you worked on this for any length of time without success, you'll kick yourself now: n4+4 can be factored. Here it is:

n^4+4n^4+4n^2+4-4n^2( n^4+4n^2+4) -4n^2((n^2+2)+2n)((n^2+2)-2n)

And since n>1 you can show that both factors are larger than one. You're done.

A similar problem was suggested to me, to prove that n^4-20n^2+4 is not prime. This one had me stumped for a while because I was trying this:

n^4- 20n^2+4n^4+4n^2+4-24n^2( n^4+4n^2+4) -24n^2((n2+2)+sqrt(24)(n))((n2+2)-sqrt(24(n))

But those factors, while real, are not integers. So it doesn't help in the effort to show compositude. Sorry, I can't help making up new words. Here's the real answer:

n4- 20n2+4n4-4n2+4-16n2( n4-4n2+4) -16n2((n2-2)+4n)((n2-2)-4n)

Here, the first factor is greater than one if n>0 -- that's the point where the n2and n terms overpower the constant. The second factor is greater than one if n>4, where the n2term overpowers the n term. So the polynomial is composite for all n>4. It's composite for n=0, 1, 2, 3, and 4, too, because its values are 4, -15, -60, -95, and -60.