Proving that if prime p>3 divides a^2 + 12, then p is congruent 2(mod 3)

Question

Prove that if prime $p>3$ divides $a^2 + 12$, then $p$ is congruent $2\pmod 3$.

I tried splitting (-12/P) to (-1/P) * (3/P) and solving 4 different cases to find when (-12/P) = 1, but I got that p is congruent to 1(mod 3). What's wrong?

Answer 1

Welcome to Mathematics Stack Exchange. I think you are correct and the statement is wrong,

because $\left(\dfrac{-12}p\right)=\left(\dfrac4p\right)\left(\dfrac{-3}p\right)=\left(\dfrac {-3}p\right)$, which is $1$ if $p\equiv1\pmod3$,

and $ -1$ if $p\equiv2\pmod3.$