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Statement

If a triangle $T$ has a tiling by congruent triangles not similar to $T$, its tile count cannot be square unless either

1. $T$ is isosceles, or
2. its angles can be labeled $(A,B,C)$ with $C=A/2+B$ and $2\sin(A/4)=M/K\in\mathbb Q$, where $M,K$ are positive integers and $2K^2-M^2$ is a square.

This is a necessary exception list for a given non-reptiling, not a claim that every isosceles triangle has a square non-reptiling.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Theorem 3, p. 2; proof p. 21. Complete rewritten deduction from the linked same-paper results and their explicit external dependencies.

Proof

Assume $T$ is not isosceles. By Theorem 11 and Proposition 13, the tiling belongs to one of the six table rows. Rows 2, 3, 4, and 6 have nonsquare counts by Propositions 26, 28, 27, and 30, respectively. Row 1 is nonsquare by Proposition 31.

Thus a square count can occur only in row 5, where $T=(2\alpha,\beta,\alpha+\beta)$ and $3\alpha+2\beta=\pi$. Proposition 29 identifies precisely its square criterion. In the notation $A=2\alpha$, $B=\beta$, it is the second exception above.

The proof does not depend on the secondary uniqueness claim in Theorem 32.

Bears on. Problem 633.