$3\times 3$ magic squares consisting of entries of a dense set $D\subseteq \mathbb{N}$

Question

Starting point. The struggle for a magic square consisting of distinct square numbers is still ongoing, but it has produced an amusing landmark result called the Parker square. One of the issues is that square numbers are very scarce - which leads to the following question.

Problem. A $3\times 3$ magic square is a $3\times 3$ integer matrix with all entries being distinct, such that all row sums, column sums, and diagonal sums are equal.

We call a set $D\subseteq \mathbb{N}$ dense if $$\lim\inf_{n\to\infty}\frac{|D\cap \{1,\ldots, n+1\}|}{n+1} > 0.$$

If $D\subseteq\mathbb{N}$ is dense, is there always a $3\times 3$ magic square with all the entries lying in $D$?

Answer 1

Yes.

By Szemerédi's theorem, your set contains an arithmetic progression of arbitrary length. In particular, it contains a progression of length 9, say it's $d_1,\ldots,d_9$. Then $$ \begin{pmatrix} d_2 & d_7 & d_6\\ d_9 & d_5 & d_1\\ d_4 & d_3 & d_8 \end{pmatrix} $$ is a magic square.