6th Brazil 1984 Problem 1 Find all solutions in positive integers to (n+1) k - 1 = n! .
n=1,k=1; n=2,k=1; n=4,k=2 Solution It is easy to check that the solutions above are the only solutions for n ≤ 4. So assume n > 4. So n! + 1 > n + 1, so k > 1. If n is odd, then n+1 is even, but n! + 1 is odd, so there are no solutions. So n is even. Hence n is composite. So n divides (n-1)! . But using the binomial theorem we have (n+1) k - 1 = n k + kC1 n k-1 + ... + kCk-2 n 2 + kn. Hence (n-1)! = n(n k-2 + n k-3 + ... + kCk-2) + k. Hence n divides k. But that means k ≥ n, and (n+1) n > n! + 1. So there are no other solutions. 6th Brazil 1984 © John Scholes jscholes@kalva.demon.co.uk 24 Oct 2003 Last corrected/updated 24 Oct 03