39th BMO 2003 Problem 1 Find all integers 0 < a < b < c such that b - a = c - b and none of a, b, c have a prime factor greater than 3.
(a, b, c) = (k, 2k, 3k), (2k, 3k, 4k) or (2k, 9k, 16k), where k = 2 m 3 n . Solution It is sufficient to find solutions (A, B, C) without any common factor k which divides A, B and C, for then the complete set of solutions is (kA, kB, kC) with k = 2 m 3 n . Put D = B - A, then C = A + 2D, which has the same parity as A, so A and B must have opposite parity. Suppose A is odd. Then C is odd and greater than 1, hence a power of 3. If A > 1, then A is also a power of 3, and hence B = (A + C)/2 is also. Contradiction. So A must be 1. Suppose C = 3 n . Then B = (A + C)/2 = (1 + (4 - 1) n )/2 = (1 + (-1) n + (-1) n-1 4n + terms in 4 2 and higher)/2. Hence n is odd and B = 2 mod 4. But B is a power of 2, so B = 2 and C = 3. Suppose A is even, then B is odd and hence a power of 3. If A is divisible by 3, then so is C. Contradiction. So A must be a power of 2. Similarly, C must be a power of 2. If A is divisible by 4, then since C is a larger power of 2, it must also be divisible by 4. Hence 2D = C - A is divisible by 4, so B is divisible by 2. Contradiction. So A must be 2. It is easy to check that (A, B, C) = (2, 3, 4) and (2, 9, 16) are solutions. It remains to show that there are no solutions with B divisible by 27. Thanks to John Pye for the next part Suppose there is such a solution, then we have A = 2, C a power of 2, B = 3 n = 2 m - 1 ( (A+C)/2). B = 27 does not work, so B ≥ 81 and hence m ≥ 4. We have 3 4 = 1 mod 16, so it is easy to check that 3 4k - 1 = 0 mod 16, 3 4k+1 - 1 = 2 mod 16, 3 4k+2 - 1 = 8 mod 16, 3 4k+3 - 1 = 10 mod 16. So n must be a multiple of 4. But since 3 4 = 1 mod 5, that means 3 n - 1 = 0 mod 5, so B cannot be a power of 2. Contradiction. 39th BMO 2003 © John Scholes jscholes@kalva.demon.co.uk 2 Mar 2003 Last corrected/updated 31 Dec 03