40th Putnam 1979 Problem A3 a n are defined by a 1 = α, a 2 = β, a n+2 = a n a n+1 /(2a n - a n+1 ). α, β are chosen so that a n+1 ≠ 2a n . For what α, β are infinitely many a n integral?
A trivial induction shows that a n+2 = αb/((n + 1)α - nβ) = αβ/(n(α - β) + α). So we must have α = β, otherwise the denominator grows without limit. But if α = β, then all a n = α. So infinitely many a n are integral iff α = β = integer. 40th Putnam 1979 © John Scholes jscholes@kalva.demon.co.uk 4 Dec 1999