36th Putnam 1975 Problem A4 m > 1 is odd. Let n = 2m and θ = e 2πi/n . Find a finite set of integers {a i } such that ∑ a i θ i = 1/(1 - θ).
We have θ m = -1. Since m is odd we have 0 = (θ m + 1) = (θ + 1)(θ m-1 - θ m-2 + ... - θ + 1). θ is not - 1, so q m-1 - θ m-2 + ... - θ + 1 = 0 (*). Since m is odd, we may write (*) as: 1 - θ(1 - θ)(1 + θ 2 + θ 4 + ... + θ m-3 = 0, or 1/(1 - θ) = θ + θ 3 + θ 5 + ... + θ m-2 . 36th Putnam 1975 © John Scholes jscholes@kalva.demon.co.uk 27 Jan 2001