IMO 2002 SL A5

Origin: IND | Category: Algebra

Problem

Let n be a positive integer that is not a perfect cube. Define real numbers a, b, c by a = 3\sqrtn, b = a −[a], c = b −[b], where [x] denotes the integer part of x. Prove that there are infinitely many such integers n with the property that there exist integers r, s, t, not all zero, such that ra + sb + tc = 0.

Solution

Write m = [ 3\sqrtn]. To simplify the calculation, we shall assume that [b] = 1. Then a = 3\sqrtn, b = 3\sqrtn−m = n−m3  m2 + m 3\sqrtn + 3\sqrt n2

, c = b−1 = u + v 3\sqrtn + w 3\sqrt n2 for certain rational numbers u, v, w. Obviously, integers r, s, t with ra + sb + tc = 0 exist if (and only if) u = m2w, i.e., if (b − 1)(m2w + v 3\sqrtn + w 3\sqrt n2) = 1 for some rational v, w. When the last equality is expanded and simplified, comparing the coeffi- cients at 1, 3\sqrtn, 3\sqrt n2 one obtains 1 : v+ ((m2 + m3 −n)m2 + m)w = n −m3, 3\sqrtn : (m2 + m3 −n)v+ (m3 + n)w = 0, 3\sqrt n2 : mv+ (2m2 + m3 −n)w = 0. (1) In order for the system (1) to have a solution v, w, we must have (2m2 + m3−n)(m2+m3−n) = m(m3+n). This quadratic equation has solutions n = m3 and n = m3 + 3m2 +m. The former is not possible, but the latter gives a −[a] > 1 2, so [b] = 1, and the system (1) in v, w is solvable. Hence every number n = m3 + 3m2 + m, m \inN, satisfies the condition of the problem.