Let α be a complex (2^{n} + 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.

**Solution**

Note that we have α + α^{2} + α^{3} + ... + α^{N} = -1 where we use N = 2^{n} (because of the shortcomings of HTML which does not allow iterated exponents). We give an inductive procedure for constructing p_{n}(x) and q_{n}(x) so that p_{n}(x)^{2} + q_{n}(x)^{2} = x^{2} + x^{4} + x^{6} + ... + x^{2N}. For x = α this becomes α^{2} + α^{4} + ... + α^{N} + α + α^{3} + ... + α^{N-1} = -1.

Evidently for n = 1, we may take p(x) = α and q(x) = α^{2}. Given p_{n}(x) and q_{n}(x), take p_{n+1}(x) = p_{n}(x) + x^{N}q_{n}(x), q_{n+1}(x) = q_{n}(x) - x^{N}p_{n}(x). Then p_{n+1}(x)^{2} + q_{n+1}(x)^{2} = (p_{n}(x)^{2} + q_{n}(x)^{2})(1 + x^{2N}) = x^{2} + x^{4} + ... + x^{4N} as required.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001