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!
 A: 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.
A: You have the $2$ values, stated to be integers, of
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{a^2c + ab^2 + bc^2}{abc} \tag{1}\label{eq1A}$$
$$\frac{a}{c} + \frac{b}{a} + \frac{c}{b} = \frac{a^2b + b^2c + ac^2}{abc} \tag{2}\label{eq2A}$$
Consider any prime $p$ where $p \mid abc$. Note the $p$-adic valuation function is $v_{p}(d) = n$, which means $p^n \mid d$ but $p^{n+1} \not\mid d$. Then have
$$v_{p}(a) = i \tag{4}\label{eq4A}$$
$$v_{p}(b) = j \tag{5}\label{eq5A}$$
$$v_{p}(c) = k \tag{6}\label{eq6A}$$
This means that for \eqref{eq1A} and \eqref{eq2A} to be integers, both numerators must have at least
$$m = i + j + k \tag{7}\label{eq7A}$$
factors of $p$. Also, with the numerator in \eqref{eq1A},
$$v_{p}(a^2c) = 2i + k \tag{8}\label{eq8A}$$
$$v_{p}(ab^2) = i + 2j \tag{9}\label{eq9A}$$
$$v_{p}(bc^2) = j + 2k \tag{10}\label{eq10A}$$
and the numerator in \eqref{eq2A} has
$$v_{p}(a^2b) = 2i + j \tag{11}\label{eq11A}$$
$$v_{p}(b^2c) = 2j + k \tag{12}\label{eq12A}$$
$$v_{p}(ac^2) = i + 2k \tag{13}\label{eq13A}$$
The sum of \eqref{eq8A}, \eqref{eq9A} and \eqref{eq10A}, as well as of \eqref{eq11A}, \eqref{eq12A} and \eqref{eq13A}, is $3i + 3j + 3k$, so their average is $i + j + k = m$. If all $3$ values being summed are the same, then $i = j = k$. Otherwise, consider them to not all be the same. Then if just one is less than $m$, the other $2$ must be greater than or equal to $m$, which is not possible since it means the numerator will have less than $m$ factors of $p$. Thus, you need to have $2$ be less than $m$, with both being the same, and one greater than $m$. For the first numerator, consider \eqref{eq10A} to be the largest, with the other $2$ of \eqref{eq8A} and \eqref{eq9A} being the same. Thus, you get that
$$2i + k = i + 2j \implies i = 2j - k \tag{14}\label{eq14A}$$
$$2i + k \lt j + 2k \implies 2i \lt j + k \tag{15}\label{eq15A}$$
Substituting \eqref{eq14A} into \eqref{eq13A} gives
$$i + 2k = 2j - k + 2k = 2j + k \tag{16}\label{eq16A}$$
This is the same as \eqref{eq12A}. The value in \eqref{eq11A} can't be the same, nor as discussed earlier, can it be less, so it must be more than \eqref{eq16A}. Thus, you have
$$2i + j \gt 2j + k \implies 2i \gt j + k \tag{17}\label{eq17A}$$
However, this contradicts \eqref{eq15A}! This means the original assumption of \eqref{eq10A} being the largest can't be true. You can similarly show that neither of \eqref{eq8A} or \eqref{eq9A} can be the largest, with the other $2$ being smaller. This means $i = j = k$. Since $p$ was any prime where $p \mid abc$, this means the $p$-adic values for all these prime factors is the same for each of $a$, $b$ and $c$, and since any primes which divide $a$, $b$ or $c$ must divide $abc$, this means you get
$$|a| = |b| = |c| \tag{18}\label{eq18A}$$
which is what was requested to be proven.
