# Prove that $abc$ is a cube of some integer.

Given three integers $a$, $b$ and $c$ such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is an integer too, prove that the product $abc$ is a cube. By the way: Merry Christmas! ;)

By dividing by a common factor if there is any, we can assume no prime number divides all of $a,b,c$. Our goal is to show that the exponent of any prime $p$ in prime decomposition of $abc$ is divisible by $3$.

Suppose $p$ divides one of the numbers, WLOG let $p\mid a$. Also, let $p^k$ be the greatest power of $p$ dividing $a$.

Our assumption on sum of fractions being integer is just saying that $abc\mid a^2c+b^2a+c^2b$. We see $p\mid abc$ and hence $p\mid a^2c+b^2a+c^2b$ and so $p\mid c^2b$ thus $p\mid b$ or $p\mid c$, but not both (as we assumed). Now we have two cases.

1) $p\mid b$. Let $p^l$ be the greatest power of $p$ dividing $b$. It's easy to see now that exponent of $p$ in $abc$ is $k+l$.

We have $p^{k+l}\mid abc$, so $p^{k+l}\mid a^2c+b^2a+c^2b$, hence $p^{k+l}\mid a^2c+c^2b$. The greatest power of $p$ dividing $c^2b$ is $p^l$ and the greatest power of $p$ dividing $a^2c$ is $p^{2k}$. If these exponents were different, then the greatest power of $p$ dividing sum of $c^2b$ and $a^2c$ would be $\min\{2k,l\}<k+l$ which is impossible. Hence $l=2k$ and $k+l=3k$, as we wanted.

2) $p\mid c$. This case is treated similarly - but now $c$ takes role of $a$ and $a$ takes role of $b$. I will leave it for you to fill in details.

These two cases in conjuction give us desired conclusion.

• $\min\{2k,l\}<k+l$? – zarathustra Dec 24 '15 at 10:09
• @zarathustra $\min\{2k,l\}\leq l<k+l$. – Wojowu Dec 24 '15 at 10:09
• ooooh good point. Nice answer! – zarathustra Dec 24 '15 at 10:16
• Great. Just great. – VanDerWarden Dec 24 '15 at 10:25