19th BMO 1983 Problem 6 Find all positive integer solutions to (n + 1) m = n! + 1.
(n,m) = (1,1), (2,1), (4,2) Solution If n = 1, then m must be 1, which gives a solution. So assume n > 1. Hence n! is even and n! + 1 is odd, so n must be even. If n = 2, then m must be 1 and we have a solution. If n = 4, then 5 m = 25, so m must be 2 which is a solution. So assume n > 4. Then 2, n/2 and n are separate terms of n!, so n 2 divides n! and hence also (n+1) m - 1. Expanding by the binomial theorem, we see that n must divide m. Hence, in particular m ≥ n. But now (n+1) m ≥ (n+1) n > n! + 1. So there are no solutions for n > 4. 19th BMO 1983 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04