Find all solutions to pn = qm ± 1, where p and q are primes and m, n ≥ 2.
Answer: 23 = 32 - 1.
pn and qm have opposite parity, so one must be even. So without loss of generality we may take p = 2. Now qm ± 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 2n has no odd factors. So m is even. Take it to be 2M.
Consider first the case 2n = q2M - 1 = (qM + 1)(qM - 1). But qM + 1 and qM - 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 23 = 32 - 1.
Finally consider 2n = q2M + 1. qM is odd, so put qM = 2N + 1, then q2M + 1 = 4N2 + 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
23 Jan 2001