Erdős' question was reported by Soifer [So09c]. It is easy to see (see for example [So09]) that any triangle can be cut into $n^2$ congruent triangles (for any $n\geq 1$). Soifer [So09b] proved that there exists at least one triangle (e.g. one with sides $\sqrt{2},\sqrt{3},\sqrt{4}$) which can only be cut into a square number of congruent triangles. (In fact Soifer proves that any triangle for which the angles and sides are both integrally independent has this property.)

It is trivial that any isosceles triangle can be cut into $2$ congruent triangles, simply by dividing down the middle.

Soifer proved [So09] that if we relax congruence to similarity then every triangle can be cut into $n$ similar triangles when $n\neq 2,3,5$ and there exists a triangle that cannot be cut into $2$, $3$, or $5$ similar triangles.

A complete classification is given by Beeson, Laczokovich, and Zhang [BLZ26]. The full classification is too complicated to be stated here, but as a consequence, the set of triangles which can be decomposed into a non-square number of congruent triangles is the set of isosceles triangles together with countably many similarity classes.

See also [634].

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This page was last edited 08 April 2026. View history

• Congrats to M. Beeson, M. Laczokovich and Y. X. Zhang for solving this problem in their latest paper: [BLZ26].

  Alfaiz — 03:35 on 07 Apr 2026

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  • As promised, the work is on the arxiv. Hopefully the introduction is fairly clear about what's done, but I'll write it here for clarity:
    
    1) We proved that the right classification is:
    
    A triangle T admits a non-square tiling if and only if it satisfies one
    of the following conditions, where (A, B, C) are the angles of T in some order:
    (1) A = B, i.e. T is an isosceles triangle (including equilateral);
    (2) $C = \pi/2$ and the legs of the right triangle T are in integer ratio $M/K$, where
    $M^2 + K^2$ is not a square;
    (3) $(A, B, C) = (\pi/6, \pi/2, \pi/3)$; (this is technically included in e.g. the next case, but I think for historical and mathematical reasons it behaves differently)
    (4) $C = \pi/3$, with $\sqrt{3} \tan(A/2)$ rational;
    (5) B = 2A, with $\sqrt{3} \tan(A/2)$ rational;
    (6) B = 2A, with $\sin(A/2)$ rational;
    (7) C = A/2 + B, with $2 \sin(A/4)$ rational, equal to $M/K$, where $2K^2 − M^2$ is not
    a square.
    (8) C = 2A + B/2, with $\sqrt{3} \tan(A/2)$ rational.
    
    2) in an upcoming revision, we prove a slight strengthening of what we have (we were almost there in the first completed draft, and as always we found an improvement after submission):
    
    Corollary. Suppose we have a tiling that is not a reptiling (that is, the triangle is similar to the tile), then the number of
    tiles cannot be a square unless one of the following holds:
    • it is isosceles;
    • it is case (2) and M^2 + K^2 is a square;
    • it is case (7) and 2K^2 − M^2 is a square.
    
    To answer the previous chain of comments on this improvement: it's amusing that in the creation of this paper the constructions from my work [Zh25] were somehow the last "piece of the puzzle" and my motivation, but we actually remove all dependence of this work to prove this stronger result (because we have to prove virtually *all* possible tilings are non-square, not just mine!).

    krzhang — 16:55 on 07 Apr 2026

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• The paper of Y. X. Zhang [Zh25] is the most recent discussion of this problem and contains a fair amount of literature in it.

  (The site has been updated to address this comment.)
  Alfaiz — 03:46 on 30 Dec 2025

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  • As far as I can see the results in this paper only apply to [634], not this problem - I'll update the remarks there.

    Thomas Bloom — 06:06 on 30 Dec 2025

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    • Actually the results in that paper [Zh25] do apply to this one in the constructive direction: for the families of (big) triangles indicated, the paper shows that (with some work not currently in the paper) that at least one non-square tiling exists. So in some sense this paper "accidentally" finishes the last piece of this Erdos problem even though, as you said, it is attempting to solve [634] instead.
      
      However, the majority of credit should be due to earlier works of Laczkovich and Beeson, who did most of the categorization. The constructions, I suspect, *could* have been done by them if they knew about this particular problem. My sense from talking to them is they did not. To take concrete action, we've started a collaboration to finish this problem that is, AFAIK, effectively done modulo a couple of small things (famous last words). Hopefully it will hit the ArXiV in under a month.

      krzhang — 18:04 on 21 Mar 2026

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      • Thanks for clarifying! I'll leave this until the paper arrives. Can you say now what form the final classification has, or is this best left until the paper stabilises into a final form?

        Thomas Bloom — 13:15 on 23 Mar 2026

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        • Probably best to wait given the complexities of collaboration (I don’t want to put extra weight on my collaborators at present). You should get one here ASAP though, given that.

          krzhang — 15:04 on 23 Mar 2026

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