TITLE: Show that $n^4-20n^2+4$ is composite when $n$ is any integer.
QUESTION [5 upvotes]: Crux Mathematicorum June 1978, 
Show that $n^4-20n^2+4$ is composite when $n$ is any integer. 

My question: What is meant by 'composite'? 
Besides, I found out that this expression could be factorised into $(n^2-4n-2)(n^2+4n-2)$. Is this related to the question? 
Hope someone can clarify what is meant by composite and what the question actually and solve this. Thanks in advance.

REPLY [3 votes]: Note that $$n^2-4n-2=(n-2)^2-6$$
and 
$$n^2+4n-2=(n+2)^2-6$$
For integer $n$, neither of $(n-2)^2$ or $(n+2)^2$ will give a value of $5,6,7$ (neither of these are perfect squares) hence neither of the factors are $-1,0,1$.  
As such, the product of both factors is a composite number.