51st Ukrainian National Mathematical Olympiad, 3rd Round

WLOG, $p\le q$, clearly $r>p$. We can write: $$p(p+1)=(r-q)(r+q+1).(1)$$ 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 $r+q+1+k=k^{2}r-k^{2}q$ 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.