49th Mathematical Olympiad in Ukraine

Let us rewrite our equation in the following way: $p(2p+1)=(m-3)(m+3)$. Since $p$ is prime, we have that $(m-3)\nmid p$ or $(m+3)\nmid p$.

$$1)(m-3)\nmid p\Rightarrow m-3=kp\Rightarrow (m+3)>kp\text{and}$$ $$3p^2>p(2p+1)=(m-3)(m+3)>k^2{p}^2,\text{therefore}3p^2>k^2{p}^2\Rightarrow k=1,\text{and thus}$$ $$\{\begin{array}{l}m-3=p\\ m+3=2p+1\end{array}\Rightarrow \{\begin{array}{l}p=5\\ m=8\end{array}\text{this is the first solution.}$$

2) $(m+3)p$, so $m+3=kp$. If $p>5$, then $m-3=kp-6>kp-k=p(k-1)$. Analogously, $3p^2>p(2p+1)=(m-3)(m+3)>(k-1)k{p}^2$, which implies that $k=1$ or $k=2$. Number $p(2p+1)=(m-3)(m+3)$ is odd, thus $k\ne 2$. For $k=1$ we have that $$\{\begin{array}{l}m+3=p\\ m-3=2p+1\end{array}$$ which is impossible.

It remains to consider the cases $p=2,3,5$. It can be easily seen that in these cases there're no other solutions.