IMO 1984 Problem A2 Find one pair of positive integers a, b such that ab(a+b) is not divisible by 7, but (a+b) 7 - a 7 - b 7 is divisible by 7 7 .
We find that (a + b) 7 - a 7 - b 7 = 7ab(a + b)(a 2 + ab + b 2 ) 2 . So we must find a, b such that a 2 + ab + b 2 is divisible by 7 3 . At this point I found a = 18, b = 1 by trial and error. A more systematic argument turns on noticing that a 2 + ab + b 2 = (a 3 - b 3 )/(a - b), so we are looking for a, b with a 3 = b 3 (mod 7 3 ). We now remember that a φ(m) = 1 (mod m). But φ(7 3 ) = 2·3·49, so a 3 = 1 (mod 343) if a = n 98 . We find 2 98 = 18 (343), which gives the solution 18, 1. This approach does not give a flood of solutions. n 98 = 0, 1, 18, or 324. So the only solutions we get are 1, 18; 18, 324; 1, 324. s are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48- X. 25th IMO 1984 © John Scholes jscholes@kalva.demon.co.uk 19 Oct 1998