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

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.

• First hint: note that $\left(\frac{a}{p}\right) = \left(\frac{2}{p}\right)^2\left(\frac{a}{p}\right) = \left(\frac{4}{p}\right)\left(\frac{a}{p}\right)= \left(\frac{4a}{p}\right).$ – Sam Streeter Oct 30 '18 at 10:07
• Quadratic reciprocity and a little bit of knowledge of how to manipulate Legendre symbols will suffice for the rest. – Sam Streeter Oct 30 '18 at 10:11
• Hint: $p$ and $q$ both leave the same remainder on division by $4a$. – Moed Pol Bollo Oct 30 '18 at 10:13
• sorry, overread that there was the tag legendre symbol. – Enkidu Oct 30 '18 at 10:13
• @AlexyVincenzo So can you solve the problem now? – Sam Streeter Oct 30 '18 at 15:31

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