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

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    $\begingroup$ I mean it in the natural combinatorial sense, with squares sides always aligned nicely etc $\endgroup$ Commented Aug 17, 2021 at 17:02
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    $\begingroup$ @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. $\endgroup$
    – Erick Wong
    Commented Aug 17, 2021 at 17:12
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    $\begingroup$ 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. $\endgroup$ Commented Aug 17, 2021 at 17:19
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    $\begingroup$ 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. $\endgroup$ Commented Aug 17, 2021 at 17:44
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    $\begingroup$ Does this answer your question? $1^2+2^2+\cdots+24^2=70^2$ and squarily squaring the torus $\endgroup$
    – RobPratt
    Commented Aug 17, 2021 at 17:58

1 Answer 1


I ran this through my tiling solver in 2019/2020. It took about 9 million seconds to answer in the negative. I fixed the unit square in place to avoid doing too much extra work. You can fix it anywhere as the toroid is 'featureless' in the sense that there are no corners or edges, every square is equivalent.

I also did a cylinder (Wrap X), Klein bottle (Wrap X, Wrap Y, Flip X), and a projective plane (Wrap X, Wrap Y, Flip X, Flip Y)

I seem to have missed out the Moebius strip (Wrap X, Flip Y). Of course it's not required, being a subset of a Klein bottle. But looking again at the projective plane tiling, I see that I stopped it well before it completed, after 460 hours. It seems it will take quite some time to complete the brute force search.

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    $\begingroup$ If there's no Klein bottle solution, shouldn't that also rule out a Möbius strip solution, since that's just a Klein bottle without one of the opposite-side identifications? $\endgroup$ Commented Aug 18, 2021 at 2:36
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    $\begingroup$ That's probably why I skipped it first time around... $\endgroup$ Commented Aug 18, 2021 at 2:42

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