Problem 983

Let x = cos -1 (1/3). If x = m/n π for some integers m, n, then cos nx = cos mπ = ±1. But we show that cos nx cannot be ±1. It follows that x/π must be irrational as required. As usual, we have cos nx = nC0 c n - nC2 c n-2 s 2 + nC4 c n-4 s 4 - ... , where c = cos x, s = sin x. We may put s 2 = 1 - c 2 to get cos nx = a polynomial of degree n in c with integer coefficients. The coefficient of c n = nC0 + nC2 + nC4 + ... = 2 n-1 . But c = 1/3, so cos nx = 2 n-1 /3 n + k/3 n-1 = (2 n-1 + 3k)/3 n for some integer k. This must be in its lowest terms since 2 n-1 is not divisible by 3. In particular, it cannot be ±1. [A variant on this is to consider cos(2 n x). By a simple induction using cos 2y = 2 cos 2 y - 1, we show that cos(2 n x) = a n /b n , where a n is not a multiple of 3 and b n is 3 to the power of 2 n . It follows that as n runs through the natural numbers, all the values cos(2 n x) are distinct. But if x/π was rational, there would only be finitely many distinct values.] 35th Putnam 1974 © John Scholes [email protected] 18 Aug 2001