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.

  • 1
    $\begingroup$ First hint: note that $\left(\frac{a}{p}\right) = \left(\frac{2}{p}\right)^2\left(\frac{a}{p}\right) = \left(\frac{4}{p}\right)\left(\frac{a}{p}\right)= \left(\frac{4a}{p}\right).$ $\endgroup$ – Sam Streeter Oct 30 '18 at 10:07
  • $\begingroup$ Quadratic reciprocity and a little bit of knowledge of how to manipulate Legendre symbols will suffice for the rest. $\endgroup$ – Sam Streeter Oct 30 '18 at 10:11
  • $\begingroup$ Hint: $p$ and $q$ both leave the same remainder on division by $4a$. $\endgroup$ – Moed Pol Bollo Oct 30 '18 at 10:13
  • $\begingroup$ sorry, overread that there was the tag legendre symbol. $\endgroup$ – Enkidu Oct 30 '18 at 10:13
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    $\begingroup$ @AlexyVincenzo So can you solve the problem now? $\endgroup$ – 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.


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