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!

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.
