$8 \cdot 7$ If $x, y$ are integers, the number of solutions to the equation $(x-8)(x-10)=2^{y}$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) More than 3.
(13th American High School Mathematics Examination, 1962)
［Solution］ The given equation can be transformed into
$$
x^{2}-18 x+80-2^{y}=0 \text {. }
$$

Considering it as an equation in $x$, we have
$$
x=9 \pm \sqrt{1+2^{y}} \text {, }
$$

where $x$ is an integer if and only if $1+2^{y}$ is a perfect square.
That is, $1+2^{y}=n^{2}$ or $n^{2}-1=2^{y}$.
Since $n^{2}-1=(n-1)(n+1)$, they must be two consecutive even numbers (their product is $2^{y}$), and each factor must be a power of 2.
The only suitable pairs are $(2,4)$ or $(-4,-2)$, i.e., $n= \pm 3$.
Substituting into equation (*) gives $y=3$; substituting back into the original equation gives
$$
x=9 \pm \sqrt{9}=9 \pm 3 \text {, i.e., } x=6 \text { or } 12 \text {. }
$$

Therefore, the original equation has only two solutions $(6,3)$ and $(12,3)$.
Hence, the answer is $(C)$.