Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$

has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition.
Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$

or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{array}\right.$
Substituting these into $3 x y+p(x-y)=p^{2}$, we get $\frac{3}{4}\left(p^{2}-1\right)+p=p^{2}$,

which simplifies to $p^{2}-4 p+3=0$.
Solving this, we get $p=3$ or $p=1$ (discard).
(2) When $\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$ or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p\end{array}\right.$, it is found through calculation that there are no prime numbers $p$ that satisfy the conditions.
Therefore, the prime number $p$ that satisfies the conditions is $p=3$.