36th Putnam 1975

m > 1 is odd. Let n = 2m and θ = e^2πi/n. Find a finite set of integers {a_i} such that ∑ a_i θⁱ = 1/(1 - θ).

Solution

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 + θ² + θ⁴ + ... + θ^m-3 = 0, or 1/(1 - θ) = θ + θ³ + θ⁵ + ... + θ^m-2.

36th Putnam 1975

© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001