37th Putnam 1976 Problem A3 Find all solutions to p n = q m ± 1, where p and q are primes and m, n ≥ 2. Solution
: 2 3 = 3 2 - 1. p n and q m have opposite parity, so one must be even. So without loss of generality we may take p = 2. Now q m ± 1 factors as (q ± 1) and a sum of m powers of q, each of which is odd. If m is odd, then the sum is odd, but that is impossible, since 2 n has no odd factors. So m is even. Take it to be 2M. Consider first the case 2 n = q 2M - 1 = (q M + 1)(q M - 1). But q M + 1 and q M - 1 are successive even numbers, so one of them has an odd factor (which is impossible) unless then are 4 and 2. Thus the only solution of this type is 2 3 = 3 2 - 1. Finally consider 2 n = q 2M + 1. q M is odd, so put q M = 2N + 1, then q 2M + 1 = 4N 2 + 4N + 2, which is not divisible by 4. So n = 1 and q = 1, which is not a solution, since 1 is not prime. 37th Putnam 1976 © John Scholes jscholes@kalva.demon.co.uk 23 Jan 2001