If there is an integer $n$ such that $n^2\equiv3\pmod p$, where $p$ is prime, prove there are integers $a$ and $b$ such that $|a^2-3b^2|=p$

Question

If there is an integer $n$ such that $n^2\equiv3\pmod p$, where $p$ is prime, prove there are integers $a$ and $b$ such that $|a^2-3b^2|=p$.

So $n^2-3 = pm$ for some integer $m$, and I know $|a^2 - 3b^2|$ is the norm map of an element in $\mathbb{Z}[\sqrt{3}]$, but I'm having difficulty piecing these pieces together.