4
$\begingroup$

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.

$\endgroup$
  • 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
  • 1
    $\begingroup$ @AlexyVincenzo So can you solve the problem now? $\endgroup$ – Sam Streeter Oct 30 '18 at 15:31
3
$\begingroup$

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

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.