T = R^2/Z^2

"If T can be tiled by N congruent squares, then N is the sum of 2 square numbers." . . .

i think this is relatively easy to prove: lift such a tiling to a tiling of R^2 . it is easy to show that any tiling of R^2 by congruent squares is invariant under some translation, with length equal to the side length of the squares. let (x, y) be this translation. since the tiling is lifted from T , it is also invariant under (1, 0) and (0, 1) translations. the group of translational invariants must be a discrete subgroup of R^2 , so this means that (x, y) is commensurate with the lattice generated by (1, 0) and (0, 1) . in other words, some (non-zero) multiple of (x, y) is in the lattice ; say M(x, y) is that multiple. now we may write x = a/M , y = b/M for some integers a, b, M . the side length of the square is 1 / sqrt(N) , because N of them tile T . thus we have (a^2 + b^2) N = M^2 . from this equation, we see that any prime = 3 mod 4 must divide N with even multiplicity. therefore N is a sum of two squares. qed mike