Belarusian Mathematical Olympiad

Answer: $(n; p) = (20; 127)$.
It is easy to see that the given equality
$$p(p-1) = 2(n^3 + 1) \qquad (1)$$
does not hold for $p=2$ and positive integer $n$. So $p \ge 3$ is an odd prime number. Then $(n+1)(n^2 - n + 1)$ is divisible by $p$.

1. If $(n+1) \nmid p$, then $n+1 = kp$ for some positive integer $k$. In particular, $n+1 \ge p$. From (1) we have $p(p-1) = 2(n+1)(n^2-n+1) \ge 2p(n^2-n+1)$, and so $p-1 \ge 2n^2-2n+2$. Then $n \ge p-1 \ge 2n^2-2n+2$ or $2n^2-3n+2 \le 0$, which is impossible.

2. Therefore $n^2 - n + 1 \nmid p$, i. e.
$$n^2 - n + 1 = kp \qquad (2)$$
for some positive integer $k$. Substituting (2) in (1), we obtain $p-1 = 2k(n+1)$ or
$$p = 2kn + 2k + 1. \qquad (3)$$
Substituting (3) in (2), we obtain $n^2 - n + 1 = 2k^2n + 2k^2 + k$ or
$$n^2 - (2k^2 + 1)n - (2k^2 + k - 1) = 0. \qquad (4)$$
The discriminant $D$ of this quadratic equation with respect to $n$ is equal to $D = (2k^2 + 1)^2 + 4(2k^2 + k - 1)$ and $D$ is evidently an odd number, $D > (2k^2 + 1)^2$; on the other hand $D < (2k^2 + 5)^2$ (indeed,
$$(2k^2+1)^2+4(2k^2+k-1) < (2k^2+5)^2 \iff 4(2k^2+k-1) < (4k^2+6) \cdot 4 \iff k-7 < 2k^2.$$
Therefore, $D = (2k^2+1)^2 + 4(2k^2+k-1) = (2k^2+3)^2$ ($D$ must be perfect square, otherwise (4) has no integer roots). Thus,
$$4(2k^2+k-1) = (2k^2+3)-(2k^2+1) = ((4k^2+4) \cdot 2 \iff 2k^2+k-1 = 2k^2+2$$
$$\iff k = 3.$$
Then (4) has the form $n^2 - 19n - 20 = 0$, so $n = 20$. From (3) we obtain $p = 2 \cdot 3 \cdot 20 + 2 \cdot 3 + 1 = 127$, and $p$ is the prime number indeed.