32nd IMO 1991 shortlist Problem 17 Find all positive integer solutions to 3 m + 4 n = 5 k .
3 2 + 4 2 = 5 2 Solution 5 k = (-1) k mod 3, and 4 n = 1 mod 3, so k must be even. Put k = 2K. Similarly 3 m = (-1) m mod 4, and 5 k = 1 mod 4, so m must be even. Put m = 2M. Now 2 2n = 4 n = 5 2K - 3 2M = (5 k + 3 M )(5 K - 3 M ). Hence for some non-negative integers a, b with (a + b) + b = 2n, we have 5 K + 3 M = 2 a+b , 5 K - 3 M = 2 b . Subtracting, we get 2·3 M = 2 b (2 a - 1), so b must be 1 and 3 M = 2 a - 1. But 3 M = 1 or 3 mod 8, so a = 1 or 2. But a + 2b is even, so a = 2. Hence M = 1 and n = 2. So we have 3 2 + 4 2 = 5 k . Hence k = 2. 32nd IMO shortlist 1991 © John Scholes jscholes@kalva.demon.co.uk 1 Jan 2003 Last corrected/updated 1 Jan 03