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](http://www.math.uconn.edu/%7Ekconrad/blurbs/ugradnumthy/3squarearithprog.pdf) and [here](http://www.math.uconn.edu/%7Ekconrad/blurbs/ugradnumthy/4squarearithprog.pdf)) 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](https://arxiv.org/abs/0912.1494), 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?
> 
> 
> 



Arithmetic Progressions of Squares
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 :-)


