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

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!

(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 $$(\frac{3}{p})$$.