For prime numbers $p, q$ and the natural number $n$,

$$
n^{2}=p^{2}+q^{2}+p^{2} q^{2}
$$

is satisfied. Determine all such triples of numbers.
By adding 1 to the equation, the right side can be factored as a product:

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

If both $p$ and $q$ are odd, then this product is divisible by four, thus $n^{2}=4 k-1$. However, a square number divided by four can only give a remainder of zero or one, so at least one of $p$ and $q$ must be even, and since they are prime, it must be equal to two. Since the roles of $p$ and $q$ are interchangeable, we can assume that $p=2$.

Then the equation becomes: $n^{2}=4+5 q^{2}$, or $(n-2)(n+2)=5 q^{2}$. The factorizations of the right side can only be $\left(1,5 q^{2}\right)$, $(q, 5 q)$, and $\left(q^{2}, 5\right)$, since $q$ is prime and $n$ is a natural number. The larger element of these pairs must equal $n+2$, while the other must equal $n-2$, so their difference is 4:

$$
5 q^{2}-1=4, \quad \text { or } \quad 5 q-q=4, \quad \text { or } \quad q^{2}-5= \pm 4
$$

From the first, $q^{2}=1$, which is not prime; from the second, $q=1$, also not prime; from the third, $q^{2}=9$ or 1, of which the only prime solution is $q=3$. Substituting this back in:

$$
p^{2}+q^{2}+p^{2} q^{2}=4+9+4 \cdot 9=49=7^{2}
$$

Thus, in this case, $p=2, q=3, n=7$ is the only solution. Adding to this the triplet obtained by swapping $p$ and $q$, $p=3, q=2, n=7$, we get all the solutions to the equation.

Based on the work of Tímea Lengyel (Kaposvár, Munkácsy M. Gymn., 1st year)