$1^2+2^2+\cdots+24^2=70^2$ 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?

It's not possible. Ian Gambini's thesis investigated squared squares on cylinders and toruses up to order 24. He found many solutions, such as the following.

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.