Prove a formula to evaluate Legendre Symbol $(\frac{-3}{p})$ using quadratic Gauss Sums (i.e. without using quadratic reciprocity)

Question

I've been trying to find a question similar to this, but have failed to find any. The full statement is let $\zeta = e^{\frac{2\pi i}{3}}$, and use the fact that $(2 \zeta + 1)^2 = -3$ and quadratic Gauss sums to prove a formula to evaluate $(\frac{-3}{p})$. I understand how to do this with reciprocity very easily, and I have a vague understanding of the proof of reciprocity using roots of unity and quadratic Gauss sums. Using reciprocity, this is what I came up with:

$$ (\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p}) $$

CASE 1: let $p \equiv 1 \text{ mod } 4$

$$ (\frac{-1}{p}) = 1 \\ (\frac{3}{p}) = (\frac{p}{3}) \\ \therefore (\frac{-1}{p})(\frac{3}{p}) = (\frac{p}{3})$$

CASE 2: let $p \equiv 3 \text{ mod } 4$

$$ (\frac{-1}{p}) = -1 \\ (\frac{3}{p}) = -(\frac{p}{3}) \\ \therefore (\frac{-1}{p})(\frac{3}{p}) = -1 \cdot -(\frac{p}{3}) = (\frac{p}{3})$$

Thus, it would be enough to perform quadratic reciprocity on $(\frac{3}{p})$ and evaluate the Legendre Symbol where $p \equiv 1 \text{ mod } 3$ and $p \equiv 2 \text{ mod } 3$. Knowing this, I try to formulate a claim using Gauss sums. Let $g(a,p)$ be the quadratic Gauss sum $\Sigma_{n=0}^{p-1} (\frac{n}{p}) \zeta_{p}^{an} $, and $\zeta_p = e^{\frac{2 \pi i}{p}}$, then

$$ g(-3,p) = (\frac{-3}{p})g(1,p) \\ \implies \frac{g(-3,p)}{g(1,p)} = (\frac{-3}{p})$$

Is this the correct way to go about proving a formula? And I am unsure when to use the statement $\zeta = e^{\frac{2\pi i}{3}}$. I know I somehow want to relate this to the fact that reciprocity shows I can evaluate this Legendre Symbol knowing if $p \equiv 1 \text{ or } 2 \text{ mod } 3$. Thanks for any help!

Answer 1

(for $p\ge 5$) $-3$ is a square iff $\zeta\in \Bbb{F}_p$ iff $3| p-1$, where $\zeta$ is a primitive 3rd root of unity, ie. a root of $x^2+x+1\in \Bbb{F}_p[x]$, either in $\Bbb{F}_p$ (if $x^2+x+1$ is reducible) or in $\Bbb{F}_{p^2}=\Bbb{F}_p[t]/(t^2+t+1)$.

In fact together with $(\frac{-1}{p})= (-1)^{(p-1)/2}$ this is how we find $(\frac{3}{p})$.