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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: \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 \(1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1\)