Every polynomial in RR[x] which takes only nonnegative values is a sum of two squares

Let p(x) \in \mathbb{R}[x] be a polynomial such that p(c) \geq 0 for all c. Prove that p(x) = a(x)^2 + b(x)^2 for some polynomials a,b \in \mathbb{R}[x].

Note that p(x) must have even degree. We proceed by induction on the degree of p(x). Note as a lemma that if h = a^2+b^2 and k = c^2 + d^2 are sums of squares, then so is hk since (evidently) hk = (ac-bd)^2 + (ad+bc)^2. We find this identity by rearranging and partially simplifying the factorization hk = (a+bi)(a-bi)(c+di)(c-di).

The base case p(x) = c is trivial; p(x) = 0^2 + 0^2 if c = 0 and p(x) = (\sqrt{c}/2)^2 + (\sqrt{c}/2)^2 if c \neq 0.

For the inductive step, suppose the result holds for all polynomials of degree n and let p(x) have degree n+2. Suppose p(x) has a real root c. Now p(x) is concave up on a sufficiently small neighborhood about c, so that p^\prime(c) = 0. In particular, c is a root of p of multiplicity at least 2. Say p(x) = q(x)(x-c)^2. Now q(x) is a sum of two squares, and (x-c)^2 = (x-c)^2 + 0^2 is as well, so that by the lemma p(x) is a sum of two squares. Suppose instead that p(x) has no real roots. Instead, we have a complex root z. Since conjugation is a ring homomorphism, \overline{z} is also a root of p. Letting z = a+bi, we see that x^2+2ax + a^2+b^2 = (x+a)^2 + b^2 is a factor of p(x); say p(x) = q(x)((x+a)^2 + b^2). Again, p(x) is a sum of two squares.

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