$1 . m$ is what integer when the equation
$$
\left(m^{2}-1\right) x^{2}-6(3 m-1) x+72=0
$$

has two distinct positive integer roots?
(Tip: $m^{2}-1 \neq 0, m \neq \pm 1$. Since $\Delta=36(m-3)^{2}>$ 0, hence $m \neq 3$. Using the quadratic formula, we get $x_{1}=\frac{6}{m-1}, x_{2}=$ $\frac{12}{m+1}$. Therefore, $(m-1)|6,(m+1)| 12$. So $m=2$, at this point, $x_{1}=6, x_{2}=4$.)