xdog:
Your proof does not work mod 4.
If A,B are both odd, then 14C^2 = 2 mod 4.
It does not follow, however, that 7C^2 = 1 mod 4. 7C^2 could also = 3 mod 4, because 1*2 = 3*2 = 2 mod 4.
There is nothing to stop A, B and C from all being 1 mod 4.
However, your approach does work mod 7.
If A and B are both multiples of 7, then C necessarily is also.
Divide both sides by 49 as many times as necessary to get
to a similar equation where A or B is not a multiple of 7.
Then the 14C^2 mod 7 = 0.
But A^2 + B^2 cannot = 0 mod 7,
because a square is = 1, 2, 4, or 0 mod 7,
but we have reduced to an equation where A^2 and B^2 cannot both be 0 mod 7.
Fixing the prior proof (spoiler)
