1. Find all integers $n$ for which $n^{4}-3 n^{2}+9$ is a prime number.
Solution. The given expression can be factored with a simple modification:

$$
n^{4}-3 n^{2}+9=n^{4}+6 n^{2}+9-9 n^{2}=\left(n^{2}+3\right)^{2}-(3 n)^{2}=\left(n^{2}+3 n+3\right)\left(n^{2}-3 n+3\right) .
$$

For the product of two integers to be a prime number $p$, one of the factors must be 1 or -1 (and the other $p$, respectively $-p$). However, the discriminants of both quadratic trinomials are $(\pm 3)^{2}-4 \cdot 4=-3$, which are negative, so both trinomials only take positive values. Therefore, it is sufficient to consider only the quadratic equations

$$
n^{2}+3 n+3=1 \quad \text { and } \quad n^{2}-3 n+3=1 \text {. }
$$

The solutions to the first equation are $n=-1$ and $n=-2$, for which the second factor takes the values 7 and 13, which are prime numbers.

The solutions to the second equation are $n=1$ and $n=2$, for which the first factor again takes prime values 7 and 13.

Answer. The given expression is a prime number precisely when $n \in\{-2,-1,1,2\}$.