Thanks, Stuart! If one relaxes the condition that all the squares be unequal sizes, but retains the condition that no points on the torus belong to more than three edges, can one find simpler solutions? I’m wondering if there’s anything like the two-unequal-squares solution but slightly more complicated. I realize that you could take the two-unequal-squares solution and replace each square by squared square, but I am wondering if there are solutions that are significantly simpler. (Stuart, can you please forward this question to Geoffrey if you think he’d know?) Thanks, Jim Propp On Saturday, July 7, 2018, Stuart Anderson <stuart.errol.anderson@gmail.com> wrote:

Geoffrey Morley has been investigating these. Here is his MathsJam presentation from 2014. The links in his paper's references to squaring.net won't work as I am updating the website.

http://www.mathsjam.com/conference/talks/2014/ GeoffreyMorley-Torus-Squaring.pdf _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun