Solution
= Solution
<Euler's criterion> implies multiplicativity of the symbol on nonzero residues and gives $(-1/p)=(-1)^{(p-1)/2}$. Apply <quadratic reciprocity> with the odd <prime> 3:
$$
\left(\frac{-3}p\right)=(-1)^{(p-1)/2}\left(\frac3p\right)=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\left(\frac p3\right)=\left(\frac p3\right).
$$
The only nonzero square modulo three is one. Since $p>3$,
$$
\boxed{\left(\frac{-3}p\right)=\begin{cases}1,&p\equiv1\pmod3,\\-1,&p\equiv2\pmod3.\end{cases}}
$$