For any odd prime $p≥5$, $(-3/p) = 1$ if $p ≡ 1,-5\pmod{12}$ and $(-3/p) = -1$ if $p ≡ -1,5 \pmod{12}$

Question

For any odd prime $p≥5$, $(-3/p) = 1$ if $p ≡ 1,-5\pmod{12}$ and $(-3/p) = -1$ if $p ≡ -1,5 \pmod{12}$.

My solution so far: if $p ≡ 1\pmod{4}$ then $(-3/p) = 1$ exactly when $p ≡ ±1\pmod{12}$ when $p ≡ 1\pmod{12}$.

And if $p ≡ -1\pmod{4}$ then $(-3/p) = 1$ exactly when $p ≡ ±5\pmod{12}$ whence $p ≡ -5\pmod{12}$. So $(-3/p) = 1$ exactly when $p ≡ 1\pmod{12}$ or $p ≡ -5\pmod{12}$.

I'm not very confident in my work. Is this correct and if it's not, how do you solve this problem?