15th Iberoamerican 2000 Problem A3 Find all solutions to (m + 1) a = m b + 1 in integers greater than 1.
(m, a, b) = (2, 2, 3). Solution Thanks to José Nelson Ramirez Taking equation mod m+1 we get (-1) b = -1, so b is odd. Hence we can divide the rhs by m+1 to get m b-1 - m b-2 + ... - m + 1. This has an odd number of terms. If m is odd, then each term is odd and so the total is odd, but (m+1) a-1 is even (note that a > 1). Contradicton, so m is even. We have m b = (m+1) a - 1. Expanding the rhs by the binomial theorem, and using b > 1, we see that m must divide a. So a is even also. Put a = 2A, m = 2M. We can factorise (m+1) a - 1 as ( (m+1) A + 1) ( (m+1) A - 1). The two factors have difference 2, so their gcd divides 2, but both factors are even, so their gcd is exactly 2. If M = 1 or a power of 2, then the smaller factor 3 A - 1 must be 2, so A = 1 and we have 3 A + 1 = 4, so (2M) b = 8. Hence M = 1 and b = 3 and we have the solution (m, a, b) = (2, 2, 3). If M is not a power of 2, then M b > 2 b , so we must have the larger factor 2·M b and the smaller factor 2 b- 1 . But the larger factor is now > 2 b+1 , so the difference between the factors is at least 3·2 b-1 > 2. Contradiction. 15th Ibero 2000 © John Scholes jscholes@kalva.demon.co.uk 9 Sep 2002 Last corrected/updated 6 July 03