The post below is unsigned, and my mailer (Gmail under Safari) alleges that it was posted from my account (it wasn't). I cannot any more locate the "view source" option on the mailer to check on its actual source. This looks like trouble! Fred Lunnon On 8/13/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred Lunnon <fred.lunnon@gmail.com> wrote:
Adam P. Goucher <apgoucher@gmx.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?
Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining.
My apologies if I over-explain, since I'm not sure quite what your question is.
Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.)
Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity.
If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage.
Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.
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