Let's determine all prime numbers $p$ and $q$ for which, for some $n>1$,

$$
\frac{p^{2 n+1}-1}{p-1}=\frac{q^{3}-1}{q-1}
$$
On the right side, simplified and subtracting 1 from both sides,

$$
\frac{p^{2 n+1}-1}{p-1}-1=\left(q^{2}+q+1\right)-1 p^{2 n+1}-p=(p-1)\left(q^{2}+q\right) p\left(p^{n}-1\right)\left(p^{n}+1\right)=(p-1) q(q+1)
$$

On the left side, one of the factors is divisible by the prime number $q$ on the right side; this factor is at least $q$. Since the largest factor is $p^{n}+1, p^{n}+1 \geq q$ always holds. We apply this estimate to the right side of (2):

$$
p\left(p^{n}-1\right)\left(p^{n}+1\right) \leq(p-1)\left(p^{n}+1\right)\left(p^{n}+2\right) p \cdot\left(p^{n}-1\right) \leq(p-1)\left(p^{n}+2\right) p^{n}+2 \leq 3 p .(3)
$$

If $p$ or $n$ is at least 3, then $p^{n}=p^{n-1} \cdot p \geq 3 p$, so (3) cannot hold. Therefore, only $p=n=2$ is possible. Substituting this into (2), we get $q(q+1)=30$, so $q=5$.

The only solution to the problem is thus $p=2, q=5$, and $n=2$.

Based on the work of Endre Csóka (Debrecen, Fazekas M. Gymnasium, 10th grade)