Prove for an integer $n = a^2+b^2$ with prime factorization $n=2^{e}p\_{1}p\_{2}...p\_rM^2$ (where $p\_1, p\_2, ...p\_r$ are distinct odd primes and e = 0 or 1), then $p\_i=1 mod 4$ for all i. My first thought is to use quadratic reciprocity but other than that i cannot think of anything else. Any hints?