Prove that (i) \(x_1x_2x_3 + 1 = 0\), (ii) \(x_4 = x_1\).
As \(x_1\) neither equals \(0\) nor \(1\), it follows that \(x_2\), \(x_3\), and \(x_4\) likewise neither equal \(0\) nor \(1\).
We can write \[\begin{align*} x_1x_2x_3 = \dfrac{x_1}{1 - x_1} \dfrac{1}{1 - x_2} &= \dfrac{x_1}{1 - x_1} \dfrac{1}{1 - \dfrac{1}{1 - x_1}} \\ &= \dfrac{x_1}{1 - x_1} \dfrac{1 - x_1}{(1 - x_1) - 1} \\ &= \dfrac{x_1}{1 - x_1} \dfrac{1 - x_1}{-x_1} \\ &= -1. \end{align*}\]from which the result immediately follows.
For the second part, note that what we have just proved for \(x_1 x_2 x_3\) applies similarly to \(x_2 x_3 x_4\). So we also have \(x_2 x_3 x_4 = -1\).
Then
\[x_2 x_3 x_4 = x_1 x_2 x_3,\]
so dividing both sides by \(x_2 x_3\) yields
\[x_4 = x_1.\]
If \(f(x) = \dfrac{1}{1 - x}\), then we have shown \(f^3(x) = x\).
We might wonder, if \(f(x) = \dfrac{p(x)}{q(x)}\), where \(p\) and \(q\) are polynomials with rational coefficients, and \(f^n(x) = x\) for (almost ) all \(x\), what are the possible values for \(n\)?
These beautiful periodic creatures are called Lyness cycles, after Robert Cranston Lyness.
