Example 9: For the equation $x^{2}-4 a x+64=0$ (where $a$ is an integer), what value of $a$ will make the equation have integer solutions?
From $x^{2}-4 a x+64=0$, we get $e=\frac{x}{4}+\frac{16}{x}$ Since $a$ is an integer, if one root is an integer, the other root is also an integer. Therefore, $x= \pm 4, \pm 8, \pm 16$. Thus, $a= \pm 5, \quad \pm 4$