Equal Legendre symbols.

Question

Any hints for proving that if $p$ and $q$ are odd primes such that $p=4a+q$ for some integer $a$ then $\left(\frac{a}{p}\right) = \left(\frac{a}{q}\right)$? I have tried using quadratic reciprocity but I don't seem to get to any answer, I suspect it has something to do with the fact that $p\equiv q \text{ } ( \text{ mod } 4) $ but I don't know how to use this fact. I can use any result given on a first number theory course. So far i have this: Since $p$ is congruent to $4a$ modulo $q$ then $\left( \frac{p}{q}\right) = \left( \frac{4a}{q}\right) = \left( \frac{4}{q}\right) \left(\frac{a}{q}\right)$. Similarly $\left(\frac{q}{p}\right) = \left(\frac{-4a}{p}\right) = \left( \frac{-1}{p}\right)\left(\frac{4}{p}\right)\left(\frac{a}{p}\right)$. Then $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = \left(\frac{4}{q}\right)\left(\frac{a}{q}\right)\left(\frac{-1}{p}\right)\left(\frac{4}{p}\right)\left(\frac{a}{p}\right)$. Thus, by quadratic reciprocity, $(-1)^{\frac{p-1}{2}\frac{q-1}{2}} = \left(\frac{4}{q}\right)\left(\frac{a}{q}\right)\left(\frac{-1}{p}\right)\left(\frac{4}{p}\right)\left(\frac{a}{p}\right)$, but $\frac{p-1}{2}\frac{q-1}{2}=2a+\left(\frac{q-1}{2}\right)^2$, so $\left(\frac{4}{q}\right)\left(\frac{a}{q}\right)\left(\frac{-1}{p}\right)\left(\frac{4}{p}\right)\left(\frac{a}{p}\right) = (-1)^{\left(\frac{q-1}{2}\right)^2}$.

Answer 1

From what I can see you have already shown that $$\left(\frac{p}{q}\right)=\left(\frac{a}{q}\right)$$ Further, $$\left(\frac{q}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)$$ Since $$\left(\frac{q}{p}\right)=\begin{cases} -\left(\frac{p}{q}\right) &\text{if }p\equiv q\equiv 3\mod 4\\ \left(\frac{p}{q}\right) & \text{if } p\equiv 1\mod 4\text{ or }q\equiv 1\mod 4 \end{cases}$$ and $$\left(\frac{-1}{p}\right)=\begin{cases} +1 & \text{if }p\equiv 1\mod 4\\ -1 & \text{if }q\equiv 3\mod 4 \end{cases}$$ The answer follows from the fact that $p\equiv q\mod 4$