Find all prime numbers $p$ and $q$ such that $2^2+p^2+q^2$ is also prime.

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1. We start by analyzing the given expression \(2^2 + p^2 + q^2\) and determining the conditions under which it is prime. Note that \(2^2 = 4\), so the expression simplifies to \(4 + p^2 + q^2\).

2. Consider the parity of \(p\) and \(q\). If both \(p\) and \(q\) are odd, then \(p^2\) and \(q^2\) are also odd, and the sum \(p^2 + q^2\) is even. Adding 4 (which is even) to an even number results in an even number. The only even prime number is 2, but \(4 + p^2 + q^2\) is greater than 2, so it cannot be prime. Therefore, \(p\) and \(q\) cannot both be odd.

3. If both \(p\) and \(q\) are even, then \(p\) and \(q\) must be 2 (since 2 is the only even prime). However, this would make the expression \(4 + 2^2 + 2^2 = 4 + 4 + 4 = 12\), which is not prime. Therefore, \(p\) and \(q\) cannot both be even.

4. Thus, one of \(p\) or \(q\) must be 2, and the other must be an odd prime. Without loss of generality, assume \(q = 2\) and \(p\) is an odd prime. The expression then becomes \(4 + p^2 + 2^2 = 4 + p^2 + 4 = p^2 + 8\).

5. We need to check if \(p^2 + 8\) can be prime for odd primes \(p\). Consider the form of \(p\). Any odd prime \(p\) can be written as \(p = 6k \pm 1\) for some integer \(k\). Let's analyze \(p^2 + 8\) for these forms:
   - If \(p = 6k + 1\), then \(p^2 = (6k + 1)^2 = 36k^2 + 12k + 1\). Thus, \(p^2 + 8 = 36k^2 + 12k + 1 + 8 = 36k^2 + 12k + 9 = 3(12k^2 + 4k + 3)\). Since \(12k^2 + 4k + 3\) is an integer, \(p^2 + 8\) is divisible by 3. For \(p^2 + 8\) to be prime, it must be equal to 3, but this is not possible since \(p^2 + 8 > 3\).
   - If \(p = 6k - 1\), then \(p^2 = (6k - 1)^2 = 36k^2 - 12k + 1\). Thus, \(p^2 + 8 = 36k^2 - 12k + 1 + 8 = 36k^2 - 12k + 9 = 3(12k^2 - 4k + 3)\). Similarly, \(p^2 + 8\) is divisible by 3 and must be equal to 3 to be prime, which is not possible.

6. The only remaining odd prime that does not fit the form \(6k \pm 1\) is \(p = 3\). Check if \(p = 3\):
   - If \(p = 3\), then \(p^2 + 8 = 3^2 + 8 = 9 + 8 = 17\), which is prime.

7. Therefore, the only solutions are \((p, q) = (3, 2)\) and \((p, q) = (2, 3)\).

The final answer is \(\boxed{(3, 2) \text{ and } (2, 3)}\).