## Friday, July 23, 2010

### a,b,c integers and cubic number

Suppose $a,b,c$ are positive integers such that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer. Prove that $abc$ is a cubic number.

Solution

We have $ab^2 + bc^2 + ca^2 = kabc$ for some k.

Let $d = \gcd(a,b,c)$. We can replace $a,b,c$ by $a/d, b/d, c/d$ respectively and the problem does not change. Thus, without loss of generality, we may assume that $d = 1$.

Let $p$ be a prime that divides $abc$, which means $p$ divides at least one of $a,b,c$. We also know that $p$ cannot divide all three, since $d = 1$.

If $p$ divides exactly one of $a,b,c$, for example $a$, then $ab^2, ca^2, kabc$ are all divisible by $p$, but not $bc^2$. Impossible. Thus, $p$ must divide exactly two of $a,b,c$.

Suppose $p$ divides $a$ and $b$. Furthermore, let $x$ be the largest integer such that $p^x$ divides $a$. Likewise, let $y$ be the largest integer such that $p^y$ divides $b$.

Since $ab^2, bc^2, kabc$ are all divisible by $b$, then so is $ca^2$. Thus $y \leq 2x$.

Since $ab^2, ca^2, kabc$ are all divisible by $a$, then so is $bc^2$, Thus $y \geq x$, which means $x \leq y \leq 2x$.

Now, since $ab^2, ca^2, kabc$ are all divisible by $p^{2x}$, then so is $bc^2$, which means $y \geq 2x$.

Therefore, $y = 2x$, which means that the degree of $p$ in the factorization of $abc$ is $x+y = 3x$.

For each prime $p$ that divides $abc$, it must occur as a cubic number in its prime factorization. Thus $abc$ is a cubic number.

A non-trivial example is $a=1,b=2,c=4$.