Round about

If $|x_1| > 2$ then $|x_2| = |2-x_1^2| > 2|x_1|-2 > |x_1| > 2$ , and hence $|x_1| > |x_4| > |x_3| > |x_2| > |x_1|$ , which is not possible. Thus we deduce that $|x_1| \le 2$ , so that we can write $x_1 = 2\cos\theta$ . But then $\begin{array}{rcl} x_2 & = & -2\cos2\theta \; = \; 2\cos(2\theta-\pi) \\ x_3 & = & 2\cos\big(2(2\theta-\pi)-\pi\big) \; = \; 2\cos(4\theta-\pi) \\ x_4 & = & 2\cos\big(2(4\theta-\pi)-\pi\big) \; = \; 2\cos(8\theta-\pi) \end{array}$ and so $\begin{array}{rcl} 2\cos\theta & = & 2\cos\big(2(8\theta-\pi) - \pi\big) \; = \; - 2\cos16\theta \\ \cos16\theta + \cos\theta & = & 0 \\ 2\cos\tfrac{17}{2}\theta \cos\tfrac{15}{2}\theta & = & 0 \end{array}$ Thus $x_1 = 2\cos\theta$ where either $\theta = \tfrac{m\pi}{15}$ or $\theta = \tfrac{m\pi}{17}$ for odd integers $m$ . Distinct possible values of $x_1$ are then $2\cos\tfrac{m\pi}{15} \; (m = 1,3,5,\ldots 15) \qquad 2\cos\tfrac{m\pi}{17} \; (m = 1,3,5,\ldots,17)$ We are told that the numbers $x_1,x_2,x_3,x_4$ must be distinct. If any two of $x_1,x_2,x_3,x_4$ are equal, then they are either all equal or else $x_1=x_3$ and $x_2=x_4$ . Either way we must have $x_1=x_3$ , in which case $2\cos\theta + 2\cos4\theta \; = \; 4\cos\tfrac52\theta \cos\tfrac32\theta \; = \; 0$ so we must exclude values for $x_1$ of the form $2\cos\tfrac{m\pi}{3}$ or $2\cos\tfrac{m\pi}{5}$ for odd integers $m$ . Thus we must exclude $2\cos\tfrac{m\pi}{15} \; (m = 3,5,9,15) \qquad 2\cos\tfrac{m\pi}{17} \; (m=17)$ from the above list of possible values of $x_1$ .

Thus there are three possible sets {x 1,x 2,x 3,x 4}, namely $\begin{array}{c} \big\{ 2\cos\tfrac{\pi}{15},2\cos\tfrac{13\pi}{15},2\cos\tfrac{11\pi}{15},2\cos\tfrac{7\pi}{15} \big\} \\ \big\{ 2\cos\tfrac{\pi}{17},2\cos\tfrac{15\pi}{17},2\cos\tfrac{13\pi}{17},2\cos\tfrac{9\pi}{17} \big\} \\ \big\{ 2\cos\tfrac{3\pi}{17},2\cos\tfrac{11\pi}{17},2\cos\tfrac{5\pi}{17},2\cos\tfrac{7\pi}{17}\big\} \end{array}$ and each of these sets has a minimum element, which must be $x_1$ . The values of $x_2,x_3,x_4$ are then fixed. Thus there are $3$ round sets.

Let $f(x) = 2 - x^2$ . Then $x_2 = 2 - x_1^2 = f(x_1)$ . Similarly, $x_3 = f(x_2)$ , $x_4 = f(x_3)$ , and $x_1 = f(x_4)$ . Thus, solutions of the given system (without the minimum condition) correspond to solutions of the equation $f^4(x) = x$ , where $f^4$ denotes the fourth iteration of $f$ . Note that $f^4(x)$ is a polynomial of degree 16, so the system has at most 16 real solutions.

Suppose that $-2 \le x_1 \le 2$ . Then there exists an angle $\theta$ such that $x_1 = -2 \cos \theta$ . Also, $$x 2 = 2 - x 1^2 = 2 - 4 \cos^2 \theta = -2 (2 \cos^2 \theta - 1) = -2 \cos 2 \theta.$$ Similarly, $x_3 = -2 \cos 4 \theta$ , $x_4 = -2 \cos 8 \theta$ , and $x_1 = -2 \cos 16 \theta$ . Thus, we want $\theta$ to satisfy $\cos 16 \theta = \cos \theta$ . From the sum-to-product formula, $$\sin \frac{15 \theta}{2} \sin \frac{17 \theta}{2} = 0.$$ Then $\theta = \frac{2k \pi}{15}$ or $\theta = \frac{2k \pi}{17}$ for some integer $k$ . Checking which values lead to distinct values of $x_1$ , we obtain the following solutions for $(x_1, x_2, x_3, x_4)$ : $(-2,-2,-2,-2)$ , $$\left( -2 \cos \frac{2k \pi}{15}, -2 \cos \frac{4k \pi}{15}, -2 \cos \frac{8k \pi}{15}, -2 \cos \frac{16k \pi}{15} \right)$$ for $1 \le k \le 7$ , and $$\left( -2 \cos \frac{2k \pi}{17}, -2 \cos \frac{4k \pi}{17}, -2 \cos \frac{8k \pi}{17}, -2 \cos \frac{16k \pi}{17} \right)$$ for $1 \le k \le 8$ . This give us $1 + 7 + 8 = 16$ distinct solutions, so these solutions are in fact all the real solutions to the given system. The important part of this argument is that the system does have 16 real solutions.

Next, we classify all 16 real solutions. It is possible that $x_1$ , $x_2$ , $x_3$ , and $x_4$ are all distinct. Otherwise, two of them are equal.

Suppose that two consecutive terms are equal, say $x_1$ and $x_2$ . From $x_2 = f(x_1)$ , we get $f(x_1) = x_1$ . Hence, $x_3 = f(x_2) = f(x_1) = x_1$ , and $x_4 = f(x_3) = f(x_1) = x_1$ , so $x_1 = x_2 = x_3 = x_4$ . It is easy to check that the only solutions to $f(x) = x$ are $x = 1$ and $x = -2$ .

Now, suppose that two non-consecutive terms are equal, say $x_1$ and $x_3$ . Then $x_4 = f(x_3) = f(x_1) = x_2$ . Also, $x_3 = f(f(x_1))$ , so $x_1$ is a root of the equation $f(f(x)) = x$ , or $$2 - (2 - x^2)^2 = x.$$ This equation factors as $$(x - 1)(x + 2)(x^2 - x - 1) = 0.$$ This factorization should make sense, because any root of $f(x) = x$ will also be a root of $f(f(x)) = x$ . Moreover, the solutions $x = 1$ and $x = -2$ lead to $x_1 = x_2$ . The roots of $x^2 - x - 1 = 0$ are $$x = \frac{1 \pm \sqrt{5}}{2}.$$

Thus, every solution to the given system falls under one of the following categories:

(1) $x_1$ , $x_2$ , $x_3$ , and $x_4$ are all distinct.

(2) $x_1 = x_2 = x_3 = x_4$ .

(3) $x_1 = x_3$ and $x_2 = x_4$ , and $x_1 \neq x_2$ .

There are two solutions in category (2) and two solutions in category (3), which leaves $16 - 2 - 2 = 12$ solutions in category (1).

In this problem, we are looking for solutions where $x_1$ , $x_2$ , $x_3$ , and $x_4$ are all distinct (category (1)), and $x_1 = \min(x_1, x_2, x_3, x_4)$ . Because each variable depends only on the previous variable via the function $f$ , any cyclic permutation of a solution gives another solution. There are 12 solutions in category (1), and in each such solution, all the $x_i$ are distinct, which means there is a unique cyclic permutation that leads to $x_1 = \min(x_1, x_2, x_3, x_4)$ . Therefore, the number of round sets is $12/4 = 3$ .