Three. (25 points) Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.

Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
Three, let the equation $x^{2}+p x+k p-1=0$ have integer roots $x_{1}$ and another root $x_{2}$.
By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=k p-1 \text {. }
$$

Thus, $x_{2}$ must also be an integer.
Assume $k>1$.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right) \\
=x_{1} x_{2}+\left(x_{1}+x_{2}\right)+1=(k-1) p .
\end{array}
$$

Since $p$ is a prime number and $k-1>0$, it follows that $p \mid\left(x_{1}+1\right)$ or $p \mid\left(x_{2}+1\right)$.
Without loss of generality, assume $p \mid\left(x_{1}+1\right)$. Then
$$
x_{1}+1= \pm m p, x_{2}+1= \pm \frac{k-1}{m} \text {, }
$$

where $m$ is a positive integer, and $m \mid(k-1)$.
Adding the two equations, we get
$$
x_{1}+x_{2}+2= \pm\left(m p+\frac{k-1}{m}\right) \text {, }
$$

which means $-p+2= \pm\left(m p+\frac{k-1}{m}\right)$.
If the right side of equation (1) is positive, then
$$
(m+1) p+\frac{k-1}{m}=2 \text {. }
$$

Clearly, the left side of equation (1) is greater than 2, which is a contradiction.
If the right side of equation (1) is negative, then
$$
(m-1) p+2+\frac{k-1}{m}=0 \text {. }
$$

Clearly, the left side of equation (1) is greater than 0, which is a contradiction.
Therefore, the assumption is false.
Since $k$ is a positive integer, the only possible value for $k$ is 1.