Question

Fermat may or may not have known that there are 3-term arithmetic progressions of squares (like $1^2, 5^2, 7^2$, and that there are no 4-term APs. Murky history aside, Keith Conrad has two pleasant expositions (here and here) providing a modern treatment of this from an algebraic viewpoint.

A natural combinatorial follow-up is: how large can a subset of $\{1^2,2^2,\dots,n^2\}$ be and still not have 3-term APs? In this paper, I showed that there are subsets of size $$\gg n c^{-\sqrt{\log\log n}},$$ where $c=2^{\sqrt{8}}$, but I don't know of an upper bound.

Is there a subset of the squares with positive relative density that is free of 3-term APs?

Answer 1

Probably not, but a proof is hopeless. Ruzsa and Gyarmati have a preprint in which they construct such a subset of size something like $N/\log \log N$.

Even the colouring version (that is, finite colour the squares, does one of the classes contain a 3-term progression) is open. A very closely-related question (Schur's theorem in the squares) is explicitly asked as Question 11 in this paper by Bergelson:

http://www.math.iupui.edu/~mmisiure/open/VB1.pdf

It is possible to show that a positive density subset of the squares contains a solution to $\frac{1}{4}(x_1 + x_2 + x_3 + x_4) = x_5$ by adapting the technique of arXiv:math/0302311. I'd have to admit this is slightly more than a back of an envelope calculation :-)