To solve this equation, we need to analyze the relationship given for prime $p$ in terms of the integers $a$ and $b$. The equation provided is:

$a(b^2 - p) + b(a^2 - p) = 5p$

Step-by-Step Solution

1. Expand and Rearrange: Start by expanding the terms in the equation:

   $ab^2 - ap + ba^2 - bp = 5p$

   Combine like terms involving $p$ on one side and constants on the other:

   $ab^2 + ba^2 = 5p + ap + bp$

2. Factor Out $p$: Factor $p$ on the right side of the equation:

   $ab^2 + ba^2 = p(5 + a + b)$

   Thus, we have:

   $p = \frac{ab^2 + ba^2}{5 + a + b}$

3. Constraints on $p$: Since $p$ is a prime number, $\frac{ab^2 + ba^2}{5 + a + b}$ must result in an integer. This condition restricts $a$ and $b$ to values that make the numerator a multiple of $5 + a + b$.

4. Testing Small Values for $a$ and $b$: Let's analyze small integer values for $a$ and $b$ and check if the result yields a prime $p$.

   • Case $a = 1$, $b = 1$: $p = \frac{1 \cdot 1^2 + 1 \cdot 1^2}{5 + 1 + 1} = \frac{1 + 1}{7} = \frac{2}{7} \quad \text{(not an integer)}$

   • Case $a = 1$, $b = 2$: $p = \frac{1 \cdot 2^2 + 2 \cdot 1^2}{5 + 1 + 2} = \frac{4 + 2}{8} = \frac{6}{8} = \frac{3}{4} \quad \text{(not an integer)}$

   • Case $a = 2$, $b = 3$: $p = \frac{2 \cdot 3^2 + 3 \cdot 2^2}{5 + 2 + 3} = \frac{2 \cdot 9 + 3 \cdot 4}{10} = \frac{18 + 12}{10} = \frac{30}{10} = 3$

     Here, $p = 3$, which is a prime number.

   • Case $a = 3$, $b = 2$: $p = \frac{3 \cdot 2^2 + 2 \cdot 3^2}{5 + 3 + 2} = \frac{3 \cdot 4 + 2 \cdot 9}{10} = \frac{12 + 18}{10} = \frac{30}{10} = 3$

     Again, $p = 3$, which is a prime number.

Conclusion

For the cases $(a, b) = (2, 3)$ and $(a, b) = (3, 2)$, the equation holds with $p = 3$, which is a prime.

Thus, the solution for $p$ is: $\boxed{p = 3}$

Would you like more detailed steps or have any questions?

Here are some related questions you might explore:

1. What would happen if we set $a$ and $b$ as non-integer values?
2. Can we generalize this approach to larger primes?
3. How would the equation change if we replaced 5 with another prime number?
4. Could there be other integer pairs $(a, b)$ for different primes $p$?
5. What if we impose additional constraints on $a$ and $b$, like parity or divisibility?

Tip: When testing equations for integer or prime solutions, try small values first to spot patterns or feasible solutions quickly.