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
$$n^2 - n + 1 = pk, \quad k \in \mathbb{N}. \qquad (1)$$
Thus $p-1 = k(n+1)$ or
$$p = kn + k + 1. \qquad (2)$$
Substituting $p$ into (1) gives
$$n^2 - (k^2 + 1)n - (k^2 + k - 1) = 0. \qquad (3)$$
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.