Suppose $p$ and $q$ are odd primes and $p = q + 4a$ for some $a$. Prove that $(\frac ap) = (\frac aq)$ holds.

Question

This problem also had me prove that $(\frac pq) = (\frac aq)$, but I've already managed to do that. I've tried messing around with the Law of Quadratic Reciprocity but can't get anything. I've also tried using the first identity by proving that $(\frac pq) = (\frac ap)$, but again I've got nothing.

Answer 1

If $p=q+4a$, then $p\equiv q$ mod $4$, in which case the Law of Quadratic Reciprocity says

$$\left({p\over q}\right)=\left({-q\over p}\right)$$ But this gives

$$\left({a\over q}\right)=\left({q+4a\over q}\right)=\left({p\over q}\right)=\left({-q\over p}\right)=\left({4a-p\over p}\right)=\left({a\over p}\right)$$

(using the fact that $4$ is a square, hence a quadratic residue for both $p$ and $q$).