2. Find all such prime numbers $p$ and $q$, for which the number $2^{2}+p^{2}+q^{2}$ is also prime.

Note. A prime number is an integer greater than one that is divisible only by one and by itself.
Solution. If the numbers $p$ and $q$ are odd, then the number $2^{2}+p^{2}+q^{2}$ is even and, obviously, greater than 2, and therefore cannot be prime. Thus, among the prime numbers $p$ and $q$, there is an even one, which is equal to 2. Suppose this is the number $p$. It remains to find all primes $q$ for which the number $q^{2}+8$ is prime.

Note that if a number is not divisible by 3, that is, it gives a remainder of 1 or 2 when divided by 3, then the square of this number gives a remainder of 1 when divided by 3, and therefore if $q$ is not divisible by 3, then $q^{2}+8$ is divisible by 3, that is, the number $q^{2}+8>3$ is divisible by 3 and is not prime. Thus, $q$ is divisible by 3, from which $q=3$. Similarly, if $q=2$, then $p=3$. It remains to substitute the pairs $(2,3)$ and $(3,2)$ into the original expression and verify that both indeed give the prime number 17.

Answer: $p=2, q=3$ or $p=3, q=2$.