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Statement

A nondegenerate Euclidean triangle $T$ admits a tiling into a positive nonsquare number of congruent triangles if and only if its angles can be labeled $(A,B,C)$ so that at least one of the following holds:

1. $A=B$; this includes the equilateral triangle.
2. $C=\pi/2$ and its legs have ratio $M/K$ for positive integers $M,K$ with $M^2+K^2$ not a square.
3. $(A,B,C)=(\pi/6,\pi/2,\pi/3)$.
4. $C=\pi/3$ and $\sqrt3\tan(A/2)\in\mathbb Q$.
5. $B=2A$ and $\sqrt3\tan(A/2)\in\mathbb Q$.
6. $B=2A$ and $\sin(A/2)\in\mathbb Q$.
7. $C=A/2+B$ and $2\sin(A/4)=M/K$ for positive integers $M,K$ with $2K^2-M^2$ not a square.
8. $C=2A+B/2$ and $\sqrt3\tan(A/2)\in\mathbb Q$.

The representation of a positive rational number as $M/K$ need not be reduced. Scaling numerator and denominator changes either quadratic expression by a rational square, so its square status is independent of the representation. A tiling covers $T$ by finitely many congruent closed triangles whose interiors are disjoint; a reptiling uses a tile similar to $T$.

Thus the answer to Problem 633 is the complement of these eight families.

Source and scope. Beeson, Laczkovich, and Zhang, Solution of Erdős Problem 633, arXiv:2604.03609v3, Theorem 1, pp. 1–2; necessity pp. 8–9; sufficiency pp. 12–17. This is a complete rewritten proof at the source's dependency boundary. The linked pages supply every essential same-paper deduction; earlier classification, existence, and counting theorems and the four elliptic-curve group data are stated and cited as external inputs, not recursively reproved.

Necessity

Suppose an $N$-tiling exists with $N$ nonsquare. If it is a reptiling, the external reptiling classification recorded in Theorem 11 gives family 2 or 3. If $T$ is isosceles, family 1 holds. We may therefore assume that $T$ is non-isosceles and the tiling is not a reptiling.

Theorem 11 gives one of six angle patterns, and Proposition 13 proves their precise rationality conditions. Its second, third, fourth, and sixth rows give families 4, 5, 6, and 8 directly. For the first row, $C=\pi/3$ and $\sqrt3\tan(A/4)$ is rational. The half-angle doubling identity gives

$$\sqrt3\tan(A/2) =\frac{2\sqrt3\tan(A/4)}{1-\tan^2(A/4)}\in\mathbb Q.$$

Here $A<\pi$, so $\tan(A/4)<1$ and the denominator is nonzero. Consequently this row also gives family 4.

For the fifth row, $T=(2\alpha,\beta,\alpha+\beta)$ and $3\alpha+2\beta=\pi$. The external tiling equation, stated precisely in Proposition 29, supplies positive integers $M,K$ with $2\sin(\alpha/2)=M/K$ and $N=2K^2-M^2$. Since $A=2\alpha$ and $N$ is nonsquare, this is family 7. These cases exhaust the possibilities. The source's p. 9 cross-reference for this equation is corrected by the explicit identification on Proposition 29's page.

Sufficiency: the three elementary families

In family 1, cut along the symmetry axis to obtain two congruent triangles. In family 2, the external reptiling construction of Golomb and Snover–Waiveris–Williams, recorded with Theorem 6 in Theorem 11, gives $M^2+K^2$ congruent similar tiles. The hypothesis makes this count nonsquare; Figure 6 on p. 25 illustrates the construction with counts $13$ and $74$.

Family 3 has a three-tile reptiling, as in Figure 7 on p. 25. To read that diagram explicitly, take the large right triangle with vertices $P=(0,0)$, $Q=(2\sqrt3,0)$, and $R=(\sqrt3/2,3/2)$. Put $D=(\sqrt3,0)$ and $S=(\sqrt3,1)$ on $PQ$ and $RQ$, respectively. The triangles $PRS$, $PDS$, and $DQS$ have side lengths $1,\sqrt3,2$. They partition the large $30$–$60$–$90$ triangle, giving three congruent tiles. These coordinates simply specify the source's diagram.

Sufficiency: excluding commensurable-angle exceptions

For families 4–8, suppose first that all angles of $T$ are rational multiples of $\pi$. In families 4, 5, and 8, Lemma 24 applies to $A/2$. Since $\sqrt3\tan(A/2)$ is rational, its possible squared tangent values are $1/3$ and $3$; the value $0$ would give $A=0$, and $1$ would give $\sqrt3\tan(A/2)=\sqrt3$. Thus $A=\pi/3$ or $2\pi/3$.

For family 6, $\cos A=1-2\sin^2(A/2)$ is rational. The cosine theorem stated with Lemma 24 forces $A\in\{\pi/3,\pi/2,2\pi/3\}$, and rationality of $\sin(A/2)$ leaves only $A=\pi/3$. In family 7, applying the same argument to $A/2$ gives $A=2\pi/3$.

In families 5 and 6, $B=2A$ would then make $C\le0$. In family 7, $A=2\pi/3$ and $C=A/2+B$ force $B=0$. In family 8, $C=2A+B/2$ with $A\ge\pi/3$ forces $B\le0$. All are impossible for a triangle. In family 4, the only nondegenerate possibility is $A=B=C=\pi/3$, already covered by family 1.

Sufficiency: the incommensurable families

It remains to treat incommensurable angles. For family 4, Proposition 10 applied to $T$ shows that $\sqrt3\tan(A/2)$ is rational for either non-$C$ angle. Label them with $A<B$; equality would make $T$ equilateral. Then Proposition 26 gives a nonsquare tiling.

Family 5 is Proposition 28 and family 6 is Proposition 27. Family 7 follows from existence and the exact square criterion in Proposition 29. Family 8 is Proposition 30. These propositions use Laczkovich's existence theorems and show that every tiling with the specified tile shape has the requisite nonsquare count. This finishes both directions.

Bears on. Problem 633.