One question about quadratic residue

Question

If a prime number $p$ is $1 \bmod 3$, how does one prove that there exists an integer $n$ such that $p$ divides $n^2+3$ ?

Help me, please.

Answer 1

As remarked in the comments, show that $(\frac{-3}{p})=1$ where the brackets are the Legendre-symbol.

Because $p \equiv 1 \mod 3$, we have $(\frac{p}{3})=1$. This and the law of quad. reciprocity gives you $$\bigg(\frac{-3}{p}\bigg)=\bigg(\frac{-1}{p}\bigg)\bigg(\frac{3}{p}\bigg)=\bigg(\frac{-1}{p}\bigg)(-1)^{\frac{p-1}{2}}\bigg(\frac{p}{3}\bigg)=\bigg(\frac{-1}{p}\bigg)(-1)^{\frac{p-1}{2}}$$ To compute the last term, distinguish the two cases $p \equiv 1 \mod 4$ and $p \not\equiv 1 \mod 4$. (This should be quite easy now.)