6th Australian 1985 Problem B2 Find all possible positive integers which have at least seven positive divisors and equal one less than the sum of the squares of their sixth and seventh divisors, when the divisors are listed in order of increasing size.
144, 1984 Solution Let the number be n and its divisors be 1 = d 1  a 2 > d 2 > cd = d 7 ), hence ac  1. Hence n = d 6 (k 2 d 6 + d 6 - 2k), so d 7 divides k 2 d 6 + d 6 - 2k and hence also d 6 - k. If d 6 - k > 0, then d 6 - k > kd 6 - 1, so (k - 1)(d 6 + 1)  |d 6 - k|, so k > 2d 6 . But d 6 is a prime or a prime square, so some d i  3. d+1 is even, so we certainly have the factors 1, 2, 4, p, 2p, 4p less than p 2 . Contradiction. The final case is d+1 a prime squared, so put d+1 = p 2 . If p = 3, then d = 8 and n = 144. Then d 1 = 1, d 2 = 2, d 3 = 3, d 4 = 4, d 5 = 6, d 6 = 8, d 7 = 9 and 144 = 82 + 92 - 1, which is a solution. If p > 3, then d is even, so we have the factors 1, 2, 4, p, 2p, 4p less than p 2 = d+1, so 4p = d 6 = p 2 - 1. Contradiction. Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered by email. However, its solution to this problem is incomplete - they do not prove that d 7 = d 6 + 1. 6th Aus 1985 © John Scholes jscholes@kalva.demon.co.uk 5 Jan 2004 Last corrected/updated 5 Jan 2004