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?
 A: 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 $\pm 1$.
A: 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.
