Ross Millikan posted this on Math StackExchange (July 26, 2018) in answer to a question about squares that are sums of three squares, giving families of such equations. Can either of these be turned into a periodic tiling of a square torus (not necessarily parallel to the edges of the square that the torus is based on)? ----- We know that (n+1)^2 − n^2 = 2n+1, so pick your favorite Pythagorean triple a^2 + b^2 = c^2 with c odd. Let c^2 = 2n+1, n = (c^2−1)/2, and (*) a^2 + b^2 + n^2 = (n+1)^2. If you pick a triple with c even, we can use (n+2)^2 − n^2 = 4n + 4, so we can let n = (c^2 − 4)/4 and have (**) a^2 + b^2 + n^2 = (n+2)^2. If c is even, c^2 is divisible by 4, so the division will come out even. ----- —Dan