5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$
5.11 .

Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$

Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rightarrow a=13, b=2 \\
\Rightarrow a-b=11
\end{array}
$$