Find all pairs $ (p, q)$ of prime numbers such that $ p$ divides $ 5^q \plus{} 1$ and $ q$ divides $ 5^p \plus{} 1$.
1. We start with the given conditions: \( p \) divides \( 5^q + 1 \) and \( q \) divides \( 5^p + 1 \). This implies:
   \[
   5^q \equiv -1 \pmod{p} \quad \text{and} \quad 5^p \equiv -1 \pmod{q}
   \]
   Squaring both sides, we get:
   \[
   5^{2q} \equiv 1 \pmod{p} \quad \text{and} \quad 5^{2p} \equiv 1 \pmod{q}
   \]

2. Let \( \text{ord}_p(5) \) be the order of 5 modulo \( p \). By definition, \( \text{ord}_p(5) \) is the smallest positive integer \( d \) such that \( 5^d \equiv 1 \pmod{p} \). Since \( 5^{2q} \equiv 1 \pmod{p} \), it follows that \( \text{ord}_p(5) \) divides \( 2q \). Similarly, \( \text{ord}_q(5) \) divides \( 2p \).

3. The possible values for \( \text{ord}_p(5) \) are \( 1, 2, q, \) or \( 2q \). Similarly, the possible values for \( \text{ord}_q(5) \) are \( 1, 2, p, \) or \( 2p \).

4. If \( \text{ord}_p(5) = 1 \), then \( 5 \equiv 1 \pmod{p} \), which is impossible since \( p \) is a prime and \( 5 \neq 1 \).

5. If \( \text{ord}_p(5) = 2 \), then \( 5^2 \equiv 1 \pmod{p} \), which implies \( p \) divides \( 24 \). The prime factors of 24 are 2 and 3. Thus, \( p = 2 \) or \( p = 3 \).

6. If \( p = 2 \), then \( 2 \) divides \( 5^q + 1 \). Since \( 5^q \) is odd, \( 5^q + 1 \) is even, and thus divisible by 2. Therefore, \( q \) can be any prime number. However, we need to check if \( q \) divides \( 5^2 + 1 = 26 \). The prime factors of 26 are 2 and 13. Thus, \( q = 2 \) or \( q = 13 \). We get the pairs \( (2, 2) \) and \( (2, 13) \).

7. If \( p = 3 \), then \( 3 \) divides \( 5^q + 1 \). Since \( 5 \equiv 2 \pmod{3} \), we have \( 5^q \equiv 2^q \pmod{3} \). For \( 3 \) to divide \( 2^q + 1 \), \( 2^q \equiv -1 \pmod{3} \). This holds if \( q \) is odd. The prime factors of \( 5^3 + 1 = 126 \) are 2, 3, and 7. Thus, \( q = 3 \) or \( q = 7 \). We get the pairs \( (3, 3) \) and \( (3, 7) \).

8. Now, consider the case where \( \text{ord}_p(5) = q \) or \( 2q \). Similarly, \( \text{ord}_q(5) = p \) or \( 2p \). This implies \( p \) divides \( q-1 \) and \( q \) divides \( p-1 \). Since \( p \) and \( q \) are primes, this leads to a contradiction unless \( p = q \).

9. Therefore, the only solutions are the pairs found earlier: \( (2, 2), (2, 13), (3, 3), (3, 7), (13, 2), (7, 3) \).

The final answer is \( \boxed{(2, 2), (2, 13), (3, 3), (3, 7), (13, 2), (7, 3)} \).