Problem:
How many ordered triples (a,b,c) of non-zero real numbers have the property that each number is the product of the other two?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
The simultaneous equations a=bc,b=ca,c=ab imply
abc=(bc)(ca)(ab)=(abc)2
so either abc=0 (ruled out) or abc=1. The same simultaneous equations above also imply
abc=a2=b2=c2
so ∣a∣=∣b∣=∣c∣=1. It cannot be that all 3 unknowns are −1, nor can exactly one be −1, for in either case a=bc is not satisfied. However, the remaining four cases,
(a,b,c)=(1,1,1),(−1,−1,1),(−1,1,−1) or (1,−1,−1)
are all solutions. Thus there are 4 solutions in all.