Problem:
Let a=π/2008. Find the smallest positive integer n such that
2[cos(a)sin(a)+cos(4a)sin(2a)+cos(9a)sin(3a)+⋯+cos(n2a)sin(na)]
is an integer.
Solution:
The product-to-sum formula for 2cos(k2a)sin(ka) yields sin(k2a+ka)− sin(k2a−ka), which equals sin(k(k+1)a)−sin((k−1)ka). Thus the given sum becomes
sin(2⋅1a)−sin(0)+sin(3⋅2a)−sin(2⋅1a)+sin(4⋅3a)−sin(3⋅2a)+⋯+sin(n(n+1)a)−sin((n−1)na).
This is a telescoping sum, which simplifies to sin(n(n+1)a)−sin(0)= sin(n(n+1)a). But sin(x) is an integer only when x is an integer multiple of π/2, so n(n+1) must be an integer multiple of 1004=22⋅251. Thus either n or n+1 is a multiple of 251, because 251 is prime. The smallest such n is 250, but 250⋅251 is not a multiple of 1004. The next smallest such n is 251, and 251⋅252 is a multiple of 1004. Hence the smallest such n is 251.
The problems on this page are the property of the MAA's American Mathematics Competitions