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Suppose q and p are odd primes and p = q+4a for some integer a. From the properties of the Legendre symbol and the Law of Quadratic Reciprocity prove that the following two identities hold.

(p/q) = (a/q), and (a/p) = (a/q)

So i first did the LAW, such that if p=q=3 (mod4) then either (p/q) = -(q/p) or (p/q) = (q/p)

with that i did some mod stuff like, p = 4a(mod q) and 4a = p (mod q) this kind of implies that (p/q) = (a/q) right?

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Since $p=q+4a$ we have $$ \left( \frac{p}{q}\right)=\left(\frac{4}{q}\right)\left(\frac{a}{q}\right)$$ but as $\left(\frac{4}{q}\right)=1$ ($4$ is a quadratic residue modulo $p$ because $4\equiv 2^2\pmod q$) then $$ \left( \frac{p}{q}\right)=\left(\frac{a}{q}\right)$$

on the other hand by the reciprocity law we have $$ \left( \frac{p}{q}\right)=\left(\frac{a}{q}\right)=(-1)^{\frac{(p-1)(q-1)}{4}}\left( \frac{q}{p}\right)= (-1)^{\frac{(p-1)(q-1)}{4}}\left( \frac{-1}{p}\right)\underbrace{\left(\frac{4}{p}\right)}_{=1}\left(\frac{a}{p}\right)=\underbrace{(-1)^{\frac{(p-1)(q-1)}{4}}\left( \frac{-1}{p}\right)}_{=1}\left(\frac{a}{p}\right)$$

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  • $\begingroup$ And for the second one: $$\left(\frac ap\right)=\pm\left(\frac {q+4a}a\right)=\pm\left(\frac qa\right)=\left(\frac aq\right)$$ $\endgroup$
    – DonAntonio
    Dec 19, 2013 at 21:50

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