4. Find all natural numbers $n$ such that $n-4$, $2n+2$, and $4n+1$ are perfect cubes.
4. Let $F=(n-4)(2 n+2)(4 n+1)=8 n^{3}-22 n^{2}-38 n-8$. According to the problem, $F$ must be a perfect cube. Since $F$ is an even number and it is clear that $F<(2 n)^{3}$, it follows that $F \leqslant(2 n-2)^{3}=8 n^{3}-24 n^{2}+24 n-8$, which reduces to $2 n^{2}-62 n \leqslant 0$, from which we get $n \leqslant 31$. We immediately see that $n=31$ is one solution $(31-4=$ $27=3^{3}, 2 \cdot 31+n=64=4^{3}$ and $4 \cdot 31+1=125=5^{3}$ ), and it is directly verified that there are no smaller solutions.