25th CanMO 1993 Problem 2 Show that the real number k is rational iff the sequence k, k + 1, k + 2, k + 3, ... contains three (distinct) terms which form a geometric progression.
Problem 947
Official solution
Suppose there are three such terms k + a, k + b, k + c. Then (k + b) 2 = (k + a)(k + c), so k(2b - a - c) = ac - b 2 . If 2b - a - c = 0, then also ac - b 2 = 0, so b is both the AM and the GM of a and c. Hence a = c. But a, b, c are assumed to be unequal, so 2b - a - c is non-zero, hence k = (ac - b 2 )/(2b - a - c), which shows that k is rational. Conversely, if k = m/n with m and n non-zero, then kmn = m 2 , so (k + m) 2 = k(k + mn + 2m), which shows that the three terms k, k + m and k + mn + 2m are a GM of distinct terms. Finally if k = 0, the terms 1, 2 and 4 are a GM. 25th CanMO 1993 © John Scholes [email protected] 18 Aug 2002