Problem source

It is given that $x,y$ and $z$ are distinct and non-zero, and that they satisfy 
	\[
	x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}.
	\]
	Show that $x^{2}y^{2}z^{2}=1$ and that the value of $x+\dfrac{1}{y}$ is either $+1$ or $-1$.

Solution source

\begin{align*}
&& x-y &= \frac1z - \frac1y \\
&& x-z &= \frac1x - \frac1y \\
&& y-z &= \frac1x - \frac1z \\
\Rightarrow && (x-y)(x-z)(y-z) &= \frac{(y-z)(y-x)(z-x)}{x^2y^2z^2} \\
\Rightarrow && x^2y^2 z^2 &= 1 \\
\end{align*}

Suppose $x + \frac1{y} =k \Rightarrow xy + 1 = ky$
Therefore $y + \frac{1}{z} = y \pm xy = k$

Therefore $1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1$