms, for all N large enough. We also give a uniform, sharp upper bound for the number of powers (with not fixed exponents) among the first N terms of arithmetic progressions. In our proofs we combine a classical result of Wigert [11] concerning the number of divisors of positive integers, a recent result of Hajdu and Tengely [6] concerning arithmetic progressions containing the most -th powers asymptotically, and a new assertion answering a question of Hajdu and Tengely from [6].

New results
Now we give our main results. We use the notation from the introduction.
Remarks. The above theorem is sharp in the sense that 1 + ε cannot be replaced by 1, and > 0 is also necessary. Indeed, Theorem 1 of [6] (see also the Remarks after it) implies that for infinitely many exponents ≥ 2 there exists a δ > 0 and an arithmetic progression It is clear that if an arithmetic progression ax + b contains an -th power then it contains infinitely many, and we have We also mention that on our way to prove Theorem 2.1, we answer a question of Hajdu and Tengely [6] (see Proposition 3.1).
We also give a uniform upper bound for the number of powers in arithmetic progressions. For this, let P a,b;N ( * ) denote the number of (arbitrary) powers among the first N terms of the arithmetic progression ax + b (x ≥ 0). Theorem 2.2 Let ax + b (x ≥ 0) be an arithmetic progression. Then for any ε > 0 there exists an N 0 such that Remark One can easily check (see also e.g. Theorem 1 of [6]) that This shows that the above result is sharp.
Further, it is also easy to see that if gcd(a, b) = 1 then there exist infinitely many exponents such that ax + b contains -th powers. Note that here the condition gcd(a, b) = 1 cannot be dropped: for example, the arithmetic progression 4x + 2 (x ≥ 0) contains no powers at all.

Proofs
To prove Theorem 2.1 we shall need some known and new assertions. The next lemma is a result of Hajdu and Tengely [6]. For its formulation, we need to introduce some new notions and notation (which will play important roles also later on). For any ≥ 2 and arithmetic progression ax + b put  Proof The statement is the first half of Theorem 1 of [6]; see also the notation in its proof on p. 970 of [6].
Remark The inequality S a,b ( ) ≤ S( ) is sharp: for any , by an appropriate choice of ax + b (given in [6]) we get equality. Observe that for odd, we have S( ) = 1.
In the proofs of Theorems 2.1 and 2.2 we shall need the following new assertion. This answers a question of Hajdu and Tengely concerning the limit of the sequence S( ) (see the 'concrete question' on p. 966 in the Remarks after Theorem 1 in [6]), and we find it of possible independent interest. Proposition 3.1 By the notation of Lemma 3.1, for any γ > 0 there exists an 1 = 1 (γ ) depending only on γ such that for > 1 we have (3) In particular, lim →∞ S( ) = 1 holds.

Remark One can easily check that (3) implies that
for large enough.
To prove the above statement, we need the next classical theorem concerning the number of divisors d(n) of a positive integer n.
Proof This is a classical result of Wigert [11]. Note that in [7], p. 56 a stronger form of this assertion is given, however, the above inequality is sufficient for our present purposes.

Proof of Proposition 3.1
As one can easily check by a direct calculation, the function (t − 1)t 1/ −1 is strictly monotone increasing for t > 0, for any fixed ≥ 3. Thus, as for ≥ 3 the product appearing in S( ) has at most d( ) terms and in every term p ≤ + 1 holds, we have Here we also used that by the condition log p log p−log( p−1) > , the terms appearing in S( ) are greater than 1. Now by Lemma 3.2 we get that hold, for any γ > 0 with > 1 , where 1 = 1 (γ ) depends only on γ . Thus the first part of the statement is proved. The second part of the claim, taking any γ with 0 < γ < 1, from this immediately follows. In view of that N 0 depends on a, b, we may assume that a(N − 1) + b ≥ 0. An -th power u belongs to the above terms if its size is 'between' b and a (N − 1) + b, and u ≡ b (mod a).
Thus we see that Here the term in brackets on the right hand side provides an upper bound for the number of (consecutive) integers (forming an interval I ) with -th power of the 'appropriate' size, the factor M a,b ( )/a is the ratio of -th powers in the residue class b (mod a), while the last term is to bound the number of possible -th powers in the progression coming from the last part of I (having less than a elements). This yields Let ε > 0 arbitrary. Clearly, there exists an N 1 = N 1 (ε, , a, b) depending on ε, , a, b such that for N > N 1 we have By Lemma 3.1 this together with (4) implies In view of Proposition 3.1 we can take an 0 such that for > 0 . This by (5)  (If u ∈ {−1, 0, 1}, then we may assume that ≤ 3.) This together with (6) and (7) gives for N large enough. This proves the statement.