Let $p \ge 2$ be a prime number and $a \ge 1$ a positive integer with $p \neq a$. Find all pairs $(a,p)$ such that: $a+p \mid a^2+p^2$
To solve the problem, we need to find all pairs \((a, p)\) such that \(a + p \mid a^2 + p^2\). Here, \(p\) is a prime number and \(a\) is a positive integer with \(p \neq a\).

1. **Initial Condition**:
   Given \(a + p \mid a^2 + p^2\), we can write this as:
   \[
   a^2 + p^2 = k(a + p) \quad \text{for some integer } k.
   \]

2. **Rewriting the Condition**:
   We can rewrite the condition as:
   \[
   a^2 + p^2 \equiv 0 \pmod{a + p}.
   \]
   This implies:
   \[
   a^2 + p^2 \equiv 0 \pmod{a + p}.
   \]

3. **Using the Euclidean Algorithm**:
   We use the Euclidean Algorithm to find the greatest common divisor (gcd):
   \[
   \gcd(a + p, a^2 + p^2) = \gcd(a + p, (a^2 + p^2) - (a + p)(a - p)) = \gcd(a + p, 2ap).
   \]
   Therefore:
   \[
   \gcd(a + p, 2ap) = a + p.
   \]

4. **Case Analysis**:
   - If \(p \nmid a\), then \(\gcd(a + p, 2ap) = a + p\) is impossible because \(a + p\) would not divide \(2ap\) unless \(a\) is a multiple of \(p\).
   - If \(a = kp\) for some integer \(k\), then:
     \[
     \gcd((k + 1)p, 2kp^2) = (k + 1)p \implies \gcd(k + 1, 2kp) = k + 1.
     \]
     This implies:
     \[
     2kp \equiv 0 \pmod{k + 1} \implies 2p \equiv 0 \pmod{k + 1}.
     \]
     Since \(p\) is prime, \(k + 1\) must be one of \(\{1, 2, p, 2p\}\).

5. **Solving for \(k\)**:
   - If \(k + 1 = 1\), then \(k = 0\), which implies \(a = 0\), but \(a\) must be a positive integer.
   - If \(k + 1 = 2\), then \(k = 1\), which implies \(a = p\). This is a valid solution.
   - If \(k + 1 = p\), then \(k = p - 1\), which implies \(a = p^2 - p\). This is also a valid solution.
   - If \(k + 1 = 2p\), then \(k = 2p - 1\), which implies \(a = 2p^2 - p\). This is another valid solution.

Conclusion:
The valid pairs \((a, p)\) are:
\[
(a, p) = (p, p), (p^2 - p, p), (2p^2 - p, p).
\]

The final answer is \( \boxed{ (p, p), (p^2 - p, p), (2p^2 - p, p) } \).