Given any real number a > 1 construct a bounded infinite sequence x_{0}, x_{1}, x_{2}, ... such that |x_{n} - x_{m}| |n - m|^{a} ≥ 1 for every pair of distinct n, m.

[An infinite sequence x_{0}, x_{1}, x_{2}, ... of real numbers is bounded if there is a constant C such that |x_{n}| < C for all n.]

**Solution**

*By Marcin Mazur, University of Illinois at Urbana-Champaign*

Let t = 1/2^{a}. Define c = 1 - t/(1 - t). Since a > 1, c > 0. Now given any integer n > 0, take the binary expansion n = ∑_{i} b_{i} 2^{i}, and define x_{n} = 1/c ∑_{bi>0} t^{i}. For example, taking n = 21 = 2^{4} + 2^{2} + 2^{0}, we have x_{21} = (t^{4} + t^{2} + t^{0})/c. We show that for any unequal n, m, |x_{n} - x_{m}| |n - m|^{a} ≥ 1. This solves the problem, since the x_{n} are all positive and bounded by (∑ t^{n} )/c = 1/(1 - 2t).

Take k to be the highest power of 2 dividing both n and m. Then |n - m| ≥ 2^{k}. Also, in the binary expansions for n and m, the coefficients of 2^{0}, 2^{1}, ... , 2^{k-1} agree, but the coefficients for 2^{k} are different. Hence c |x_{n} - x_{m}| = t^{k} + ∑_{i>k} y_{i}, where y_{i} = 0, t^{i} or - t^{i}. Certainly ∑_{i>k} y_{i} > - ∑_{i>k} t^{i} = t^{k+1}/(1 - t), so c |x_{n} - x_{m}| > t^{k}(1 - t/(1 - t)) = c t^{k}. Hence |x_{n} - x_{m}| |n - m|^{a} > t^{k} 2^{ak} = 1.

Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

14 Sep 1999