Not quartics again

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

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

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

$n^4 + 4 = ((n+1)^2 +1)((n-1)^2+1)$

Clearly $((n+1)^2 +1)>1$ if $n$ is a positve integer. So the other factor must be equal to 1 to make it prime:

$(n-1)^2+1 = 1 \rightarrow n=1$

$\therefore n^4+4$ iff $\boxed{n=1}$

n^4 + 4 = n^4 + 4n^2 + 4 - 4n^2 = (n^2 + 2)^2 - (2n)^2 = (n^2 + 2 - 2n)*(n^2 +2 + 2n). If n^4 + 4 is prime, then (n^2 + 2 - 2n) must equal 1; So n = 1. Ed Gray