5. Find all natural values of $n$ such that the number $n^{4}+64^{n}$ is composite.
5.I.5. Let's use the identity

$$
x^{4}+4 y^{4}=\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2}=\left(x^{2}-2 x y+2 y^{2}\right)\left(x^{2}+2 x y+2 y^{2}\right) .
$$

If the numbers $x$ and $y$ are natural, then the number $x^{2}-2 x y+2 y^{2}=(x-y)^{2}+y^{2}$ can be equal to 1 only if $x=y=1$. Therefore, for all other values of $x$ and $y$, the number $x^{4}+4 y^{4}$ is composite.

If $n$ is an even number, then $n^{4}+64^{n}$ is an even number, and therefore it is not prime. Let $n=2 k+1$. Then $64^{n}=64 \cdot 64^{2 k}=64 \cdot 8^{4 k}=4 \cdot\left(2 \cdot 8^{k}\right)^{4}=4 m^{4}$. As already proven, the number $n^{4}+64^{n}=n^{4}+4 m^{4}$ is composite.

Answer: $n$ is any natural number.