21st VMO 1983 Problem A1 For which positive integers m, n with n > 1 does 2 n - 1 divides 2 m + 1?
(m,n) = (k,1) or (2k+1,2) Solution Suppose n > 2. The residues 1, 2, 4, 8, ... , 2 n-1 are obviously distinct mod 2 n -1. But 2 n = 1 mod 2 n -1, so all power of 2 must equal one of these residues mod 2 n -1. Hence for any m we have 2 m +1 = one of 2, 3, 5, 9, ... , 2 n-1 +1 mod 2 n -1. Since 2 n-1 +1 < 2 n -1, these are all non-zero and so there are no solutions. If n = 1, then obviously any value of m works. If n = 2, then we need 2 m +1 = (3-1) m +1 to be a multiple of 3 and hence m odd. Thanks to Suat Namli 21st VMO 1983 © John Scholes jscholes@kalva.demon.co.uk 12 February 2004 Last corrected/updated 12 Feb 04