Prove that we can find a real polynomial p(y) such that p(x - 1/x) = x^{n} - 1/x^{n} (where n is a positive integer) iff n is odd.

**Solution**

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} = (x^{2m+1} - x^{-(2m+1)}) - (2m+1)C1 (x^{2m-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 x^{2m} - 1/x^{2m} is negative for x = -1/2 and positive for x = 2, so we cannot express x^{2m} - 1/x^{2m} as a function of (x - 1/x), polynomial or otherwise.

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002