I'm trying to determine for which primes $p$ we have $\displaystyle\binom{6}{p}=1$, where $\displaystyle\binom{6}{p}$ is the the Legendre symbol. I know how to do this, but my question what is the quickest way? Breaking the problem down we have $$\binom{3}{p}\binom{2}{p}=1$$ so let's suppose they are both equal to $1$. Then $\displaystyle\binom{2}{p}=1\iff p\equiv\pm1\mod8$ and $\displaystyle\binom{3}{p}=1\iff\binom{p}{3}(-1)^{\frac{p-1}{2}}=1\iff p\equiv1\mod 3\wedge p\equiv1\mod 4$ OR $p\equiv2\mod 3\wedge p\equiv3\mod 4$
So that's half of the problem. My question is, is there a more efficient way to do this rather than ending up with $8$ systems of congruence equations? Or any tricks to do it more quickly?