Show that for some fixed positive n we can always express a polynomial with real coefficients which is nowhere negative as a sum of the squares of n polynomials.

**Solution**

We consider what polynomials are nowhere negative. If p(x) has a root of odd order, then it changes sign at the root. So any real roots of the polynomial must be of even order. If x = α + i β is a complex root, then so is x = α - i β, since the coefficients of the polynomial are real.

Collecting all the real roots together gives us a square factor q(x)^{2}. If the complex roots are α_{1} ± i β_{1}, α_{2} ± i β_{2}, α_{3} ± i β_{3}, ... , α_{m} ± i β_{m}, then let a(x) = ∏ (x - α_{i} - i β_{i}) and b(x) = ∏ (x - α_{i} + i β_{i}), so that p(x) = q(x)^{2}a(x)b(x). Then we can find real polynomials, r(x) and s(x), such that a(x) = r(x) + i s(x). Substituting -i for i, we must have b(x) = r(x) - i s(x) and hence a(x) b(x) = r(x)^{2} + s(x)^{2}. So the original polynomial p(x) = q(x)^{2}(r(x)^{2} + s(x)^{2}), which is the sum of two squares.

© John Scholes

jscholes@kalva.demon.co.uk

14 Dec 1999