Let's determine those prime numbers $p$ and $q$ for which $p+q$ and $p^{2}+q^{2}-q$ are both prime numbers.
If the sum of primes $p$ and $q$ is also a prime, then one of them must be 2, otherwise the sum would be a composite number greater than 2 and thus could not be a prime. Since the other prime in the condition is $p^{2}+q^{2}-q$, and here $\left(q^{2}-q\right)=q(q-1)$ is an even positive number, $p^{2}$ cannot be even; thus, only $q=2$ is possible. In this case, $p+q=p+2, p^{2}+q^{2}-q=p^{2}+2$, so we are looking for a prime $p$ such that $p+2$ and $p^{2}+2$ are also primes.

$$
p^{2}+2=\left(p^{2}-1\right)+3=(p-1)(p+1)+3
$$

Thus, if $p$ is not divisible by 3, then one of the numbers $(p-1),(p+1)$ is divisible by 3, and thus on the right side of (1) there stands a multiple of 3 greater than 3, which cannot be a prime number.

The only remaining possibility is $p=3$. In this case, $p+2=5 ; p^{2}+2=11$, so we have found a solution to the problem, and as we have seen, there is no other solution.

The primes $p$ and $q$ in the problem are thus 3 and 2.

Zoltán Kovács (Szeged, Radnóti M. Experimental Gymnasium, 8th grade)