Posted: http://www.artofproblemsolving.com/Forum/topic-2592.html

we can slove the problem generaly $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$is enteger prove that abc is acube

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As proved in the given link, the result still holds even if $3$ is replaced by any integer. I was wondering why the statement asked for that particuliar case, but I think I know now Pierre.

Pascual2005 wrote: Let $a,b,c$ be integers such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=3$ prove that $abc$ is a perfect cube!

am i right?

@above How do you know that $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ has to be integers. 6

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@Vukch $a, b, c$ can be negative.

@above, would you enlighten me how to prove it with negative numbers?

Vulch wrote: @above, would you enlighten me how to prove it with negative numbers? It has something to do with $p$-adic valuations if I remember correctly.

Simply note that by the AM-GM Inequality we know \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{abc}{abc}} = 3\]with equality holding if and only if $a=b=c$. Thus, $abc=a^3$ which is most clearly a perfect cube. @below Good Point. Hmm lemme see.

However, $a, b, c$ can be negative, so I don't think the above solution works as AM-GM is only for positive numbers?

The following proof lets the sum equal any integer not just 3: We can divide out the gcd of all 3 numbers without changing the value of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, so WLOG let $\gcd(a,b,c)=1$. Note the sum is an integer iff $abc|a^2c+b^2a+c^2b$. To show $abc$ is a cube, we can show for all primes $p$, $3|v_p(abc)$ (Note that $v_p(a)$ is defined to be the maximal number of times one can divide $a$ by $p$ e.g. $v_2(96)=5$). So let $p$ be an arbitrary prime. If $p$ doesn't divide any of the 3 integers, then $v_p(abc)=0$ which is a multiple of $3$, done. Otherwise, WLOG let it divide $a$. Let $v_p(a)=k$. Then we have $a|c^2b$, thus either $p|b$ or $p|c$, but note that by our gcd=1 assumption we can't have both. So we split into cases: Case 1: $p|b$ Then since $p$ doesn't divide $c$, we must have $p^k|b$. But then $p^{2k}|abc$ thus $p^{2k}|a^2c+b^2a+c^2b$. Note $a^2c$ and $b^2a$ must be multiples of $p^{2k}$ so we have $p^{2k}|c^2b$ thus $p^{2k}|b$ so $v_p(b)\geq 2k$. THIS implies $p^{3k}|a^2c+b^2a+c^2b$ which means $p^{3k}|a^2c+c^2b$. If $v_p(b)>2k$, note that $v_p(a^2c)=2k$ and $v_p(c^2b)>2k$, thus $v_p(a^2c+c^2b)=2k<3k$ contradiction! Thus $v_p(b)=2k$ implying that $v_p(abc)=3k$ which is a multiple of $3$. Case 2: $p|c$ Then since $p$ doesn't divide $b$, we must have $p^k|c^2$. But then $p^{\lceil\frac{3k}{2}\rceil}|abc$ thus $p^{\lceil\frac{3k}{2}\rceil}|a^2c+b^2a+c^2b$. Note $a^2c$ is a multiple of $p^{\lceil\frac{3k}{2}\rceil}$ so we have $p^{\lceil\frac{3k}{2}\rceil}|b^2a+c^2b$. If $v_p(c^2)>k$, note that $v_p(b^2a)=k$ and $v_p(c^2b)>k$, thus $v_p(b^2a+c^2b)=k<\lceil\frac{3k}{2}\rceil$ contradiction! Thus $v_p(c^2)=k$ implying $k$ is even and that $v_p(abc)=\frac{3k}{2}$ which is a multiple of $3$. It's funny how I'm pretty sure the problem writer made the sum equal $3$ so that people would think AM-GM had something to do with solving it...