# 1991 IMO Problems/Problem 6

## Problem

An infinite sequence $x_0, x_1, x_2,\ldots$ of real numbers is said to be bounded if there is a constant $C$ such that $|x_i| \leq C$ for every $i \geq 0$.

Given any real number $a > 1$, construct a bounded infinite sequence $x_0,x_1,x_2,\ldots$ such that $|x_i-x_j|\cdot |i-j|^a\geq 1$ for every pair of distinct nonnegative integers $i,j$.

## Solution 1

Since $a>1$, the series $\sum_{k=1}^\infty\frac{1}{k^a}$ is convergent; let $L$ be the sum of this convergent series. Let $I\subset \mathbb{R}$ be the interval $[-L,L]$ (or any bounded subset of measure $\geq 2L$).

Suppose that we have chosen points $x_0,x_1,\ldots,x_{m-1}$ satisfying

$\qquad(\ast)\quad |x_i-x_j|\cdot |i-j|^a\geq 1$

for all distinct $i,j. We show that we can choose $x_m\in I$ such that $(\ast)$ holds for all distinct $i,j\leq m$. The only new cases are when one number (WLOG $i$) is equal to $m$, so we must guarantee that $|x_m-x_j|\cdot |m-j|^a\geq 1$ for all $0\leq j.

Let $U_j$ be the interval $(x_j-\frac{1}{|m-j|^a},x_j+\frac{1}{|m-j|^a})$, of length $\frac{2}{|m-j|^a}$. The points that are valid choices for $x_m$ are precisely the points of $I\setminus(U_0\cup \cdots \cup U_{m-1})$, so we must show that this set is nonempty. The total length $\mu(U_0\cup \cdot\cup U_{m-1})$ is at most the sum of the lengths $\mu(U_0)+\cdots+\mu(U_{m-1})=\frac{2}{m^a}+\frac{2}{(m-1)^a}+\cdots+\frac{2}{2^a}+\frac{2}{1^a}$. This is $2\sum_{k=1}^m\frac{1}{k^a}<2\sum_{k=1}^\infty \frac{1}{k^a}=2L$.

Therefore the total measure of $U_0\cup \cdots \cup U_{m-1}$ is $<2L=\mu(I)$, so $I\setminus(U_0\cup \cdots \cup U_{m-1})$ has positive measure and thus is nonempty. Choosing any $x_m\in I\setminus(U_0\cup \cdots \cup U_{m-1})$ and continuing by induction constructs the desired sequence.

## Solution 2

The argument above would not work for $a=1$, since ${\textstyle\sum \frac{1}{k^a}}$ only converges for $a>1$. But Osmun Nal argues in this video that $x_k=k\sqrt{2}-\lfloor k\sqrt{2}\rfloor$ satisfies the stronger inequality $|x_i-x_j|\cdot |i-j|\geq 1$ for all distinct $i,j$; in other words, this sequence simultaneously solves the problem for all $a\geq 1$ simultaneously.