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Sep 9, 2020 at 3:06 comment added Gerry Myerson The Erdos conjecture is that $A$ contains arithmetic progressions of arbitrary length. That $A$ contains a 3-term AP (or infinitely many 3-term APs) is the simplest special case. And this special case is what's proved at the arxiv reference, without resolving the general case.
Sep 5, 2020 at 3:38 comment added Antoine Labelle This is now solved! Seems like this wasn't that much out of reach. arxiv.org/abs/2007.03528
May 8, 2018 at 13:38 comment added C.S. @Kevin: What doing mean by saying $A$ contains a 3-term arithmetic progession?
Nov 7, 2013 at 4:31 comment added Suvrit wow, in my ignorance, I find it hard to believe that this question is within the "completely out of reach" category! really awesome question though!
Jul 1, 2012 at 15:12 history answered Kevin O'Bryant CC BY-SA 3.0