2019/014) Find the values of n such that $n^4+4$ is a prime

We have $n^4+4 = n^4+4n^2 + 4 - 4n^2 = (n^2+2)^2 -(2n)^2 = (n^2+2n+2)(n^2-2n +2)$
$n^4+4$ is a prime iff $n^2+2n+2$ is a prime and $n^2-2n+2=1$
$n^2-2n+2=1=>(n-1)^2 = 0$  or n = 1
And for n= 1 $n^2+2n+2=5$ which is a prime
we could also compute $n^4+4= 1 + 4 =5$
So 1 is the only choice  for n