For which $n$ can some triangle be cut into $n$ mutually congruent triangles? (Erdős #634)
Statement
Find all positive integers $n$ for which there exists at least one triangle that can be dissected into $n$ mutually congruent triangles. Here the $n$ pieces must be congruent to one another (but need not be similar to the original triangle), and they must tile the original triangle exactly, with no gaps or overlaps.
Acceptance. FULLY RESOLVES: a complete classification of the set of $n$ having the property, with proof — both the constructions for the $n$ that qualify and the impossibility proofs for the $n$ that do not. Machine-checkable proof preferred, else a full written proof. ADVANCES: settle the property for a specific $n$ whose status is currently open (for example $n=19$): either (a) exhibit an explicit dissection of a named triangle into $n$ mutually congruent triangles — give the vertices of the outer triangle and of every piece, plus a machine-checkable certificate that the pieces are pairwise congruent and tile the triangle exactly with no gaps or overlaps — or (b) prove that no triangle can be cut into $n$ congruent triangles; OR prove that a new infinite family of $n$ has (or lacks) the property, strengthening the families stated in the background. Deliver the witness dissection (coordinates plus congruence/tiling certificate), the impossibility proof, or the new family with its proof.
Background
This question of Erdős was reported by Soifer [So09c]; Erdős offered $25 for a solution. Listed as open on erdosproblems.com/634 (fetched 2026-07-13, status 'open', tagged 'geometry'). Every perfect square $n$ works — indeed any triangle can be cut into $n^2$ congruent copies by the standard barycentric grid subdivision. Soifer [So09c] showed that $n$ of the forms $2m^2,\ 3m^2,\ 6m^2,\ m^2+\ell^2$ also have the property. On the negative side, Beeson showed that $n=7$ and $n=11$ do NOT have the property, and it is conjectured that no prime $\equiv 3\pmod 4$ has it; the smallest value whose status is currently unknown is $n=19$. Two relaxations are understood: if the small triangles need only be similar (not congruent), Soifer [So09] proved every triangle can be cut into $N$ similar triangles for all $N\neq 2,3,5$; if the pieces must be similar to the original triangle, Snover–Waiveris–Williams [SWW91] proved the only possible $N$ are $m^2$, $m^2+\ell^2$, and $3m^2$. Most recently Zhang [Zh25] proved that $n^2ab$ has the property whenever $a\ge b\ge1$ are integers and $n\ge 3\lceil (a^2+b^2+ab-a-b)/(ab)\rceil$, via an explicit family of tilings of an equilateral triangle by congruent triangles with a $120^\circ$ angle. Attacker's tool: for a fixed open $n$ (such as $19$) a positive answer is a finite, fully checkable witness — an explicit dissection of a specific triangle into $n$ congruent pieces given by coordinates — and computer search over candidate tilings can locate such a witness; ruling $n$ out instead requires an impossibility argument in the style of Beeson.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #634 (T. F. Bloom) | website |
| REF-02 | M. Beeson — Tiling triangles with congruent triangles (talk slides; non-existence for n=7, 11) | paper |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.