24
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

24
$\begingroup$

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 :-)

$\endgroup$
3
  • $\begingroup$ Can you say something about the Ruzsa&Gyarmati construction? $\endgroup$ Oct 25, 2011 at 15:46
  • $\begingroup$ Kevin - I'd suggest asking one of them. I think there is a preprint. $\endgroup$
    – Ben Green
    Oct 25, 2011 at 22:57
  • 4
    $\begingroup$ They got $c N /\sqrt{\log\log N}$. To appear in Acta Arithmetica. $\endgroup$ Apr 18, 2012 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.