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OPEN - $25 Classify those triangles which can only be cut into a square number of congruent triangles. Erdős' question was reported by Soifer [So09c]. It is easy to see (see for example [So09]) that any triangle can be cut into$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}$) which can only be cut into a square number of congruent triangles. (In fact Soifer proves that any triangle for which the angles and sides are both integrally independent has this property.) Soifer proved [So09] that if we relax congruence to similarity then every triangle can be cut into$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. See also [634]. Additional thanks to: Boris Alexeev, Yan Zhang OPEN -$25
Find all $N$ such that there is at least one triangle which can be cut into $n$ congruent triangles.
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 than 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].