Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:
$\frac{4a + p}{b}+\frac{4b + p}{a}$ and  $ \frac{a^2}{b}+\frac{b^2}{a}$
are integers.
1. We start with the given conditions:
   \[
   \frac{4a + p}{b} + \frac{4b + p}{a} \quad \text{and} \quad \frac{a^2}{b} + \frac{b^2}{a} \quad \text{are integers}.
   \]

2. We need to show that \(ab\) divides both \(4(a^2 + b^2) + p(a + b)\) and \(a^3 + b^3\). 

3. First, consider the expression \(\frac{a^2}{b} + \frac{b^2}{a}\):
   \[
   \frac{a^2}{b} + \frac{b^2}{a} = \frac{a^3 + b^3}{ab}.
   \]
   For this to be an integer, \(ab\) must divide \(a^3 + b^3\).

4. We know that:
   \[
   a^3 + b^3 = (a + b)(a^2 - ab + b^2).
   \]
   Therefore, \(ab\) must divide \((a + b)(a^2 - ab + b^2)\).

5. Next, consider the expression \(\frac{4a + p}{b} + \frac{4b + p}{a}\):
   \[
   \frac{4a + p}{b} + \frac{4b + p}{a} = \frac{4a^2 + pa + 4b^2 + pb}{ab} = \frac{4(a^2 + b^2) + p(a + b)}{ab}.
   \]
   For this to be an integer, \(ab\) must divide \(4(a^2 + b^2) + p(a + b)\).

6. Combining these results, we have:
   \[
   ab \mid (a + b)(a^2 - ab + b^2) \quad \text{and} \quad ab \mid 4(a^2 + b^2) + p(a + b).
   \]

7. Let \(d = \gcd(a, b)\). Then we can write \(a = dx\) and \(b = dy\) where \(\gcd(x, y) = 1\).

8. Substituting \(a = dx\) and \(b = dy\) into the divisibility conditions, we get:
   \[
   d^2xy \mid p(x^2 + y^2).
   \]

9. Since \(\gcd(x, y) = 1\), \(xy \mid p(x^2 + y^2)\). This implies \(x \mid p(x^2 + y^2)\) and \(y \mid p(x^2 + y^2)\).

10. Let \(q\) be a prime that divides \(x\). Then \(q \mid p(x^2 + y^2)\). Since \(q \mid x\), \(q\) must divide \(p\) or \(y^2\). But \(\gcd(x, y) = 1\), so \(q\) cannot divide \(y\). Therefore, \(q = p\).

11. Similarly, if \(r\) is a prime that divides \(y\), then \(r = p\).

12. This implies that \(x\) and \(y\) are either 1 or \(p\). Without loss of generality, assume \(y = 1\). Then \(x \mid p\), so \(x = 1\) or \(x = p\).

13. If \(x = 1\), then \(a = d\) and \(b = d\). If \(x = p\), then \(a = dp\) and \(b = d\).

14. Therefore, the possible solutions are \(a = b\) or \(a = pb\).

The final answer is \( \boxed{ a = b } \) or \(a = pb\).