Let α be a complex (2n + 1)th root of unity. Prove that there always exist polynomials p(x), q(x) with integer coefficients, such that p(α)2 + q(α)2 = -1.
Note that we have α + α2 + α3 + ... + αN = -1 where we use N = 2n (because of the shortcomings of HTML which does not allow iterated exponents). We give an inductive procedure for constructing pn(x) and qn(x) so that pn(x)2 + qn(x)2 = x2 + x4 + x6 + ... + x2N. For x = α this becomes α2 + α4 + ... + αN + α + α3 + ... + αN-1 = -1.
Evidently for n = 1, we may take p(x) = α and q(x) = α2. Given pn(x) and qn(x), take pn+1(x) = pn(x) + xNqn(x), qn+1(x) = qn(x) - xNpn(x). Then pn+1(x)2 + qn+1(x)2 = (pn(x)2 + qn(x)2)(1 + x2N) = x2 + x4 + ... + x4N as required.
44th Putnam 1983
© John Scholes
16 Jan 2001