# 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. [duplicate]

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.

• Nice approach!! Oct 4, 2020 at 6:14

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{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}\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{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}\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.