# Is it possible to cover a $70\times70$ torus (or klein bottle/projective plane) with $24$ squares with side length $1,2,3\ldots,24$?

Is it possible to cover a $$70\times70$$ torus with $$24$$ squares with side length $$1,2,3\ldots,24$$? It is known it cannot be done in a $$70\times70$$ square, which is a shame as the identity $$1^2+2^2+3^2+\cdots+24^2=70^2$$ is so nice, but perhaps there's still something good to be found in this vein.

As a bonus, attempting this with a $$70\times70$$ Klein bottle or projective plane would be interesting, too.

edit: The torus case was resolved in the negative in this question, which leaves just the Klein bottle and projective plane identifications left. I think those are still sufficiently interesting questions that I'll wait to see if they can be answered

• I mean it in the natural combinatorial sense, with squares sides always aligned nicely etc Aug 17, 2021 at 17:02
• @BerenGunsolus I think Thomas is just asking you to state explicitly that when you say torus, Klein bottle, projective plane, you're specifically referring to the familar unit square with various sides glued together, rather just than the generic topological spaces. Aug 17, 2021 at 17:12
• Another, related question might be a tiling of a $70m\times 70n$ rectangle with equal numbers of squares of size $1,\dots,24,$ and similar for torus and klein bottle. Aug 17, 2021 at 17:19
• Re: The larger rectangle case: We can tile a $70m\times 70$ when $m$ is a multiple of $\operatorname{lcm}(1,2,\dots,24),$ with equal numbers of each square, simply by assembling $70m\times k$ rectangles out of $k\times k$ squares. So we know that that case can be done, but the question is whether there are smaller examples. Aug 17, 2021 at 17:44
• Does this answer your question? $1^2+2^2+\cdots+24^2=70^2$ and squarily squaring the torus Aug 17, 2021 at 17:58