Prove that we can find a real polynomial p(y) such that p(x - 1/x) = xn - 1/xn (where n is a positive integer) iff n is odd.
Take n odd. For n = 1 the result is obvious. Now suppose the result is true for odd n ≤ 2m - 1. Expand (x - 1/x)2m+1 by the binomial theorem. Since (2m+1)C0 = (2m+1)C(2m+1), (2m+1)C1 = (2m+1)C2m, (2m+1)C2 = (2m+1)C(2m-1), ... , (2m+1)Cm = (2m+1)C(m+1), we may group these pairs of terms together to get: (x - 1/x)2m+1 = (x2m+1 - x-(2m+1)) - (2m+1)C1 (x2m-1 - x-(2m-1)) + ... + (-1)m (2m+1)Cm (x - 1/x). This gives the polynomial for n = 2m+1 in terms of the lower order polynomials.
Note that x - 1/x has the same value 3/2 for x = -1/2 and x = 2. But x2m - 1/x2m is negative for x = -1/2 and positive for x = 2, so we cannot express x2m - 1/x2m as a function of (x - 1/x), polynomial or otherwise.
20th Putnam 1959
© John Scholes
15 Feb 2002