Arithmetic Progressions of Squares - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:03:41Z http://mathoverflow.net/feeds/question/78949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78949/arithmetic-progressions-of-squares Arithmetic Progressions of Squares Kevin O'Bryant 2011-10-24T03:31:35Z 2011-10-24T13:13:14Z <p>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 (<a href="http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/3squarearithprog.pdf" rel="nofollow">here</a> and <a href="http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/4squarearithprog.pdf" rel="nofollow">here</a>) providing a modern treatment of this from an algebraic viewpoint.</p> <p>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 <a href="http://front.math.ucdavis.edu/0912.1494" rel="nofollow">this paper</a>, 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. </p> <blockquote> <p>Is there a subset of the squares with positive relative density that is free of 3-term APs?</p> </blockquote> http://mathoverflow.net/questions/78949/arithmetic-progressions-of-squares/78977#78977 Answer by Ben Green for Arithmetic Progressions of Squares Ben Green 2011-10-24T13:08:09Z 2011-10-24T13:13:14Z <p>Probably not, but a proof is hopeless. Ruzsa and Gyarmati have a preprint in which they construct such a subset of size something like $N/\log \log N$.</p> <p>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:</p> <p><a href="http://www.math.iupui.edu/~mmisiure/open/VB1.pdf" rel="nofollow">http://www.math.iupui.edu/~mmisiure/open/VB1.pdf</a></p> <p>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 :-)</p>