Continuing to hare off in my own direction, I just found a toroidal tiling with the squares of 1, 2, 3, and 4. The fundamental region isn't even a rectangle; it's a parallelogram with sides (2,-4) and (7,1). (2D cross product is 30 = 1 + 4 + 9 + 16.). I can see how to do the same thing with the squares from 1 to 5, but I don't see 1 to 6 offhand. On Fri, Jul 3, 2020 at 5:07 PM Dan Asimov <dasimov@earthlink.net> wrote:
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
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