367. Find all natural $n$, for which the number $n^{4}+4$ is composite.
$\triangle$ Let's try to factor the expression $n^{4}+4$ into factors with integer coefficients. We are accustomed to the fact that the sum of squares does not factor into factors with integer coefficients. It turns out that sometimes it does. In this case, it is done using the "plus-minus" trick as follows:

$$
n^{4}+4=\left(n^{4}+4+4 n^{2}\right)-4 n^{2}=\left(n^{2}+2\right)^{2}-4 n^{2}=\left(n^{2}+2 n+2\right)\left(n^{2}-2 n+2\right)
$$

It is clear that the factor $n^{2}+2 n+2$ is always greater than 1. The second factor $n^{2}-2 n+2$ can be equal to 1:

$$
n^{2}-2 n+2=1, \quad n^{2}-2 n+1=0, \quad(n-1)^{2}=0, \quad n=1
$$

Since when $n=1$, the factor $n^{2}+2 n+2$ takes the value 5, which is a prime number, the value $n=1$ must be discarded.

Answer: all $n \neq 1$.