<s>[INST] 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?
 [/INST] You can use divisibility as I show here. First, let
$$\frac{ab}{c} = \frac{d_1}{e_1} \tag{1}\label{eq1A}$$
$$\frac{bc}{a} = \frac{d_2}{e_2} \tag{2}\label{eq2A}$$
$$\frac{ac}{b} = \frac{d_3}{e_3} \tag{3}\label{eq3A}$$
where each fraction $\frac{d_i}{e_i}$ for $1 \le i \le 3$ is in lowest terms, i.e., $\gcd(d_i, e_i) = 1$. Since the sum of these fractions is an integer, say $n$, we have
$$\begin{equation}\begin{aligned}
\frac{d_1}{e_1} + \frac{d_2}{e_2} + \frac{d_3}{e_3} & = n \\
d_1(e_2)(e_3) + d_2(e_1)(e_3) + d_3(e_1)(e_2) & = n(e_1)(e_2)(e_3)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Consider one of the fractions in the first $3$ equations to not be an integer, say WLOG in \eqref{eq1A}, then there exists a prime $p \mid e_1$, so $p \not\mid d_1$. Using the $p$-adic order function, i.e., which gives the highest power of $p$ which divides a given value, we have
$$\nu_p(e_1) \gt 0 \implies \nu_p(c) \gt \nu_p(a) + \nu_p(b) \tag{5}\label{eq5A}$$
If $p \not\mid e_2$ and $p \not\mid e_3$, then in \eqref{eq4A} on the left side, $p$ doesn't divide the first term, but it divides the second & third terms, plus it divides the right side term, which is not possible. Thus, $p \mid e_2$ and/or $p \mid e_3$, say WLOG we have $p \mid e_2$. This gives
$$\nu_p(e_2) \gt 0 \implies \nu_p(a) \gt \nu_p(b) + \nu_p(c) \tag{6}\label{eq6A}$$
Substituting this into \eqref{eq5A} gives
$$\begin{equation}\begin{aligned}
\nu_p(c) & \gt (\nu_p(b) + \nu_p(c)) + \nu_p(b) \\
\nu_p(c) & \gt 2\nu_p(b) + \nu_p(c) \\
0 & \gt 2\nu_p(b)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
which is not possible since $\nu_p(b) \ge 0$. Thus, the original assumption that one of the fractions in \eqref{eq1A}, \eqref{eq2A} or \eqref{eq3A} is not an integer must be false, i.e., they are actually all integers instead.
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