On Tuesday 15 May 2007 20:21, James Propp wrote:

This is a really nice question; has anyone made progress with it?

Mike Reid gave (in mail to math-fun) what purports to be a proof of Dan's conjecture; it looks plausible to me. Here it is again, since apparently it didn't reach you the first time around. | 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 -- g