5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.

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5.4.

If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$

Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (corresponding to $n^{4}-3 n^{2}+9=7$ or 13, both of which are prime numbers).
If $n<0$, then $n=-1$ or -2.
In summary, there are 4 integers $n$ that satisfy the condition.