An improved lower bound related to the Furstenberg-Sárközy theorem

Let D(n) denote the cardinality of the largest subset of the set {1, 2, . . . , n} such that the difference of no pair of elements is a square. A well-known theorem of Furstenberg and Sárközy states that D(n) = o(n). In the other direction, Ruzsa has proven that D(n) & nγ for γ = 1 2 ( 1 + log 7 log 65 ) ≈ 0.733077. We improve this to γ = 1 2 ( 1 + log 12 log 205 ) ≈ 0.733412.


Introduction
The following theorem, first conjectured by Lovász, was proven independently by Furstenberg [5] and Sárközy [14], [15], [16] around 1977: Theorem 1.Every subset of the natural numbers of positive upper density contains two distinct elements whose difference is a square.
One can reformulate this theorem as follows.Let D(n) be the cardinality of the largest subset of the set {1, 2, . . ., n} such that the difference of no pair of elements is a square.Then: The best known quantitative form of (1) is the following bound of Pintz, Steiger, and Szemerédi [11] from 1988: Erdős originally conjectured that D(n) x 1/2 log c (x) for some positive constant c.This was disproved by Sárközy [15], who put forth the weaker conjecture that D(n) n 1/2+ for every > 0. This was disproved by Ruzsa [12] in 1984.More precisely, he proved the following: Theorem 2. In the notation above, D(n) n γ where γ = 1 2 1 + log 7 log 65 ≈ 0.733077.
It is perhaps insightful to compare these results with the known progress on Roth's theorem.Let R(n) denote the largest subset of {1, 2, . . ., n} that does not contain a (nontrivial) three term arithemetic progression.Roth's theorem states that R(n) = o(n).The known proofs of Roth's and the Furstenberg-Sárköky theorem are based on very similar considerations, although the later case is somewhat simpler.For instance, see the recent proofs of Layla [8] and Tao, Green and Ziegler [17].The best known quantitative form of Roth's theorem is due to Bloom [2] (see also Sanders [13]) and states that: In the other direction, an example of Behrend from 1946 shows that for some universal c.The nature of the constant c has been recently refined by Elkin [3] and Green and Wolf [7].Note that both the known upper and lower bounds are considerably smaller for D(n) than R(n).Indeed, while Behrend's example rules out the possibility of extending Roth's theorem to polynomial sparse sets (in other words, obtaining a power savings in the estimate (3)), such a possibility has not been ruled out in the context of the Furstenberg-Sárköky theorem (1).The purpose of the current work is to obtain a slight improvement to Ruzsa's Theorem 2.More specifically, we prove the following theorem: Theorem 3. In the notation above, More generally, let D k (n) denote the cardinality of the largest subset of {1, 2, . . ., n} such that the difference of no pair of elements is a k-th power.Thus With this notation, Ruzsa also obtained D 3 (n) n γ 3 for γ 3 = 2 3 + log 3 3 log 7 ≈ 0.854858.We are also able to slightly improve this as follows.
the electronic journal of combinatorics 22(1) (2015), #P1.32 The proofs of Theorems 2, 3 and 4 are based on an observation of Ruzsa which states roughly (see Lemma 5) that if one can find a square-free natural number m and a large subset A of the residue mod m such that the difference of no two distinct elements in A is a square mod m, then one can use this example to construct a large subset B of {1, 2, . . ., N } such that the difference of no two distinct elements of B is an integral square.Using this lemma, Ruzsa proved Theorem 2 by exhibiting an explicit set of 7 residues mod 65 with the desired property.Our proof of Theorem 3 will follow by exhibiting an explicit set of 12 residues mod 205.Similarly, our proof of Theorem 4 is based on a set of 14 residues mod 91.

Ruzsa's Lemma
Our starting point is the following result of Ruzsa [12].
Lemma 5. Let m denote a square-free positive integer, and let r k (m) denote the maximal number of residues mod m such that the difference of no two such elements is a k-th power residue.Moreover, define Then, 3 Proof of Theorem 3 Given lemma 5, it suffices to demonstrate a set A of 12 residues mod 205 whose difference set (the set of pairwise differences, A − A) contains no square.We claim that {7, 21, 50, 64, 76, 83, 106, 120, 139, 182, 193, 199} is such a set.Indeed, to enable the reader to independently verify this for herself, Table 1 shows the difference set of A. More specifically, the element in the row labeled i and column labeled j is the value i − j mod 205.One may the check this against the following list of squares mod 205:

Further remarks
In light of lemma 5, it is tempting to computationally search of favorable sets of residues.Indeed, this is how we found those presented above.We were able to exhaustively check all moduli (strictly) less than 533 with a brute-force search.This took about a month of computing time on a modern desktop computer.Given a square-free modulus m the problem of finding the largest subset whose difference set contains no squares is equivalent to the problem of finding the maximal clique in a dense graph.In complete generality, this problem is known to be NP complete.There are, however, non-trivial algorithms available for this problem.Using the graph algorithm of Konc and Janezic [9] implemented in C by Konc [10] we were able to extend this range up to m 733.Indeed, the set of 12 residues mod 205 given below gives the optimal result among sets within this range.
It is natural to ask what the limitations of this method are.As above, let r 2 (m) denote the cardinality of the largest set of residues mod m whose difference set does not contain a square.It is well known (see [4], for instance) that r 2 (p) p 1/2 for all primes p. Ruzsa has conjectured that r 2 (m) m 1/2 for square-free m.This would imply that γ 3 4 would be the limitation of this method.Ruzsa's conjecture remains open; indeed it does not seem to be known if there exists a δ > 0 such that r 2 (m) m 1−δ for all square-free m.We note that the paper [5] claims the inequality r 2 (m) m 1/2+ for square-free m, however the referee of this note has pointed out that the proof presented there contains a serious error.
It is unclear, at least to the author, what one should expect the true order of D(n) to be.

Table 1 :
The difference set of A Again using lemma 5, it suffices to demonstrate a set A of 14 residues mod 91 whose difference set (the set of pairwise differences, A − A) contains no cube.We claim that {3, 19, 23, 25, 29, 35, 41, 47, 66, 72, 78, 84, 88, 90} is such a set.The difference set mod 91 is as follows.