WakeUp wrote: For a given real number $a$ and a positive integer $n$, prove that: i) there exists exactly one sequence of real numbers $x_0,x_1,\ldots,x_n,x_{n+1}$ such that \[\begin{cases} x_0=x_{n+1}=0,\\ \frac{1}{2}(x_i+x_{i+1})=x_i+x_i^3-a^3,\ i=1,2,\ldots,n.\end{cases}\] ii) the sequence $x_0,x_1,\ldots,x_n,x_{n+1}$ in i) satisfies $|x_i|\le |a|$ where $i=0,1,\ldots,n+1$. Liang Yengde i) Since $\frac{1}{2}(x_i+x_{i+1})=x_i+x_i^3-a^3$, \[2x_i^3+x_i=x_{i+1}+2a^3\quad (1).\] And since $x_{n+1}=0$, $2x_n^3+x_n=2a^3\quad (2)$. Now, let consider the function $f(t)=2t^3+t$. It is clearly $f(t)$ is a increasing function in $\mathbb{R}$, (2) has only real root. So $x_n$ exists and only! Now let consider (1). It has only real root $x_i$ for a fix $x_{i+1}$. Then, by induction hypothesys, there exists exactly one sequence of real numbers $x_0,x_1,\ldots,x_n,x_{n+1}$. ii). We will prove that the such sequence in i) satisfies the condition $|x_i|\le |a|$. Since $2x_n^3+x_n=2a^3$, $|2x_n^3+x_n|=|2a^3|$. For a contradiction, if $|x_n|> |a|$, $|2x_n^3+x_n|=|x_n|(2|x_n|^2+1)>2|a^3|$. Then $|x_n|\le |a|$. Since (1), $|x_{n-1}|(2|x_{n-1}|^2+1)=|2x_{n-1}^3+x_{n-1}|=|x_{n}+2a^3|\le |x_{n}|+2|a^3|\le|a|+2|a^3|$, implies $2|x_{n-1}^3|+|x_{n-1}|-2|a^3|-|a|\le 0$, or $(|x_{n-1}|-|a|)(2|x_{n-1}|^2+2|x_{n-1}||a|+2|a|^2+1)\le 0$, then $|x_{n-1}|\le |a|$. By (1) and by induction hypothesys, $|x_i|\le |a|\forall i\1,2,\cdots,n$. And since it is clearly $|x_0|=|x_{n+1}|=0\le |a|$, we are done.

This problem should have \[\begin{cases} x_0=x_{n+1}=0,\\ \frac{1}{2}(x_{i-1}+x_{i+1})=x_i+x_i^3-a^3,\ i=1,2,\ldots,n.\end{cases}\]