Quote:
Originally Posted by Charles Kusniec
Now, the challenge is to find at least one other form of proof where you use the sum of two squares theorem.

I forgot to point out that the thing is also trivial when n = 1, y^2 = 0 (mod 1).
What I proved was that your given hypotheses imply that
2*n  1 divides 4*y^2 + 1 if n > 1
2*n + 1 divides 4*y^2 + 1, if n < 0.
The fact that every prime divisor p of 4*y^2 + 1 is congruent to 1 (mod 4) is elementary, and does not require the twosquares theorem. Using Fermat's "little theorem,"
p divides (2*y)^(p1)  1. From the above,
p divides (2*y)^4  1, but p does
not divide (2*y)^2  1.
Therefore, 2*y has multiplicative order 4 (mod p), from which it follows that p1 is divisible by 4.