Prime or Composite

By the Sophie Germain identity, $n^4+4=(n^2+2n+2)(n^2-2n+2)$ . For it to be a prime number, one of the two terms must be 1. So either $n^2+2n+2=1$ or $n^2-2n+2=1$ . The former case is $(n+1)^2=0$ so $n=-1$ and the latter case is $(n-1)^2=0$ so $n=1$ . Therefore, sum of all integer values of n is $\color{#D61F06}{\boxed{0}}$ .

$n^{4}$ + 4

$n^{4}$ + 4 $n^{2}$ + 4 - 4 $n^{2}$

[( $n^{2}$ + 2)^{2}]- 4 $n^{2}$

( $n^{2}$ +2n+2)( $n^{2}$ -2n+2)

.•. for +ve integers, for $n^{4}$ + 4 to be prime,

( $n^{2}$ -2n+2) = 1

.•. $n^{2}$ -2n+1 = 0

.•. $(n-1)^{2}$ = 0

.•. n-1 = 0

.•. $\boxed{n = 1}$

Similarly, for -ve integers,

( $n^{2}$ +2n+2) = 1

$(n+1)^{2}$ = 0

$\boxed{n = -1}$

.•. 1 + (-1) = 0

Hence, $\boxed{0}$