IMC 2012 · Day 2 · P9

Throughout the solution we will use the notation $x_{n+1} = x_1$.

We prove that the set of possible values of $a$ is $$\left( -\infty, \frac{1}{4} \right] \cup \left\{ \frac{1}{4 \cos^2 \frac{k\pi}{n}}\ ;\ k \in \mathbb{N},\ 1 \le k < \frac{n}{2} \right\}.$$ In the case $a \le \frac{1}{4}$ we can choose $x_1$ such that $x_1 (1 - x_1) = a$ and set $x_1 = x_2 = \dots = x_n$. Hence we will now suppose that $a > \frac{1}{4}$.

The system (1) gives the recurrence formula $$x_{i+1} = \varphi(x_i) = 1 - \frac{a}{x_i} = \frac{x_i - a}{x_i}, \quad i = 1, \dots, n.$$ The fractional linear transform $\varphi$ can be interpreted as a projective transform of the real projective line $\mathbb{R} \cup \{\infty\}$; the map $\varphi$ is an element of the group $PGL_2(\mathbb{R})$, represented by the linear transform $M = \begin{pmatrix} 1 & -a \\ 1 & 0 \end{pmatrix}$. (Note that $\det M \ne 0$ since $a \ne 0$.) The transform $\varphi^n$ can be represented by $M^n$. A point $[u, v]$ (written in homogenous

coordinates) is a fixed point of this transform if and only if $(u, v)^T$ is an eigenvector of $M^n$. Since the entries of $M^n$ and the coordinates $u, v$ are real, the corresponding eigenvalue is real, too.

The characteristic polynomial of $M$ is $x^2 - x + a$, which has no real root for $a > \frac{1}{4}$. So $M$ has two conjugate complex eigenvalues $\lambda_{1,2} = \frac{1}{2} \left( 1 \pm \sqrt{4a - 1}\, i \right)$. The eigenvalues of $M^n$ are $\lambda_{1,2}^n$, they are real if and only if $\arg \lambda_{1,2} = \pm \frac{k\pi}{n}$ with some integer $k$; this is equivalent with $$\pm \sqrt{4a - 1} = \tan \frac{k\pi}{n},$$ $$a = \frac{1}{4} \left( 1 + \tan^2 \frac{k\pi}{n} \right) = \frac{1}{4 \cos^2 \frac{k\pi}{n}}.$$ If $\arg \lambda_1 = \frac{k\pi}{n}$ then $\lambda_1^n = \overline{\lambda_2^n}$, so the eigenvalues of $M^n$ are equal. The eigenvalues of $M$ are distinct, so $M$ and $M^n$ have two linearly independent eigenvectors. Hence, $M^n$ is a multiple of the identity. This means that the projective transform $\varphi^n$ is the identity; starting from an arbitrary point $x_1 \in \mathbb{R} \cup \{\infty\}$, the cycle $x_1, x_2, \dots, x_n$ closes at $x_{n+1} = x_1$. There are only finitely many cycles $x_1, x_2, \dots, x_n$ containing the point $\infty$; all other cycles are solutions for (1).

Remark. If we write $x_j = P + Q \tan t_j$ where $P, Q$ and $t_1, \dots, t_n$ are real numbers, the recurrence relation re-writes as $$(P + Q \tan t_j)(1 - P - Q \tan t_{j+1}) = a$$ $$(1 - P) Q \tan t_j - P Q \tan t_{j+1} = a + P(P - 1) + Q^2 \tan t_j \tan t_{j+1} \quad (j = 1, 2, \dots, n).$$ In view of the identity $\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}$, it is reasonable to choose $P = \frac{1}{2}$, and $Q = \sqrt{a - \frac{1}{4}}$. Then the recurrence leads to $$t_j - t_{j+1} \equiv -\arctan \sqrt{4a - 1} \pmod{\pi}.$$