IMO 1991 SL 28

Origin: NET

Problem

Given a real number a > 1, construct an infinite and bounded sequence x0, x1, x2, . . . such that for all natural numbers i and j, i ̸= j, the following inequality holds: |xi −xj||i −j|a \geq1.

Solution

Let xn = c(n \sqrt 2 −[n \sqrt 2]) for some constant c > 0. For i > j, putting p = [i \sqrt 2] −[j \sqrt 2], we have |xi−xj| = c|(i−j) \sqrt 2−p| = |2(i −j)2 −p2|c (i −j) \sqrt 2 + p \geq c (i −j) \sqrt 2 + p \geq c 4(i −j), because p < (i −j) \sqrt 2 + 1. Taking c = 4, we obtain that for any i > j, (i −j)|xi −xj| \geq1. Of course, this implies (i −j)a|xi −xj| \geq1 for any a > 1. Remark. The constant 4 can be replaced with 3/2 + \sqrt 2. Second solution. Another example of a sequence {xn} is constructed in the following way: x1 = 0, x2 = 1, x3 = 2 and x3ki+m = xm + i 3k for i = 1, 2 and 1 \leqm \leq3k. It is easily shown that |i −j| \cdot |xi −xj| \geq1/3 for any i ̸= j. Third solution. If n = b0+2b1+\cdot \cdot \cdot+2kbk, bi \in{0, 1}, then one can set xn to be = b0+2−ab1+\cdot \cdot \cdot+2−kabk. In this case it holds that |i−j|a|xi−xj| \geq 2a−2 2a−1.