Prove that, for $p > 3$, $(\frac 3p) = 1$ when $p \equiv 1,11 \pmod{12}$ and $(\frac 3p) = -1$ when $p \equiv 5,7 \pmod{12}$

$(\frac 3p)$ is the Legendre symbol here, not a fraction, if that wasn’t clear.

This is what I have so far:

We know by Gauss’s Lemma that $(\frac qp) = (-1)^v$ where
$$v = \#\left\{1 \le a \le \frac{p-1}{2} \mid aq \equiv k_a \pmod{12}),\space -p/2 < k_a < 0\right\}$$

And we also know that multiples of $(\dfrac{p-1}{2}) \times 3$ must be between $0$ and $p/2$, between $p/2$ and $p$, or between $p$ and $3p/2$.

Representing $p$ as $12k + r$ seems like the next step, but I’m not sure where to go from there.

Solutions Collecting From Web of "Prove that, for $p > 3$, $(\frac 3p) = 1$ when $p \equiv 1,11 \pmod{12}$ and $(\frac 3p) = -1$ when $p \equiv 5,7 \pmod{12}$"

By Quadratic Reciprocity $$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)(-1)^{(p-1)(3-1)/4}=\left(\frac{p}{3}\right)(-1)^{(p-1)/2}=\left(\frac{p}{3}\right)\left(\frac{-1}{p}\right).$$ We will use the fact that $-1$ is a quadratic residue modulo $p$ if $p\equiv 1\pmod{4}$ and is not if $p\equiv -1\pmod{4}$. Since $3\nmid p$, $p\equiv 1,2\pmod{3}$. Thus, $\left(\frac{p}{3}\right)=\left(\frac{1}{3}\right)=1$ or $\left(\frac{p}{3}\right)=\left(\frac{2}{3}\right)=-1$. Now, for $\left(\frac{3}{p}\right)=1$ either $\left(\frac{p}{3}\right)=1$ and $\left(\frac{-1}{p}\right)=1$, or they are both equal to $-1$. In the former case, $p\equiv 1\pmod{4}$ and $p\equiv 1\pmod{3}$, so by CRT, $p\equiv 1\pmod{12}$. In the latter case $p\equiv 3\pmod{4}$ and $p\equiv 2\pmod{3}$, so again by CRT $p\equiv 11\equiv -1\pmod{12}$. The result follows from here.