Are there any positive integers such that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1$?

Question

Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.

I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.

Answer 1

By the AM-GM inequality, one has $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3\sqrt[3]{\frac{x}{y} \cdot\frac{y}{z} \cdot\frac{z}{x} }=3>1 $$ and hence it is impossible.

Answer 2

Rearrange the equality

$x^2z+y^2x+z^2y=xyz$

Since $x,y,z>0$, we have

$x^2z+y^2x+z^2y>x^2z$

Multiply both sides with $\frac{y}{x}$

$y^2z>xyz$

And $y>x$

On the other hand,

$x^2z+y^2x+z^2y>y^2x$

Multiply both sides with $\frac{z}{y}$ and you get

$xz^2>xyz$, and $z>y$

Lastly,

$x^2z+y^2x+z^2y>z^2y$ Multiply both sides with $\frac{x}{z}$ and get

$x^2y>xyz$ and $x>z$

So we got, $y>x, z>y, x>z$ which is clearly not possible when $x,y,z$ are positive.

Answer 3

Here is an alternative.

Multiply both sides by $xyz$

$$x^2z + xy^2 + yz^2 = xyz$$

Subtract the third term $yz^2$ from both sides

$$x^2z + xy^2 = (x - z)yz$$

The LHS is clearly positive so the RHS must be as well. Hence $x > z$.

Do the same for the other two terms gives: $z > y$ and $y > x$.

These cannot all be true.

Answer 4

Assume, that a solution exists. Notice, that in each of the summands in $$ \frac xy + \frac yz + \frac zx = 1 $$ has to lie in the open intervall $(0,1)$, because $x,y,z$ are required to be positive. Therefore, we have $\frac xy < 1$, i. e. $x < y$. Similarly, we conclude $y < z$ and $z < x$. Hence $$ x < y < z < x, $$ a contradiction. Thus, no solution exists.

Answer 5

Clearly, 2 among $x,y,z$ cannot be equal. Without loss of generality, we suppose: $$z>y>x>0$$ Therefore, $\dfrac{z}{x}>1$, which contradicts $$\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=1$$