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A triplet (a,b,c) is called friendly if product of any two is equal to third number . Find number of such ordered triplet. a,b,c are real numbers

maveric
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    To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$. – JMoravitz Nov 14 '18 at 16:24
  • how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible. – maveric Nov 14 '18 at 16:27

1 Answers1

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Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get

$$ab\cdot bc \cdot ca = abc$$ $$(abc)^2=abc$$

The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.

If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.

If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=\pm 1$, $b=\pm 1$ and $c=\pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.

lisyarus
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