Show that $n^{4}-20n^{2}+4$ is a composite whenever $n$ is an integer

Question

Show that $n^{4}-20n^{2}+4$ is a composite whenever $n$ is an integer. I'm having trouble putting together a proof for this problem.

Answer 1

You can factor this as $(n^2-4n-2)(n^2+4n-2)$. Now you just have to show that neither factor is $1$.

Answer 2

I wanted to explain where the factorization in Ross Milikan's answer comes from, as it may not be evident at first sight.

Looking at $n^4-20n^2+4$, it doesn't seem like the quadratic $x^2-20x+4$ factors nicely. However, it looks fairly close to $x^2-4x+4$ which we know is $(x-2)^2$. This leads us to write $$n^4-20n^2+4=\left(n^4-4n^2+4\right)-\left(16n^2\right).$$ This becomes $$\left(n^2-2\right)^2 -(4n)^2,$$ which is a difference of squares, and factors into $$(n^2-4n-2)(n^2+4n-2).$$