Let a, b, c be ints. $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an int, show that each of $\frac{ab}{c}, \frac{bc}{a}, \frac{ac}{b}$ is an int. Let a, b, c $\in \mathbb{Z}$ . If $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an integer, prove that each of $\frac{ab}{c}, \ \frac{bc}{a}, \ \frac{ac}{b}$ is an integer.
I've tried to solve this problem but still got no solution. All I think is divisibility and GCD
$\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b} \\ = \frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2}}{abc}$
Note that $2a^{2}bc + 2ab^{2}c + 2abc^{2}$ is divisible by abc. Put those in, we get:
$\frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} + 2a^{2}bc + 2ab^{2}c + 2abc^{2}}{abc} \\ = \frac{(ab + bc + ac)^{2}}{abc}$
Because it's an integer, thus $abc \mid (ab + bc + ac)^{2}$
Assume $GCD(ab + bc + ac, abc) = d$, then $ab + bc + ac = dk_1$ and $abc = dk_2$ for an integer d where $GCD(k_1, k_2) = 1$
$\frac{(ab + bc + ac)^{2}}{abc} = \frac{d^{2}{k_1}^{2}}{dk_2} = \frac{d{k_1}^2}{k_2}$
Because $GCD(k_1, k_2) = 1$, thus the only possibility is $k_2 \mid d$. Let d = $k_{2}p$ where p is an integer, thus it implies that $abc = dk_2 = {k_2}^{2}p$
I got stuck here, I probably used the wrong method to solve this problem, does anyone know how to solve this?

Let $x=\frac{bc}{a}, y=\frac{ca}{b}, z=\frac{ab}{c}$, then $x, y, z \in \Bbb{Q}$ and by condition $x+y+z=\alpha \in \Bbb{Z}$. It is easy to verify that $yz+zx+xy=a^2+b^2+c^2=\beta \in \Bbb{Z}$, $xyz=abc=\gamma \in \Bbb{Z}$. So $x, y, z$ are the rational roots of the monic polynomial $t^3-\alpha t^2+\beta t-\gamma=0$ whose coefficients are all integer, hence $x, y, z$ must be integer.
