$x^4+4$ is composite for $x>1$ 
$x^4+4$ is composite for $x>1$ 

I know the Sophie Germain indentity and the  get the factorization $$x^4+4 = (x^2+2-2x)(x^2+2+2x)$$
But I am stuck here. I cannot see any general factor here. 

A version in two variables is
$$ ( x^2 + 2 x y + 2 y^2) (x^2 - 2 x y + 2 y^2) = x^4 + 4 y^4 $$
where both quadratic forms are positive definite, so that there are only a finite number of $(x,y)$ pairs such that either factor is $1.$
