# $1^2+2^2+\cdots+24^2=70^2$ and squarily squaring the torus

The unique nontrivial solution to $1^2+2^2+\cdots+n^2=m^2$ is $(n,m)=(24,70)$. (This fact has connections to modular forms, special functions, lattices and string theory.) Martin Gardner, in the September 1966 issue of Scientific American, attributed the following question to someone named Richard Britton: can we tile a $70\times70$ square with $1\times1$, $2\times2$, $\cdots$, $24\times24$ squares? The answer, after some computer analysis, has turned out to be negative. (If I understand correctly it wasn't quite brute force but it was something of an exhaustive search. I am also curious about the potential for a paper-and-pen combinatorial disproof.) So I will relax the question a bit: is it possible to tile the $70\times70\,$ torus with these squares? Equivalently, can we tile the $70\times70$ square with them if we allow wrapping across both pairs of opposite sides?

• without cutting squares, I guess.
– mike
Commented Sep 12, 2014 at 19:06
• it's really not a problem with cutting allowed, @mike. That's what the word "tile" implies, anyway. Commented Sep 12, 2014 at 19:16
• when my contractor tiled my bathroom recently, they just cut them when the squares did not fit ;-)
– mike
Commented Sep 12, 2014 at 19:26
• Ah, sorry - missed the word negative. Now found this paper. aaai.org/Papers/ICAPS/2004/ICAPS04-019.pdf Commented Sep 12, 2014 at 19:34
• This discussion deals briefly with the original problem mathoverflow.net/questions/133348/… Commented Sep 12, 2014 at 19:40

More than that, he enumerated all possible solutions of order 24. He also looked at squared cylinders. On the last page (87) of his thesis, Gambini verifies that the $70 \times 70$ torus is impossible.