# If $a,b,c, \frac{a}{b}+\frac{b}{c}+\frac{c}{a}, \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \in \mathbb{Z}$ prove that $\displaystyle |a|=|b|=|c|$.

If $\displaystyle a,b,c \in \mathbb{Z}$ and $\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\displaystyle \frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ are also integers then prove that $\displaystyle |a|=|b|=|c|$.

This is one of those things that seems like it should be fairly obvious, yet I've not found a nice way of doing it after a few days. My best attempt has been to say the following, which I will only sketch as it is very messy.

Let $\displaystyle \alpha=\frac{a}{b},\beta=\frac{b}{c},\gamma=\frac{c}{a}=\frac{1}{\alpha\beta}$. Let $\displaystyle \alpha+\beta+\gamma=N_1 \in \mathbb{Z}, \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=N_2 \in \mathbb{Z}$.

Treating $N_1,N_2, \alpha$ as given we can write a quadratic for $\displaystyle \gamma=\frac{c}{a}$. $\gamma$ must be rational and hence we get the restriction (after algebra) that $(N_1-\alpha)^2(N_2-\alpha)^2-4(N_1-\alpha)$ must be a perfect square, as well as $5$ more corresponding to formulating different quadratics. You can eventually get the required answer $|\alpha|=|\beta|=|\gamma|=1$ to drop out.

Any better ideas?

• $N_1-N_2=-\frac{(a-b) (a-c) (b-c)}{a b c}$, $N_1^2+N_2^2-N_1N_2=\frac{(a^2+a b+b^2) (a^2+a c+c^2) (b^2+b c+c^2)}{a^2 b^2 c^2}$ , but doesn't help much... May 14, 2015 at 22:45

Of course we may assume $a,b,c \ne 0$. Let $S(a,b,c) = \{a/b+b/c+c/a, b/a+c/b+a/c\}$. Note that $S$ is invariant under permutations of $a,b,c$. Also $S$ is invariant under scaling: $S(ta,tb,tc) = S(a,b,c)$ for any $t \ne 0$, so WLOG we may assume $\gcd(a,b,c) = 1$, and under this assumption show that all $a,b,c$ are $\pm 1$.

Now let $p$ be a prime that divides at least one of the numbers (but not all three). By permutation invariance we may assume its largest power in $a,b,c$ is in $a$, with second-largest in $b$. Thus $a = p^m a'$, $b = p^n b'$, with $a', b', c$ coprime to $p$, and $0 \le n \le m$, $m \ge 1$. But then $a/b + b/c + c/a = p^{n-m} a'/b' + p^m b'/c + c/(p^n a')$; the first two terms have no $p$ in their denominators, but the third does, so the result can't be an integer. We conclude there is no such $p$, which means $a,b,c$ must all be $\pm 1$.

• This is a nice answer, right from basic principles. Should have thought of this. May 15, 2015 at 6:42

Using your notation, we see that $\alpha \beta \gamma = 1$ and then $N_2 = \alpha \beta + \alpha \gamma + \beta \gamma$.

So, $\alpha , \beta , \gamma$ are the rational roots of the polynomial $x^3 - N_1x^2 + N_2x -1 \in \mathbb{Z}[x]$.

Applying the Rational Root Test, we conclude that $\alpha , \beta , \gamma = \pm 1$ and the result follows.

• So slick, my personal favourite. May 15, 2015 at 6:42