The Graham conjecture implies the Erdos-Turan conjecture

TL;DR

This paper demonstrates that proving the Graham conjecture for a specific grid size implies the truth of the Erdős-Turán conjecture for certain progression lengths, linking two significant conjectures in number theory.

Contribution

It establishes a conditional implication from the Graham conjecture to the Erdős-Turán conjecture, connecting progressions and grid structures in number theory.

Findings

Proves that Graham conjecture for s implies Erdős-Turán for 2s-1

Links grid patterns in extbf{N}× extbf{N} to arithmetic progressions in extbf{N}

Provides a new approach to relate two longstanding conjectures

Abstract

Erd\"{o}s and Tur\'{a}n once conjectured that any set $A\subset\mathbb{N}$ with $\sum_{a\in A}{1}/{a}=\infty$ should contain infinitely many progressions of arbitrary length $k\geq3$. For the two-dimensional case Graham conjectured that if $B\subset \mathbb{N}\times\mathbb{N}$ satisfies $$\sum\limits_{(x,y)\in B}\frac{1}{x^2+y^2}=\infty,$$ then for any $s\geq2$, $B$ contains an $s\times s$ axes-parallel grid. In this paper it is shown that if the Graham conjecture is true for some $s\geq2$, then the Erd\"{o}s-Tur\'{a}n conjecture is true for $k=2s-1$.

Equations

$$\sum _{(x,y)\in B}\frac{1}{x^2+y^2}=\infty ,$$

$$A=\{a_1<a_2<a_3<\cdots \}\subset \mathbb{N}$$

$$B=\Big{\{}(a_{n}+m,m):n\in\mathbb{N},m\in\mathbb{N}\Big{\}}.$$

$$\sum _{(x,y)\in B}\frac{1}{x^2+y^2}$$

$$\ge \sum _{n\in \mathbb{N}}\sum _{m=1}^{a_n}\frac{1}{(a_n+m{)}^2+m^2}$$

$$\ge \sum _{n\in \mathbb{N}}\frac{a_n}{(a_n+a_n{)}^2+a_n^2}$$

$$=\sum _{n\in \mathbb{N}}\frac{1}{5a_n}$$

$$=\infty .$$

$$\displaystyle(a_{n}+m,m+l),\ \$$

$$(a_n+m,m),$$

$$r(k,N)\ge N\cdot \exp (-c(\log N{)}^{1/(k-1)}).$$

$$\Theta (A)=\{(a+m,m):a\in A,m=1,2,\dots ,N\}\subset \{1,2,\dots ,2N{\}}^2.$$

$$\tilde{r}(s,2N)$$

$$\ge N^2\exp (-c(\log N{)}^{1/(2s-2)}).$$

Full text

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then for any $s\geq 2$, $B$ contains an $s\times s$ axes-parallel grid. In this paper it is shown that if the Graham conjecture is true for some $s\geq 2$, then the Erdös-Turán conjecture is true for $k=2s-1$.

1. Introduction

One famous conjecture of Erdös and Turán [2] asserts that any set $A\subset\mathbb{N}$ with $\sum_{a\in A}{1}/{a}=\infty$ should contain infinitely many progressions of arbitrary length $k\geq 3$. There are two important progresses towards this direction due to Szemerédi [7] and Green and Tao [5] respectively, which assert that if $A$ has positive upper density or $A$ is the set of the prime numbers, then $A$ contains infinitely many progressions of arbitrary length.

If one considers the similar question in the two-dimensional plane, Graham [4] conjectured that if $B\subset\mathbb{N}\times\mathbb{N}$ satisfies

[TABLE]

then $B$ contains the four vertices of an axes-parallel square. More generally, for any $s\geq 2$ it should be true that $B$ contains an $s\times s$ axes-parallel grid. Furstenberg and Katznelson [3] proved the two-dimensional Szemerédi theorem, that is, any set $B\subset\mathbb{N}\times\mathbb{N}$ with positive upper density contains an $s\times s$ axes-parallel grid. In another words, such a set $B$ contains any finite pattern.

The purpose of this paper is to show that if the Graham conjecture is true, then the Erdös-Turán conjecture is also true.

2. The Graham conjecture implies the Erdös-Turán conjecture

Suppose that the Erdös-Turán conjecture is false for $k=3$. Then there exists a set

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with $\sum_{n\in\mathbb{N}}{1}/{a_{n}}=\infty$ such that $A$ contains no arithmetic progression of length 3. Define a set $B\subset\mathbb{N}\times\mathbb{N}$ by

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Then

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In the sequel we indicate that $B$ contains no square and argue it by contradiction. This would mean that the Graham conjecture is false for $s=2$. Suppose that for some $n,m,l\in\mathbb{N}$, $B$ contains a square of the following form:

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It follows easily from the construction of $B$ that $a_{n}-l,a_{n},a_{n}+l\in A$, which yields a contradiction since $A$ contains no arithmetic progression of length 3 according to the initial assumption.

Similarly, if the Graham conjecture is true for some $s\geq 2$, then the Erdös-Turán conjecture is true for $k=2s-1$. The interested reader can easily provide a proof.

3. Concluding Remarks

Let $r(k,N)$ be the maximal cardinality of a subset $A$ of $\{1,2,\ldots,N\}$ which is free of $k$-term arithmetic progressions. Behrend [1] and Rankin [6] had shown that

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Similarly, let $\widetilde{r}(s,N)$ be the maximal cardinality of a subset $B$ of $\{1,2,\ldots,N\}^{2}$ which is free of $s\times s$ axes-parallel grids. For any set $A\subset\{1,2,\ldots,N\}$, define

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Following the discussion in Section 2, one can easily deduce that if $A$ is free of $2s-1$ term of arithmetic progression, then $\Theta(A)$ is free of $s\times s$ axes-parallel grid. Hence

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We end this paper with a question. Does the Erdös-Turán conjecture imply the Graham conjecture?