If $p>3$ and $x^2+3y^2=p$, prove $\left(\frac{-3}{p}\right)=1$ and $p\equiv 1\pmod{6}$.
$$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)$$
We know that $$\left(\frac{3}{p}\right)= \left(\frac{p}{3}\right)\times (-1)^{(p-1)(3-1)/4}= \left(\frac{p}{3}\right) \times (-1)^{(p-1)/2}$$
Can anyone explain how we get from:
$$\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)=(-1)^{(p-1)(q-1)/4}$$ to
$$\left(\frac{q}{p}\right)=\left(\frac{p}{q}\right)\times (-1)^{(p-1)(q-1)/4}$$
Using 3/p Legendre Calculation
I see that they get answer modulo $12$, not $6$.
I can see that $x^2\equiv -3y^2 \pmod{p} \Rightarrow (xy^{-1})^2\equiv -3 \pmod{p} \Rightarrow -3$ is Q.R. (since $y \equiv 0 \pmod{p}$ is not true)
It's because if $p\neq q$, then $\left(\frac{p}{q}\right)=\frac{1}{\left(\frac{p}{q}\right)}$.
If $p>3$ and $x^2+3y^2=p$, then, for contradiction, let $p\mid y$. But then $p\mid p-3y^2=x$, so $p^2\mid x^2,y^2$, so $p^2\mid x^2+3y^2=p$, contradiction. Therefore $\gcd(y,p)=1$, so $$x^2\equiv -3y^2\pmod{p}\iff \left(xy^{-1}\right)^2\equiv -3\pmod{p}\implies \left(\frac{-3}{p}\right)=1$$
By Quadratic Reciprocity (QR) Lemma (see below) this is equivalent to $p\equiv 1\pmod{3}$, and since $p>3$ and $p$ is odd, we get $p\equiv 1\pmod{6}$.
The QR Lemma: If $p>3$, then $\left(\frac{-3}{p}\right)=1\iff p\equiv 1\pmod{3}$.
Proof: If $p\equiv 1\pmod{4}$, then $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=1\iff p\equiv 1\pmod{3}$$
If $p\equiv 3\pmod{4}$, then $$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=-\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)=1\iff p\equiv 1\pmod{3}$$