$33 \cdot 65$ If the number $121_{b}$ (i.e., the number in base $b$) is the square of an integer, then $b$ is
(A) only $b=10$.
(B) only $b=10, b=5$.
(C) $2 \leqslant b \leqslant 10$.
(D) $b>2$.
(E) no such $b$ exists.
(13th American High School Mathematics Examination, 1962)
［Solution］From the given and the definition of base $b$, we have
$121_{b}=1 \cdot b^{2}+2 \cdot b+1=(b+1)^{2}$, which is a perfect square.
However, in base $b$, the largest digit can only be $b-1$, so $b$ cannot be 1 or 2, i.e., $b>2$.
Therefore, the answer is $(D)$.