$29 \cdot 15$ Given that $2^{96}-1$ can be divided by two integers between 60 and 70, these two numbers are
(A) 61,63 .
(B) 61,65 .
(C) 63,65 .
(D) 63,67 .
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational Competition, 1987)
[Solution] $2^{96}-1=2^{6 \cdot 16}-1=\left(2^{6 \cdot 8}\right)^{2}-1$
$$
\begin{aligned}
& =\left(2^{6 \cdot 8}-1\right)\left(2^{6 \cdot 8}+1\right) \\
& =\left(2^{6 \cdot 4}-1\right)\left(2^{6 \cdot 4}+1\right)\left(2^{6 \cdot 8}+1\right) \\
& =\left(2^{6}-1\right)\left(2^{6}+1\right)\left(2^{6 \cdot 2}+1\right)\left(2^{6 \cdot 4}+1\right)\left(2^{6 \cdot 8}+1\right), \\
& 2^{6}-1=63, \quad 2^{6}+1=65 .
\end{aligned}
$$
Therefore, the answer is $(C)$.