Prove the non-existence of any triplet satisfying $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=1$

Question

Prove that there do not exist any positive integers $x,y$ and $z$ that satisfy $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=1$

My attempt:- Since $x,y $ and $z$ are positive integers, we can assume without loss of generality that $x\le y\le z$ . Hence we have $\frac{z}{y}$ and $\frac{y}{x} \ge 1$. This implies, $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\gt 2$$. Therefore, there do not exist any integers $x,y $ and $z$ satisfying the above equation. Does this look good? Can this be solved using divisibility or any other technique ?

Answer 1

You can use arithmetic - geometric mean inequality:

$$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq 3\sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}}=3$$

So the smallest value of $\frac{x}{y} + \frac{y}{z} + \frac{z}{x}$ is $3$ for $x=y=z$.

Answer 2

Let your fractions be $a,b,c,$ then note that also, $abc=1.$ Now, letting $ab+bc+ca=k,$ we have that $a,b,c$ are the roots of $$x^3-x^2+kx-1.$$

Since this is monic, it is clear that the three roots cannot all be integers, since the only integers that divide $1$ are $\pm1,$ and we are done.