$10 \cdot 42$ If $a_{1}, a_{2}, a_{3}, \cdots$ is a sequence of positive numbers and satisfies $a_{n+2}=a_{n} a_{n+1}$ for all positive integers $n$, then the sequence $a_{1}, a_{2}, a_{3}, \cdots$ is a geometric sequence if and only if
(A) $a_{1}$ and $a_{2}$ are any positive numbers.
(B) $a_{1}=a_{2}$.
(C) $a_{1}=1$.
(D) $a_{2}=1$.
(E) $a_{1}=a_{2}=1$.
(28th American High School Mathematics Examination, 1977)
[Solution] According to the problem, the second to fifth terms of the sequence $\left\{a_{n}\right\}$ are
$$
a_{2}, a_{1} a_{2}, a_{1} a_{2}^{2}, a_{1}^{2} a_{2}^{3} .
$$

If they form a geometric sequence, the ratio of consecutive terms must be equal:
$$
\frac{a_{1} a_{2}}{a_{2}}=\frac{a_{1} a_{2}^{2}}{a_{1} a_{2}}=\frac{a_{1}^{2} a_{2}^{3}}{a_{1} a_{2}^{2}}, \text { i.e., } a_{1}=a_{2}=a_{1} a_{2} \text {. }
$$

Since $a_{1}$ and $a_{2}$ are positive numbers, it must be that $a_{1}=a_{2}=1$.
On the other hand, if $a_{1}=a_{2}=1$, then $\left\{a_{n}\right\}$ is the geometric sequence $1,1,1, \cdots$ Therefore, the answer is $(E)$.