= Solution
<Quadratic reciprocity> with the <prime> three gives
$$
\left(\frac3p\right)=(-1)^{(p-1)/2}\left(\frac p3\right).
$$
The latter symbol is $+1$ when $p\equiv1\pmod3$ and $-1$ when $p\equiv2\pmod3$. A <prime> greater than three occupies one of the classes $1,5,7,11$ modulo twelve. Combining its residues modulo three and four gives signs $+1,-1,-1,+1$ respectively. Hence
$$
\boxed{\left(\frac3p\right)=1\text{ for }p\equiv\pm1\pmod{12},\qquad
\left(\frac3p\right)=-1\text{ for }p\equiv\pm5\pmod{12}.}
$$
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