TITLE: If a prime $p$ satisfies $p \equiv 5 \mod 12$, then $p \in \mathbb{Z}(\sqrt{3})$ is irreducible.
QUESTION [7 upvotes]: I would like to show that if a prime  $p$ satisfies $p \equiv 5 \mod 12$, then $p \in \mathbb{Z}(\sqrt{3})$ is irreducible.   
The following fact, which I have proven, seems to be helpful:
Let $N$ be the norm map given by $N(a + b\sqrt{3})  = a^2 - 3b^2$.   If $p \in \mathbb{N}$ is a prime number such that there is no $x \in \mathbb{Z}(\sqrt{3})$ such that $N(x) = p$, then $p$ is irreducible in $\mathbb{Z}(\sqrt{3})$.
So I have to show that $p$ is not in the image of a norm map.  I said suppose it is ($N(x) = p$). Then we have $N(x^2) = N(x)N(x) = p^2 \equiv 1 \mod 12$.
But I don't really know where to proceed from there.  Any help is appreciated.

REPLY [6 votes]: Hint: We show that if $a^2-3b^2$ is divisible by $p$, then $a$ and $b$ are divisible by $p$. Suppose that $b$ is not divisible by $p$. From $3b^2\equiv a^2\pmod{p}$ we conclude that $3$ is a quadratic residue of $p$. However, by Reciprocity, $(3/p)=(p/3)=(2/3)=-1$.