ThEdArKlOrD

Famous example of Lagrange Interpolation Polynomials

See here https://www.math.ust...ibur/v15_n2.pdf (Example $6$)

Suppose to the contradiction that $x_{i+1}-x_i-1>3\sqrt[3]{x_ix_{i+1}}$ for all $i\in \mathbb{1,2,...,10}$

This give us $(\sqrt[3]{x_{i+1}}-\sqrt[3]{x_i}-1)(\sqrt[3]{x_{i+1}}^2+\sqrt[3]{x_i}\sqrt[3]{x_{i+1}}+\sqrt[3]{x_i}^2-\sqrt[3]{x_{i}}+\sqrt[3]{x_{i+1}}+1)>0$, so $\sqrt[3]{x_{i+1}} >\sqrt[3]{x_i}+1$, for all $i\in \mathbb{1,2,...,10}$

So $\sqrt[3]{x_{11}} >\sqrt[3]{x_1} +10\geq 11$, contradict with $x_{11}\leq 1000$

This is equivalent to APMO $1989$

See here https://www.artofpro...c6h78608p450334

If you mean $f^2(x)=f(x)^2$

Let $P(x,y)$ denote $f(x)^2=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(x)^2=2f(x)f(-x)$,so $f(-x)^2=2f(x)f(-x)$ give us $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$,

From $f(x)^2=2f(x)f(-x)$, suppose there exist $t$ such that $f(t)\neq 0$, we get $f(t)=2f(-t)$, but since $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$, give $f(t)=f(-t)=0$, contradiction, so $f(x)=0$ for all $x\in \mathbb{R}$

If you mean $f^2(x)=f(f(x))$

Let $P(x,y)$ denote $f(f(x))=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(f(x))=2f(x)f(-x)=f(f(-x))$ for all $x\in \mathbb{R}$ and $P(x,x)$ give us $f(f(x))=f(0)^2+f(4x)f(2x)$ for all $x\in \mathbb{R}$

Then $P(-y,y)$ give $f(f(-y))=f(-2y)f(2y)+f(2y)f(4y)$ for all $y\in \mathbb{R}$, so $f(0)^2+f(4x)f(2x)=f(f(x))=f(f(-x))=f(-2x)f(2x)+f(2x)f(4x)$ for all $x\in \mathbb{R}$

So $f(0)^2=f(-2x)f(2x)$ for all $x\in \mathbb{R}$, this give us $f(f(x))=f(f(-x))=2f(0)^2$ for all $x\in \mathbb{R}$

Then, $P(x,y)$ change to $2f(0)^2=f(0)^2+f(3y+x)f(3y-x)$ for all $x,y\in \mathbb{R}$, so $f(0)^2=f(3y)^2=f(0)f(6y)$ for all $y\in \mathbb{R}$

So $f(0)=0$ give $f(y)=0$ for all $y\in \mathbb{R}$ or $f(0)\neq 0$ give us $f(6y)=f(0)$ for all $y\in \mathbb{R}$

So $f$ is constant function and the rest is easy

Suppose there are $t$ elements in $S$

For each element $p\in S$, we write down $p$ in a line, under word $p$, we write down all word different from $p$ by exactly $1$ letter, note that there are total $15+1$ words from each $p$ (including $p$)

Note that among any two words differ at least $1$ letter

So $(15+1)\times t \leq 2^{15}$ give us $t\leq 2^{11}$