$12 \cdot 18 \quad$ The equation $(x-a)(x-8)-1=0$ has two integer roots, find $a$.
(China Junior High School Mathematics League, 1990)
［Solution］Let the two integer roots of the equation be $x_{1}, x_{2}$, then
$$
(x-a)(x-8)-1=\left(x-x_{1}\right)\left(x-x_{2}\right) \text {. }
$$

Let $x=8$, then the above equation becomes
$$
\left(8-x_{1}\right)\left(8-x_{2}\right)=-1 .
$$

So $x_{1}=7, x_{2}=9$ or $x_{1}=9, x_{2}=7$.
Therefore, we have $(x-a)(x-8)-1=(x-7)(x-9)$.
Let $x=9$, then the above equation becomes
$$
(9-a)-1=0 \text {. }
$$

Thus,
$$
a=8 \text {. }
$$