$p > 3$  is a prime. Find all integers $a$, $b$, such that $a^2 + 3ab + 2p(a+b) + p^2 = 0$.
Given the equation \(a^2 + 3ab + 2p(a+b) + p^2 = 0\), we need to find all integers \(a\) and \(b\) such that the equation holds for a prime \(p > 3\).

1. **Rewrite the equation:**
   \[
   a^2 + 3ab + 2p(a+b) + p^2 = 0
   \]
   We can rearrange this equation as:
   \[
   b(3a + 2p) + (a + p)^2 = 0
   \]
   Since \(p > 3\), \(3a + 2p \neq 0\). Even if \(p = 3\), \(3a + 2p = 0\) would imply \(a = -2\), which does not satisfy the equation.

2. **Divisibility condition:**
   \[
   3a + 2p \mid (a + p)^2
   \]
   We need to find \(a\) and \(b\) such that \(3a + 2p\) divides \((a + p)^2\).

3. **Case 1: \(3a + 2p = \pm 1\):**
   \[
   3a + 2p = 1 \quad \text{or} \quad 3a + 2p = -1
   \]
   Solving for \(a\):
   \[
   a = \frac{-2p \pm 1}{3}
   \]
   We need to check if \(a\) is an integer for different values of \(p\) modulo 3.

4. **Case 2: \(|3a + 2p| > 1\):**
   Let \(q\) be a prime such that \(q \mid \gcd(3a + 2p, (a + p)^2)\). Then \(q \mid a\) and \(q \mid p\), implying \(q = p\). Thus, \(a = kp\) for some integer \(k\).

   We need:
   \[
   3k + 2 \mid (k + 1)^2p
   \]
   Since \(\gcd(3k + 2, k + 1) = 1\), it follows that \(3k + 2 \mid p\).

5. **Subcase: \(3k + 2 = -1\):**
   \[
   3k + 2 = -1 \implies k = -1
   \]
   Thus, \(a = -p\) and substituting \(a = -p\) into the original equation, we get:
   \[
   b(3(-p) + 2p) + (-p + p)^2 = 0 \implies b(-p) = 0 \implies b = 0
   \]
   This gives the solution \((a, b) = (-p, 0)\).

6. **Subcase: \(3k + 2 = \pm p\):**
   \[
   3k + 2 = p \quad \text{or} \quad 3k + 2 = -p
   \]
   Solving for \(k\):
   \[
   k = \frac{p - 2}{3} \quad \text{or} \quad k = \frac{-p - 2}{3}
   \]
   Thus:
   \[
   a = \frac{p(p - 2)}{3} \quad \text{or} \quad a = \frac{p(-p - 2)}{3}
   \]

   These solutions are valid if \(a\) is an integer, which depends on the residue of \(p\) modulo 3.

Conclusion:
The solutions are:
1. \(a = \frac{-2p \pm 1}{3}\)
2. \(a = -p\) with \(b = 0\)
3. \(a = \frac{p(p - 2)}{3}\)
4. \(a = \frac{p(-p - 2)}{3}\)

The final answer is \( \boxed{ (a, b) = (-p, 0) } \) and other solutions depending on the residue of \(p\) modulo 3.