Solutions to $a,\ b,\ c,\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}$ I came across a puzzle in a Maths Calendar I own. Most of them I can do fairly easily, but this one has me stumped, and I was hoping for a hint or solution. The question is:
What are the solutions to
$$\left \{ a,\ b,\ c,\ \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a},\ \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \right \} \subset \mathbb{Z}$$
I've tried a few things, but don't think I've made any meaningful progress, besides determining that $a = \pm b = \pm c \ $ are the only obvious possible solutions. My hope is to prove that no other solution can exist.

I don't know if it helps, but I also did a brute force search for coprime numbers $a,b,c$ for which $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \in \mathbb{Z}$, with $1 \leq a \leq b \leq c$, and $a \leq 100, b \leq 1000, c \leq 10000$. 
The reason for coprimality is that if a solution has a common factor, we can divide through by the common factor and have another solution that satisfies the conditions.
The triplets I found which satisfy this are:
$(a, b, c) = (1, 1, 1), (1,2,4), (2, 36, 81), (3, 126, 196), (4, 9, 162), (9, 14, 588), (12, 63, 98), (18, 28, 147), (98, 108, 5103)$
None of these except the first satisfy $\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{a}{c} \in \mathbb{Z}$.

Suppose that $\displaystyle a,b,c,\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \in \mathbb Z$.
Consider polynomial 
$$P(x)=\left(x-\frac{a}{b}\right)\left(x-\frac{b}{c}\right)\left(x-\frac{c}{a}\right) = x^3-\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)x^2+\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)x-1.$$
Its coefficients are integers. Since the leading coefficient is $1$, all rational roots of $P$ are integers. Since the constant term is $-1$, it follows that all integer roots of $P$ are $1$ or $-1$ (they must divide the constant term). Since $\dfrac ab, \dfrac bc, \dfrac ca$ are rational roots of $P$, it follows that $\dfrac ab, \dfrac bc, \dfrac ca \in \{-1,1\}$.
