Three, (12 points) The equation $x^{2}+k x+4-k$ $=0$ has two integer roots. Find the value of $k$.

 untranslated part:
关于 $x$ 的方程 $x^{2}+k x+4-k$ $=0$ 有两个整数根. 试求 $k$ 的值.

 translated part:
The equation $x^{2}+k x+4-k$ $=0$ has two integer roots. Find the value of $k$.
Three, let $x_{1}, x_{2}$ be the roots of the equation, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} \cdot x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$, so
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$

By the problem statement,
$\left\{\begin{array}{l}x_{1}-1=1, \\ x_{2}-1=5\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=5, \\ x_{2}-1=1\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=-1 . \\ x_{2}-1=-5\end{array}\right.$ or
$$
\left\{\begin{array}{l}
x_{1}-1=-5 \\
x_{2}-1=-1
\end{array}\right.
$$

That is, $x_{1}+x_{2}=8$ or $x_{1}+x_{2}=4$.
Thus, $k_{1}=-8$ or $k_{2}=4$.