China Mathematical Olympiad

(1) Proof of existence: From $x_{i+1} = 2x_i + 2x_i^3 - 2a^3 - x_{i-1}$, $i = 1, 2, \dots, n$ and $x_0 = 0$, we get that $x_i$ is a polynomial of $x_1$ with degree $3^{i-1}$ and real coefficients, for $1 \le i \le n+1$. Specifically, $x_{n+1}$ is a polynomial of $x_1$ with degree $3^n$ and real coefficients. As $3^n$ is an odd number, there exists a real number $x_1$ such that $x_{n+1} = 0$. Then from this $x_1$ and $x_0 = 0$ we can calculate $x_i$. The sequence $x_0, x_1, \dots, x_n, x_{n+1}$ obtained in this way satisfies the required condition.

Proof of uniqueness: Suppose there are two sequences $w_0, w_1, \dots, w_n, w_{n+1}$ and $v_0, v_1, \dots, v_n, v_{n+1}$, both of which satisfy Condition (1). Then $\frac{1}{2}(w_{i+1} + w_{i-1}) = w_i + w_i^3 - a^3$ and $\frac{1}{2}(v_{i+1} + v_{i-1}) = v_i + v_i^3 - a^3$. Thus,
$$\frac{1}{2}(w_{i+1} - v_{i+1} + w_{i-1} - v_{i-1}) = (w_i - v_i)(1 + w_i^2 + w_i v_i + v_i^2).$$
Suppose $|w_{i_0} - v_{i_0}|$ is the greatest. Then
$$\begin{aligned} |w_{i_0} - v_{i_0}| &\le |w_{i_0} - v_{i_0}| (1 + w_{i_0}^2 + w_{i_0} v_{i_0} + v_{i_0}^2) \\ &\le \frac{1}{2} |w_{i_0+1} - v_{i_0+1}| + \frac{1}{2} |w_{i_0-1} - v_{i_0-1}| \\ &\le |w_{i_0} - v_{i_0}|. \end{aligned}$$
So, either $|w_{i_0} - v_{i_0}| = 0$ or $(1+w_{i_0}^2+w_{i_0}v_{i_0}+v_{i_0}^2) = 0$, that is, either $|w_{i_0} - v_{i_0}| = 0$ or $w_{i_0}^2+v_{i_0}^2+(w_{i_0}+v_{i_0})^2 = 0$. However it must be $|w_{i_0} - v_{i_0}| = 0$. Since $|w_{i_0} - v_{i_0}|$ is the greatest. Then $|w_i - v_i| = 0$ for every $i = 1, 2, \dots, n$. This completed the proof of (1).

(2) Suppose that $|x_{i_0}|$ is the greatest. Then we have
$$\begin{aligned} |x_{i_0}| + |x_{i_0}|^3 &= |x_{i_0}| (1 + |x_{i_0}|^2) \\ &= \left| \frac{1}{2}(x_{i_0+1} + x_{i_0-1}) + a^3 \right| \\ &\le \frac{1}{2} |x_{i_0+1}| + \frac{1}{2} |x_{i_0-1}| + |a|^3 \end{aligned}$$
$$\leq |x_{i_0}| + |a|^3.$$
That means $|x_{i_0}| \leq |a|$. So $|x_i| \leq |a|$, $i = 0, 1, \dots, n+1$.