<s>[INST] Solving $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$, $\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$ Whether non-zero integers $a, b, c$ with the property that $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$$ and $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$$ Calculate all possible values for $m + n$. 
 [/INST] let integer $a,b,c$ such $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$ and $\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$ is also integer, then we have
$$|a|=|b|=|c|$$
proof: WLOG:we may assume $gcd(a,b,c)=1$,Indeed,we can consider$\dfrac{a}{d},\dfrac{b}{d},\dfrac{c}{d}$
where $\gcd{(a,b,c)}=d$
a contradiction,then $a,b,c$ least one $\neq\pm 1$,WLOG $a\neq\pm 1$,and let prime number $p|a$,then
$$p|abc\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)=a^2c+b^2a+c^2b$$
so
$p|c^2b$
so $p|b$ or $p|c$
WLOG: $p|b,p\nmid c$,and let $v_{p}(a)=r,v_{p}(b)=s,r\le s$,then we have
$$p^{r+s}|abc\left(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\right)=a^2b+c^2a+b^2c$$
But $$p^{s+r}\nmid c^2a,p^{r+s}\nmid b^2c,p^{r+s}|a^2b$$
a contradiction.
so
$$|a|=|b|=|c|$$
then easy to find $m,n$
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