Belorusija 2012

We have $p^2-p-1=n^3$. First, note that $p^2>n^3\ge n^2\Rightarrow p\ge n+1$. If $p=n+1$, then the equation becomes $n^3-n^2-n+1=0\Rightarrow n=1,p=2$, which is a solution of the problem.

Let now $p>n+1$. Rewrite the equation as $p(p-1)=(n+1)(n^2-n+1)$. So, $(n+1)(n^2-n+1)\nmid p$. Since $n+1<p$, we have $n^2-n+1\nmid p$, hence $$n^2-n+1=pk,k\in \mathbb{N}.(1)$$ Thus $p-1=k(n+1)$ or $$p=kn+k+1.(2)$$ Substituting $p$ into (1) gives $$n^2-(k^2+1)n-(k^2+k-1)=0.(3)$$ The discriminant of this equation is $D=k^4+6k^2+4k-3$ must be a square of an integer. But if $k\ge 4$, then it is easy to see that $(k^2+3{)}^2<D<(k^2+4{)}^2$, a contradiction.

It remains to inspect the cases $k=1,2,3$, which gives the answer.