Problem #634: Find all $n$ such that there is at least one triangle which...

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Progress

Erdős' question was reported by Soifer [So09c]. It is easy to see that all square numbers have this property (in fact for square numbers any triangle will do). Soifer [So09c] has shown that numbers of the form $2n^2,3n^2,6n^2,n^2+m^2$ also have this property. Beeson has shown (see the slides below) that $7$ and $11$ do not have this property. It is possible that any prime of the form $4n+3$ does not have this property.

In particular, it is not known if $19$ has this property (i.e. are there $19$ congruent triangles which can be assembled into a triangle?).

For more on this problem see these slides from a talk by Michael Beeson. As a demonstration of this problem we include a picture of a cutting of an equilateral triangle into $27$ congruent triangles from these slides.

Soifer proved [So09] that if we relax congruence to similarity then every triangle can be cut into $N$ similar triangles when $N\neq 2,3,5$.

If one requires the smaller triangles to be similar to the larger triangle then the only possible values of $N$ are $n^2,n^2+m^2,3n^2$, proved by Snover, Waiveris, and Williams [SWW91].

Zhang [Zh25], among other results, has proved that for any integers $a \geq b$, if\[n\geq 3\left\lceil \frac{a^2+b^2+ab-a-b}{ab}\right\rceil\]then $n^2ab$ has this property (indeed, they explicitly show that an equilateral triangle can be tiled with $n^2ab$ many triangles of side lengths $a,b,\sqrt{a^2+b^2+2+ab}$).

See also [633].