Translate w to the origin. Then L = |v|^2 = x^2 + y^2 is even, and so x and y are either both even or both odd. Then x + iy = (x + iy) ((1 - i)/2) (1 + i) = (((x + y) + i(y - x)) / 2) (1 + i). Since x + y and y - x are both even, the cofactor of 1 + i is an integer. -- Gene On Sunday, August 13, 2017, 10:41:05 AM PDT, Fred Lunnon <fred.lunnon@gmail.com> wrote: << Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i. >> C'mon, you young hotshots --- spare a thought for the feeble-minded pensioners among us! Why is that? WFL On 8/13/17, James Propp <jamespropp@gmail.com> wrote:
Very nice!
Jim
On Sun, Aug 13, 2017 at 9:13 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
What about equilateral polygons in the integer grid? A dodecagon is possible if it's in the form of a Greek cross (aka X-pentomino). Are all even-N-gons possible? I suspect that no odd-N-gon is possible. I'm searching for a proof.
Suppose we have an equilateral polygon with vertices in the Gaussian integers and common side-length sqrt(L), where L is an integer. Without loss of generality, assume that one of the vertices is at the origin.
Now, if L is even, then for any pair of adjacent vertices v, w in Z[i], we have that (v - w) is divisible by 1 + i. Hence, by induction, every vertex is divisible by 1 + i, so we can divide throughout and obtain an equilateral polygon with common side-length sqrt(L/2).
Hence, we can assume without loss of generality that L is odd. Now colour the vertices of the integer grid red and blue in a checkerboard fashion. Every pair of adjacent vertices in an equilateral polygon of common side length L have opposite colours, so no odd-sided equilateral N-gons can exist. QED.
Best wishes,
Adam P. Goucher