4. Given that the integer $n$ is not a multiple of 5. Then the remainder when $n^{4}+4$ is divided by 5 is $\qquad$ .
4. 0 .
$$
\text { Given } \begin{aligned}
& n^{4}+4=\left(n^{4}-1\right)+5 \\
= & \left(n^{2}+1\right)\left(n^{2}-1\right)+5 \\
= & \left(n^{2}-4\right)\left(n^{2}-1\right)+5\left(n^{2}-1\right)+5 \\
= & (n+2)(n-2)(n+1)(n-1) \\
& +5\left(n^{2}-1\right)+5,
\end{aligned}
$$

and $n$ is not a multiple of 5, so among $n+2, n-2, n+1, n-1$, there must be one that is a multiple of 5, thus $n^{4}+4$ is also a multiple of 5. Therefore, $n^{4}+4 \equiv 0(\bmod 5)$