Find all integers $n$ for which $81\frac{n^4}{4}-2017n^2+81$ is a prime.
I know completing the square helps with this problem, but I'm not how completing the square is going to get me to the right answer.
When I complete the square I get: $(\frac{9}{2}n^2-9)^2-\frac{6609}{4}n^2$
HINT
Following the hint given in the comments, we need that $n=2k$ and therefore
$$81\frac{n^4}{4}-2017n^2+81=324k^4-8068k^2+81$$
and from $(2n-m)^2=4n^2-4nm+m^2$ we have
$$324k^4-8068k^2+81=[4(9k^2)^2-4\cdot 9k^2\cdot 9+9^2]-88^2k^2$$