$31 \cdot 43$ The number of ordered triples $(a, b, c)$ of nonzero real numbers that have the property that each number is the product of the other two is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(E) 5 .
(38th American High School Mathematics Examination, 1987)
［Solution］Given $a=b c, b=a c, c=a b$,
$\therefore a=b c=(a c)(a b)=a^{2} c b$, i.e., $b c=a^{2} b c$,
$\because b \neq 0, \quad c \neq 0$, thus $a^{2}=1, \quad a= \pm 1$.
Similarly, $b= \pm 1, c= \pm 1$. Note that $(-1,-1,-1)$ does not satisfy the conditions.
Therefore, $(a, b, c)=(1,1,1)$ or $(1,-1,-1)$ or $(-1,1,-1)$ or $(-1,-1,1)$, a total of four sets.
Hence, the answer is $(D)$.