21st BMO 1985 Problem 6 Find all non-negative integer solutions to 5 a 7 b + 4 = 3 c .
5 1 7 0 + 4 = 3 2 Solution Note that lhs > 4, so c ≥ 2. If a = 0, then 7 b = 4 = 1 mod 3, so lhs = 2 mod 3. Contradiction. Hence a ≥ 1. So 3 c = 4 mod 5, so c = 4k+2. In particular c must be even , so 5 a 7 b = (3 2k+1 +2)(3 2k+1 -2). The gcd of (3 2k+1 +2) and (3 2k+1 -2) must divide their sum 4, but they are odd, so they are coprime. Hence one is 5 a and the other is 7 b . If k = 0, then we have the solution shown above. So assume k > 0. Hence (3 2k+1 +2) and (3 2k+1 -2) are both ≥ 25, so a ≥ 2 and b ≥ 1, and we have 5 a - 7 b = ± 4. So 7 b = ± 4 mod 25. Contradiction, because powers of 7 must be ±1 or ±7 mod 25. So there are no solutions with k > 0. 21st BMO 1985 © John Scholes jscholes@kalva.demon.co.uk 11 February 2004 Last corrected/updated 11 Feb 04