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.