Find all integers $n$ such that $n^{3}-3 n^{2}+n+2$ is a power of 5.

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We then have: $(n-2)\left(n^{2}-n-1\right)=5^{a}$. Thus, we have: $n-2=5^{x}$ and $n^{2}-n-1=5^{y}$. It follows that $5^{y}-5^{2 x}-3 \cdot 5^{x}=1$. Therefore, if $x, y \geqslant 1$ then $0 \equiv 1[5]$, which is absurd. Therefore, $x=0$ or $y=0$. If $x=0$, then $y=1$, which gives $n=3$. If $y=0$, then $5^{2 x}+3 \cdot 5^{x}=0$, which is absurd. The only solution is thus $n=3$.