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 $$\begin{array}{rl}|w_{i_0}-v_{i_0}|& \le |w_{i_0}-v_{i_0}|(1+w_{i_0}^2+w_{i_0}{v}_{i_0}+v_{i_0}^2)\\ & \le \frac{1}{2}|w_{i_0+1}-v_{i_0+1}|+\frac{1}{2}|w_{i_0-1}-v_{i_0-1}|\\ & \le |w_{i_0}-v_{i_0}|.\end{array}$$ 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 $$\begin{array}{rl}|x_{i_0}|+|x_{i_0}{|}^3& =|x_{i_0}|(1+|x_{i_0}{|}^2)\\ & =\left|\frac{1}{2}(x_{i_0+1}+x_{i_0-1})+a^3\right|\\ & \le \frac{1}{2}|x_{i_0+1}|+\frac{1}{2}|x_{i_0-1}|+|a{|}^3\end{array}$$ $$\le |x_{i_0}|+|a{|}^3.$$ That means $|x_{i_0}|\le |a|$. So $|x_i|\le |a|$, $i=0,1,\dots ,n+1$.