32nd BMO 1996 Problem 1 Find all non-negative integer solutions to 2 m + 3 n = k 2 . Solution
: 2 0 + 3 1 = 2 2 , 2 3 + 3 0 = 3 2 , 2 4 + 3 2 = 5 2 . If m = 0, then 3 n = k 2 - 1 = (k + 1)(k - 1). But if 3 divides both k + 1 and k - 1, then it divides their difference 2, which is impossible, so k - 1 = 1 and we have the solution m = 0, n = 1, k = 2. If m = 1, then k 2 - 2 = 3 n . But squares are 0 or 1 mod 3, so k 2 - 2 is 1 or 2 mod 3, so n must be 0. But that does not give a solution. So we may assume m > 1. Hence k 2 = 3 n = (4 - 1) n = (-1) n mod 4. But squares are 0 or 1 mod 4, so n must be even. Put n = 2s. Then 2 m = k 2 - 3 2s = (k + 3 s )(k - 3 s ). So both k + 3 s and k - 3 s are powers of 2. Suppose k - 3 s = 2 a and k + 3 s = 2 b , so that m = a + b. Then 2.3 s = 2 b - 2 a . So a = 1 and 3 s = 2 b-1 - 1. But 2 b-1 - 1 = (-1) b-1 - 1 mod 3, so b - 1 must be even or s = 0. If s = 0, then b - 1 = 1, so m = 3 and we have the solution 2 3 + 3 0 = 3 2 . So suppose b -1 is even. Put b - 1 = 2r, then 3 s = (2 r - 1)(2 r + 1). But 3 cannot divide both 2 r - 1 and 2 r + 1 or it divides their difference, which is 2. So we must have 2 r - 1 = 1 and hence r = s = 1, b = 2r+1 = 3, n = 2s = 2, m = a + b = 4 and we have the solution 2 4 + 3 2 = 5 2 . 32nd BMO 1996 © John Scholes jscholes@kalva.demon.co.uk 14 Aug 2002 Last corrected/updated 22 Aug 03