60th Putnam 1999

Problem A2

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.



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 α1 ± i β1, α2 ± i β2, α3 ± i β3, ... , αm ± i βm, then let a(x) = ∏ (x - αi - i βi) and b(x) = ∏ (x - αi + i βi), so that p(x) = q(x)2a(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.



60th Putnam 1999

© John Scholes
14 Dec 1999