1. Determine all prime numbers $p$ and $q$ for which the number $2^{2}+p^{2}+q^{2}$ is also a prime number.
I/1. If $(p, q)$ is a solution to the problem, then $(q, p)$ is also a solution. Therefore, it is sufficient to consider the case $p \leq q$. Clearly, $p=q=2$ is not a solution. If $p$ and $q$ are odd prime numbers, the number $2^{2}+p^{2}+q^{2}$ is even and greater than 2, so it is not a prime number. Therefore, $p=2$.

Determine when the number $8+q^{2}$ is a prime number. If $q=3$, then $8+q^{2}=17$ is a prime number. Otherwise, one of the numbers $q-1$ or $q+1$ is divisible by 3, so 3 divides $9+(q-1)(q+1)=8+q^{2}$, and it is not a prime number.

The number $2^{2}+p^{2}+q^{2}$ is a prime number only if $p=2$ and $q=3$ or $p=3$ and $q=2$.

The observation that when $p$ and $q$ are odd, $2^{2}+p^{2}+q^{2}$ is not a prime number $\ldots 1$ point

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The conclusion that $3 \mid 8+q^{2}$ when the prime number $q$ is not divisible by 3 $\ldots \ldots \ldots \ldots \ldots \ldots \ldots 2$ points