For which positive integers m, n with n > 1 does 2n - 1 divides 2m + 1?
Answer
(m,n) = (k,1) or (2k+1,2)
Solution
Suppose n > 2. The residues 1, 2, 4, 8, ... , 2n-1 are obviously distinct mod 2n-1. But 2n = 1 mod 2n-1, so all power of 2 must equal one of these residues mod 2n-1. Hence for any m we have 2m+1 = one of 2, 3, 5, 9, ... , 2n-1+1 mod 2n-1. Since 2n-1+1 < 2n-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 2m+1 = (3-1)m+1 to be a multiple of 3 and hence m odd.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
12 February 2004
Last corrected/updated 12 Feb 04