Five. (Full marks 25 points) Let the quadratic equation in $x$ be $\left(k^{2}-6 k+8\right) x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}=4$ with both roots being integers. Find all real values of $k$ that satisfy the condition.

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$$
\begin{array}{l}
(k-4)(k-2) x^{2}+\left(2 k^{2}-6 k-4\right) x \\
+(k-2)(k+2)=0, \\
(k-4) x+(k-2)][(k-2) x+(k+2)]=0, \\
\because(k-4)(k-2) \neq 0, \\
\therefore x_{1}=-\frac{k-2}{k-4}=-1-\frac{2}{k-4}, \\
x_{2}=-\frac{k+2}{k-2}=-1-\frac{4}{k-2} . \\
\therefore k-4=-\frac{2}{x_{1}+1}, k-2=-\frac{2}{x_{2}+1} . \\
\left(x_{1} \neq-1, x_{2} \neq-1\right)
\end{array}
$$

Eliminating $k$, we get
$$
\begin{array}{l}
x_{1} x_{2}+3 x_{1}+2=0, \\
\therefore x_{1}\left(x_{2}+3\right)=-2 .
\end{array}
$$

Since $x_{1} 、 x_{2}$ are both integers,
$$
\begin{array}{l}
\therefore\left\{\begin{array} { l } 
{ x _ { 1 } = - 2 , } \\
{ x _ { 2 } + 3 = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=1, \\
x_{2}+3=-2 ;
\end{array} ;\left\{\begin{array}{l}
x_{1}=2, \\
x_{2}+3=-1 .
\end{array}\right.\right.\right. \\
\therefore\left\{\begin{array} { l } 
{ x _ { 1 } = - 2 , } \\
{ x _ { 2 } = - 2 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x _ { 1 } = 1 , } \\
{ x _ { 2 } = - 5 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=2, \\
x_{2}=-4 .
\end{array}\right.\right.\right. \\
\therefore k=6,3, \frac{10}{3} .
\end{array}
$$

Upon verification, $k=6,3, \frac{10}{3}$ satisfy the conditions.