Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.

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The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]

To find the points where the graph intersects the $x$-axis, we set $y = 0$:
\[ x^2 + px + (k+1)p - 4 = 0 \]

This is a quadratic equation in the form:
\[ x^2 + px + (k+1)p - 4 = 0 \]

The roots of this quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = p \), and \( c = (k+1)p - 4 \).

Substituting these values into the quadratic formula, we get:
\[ x = \frac{-p \pm \sqrt{p^2 - 4 \cdot 1 \cdot ((k+1)p - 4)}}{2 \cdot 1} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4((k+1)p - 4)}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4(k+1)p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4p(k + 1) + 16}}{2} \]

For the quadratic equation to have at least one integer root, the discriminant must be a perfect square. Let's denote the discriminant by \( D \):
\[ D = p^2 - 4p(k + 1) + 16 \]

We need \( D \) to be a perfect square. Let \( D = m^2 \) for some integer \( m \):
\[ p^2 - 4p(k + 1) + 16 = m^2 \]

Rearranging the equation, we get:
\[ p^2 - 4p(k + 1) + 16 - m^2 = 0 \]

This is a quadratic equation in \( p \):
\[ p^2 - 4p(k + 1) + (16 - m^2) = 0 \]

For \( p \) to be a prime number greater than 2, the discriminant of this quadratic equation must be a perfect square. The discriminant of this quadratic equation is:
\[ \Delta = (4(k + 1))^2 - 4 \cdot 1 \cdot (16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 4(16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 64 + 4m^2 \]
\[ \Delta = 16(k + 1)^2 + 4m^2 - 64 \]

For \( \Delta \) to be a perfect square, we need:
\[ 16(k + 1)^2 + 4m^2 - 64 = n^2 \]
for some integer \( n \).

Simplifying, we get:
\[ 4(4(k + 1)^2 + m^2 - 16) = n^2 \]
\[ 4(k + 1)^2 + m^2 - 16 = \left(\frac{n}{2}\right)^2 \]

Let \( \frac{n}{2} = t \), then:
\[ 4(k + 1)^2 + m^2 - 16 = t^2 \]

We need to find integer solutions for \( k \) and \( m \) such that the above equation holds. Testing small values of \( k \):

For \( k = 1 \):
\[ 4(1 + 1)^2 + m^2 - 16 = t^2 \]
\[ 4 \cdot 4 + m^2 - 16 = t^2 \]
\[ 16 + m^2 - 16 = t^2 \]
\[ m^2 = t^2 \]

This is true for \( m = t \). Therefore, \( k = 1 \) is a solution.

Thus, the value of \( k \) is:
\[ \boxed{1} \]
From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$

has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\
=x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4 \\
=(k-1) p .
\end{array}
$$
(1) If $k=1$, then the equation becomes
$$
x^{2}+p x+2(p-2)=0 \text {, }
$$

which has two integer roots -2 and $2-p$.
(2) If $k>1$, then $k-1>0$.
Since $x_{1}+x_{2}=-p$ is an integer, and at least one of $x_{1}, x_{2}$ is an integer, it follows that $x_{1}, x_{2}$ are both integers.
Also, since $p$ is a prime number, by equation (1), we know that $p \mid (x_{1}+2)$ or $p \mid (x_{2}+2)$.
Without loss of generality, assume $p \mid (x_{1}+2)$. Then we can set $x_{1}+2=m p$ (where $m$ is a non-zero integer).
Thus, from equation (1), we get $x_{2}+2=\frac{k-1}{m}$.
Therefore, $\left(x_{1}+2\right)+\left(x_{2}+2\right)=m p+\frac{k-1}{m}$, which means
$x_{1}+x_{2}+4=m p+\frac{k-1}{m}$.
Since $x_{1}+x_{2}=-p$, we have
$-p+4=m p+\frac{k-1}{m}$
$\Rightarrow (m+1) p+\frac{k-1}{m}=4$.
If $m$ is a positive integer, then
$(m+1) p \geqslant (1+1) \times 3=6, \frac{k-1}{m}>0$.
Thus, $(m+1) p+\frac{k-1}{m}>6$, which contradicts equation (2).
If $m$ is a negative integer, then
$(m+1) p \leqslant 0, \frac{k-1}{m}<0$.
Thus, $(m+1) p+\frac{k-1}{m}<0$, which also contradicts equation (2).
Therefore, when $k>1$, the equation
$$
x^{2}+p x+(k+1) p-4=0
$$

cannot have integer roots.
In summary, $k=1$.
(Provided by Xu Shenglin)