The well-known Erdos-Turan conjecture is the following.
Let $V \subset \mathbb{N}$ be such that $\sum_V k^{-1}$ diverges. Then $V$ contains arithmetic progressions of every possible length.
A recent result showed such a set $V$ must contain infinitely many length-$3$ progressions.
I am wondering: if $V \subset \mathbb{N}$ is such that $\sum_V k^{-1}=\infty$, what are some properties of $V$? This Wikipedia page gives a small number of properties, but I am curious whether there are others (being non-trivial).
Best Answer
Such a 'large' set must have infinitely many distinct pairs that differ by perfect squares.
This follows from a quantitative strengthening of Sárközy's theorem proved by Pintz-Steiger-Szemerédi. The same is true if we replace squares by any other power (so perfect cubes, fourth powers, and so on).
In general, the following is true. We say that $h\in\mathbb{Z}[x]$ is an intersective polynomial if it is non-zero and for every $q$ there is some $n$ such that $q\mid h(n)$ (for example, $h(x)=x^k$ is intersective). If $A\subset \mathbb{N}$ is such that $\sum_{a\in A}\frac{1}{a}=\infty$ then for any fixed intersective polynomial $h$ there are infinitely many pairs $a\neq b\in A$ such that $a-b=h(n)$ for some $n$. This generalisation of the Pintz-Steiger-Szemerédi result was proved by Alex Rice (https://arxiv.org/pdf/1612.01760.pdf) (the case when $h$ is a perfect power was already proved by Balog-Pelikan-Pintz-Szemerédi).
Note being intersective is also necessary for such a property to hold, otherwise if $q$ is such that $h(n)\equiv 0\pmod{q}$ has no solutions then $A$ being the set of all multiples of $q$ is a 'large' set that has no solutions to $a-b=h(n)$.