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Created December 11, 2021 01:33
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Proof thingy
All pythagorean triples can be generated from two integers m and n with the following:
a = m^2 - n^2
b = 2mn
c = m^2 + n^2
If either m or n is divisible by 3, then b is divisible by three. Else, there are three possibilities for m and n (modulo 3):
(2, 1) => 2^2 - 1^2 = 3, divisible by 3
(2, 2) => 2^2 - 2^2 = 0, divisible by 3
(1, 1) => 1^2 - 1^2 = 0, divisible by 3
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