If $x_2 = 1/(1 - x_1), x_3 = 1/(1 - x_2)$, can we show $x_1x_2x_3+1=0$?

If $x_2 = 1/(1 - x_1), x_3 = 1/(1 - x_2)$, can we show $x_1x_2x_3+1=0$?

Question

It is given that

$$\begin{equation*} x_2 = \frac{1}{1 - x_1}, \quad x_3 = \frac{1}{1 - x_2}, \quad x_4 = \frac{1}{1 - x_3}, \quad\text{where }x_1 \ne 0\text{ or }1\text{.} \end{equation*}$$

Prove that (i) $x_1x_2x_3 + 1 = 0$, (ii) $x_4 = x_1$.

Solution

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.