$31 \cdot 40$ The number of integer solutions to the equation $2^{2 x}-3^{2 y}=55$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) Finite, but more than 3.
(12th American High School Mathematics Examination, 1961)
[Solution] From $2^{2 x}-3^{2 y}=\left(2^{x}-3^{y}\right)\left(2^{x}+3^{y}\right)$, and
$$
55=5 \cdot 11=1 \cdot 55 \text {. }
$$

then $\left(2^{x}-3^{y}\right)\left(2^{x}+3^{y}\right)=5 \cdot 11$, noting that $2^{x}+3^{y}>2^{x}-3^{y}$,
we have $\quad\left\{\begin{array}{l}2^{x}-3^{y}=5, \\ 2^{x}+3^{y}=11 ;\end{array}\right.$ or $\left\{\begin{array}{l}2^{x}-3^{y}=1, \\ 2^{x}+3^{y}=55 .\end{array}\right.$
From the first set of equations, we get the solution $x=3, y=1$; the second set of equations has no solution.
Therefore, the answer is $(B)$.