<s>[INST] Legendre symbol of $(-3/p)$. Question. Show $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1\bmod 3,\\
-1 & p\equiv 2\bmod 3. \end{cases}$$
Attempt. So, using the established results of $$\Big(\frac{3}{p} \Big) = \begin{cases}+1 & p\equiv\pm1\bmod 12,\\
-1 & p\equiv\pm5\bmod 12,\end{cases}$$ $$\Big(\frac{-1}{p}\Big)=\begin{cases}+1 & p\equiv1\bmod 4,\\
-1 & p\equiv3\bmod 4,\end{cases}$$ and observing that $$\Big(\frac{-1}{p} \Big) = \begin{cases}+1 & p\equiv1,5\bmod 12,\\
-1 & p\equiv7,11\bmod 12,\end{cases}$$ we get $$\Big(\frac{-3}{p}\Big)=\begin{cases}+1 & p\equiv 1,7\bmod 12,\\
-1 & p\equiv 5,11\bmod 12. \end{cases}$$ Providing I've carried this out right;  how can I reduce this further so it is in the form of modulo $3$?
 [/INST] You have that $p\equiv 1,7\pmod{12}$ if and only if $p\equiv 1\pmod{3}$ and $p$ odd.
Similarly for $p\equiv 5,11\pmod{12}$ and $p$ odd with $p\equiv 2\pmod{3}.$\
There is an easier approach if you know quadratic reciprocity:
$$\begin{align}\left(\frac{-3}{p}\right)&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\
&=\left(-1\right)^{\frac{p-1}{2}}\left(-1\right)^{\frac{p-1}{2}\frac{3-1}{2}}\left(\frac{p}{3}\right)\\
&=(-1)^{p-1}\left(\frac p3\right)\\&=\left(\frac{p}{3}\right)
\end{align}$$
More generally, when $q\equiv 3\pmod{4}$ is a prime you get that:
$$\left(\frac{-q}{p}\right)=\left(\frac{p}{q}\right)$$
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