If $a$, $b$, $c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are all integers, then $|a|=|b|=|c|$ 
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.

Well this is what I have done so far:
From the fact that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, we get 
$$abc \mid ab^2 + bc^2 + ca^2 \tag{1}$$
In the same way, we also have 
$$abc \mid a^2b + b^2c + c^2a \tag{2}$$ 
so 
$$\begin{align}
abc \mid a(ab^2 + bc^2 + ca^2) - b(a^2b + b^2c + c^2a) &\Rightarrow abc \mid c(a^3-b^3) \tag{3}\\
&\Rightarrow ab \mid a^3-b^3 \tag{4}\\
&\Rightarrow ab \mid b^3(a^3-b^3) \tag{5}
\end{align}$$ and so $a \mid b^6$. In the same way, we can also get $b \mid c^6$ and $c \mid a^6$. 

But what should I do after this?

Any help is surely appreciated! Thanks!

I think what you've actually shown is that $a \mid b^5$, $b \mid c^5$ and $c \mid a^5$.
From this
you can argue as follows. Suppose $p$ is a prime factor of $a$. Then $a \mid b^5$ implies that
$p \mid b$, and similarly $b \mid c^5$ now implies that $p \mid c$.
Thus $p$ divides all of $a$, $b$, and $c$. Similarly, any prime factor of $b$ or $c$ divides all of $a$, $b$, and $c$.
So in the above situation, you can replace  $a$ by $A = a/p$, $b$ by $B = b/p$, and $c$ by $C = c/p$ for any prime factor of $a,b,$ or $c$, and $A$, $B$, and $C$ will satisfy the conditions of the original problem.
Now repeat the above as many times as needed, as long as there are any prime factors left. In the end you are left with your numbers being $\pm 1$. Looking backwards this means $|a| = |b| = |c|$ as desired.
