For what integer $2 m$ does the equation $\left(m^{2}-1\right) x^{2}$ $-6(3 m-1) x+72=0$ have two distinct positive integer roots?
Solution: Clearly, $m \neq \pm 1$. The original equation can be factored as
$$
[(m-1) x-6][(m+1) x-12]=0 \text {. }
$$

Thus, $x_{1}=\frac{6}{m-1}, x_{2}=\frac{12}{m+1}$.
Since $x_{1}$ and $x_{2}$ are positive integers,
$$
\therefore m-1=1,2,3,6 \text { and } m+1=1,2,3,4 \text {, }
$$
6,12.
Solving, we get $m=2$ or $m=3$.
However, when $m=3$, $x_{1}=x_{2}$, which should be discarded.
Therefore, $m=2$ is the solution.