Solving $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$, $\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$

Question

Whether non-zero integers $a, b, c$ with the property that $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = m \in \mathbb{Z}$$ and $$\frac{a}{c} + \frac{c}{b} + \frac{b}{a }= n \in \mathbb{Z}$$ Calculate all possible values for $m + n$.

Answer 1

Let $\frac{a}{b}=x$ , $\frac{b}{c}=y$ , $\frac{c}{a}=z$

$x+y+z=m$, $xy+yz+zx=n$

$m+n+xyz+1=1+x+y+z+xy+yz+zx+xyz$

$m+n+2=(x+1)(y+1)(z+1)\geqslant 2^3\sqrt{xyz}=2^3$ AM-GM

$m+n+2\geqslant 8$

$m+n \geqslant 6 $

I'm not sure about my solution

Answer 2

We can assume that the three numbers do not a share a prime in their factorization. Rearrange the equations a bit:

$$a^2c + b^2a+c^2b = mabc \\ a^2b + c^2a + b^2c = nabc$$

Pick your favorite prime $p$. Suppose $N$ is largest so that $p^N\mid gcd(a,b)$.

If $p^{N+1}\nmid a$, then examine the second equation to see that $p^{2N}\mid a^2b,b^2c, nabc$ and so $p^{2N}\mid c^2a$, and so $p \mid c$.

If $p^{N+1}\nmid b$, then examine the first equation to see that $p^{2N}\mid a^2c,b^2a,mabc$ and so $p^{2N}\mid c^2b$, and so $p \mid c$.

This implies that a prime shared by 2 of the numbers is shared by the third, contradicting our initial assumption.

This is a hint to get you started. I'll add the rest of the solution later.

Answer 3

let integer $a,b,c$ such $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$ and $\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}$ is also integer, then we have $$|a|=|b|=|c|$$

proof: WLOG:we may assume $gcd(a,b,c)=1$,Indeed,we can consider$\dfrac{a}{d},\dfrac{b}{d},\dfrac{c}{d}$ where $\gcd{(a,b,c)}=d$

a contradiction,then $a,b,c$ least one $\neq\pm 1$,WLOG $a\neq\pm 1$,and let prime number $p|a$,then $$p|abc\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)=a^2c+b^2a+c^2b$$ so $p|c^2b$ so $p|b$ or $p|c$ WLOG: $p|b,p\nmid c$,and let $v_{p}(a)=r,v_{p}(b)=s,r\le s$,then we have $$p^{r+s}|abc\left(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\right)=a^2b+c^2a+b^2c$$ But $$p^{s+r}\nmid c^2a,p^{r+s}\nmid b^2c,p^{r+s}|a^2b$$ a contradiction.

so $$|a|=|b|=|c|$$ then easy to find $m,n$