II. (25 points) Does there exist a prime number $p$ such that the equation
$$
x^{2}-4 p(x-p)-5 p-1=0
$$

has two integer roots? If it exists, find all possible values of $p$ and the roots of the equation; if not, explain why.
There exists a prime number $p$ that satisfies the condition. The reason is as follows.

The original equation is equivalent to $(x-2 p)^{2}=5 p+1$. According to the problem, $5 p+1$ must be a perfect square. Let $5 p+1=n^{2}$.
Note that, $p \geqslant 2$. Then $n \geqslant 4$, and $n$ is an integer. Therefore, $5 p=n^{2}-1=(n+1)(n-1)$.
It is easy to see that at least one of $n+1$ and $n-1$ is a multiple of 5, i.e., $n=5 k \pm 1$ (where $k$ is a positive integer). Thus,
$$
5 p+1=25 k^{2} \pm 10 k+1 \Rightarrow p=k(5 k \pm 2) \text {. }
$$

Since $p$ is a prime number and $5 k \pm 2>1$, it follows that $k=1$, $p=3$ or 7.
When $p=3$, the original equation becomes
$$
(x-6)^{2}=16 \Rightarrow x_{1}=2, x_{2}=10 \text {; }
$$

When $p=7$, the original equation becomes
$$
(x-14)^{2}=36 \Rightarrow x_{1}=8, x_{2}=20 .
$$