, 2023

Solution:
Answer: The only possible integer value is $3$.

If both $p$ and $q$ are odd, then the numerator is odd while the denominator is even. Since an even number never divides an odd number, this does not lead to an integer value. Hence we can assume that one of our primes is even and therefore equal to $2$. Since the expression is symmetric in $p$ and $q$, we can assume without loss of generality that $q = 2$.

Substituting $q = 2$, it remains to determine all integer values taken by the expression
$$\frac{2p + p^{p} + 4}{p + 2} = \frac{2(p + 2) + p^{p}}{p + 2} = 2 + \frac{p^{p}}{p + 2}$$
In order for this to be an integer, we must have that $p + 2$ is a divisor of $p^{p}$. But since $p$ is prime, the only positive divisors of $p^{p}$ are $1, p, p^{2}, \ldots, p^{p-1}$ and $p^{p}$. If $p > 2$, then we have
$$p < p + 2 < p + p = 2p < p^{2}$$
and $p + 2$ is strictly squeezed between two consecutive divisors of $p^{p}$. It follows that for $p > 2$ the expression $p + 2$ never divides $p^{p}$ and we don't get integer values. The only case remaining is $p = q = 2$, making the original expression equal to $3$.