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Here is my solution; can someone check it out?

The sum $$ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} $$ is just
$$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} $$ but 'reversed'. No element of this expression can be a fraction. I prove that by substituting $\frac{1}{3}$ under $\frac{a}{b}$. Therefore $a,b,c=1||0||-1$.

Daniel W. Farlow
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I I
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    How about $a=b=-c=1$? – Jaakko Seppälä Jan 30 '15 at 22:46
  • Can you show how you "prove" that no element of the expression can be a fraction? I don't see how that works at all. ~~~~~~ @user2219896 positive integers I guess. – Joffan Jan 30 '15 at 22:56
  • @user2219896 yeah, $\frac{1}{3}+\frac{3}{c}+\frac{c}{1}=x$ after solving we get $c+\frac{10c}{3c}=x$ – I I Jan 30 '15 at 22:59
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    see http://math.stackexchange.com/questions/1102537/solving-fracab-fracbc-fracca-m-in-mathbbz-fraca/1102644#1102644 – math110 Jan 31 '15 at 04:20

1 Answers1

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We assume that $a,b,c>0$.

By cancelling any common factors, positive integers $a,b,c$ satisfying the conditions can be replaced by positive integers $a',b',c'$ with no common factors, which still satisfy the conditions. Moreover $a=b=c$ iff $a'=b'=c'$. Therefore, it suffices to prove that the problem holds when $a,b,c$ have no common factors.

For a prime $p$, write $a=p^{\alpha}A$, $b=p^{\beta}B$ and $c=p^{\gamma}C$, where $A,B,C$ are not divisible by $p$. Since $a,b,c$ share no common factor, one of $\alpha,\beta,\gamma$ must be zero. Since the problem doesn't change under any permutation of $a,b,c$, we can assume that $\gamma=0$ and $\alpha\geq\beta$.

Now \begin{align*} \frac{a}{b}+\frac{b}{c}+\frac{c}{a} &= p^{\alpha-\beta}\frac{A}{B}+p^{\beta}\frac{B}{C}+p^{-\alpha}\frac{C}{A} \\ &= \frac{p^{2\alpha-\beta}A^2C+p^{\alpha+\beta}AB^2+BC^2}{p^{\alpha}ABC} \end{align*}

If $\alpha\neq0$, then $2\alpha-\beta,\alpha+\beta>0$. This means that the first two terms in the numerator are divisible by $p$, and the third is not. Therefore the numerator is not divisible by $p$. However there is a factor of $p$ in the denominator, contradicting the fact that this is an integer. Therefore $\alpha=0\implies\beta=0$.

This implies that none of $a,b,c$ are divisible by any prime, so they are all equal to 1. Hence $a=b=c$.

Peter Huxford
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