
1. Find all integers $n$ such that $n^{4}-3 n^{2}+9$ is prime.

(Aleš Kobza)


Solution. We factorize the given expression:

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

Should the product be a prime $p$, one of the factors should be 1 or -1 and the other one then $p$ or $-p$. But both of the factors are only positive (they have negative discriminants), thus the only possibilities are

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

The roots of the first equation are $n=-1$ and $n=-2$, the second equation has roots $n=1$ and $n=2$. In both cases the values of the other factor are 7 or 13 which are primes.

Conclusion. The number $n^{4}-3 n^{2}+9$ is prime iff $n \in\{-2,-1,1,2\}$.
