Ukrainian National Mathematical Olympiad, 3rd Round

WLOG, $p \le q$, clearly $r > p$. We can write:
$$p(p+1) = (r-q)(r+q+1). \quad (1)$$
Since $p$ is prime, then either $r-q$ or $r+q+1$ is divisible by $p$. If $p$ divides $r-q$, then $r-q \ge p$,
$$p(p+1) \le (r-q)(r-q+1) < (r-q)(r+q+1),$$
which contradicts to (1).

Let suppose that $r+q+1$ is divisible by $p$. If $p=2$, then $r+q+1$ has to be even, which implies that $r$ is odd, hence $q$ is even, thus $q=2$ and $r=3$.

If $p > 2$, let $r+q+1 = kp$. Then $r$ and $q$ are odd and $k$ is odd, $k > 1$. Then $p+1 = k(r-q)$ and we get $k^2(r-q) = kp+k$ or $r+q+1+k = k^2r-k^2q$ which is equivalent to
$$(k^2+1)q = (k^2-1)r - (k+1).$$
RHS is divisible by $(k+1)$, thus LHS is divisible by the same number, since $k$ is odd, then $\frac{k^2+1}{2}$ is divisible by $\frac{k+1}{2}$. Observe, that
$$\left(\frac{k^2+1}{2}, \frac{k+1}{2}\right) = \left(\frac{k^2+1}{2} - \frac{(k-1)(k+1)}{2}, \frac{k+1}{2}\right) = \left(1, \frac{k+1}{2}\right) = 1.$$
$\frac{k^2+1}{2}$ and $\frac{k+1}{2}$ are coprime, so $\frac{k+1}{2}$ is divisible by $q$ and since $k > 1$, then $\frac{k+1}{2} > 1$ and $q = \frac{k+1}{2}$.

We get $r = kp - q - 1$ and $q = \frac{k+1}{2}$. Plugging in this into $p+1 = k(r-q)$ we get $p+1 = (kp - k - 2)k$. But the last equality is impossible because $k \ge 3$, $p \ge 3$ implies
$$(kp - k - 2)k > kp - k - 2 = k(p-1) - 2 \ge 3(p-1) - 2 = p + (2p-5) \ge p+1.$$
This completes the proof.