I am studying Ireland & Rosen's Number Theory text, and I am asked to find the quadratic character of $-3$ modulo a prime $p$ via the method presented in $6.2$ of the text (in which they find the quadratic character of $2$). Here is my work:
As a start, let $\omega = e^{2\pi i /3}$ be a root of $x^3 - 1$. Let $\tau = 2\omega + 1$. Observe that $(2 \omega + 1)^2 = -3$. Raising both sides to the power $(p-1)/2$, I get
$$ (2 \omega + 1)^{p-1} \equiv (-3)^{(p-1)/2} \equiv \Big{(}\frac{-3}{p}\Big{)} \pmod{p}. $$
Next, I multiply by $(2 \omega + 1)$ to get
$$ (2 \omega + 1)\Big{(}\frac{-3}{p}\Big{)} \equiv (2 \omega + 1)^p \equiv (2\omega)^p + 1 \equiv 2\omega^{p} + 1\pmod{p}. $$
From here, I am at a stand-still. At this point in the text for finding the quadratic character of $2$, the authors consider cases for $p$ modulo $8$, but they don't explain this choice. Logically, I thought it wise to consider $p \equiv \pm 1 \pmod{3}$, which case $p \equiv 1 \pmod{3}$ implies $2\omega^p + 1 = \tau$, and $p \equiv -1 \pmod{3}$ implies $2\omega^p + 1 = -\tau$, but I can't seem to see the horizon still.
I would prefer a nudge, instead of a full answer, as I would like to continue this problem myself.