Saudi Arabian Mathematical Competitions, 2015

Let $P_{1}(X)=X$, $P_{2}(X)=8 X^{3}+4 X-1$ and define by induction $P_{k+1}(X)$ by $P_{k+1}(X)=8 P_{k}(X)^{3}+4 P_{k}(X)-3 P_{k-1}(X)-1$ for all integer $k \geq 2$. Clearly, $P_{k}(X)$ is a polynomial of odd degree for all $k \geq 1$.

Let $n \geq 2$ be an integer and $a$ a real zero of the polynomial $P_{n}(X)$. The real $a$ exists since the degree of $P_{n}(X)$ is odd.

Consider the finite sequence $x_{0}, x_{1}, \ldots, x_{n}$ of real numbers defined by $x_{0}=0$, $x_{1}=a$ and $x_{i+1}=8 x_{i}^{3}+4 x_{i}-3 x_{i-1}-1$, for all $1 \leq i \leq n-1$. Clearly, $x_{1}=P_{1}(a)$ and $x_{2}=P_{2}(a)$. Assume that for $2 \leq k \leq n-1$, $x_{k-1}=P_{k-1}(a)$ and $x_{k}=P_{k}(a)$. We have

$x_{k+1}=8 x_{k}^{3}+4 x_{k}-3 x_{k-1}-1=8 P_{k}(a)^{3}+4 P_{k}(a)-3 P_{k-1}(a)-1=P_{k+1}(a)$.

This proves that $x_{k}=P_{k}(a)$ for all $1 \leq k \leq n$ and in particular $x_{n}= P_{n}(a)=0$. This proves the existence of the sequence $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $x_{0}=x_{n}=0$ and $x_{i+1}-8 x_{i}^{3}-4 x_{i}+3 x_{i-1}+1=0$ for all $i=1,2, \ldots, n-1$.

Conversely, if such a sequence $x_{0}, x_{1}, \ldots, x_{n}$ exists then $x_{1}$ is a real zero of the polynomial $P_{n}(X)$. Therefore, proving the uniqueness of the sequence is equivalent to proving that $P_{n}(X)$ has a unique real zero.

Let $x, y$ be two real numbers. We have $\left|P_{2}(x)-P_{2}(y)\right|=4|x-y|\left|2\left(x^{2}+x y+y^{2}\right)+1\right| \geq |x-y|=\left|P_{1}(x)-P_{1}(y)\right|$, since $x^{2}+x y+y^{2} \geq 0$. Assume that $\left|P_{k}(x)-P_{k}(y)\right| \geq \left|P_{k-1}(x)-P_{k-1}(y)\right|$, for some integer $k \geq 2$. We have

$$\begin{gathered} \left|P_{k+1}(x)-P_{k+1}(y)\right| \geq \\ \geq 4\left|P_{k}(x)-P_{k}(y)\right| \cdot \left|2\left(P_{k}^{2}(x)+P_{k}(x) P_{k}(y)+P_{k}^{2}(y)\right)+1\right| \\ -3\left|P_{k-1}(x)-P_{k-1}(y)\right| \\ \geq 4\left|P_{k}(x)-P_{k}(y)\right|-3\left|P_{k-1}(x)-P_{k-1}(y)\right| \geq \left|P_{k}(x)-P_{k}(y)\right| \end{gathered}$$

Hence, the sequence $\left(\left|P_{k}(x)-P_{k}(y)\right|\right)_{k \geq 1}$ is non-decreasing and if $x \neq y$ then $P_{k}(x) \neq P_{k}(y)$ for all $k \geq 1$. We deduce that $P_{n}(X)$ has a unique real zero, and therefore the sequence $x_{0}, x_{1}, \ldots, x_{n}$ is unique.

Let $c=\max \left\{\left|x_{i}\right|: 1 \leq i \leq n-1\right\}$. There exists $k \in \{1,2, \ldots, n-1\}$ such that $c=\left|x_{k}\right|$. Because $x_{k}$ and $x_{k}^{3}$ have the same sign, it follows that

$$\begin{aligned} c+2 c^{3} & =\left|x_{k}\right|+2\left|x_{k}\right|^{3}=\left|x_{k}+2 x_{k}^{3}\right|=\frac{1}{4}\left|1+x_{k+1}+3 x_{k-1}\right| \\ & \leq \frac{1}{4}+\frac{1}{4}\left(\left|x_{k+1}\right|+3\left|x_{k-1}\right|\right) \leq \frac{1}{4}+c, \end{aligned}$$

which implies that $c \leq \frac{1}{2}$.