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How many positive integers $p$, $q$, $r$ exists satisfying the relation below?

$$p/q + q/r + r/p = 2$$

Thank in advance. I tried so much but could not think how to proceed.

I just need a hint from where to begin.

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The following link gives not any help to find the answer to the problem but shows a related number theoretic problem math.stackexchange.com/questions/113546/… –  miracle173 Mar 17 '12 at 6:50
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1 Answer

up vote 1 down vote accepted

HINT : Note that from AM-GM inequality, we have that $a+b+c \geq 3 \sqrt[3]{abc}$. If you wish to view the answer, hover your cursor over the grey region directly below.

Answer: Since $p$, $q$ and $r$ are positive, so are $\frac{p}{q}$, $\frac{q}{r}$, $\frac{r}{p}$. From AM-GM inequality, we have that $$\displaystyle \frac{p}{q} + \frac{q}{r} + \frac{r}{p} \geq 3 \sqrt[3]{\frac{p}{q} \times \frac{q}{r} \times \frac{r}{p}} = 3.$$ Hence, there doesn't exists any $p$, $q$ and $r$ satisfying $\displaystyle \frac{p}{q} + \frac{q}{r} + \frac{r}{p} = 2$.

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Beat me to it by 30 seconds. :-) (I +1'd it. ) –  user21436 Mar 17 '12 at 6:00
    
@Sivaram what if the condition of p,q,r being integer is removed.will the answer remains the same. –  vikiiii Mar 17 '12 at 6:15
    
@vikiiii Yes. The proof only needs $a$,$b$,$c$ are positive. –  user17762 Mar 17 '12 at 6:16
    
@Sivaram : thanks.Idea of giving answer in grey region is awesum. –  vikiiii Mar 17 '12 at 6:23
    
I got that idea from one of @KannappanSampath answers. –  user17762 Mar 17 '12 at 6:24
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