Find all pairs $ (p, q)$ of primes such that $ {p}^{p}\plus{}{q}^{q}\plus{}1$ is divisible by $ pq$.
1. We need to find all pairs \((p, q)\) of primes such that \(pq \mid p^p + q^q + 1\). This means that \(p^p + q^q + 1\) is divisible by \(pq\).

2. Since \(p\) and \(q\) are primes, \(\gcd(p, q) = 1\).

3. Without loss of generality, assume \(p > q\).

4. Consider the case \(q = 2\):
   - We need \(pq \mid p^p + 2^2 + 1\).
   - This simplifies to \(2p \mid p^p + 5\).
   - Check \(p = 5\):
     - \(2 \cdot 5 = 10\)
     - \(5^5 + 5 = 3125 + 5 = 3130\)
     - \(3130 \div 10 = 313\), which is an integer.
   - Therefore, \((p, q) = (5, 2)\) is a solution.

5. Now, consider \(p > q > 2\):
   - Since \(p \mid q^q + 1\), we have \(q^q \equiv -1 \pmod{p}\).
   - This implies \(q^{2q} \equiv 1 \pmod{p}\).
   - Let \(\text{ord}_p(q) = a\), where \(\text{ord}_p(q)\) is the order of \(q\) modulo \(p\). Then \(a \mid 2q\).

6. Analyze possible values of \(a\):
   - If \(a = 1\), then \(q \equiv 1 \pmod{p}\), which is impossible since \(p > q\).
   - If \(a = 2\), then \(q^2 \equiv 1 \pmod{p}\), implying \(p \mid q^2 - 1\). This means \(p \mid (q-1)(q+1)\).
     - Since \(p\) is a prime and \(p > q\), \(p\) must divide either \(q-1\) or \(q+1\).
     - If \(p \mid q-1\), then \(p \leq q-1\), which is impossible.
     - If \(p \mid q+1\), then \(p = q+1\), which is also impossible since both \(p\) and \(q\) are primes and \(p > q\).
   - If \(a = q\), then \(q^q \equiv 1 \pmod{p}\), implying \(p \mid q^q - 1\). This means \(p \mid 2\), which is impossible since \(p > 2\).
   - If \(a = 2q\), then \(2q \mid p-1\), implying \(q \mid p-1\). This means \(p = kq + 1\) for some integer \(k\).
     - Since \(q \mid p-1\), \(q \mid p^p - 1\), implying \(q \mid 2\), which is impossible since \(q > 2\).

7. Therefore, the only possible solutions are \((p, q) = (2, 5)\) and \((p, q) = (5, 2)\).

The final answer is \(\boxed{(2, 5)}\) and \(\boxed{(5, 2)}\).