Find all positive integers $n$ such that $n^{4}-27 n^{2}+121$ is a prime positive integer.
We can rewrite $n^{4}-27 n^{2}+121$ as

$$
\left(n^{4}+22 n^{2}-121\right)-49 n^{2}=\left(n^{2}+11\right)^{2}-(7 n)^{2}=\left(n^{2}+7 n+11\right)\left(n^{2}-7 n+11\right) .
$$

For this to be prime, we would need $n^{2}-7 n+11=1$. Rearranging and factoring gives $(n-2)(n-5)=0$, so $n=2$ or $n=5$. If $n=2$, the expression equals 29 , which is indeed prime, and if $n=5$, the expression equals 71 , which is also prime. Thus, our answer is $n=2,5$.