Let $p,q$ be odd primes such that $p-q=4a.$ Prove that $\Bigg(\dfrac{a}{p}\Bigg)=\Bigg(\dfrac{a}{q}\Bigg).$

Question

Let $p,q$ be odd primes such that $p-q=4a.$ Prove that $\Bigg(\dfrac{a}{p}\Bigg)=\Bigg(\dfrac{a}{q}\Bigg).$

Could anyone advise on how to prove the equality? Hints will suffice, thank you.

Answer 1

$\Big(\dfrac{a}{p}\Big)= \Big(\dfrac{4a}{p}\Big)=\Big(\dfrac{p-q}{p}\Big)=\Big(\dfrac{-q}{p}\Big) =\Big(\dfrac{-1}{p}\Big)\Big(\dfrac{q}{p}\Big)=(-1)^{\frac{p-1}{2}}\Big(\dfrac{q}{p}\Big)$

$\Big(\dfrac{a}{q}\Big)= \Big(\dfrac{p}{q}\Big)=(-1)^{\frac{p-1}{2} \cdot\frac{q-1}{2}}\Big(\dfrac{q}{p}\Big).$

If $\dfrac{p-1}{2}$ is even, then we are done. If $\dfrac{p-1}{2}$ is odd, then $p \equiv 3 \ (\text{mod} \ 4) \implies q+4a \equiv 3 \ (\text{mod} \ 4) \implies q-1 \equiv 2 \ (\text{mod} \ 4) \implies \dfrac{q-1}{2}$ is odd.