a, b, c, ... , k are positive integers such that a divides 2b - 1, b divides 2c - 1, ... , k divides 2a - 1. Show that a = b = c = ... = k = 1.
Solution
Suppose a prime p divides 2b - 1. Then 2b = 1 mod p and (by Fermat's theorem) 2p-1 = 1 mod p. Hence if d is the greatest common divisor of b and p-1, then 2d = 1 mod p. Evidently d > 1 (since 2 is not 1 mod p). Since d divides p-1, we have d < p. But d divides b, so there is a prime q smaller than p which divides b.
But if b > 1, then we may take p to be the smallest prime which divides 2b - 1. Then we find q < p which divides b and hence 2c - 1. Continuing in this way we find a prime strictly smaller than p which divides 2b - 1. Contradiction. So b must be 1. Similarly for the other integers.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 2002