1.50 For what integer $k$ does the equation
$$
\left(k^{2}-1\right) x^{2}-6(3 k-1) x+72=0
$$

have two distinct positive integer roots?
(Xining, Qinghai Province, China Junior High School Mathematics Competition, 1983)
［Solution］Since the equation has two distinct positive integer roots, then
$$
\begin{array}{c}
k^{2}-1 \neq 0, \quad k \neq \pm 1 . \\
\Delta=6^{2}(3 k-1)^{2}-4 \cdot 72\left(k^{2}-1\right)=[6(k-3)]^{2} .
\end{array}
$$

By the quadratic formula, we get
$$
\begin{array}{l}
x=\frac{6(3 k-1) \pm 6(k-3)}{2\left(k^{2}-1\right)}, \\
x_{1}=\frac{12}{k+1}, \quad x_{2}=\frac{6}{k-1} .
\end{array}
$$

For $x_{1}, x_{2}$ to both be positive integers, $k+1$ must be a positive divisor of 12, and $k-1$ must be a positive divisor of 6, i.e.,
$$
\left\{\begin{array}{l}
k+1=1,2,3,4,6,12 \\
k-1=1,2,3,6 .
\end{array}\right.
$$

Thus, $k=2,3$.
When $k=3$, $x_{1}=x_{2}=3$, which are two equal positive integer roots.
Therefore, when $k=2$, $x_{1}=4, x_{2}=6$ meets the requirements of the problem.