Indian National Mathematical Olympiad

Solution:
Obviously $p \neq q$. We write this in the form
$$p\left(p^{n-1}+p^{n-2}+\cdots+1\right)=q(q+1)$$
If $q \leq p^{n / 2}-1$, then $q<p^{n / 2}$ and hence we see that $q^{2}<p^{n}$. Thus we obtain
$$q^{2}+q<p^{n}+p^{n / 2}<p^{n}+p^{n-1}+\cdots+p$$
since $n>2$. It follows that $q \geq p^{n / 2}$. Since $n>2$ and is an even number, $n / 2$ is a natural number larger than 1. This implies that $q \neq p^{n / 2}$ by the given condition that $q$ is a prime. We conclude that $q \geq p^{n / 2}+1$. We may also write the above relation in the form
$$p\left(p^{n / 2}-1\right)\left(p^{n / 2}+1\right)=(p-1) q(q+1)$$
This shows that $q$ divides $\left(p^{n / 2}-1\right)\left(p^{n / 2}+1\right)$. But $q \geq p^{n / 2}+1$ and $q$ is a prime. Hence the only possibility is $q=p^{n / 2}+1$. This gives
$$p\left(p^{n / 2}-1\right)=(p-1)(q+1)=(p-1)\left(p^{n / 2}+2\right)$$
Simplification leads to $3 p=p^{n / 2}+2$. This shows that $p$ divides $2$. Thus $p=2$ and hence $q=5, n=4$. It is easy to verify that these indeed satisfy the given equation.