N1) Find all integer values that the expression

$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$

can take, where $p$ and $q$ are prime numbers.
Answer: The only attainable relative integer is 3. Solution: If $p$ and $q$ are odd, the numerator is odd but the denominator is even. Since an even number never divides an odd number, the expression is not a natural integer. Therefore, we can assume that at least one of the two primes is even. Since the expression is symmetric in $p$ and $q$, we can assume without loss of generality that $q=2$.

With $q=2$, it suffices to determine all the relative integers the following expression can take:

$$
\frac{2 p+p^{p}+4}{p+2}=\frac{2(p+2)+p^{p}}{p+2}=2+\frac{p^{p}}{p+2}
$$

For this expression to be an integer, it is necessary 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 $p>2$, we have

$$
p+2 > p \geq p^{1}
$$

Thus, for $p>2$, the expression $p+2$ never divides $p^{p}$, and we do not obtain any integer values. The only remaining case is when $p=q=2$, making the initial fraction equal to 3.