MHB Can a nonnegative polynomial be expressed as a sum of squares of polynomials?

[Ackbach]

Here is this week's POTW:

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Let $p(x)$ be a polynomial that is nonnegative for all real $x$. Prove that for some $k$, there are polynomials $f_1(x),\dots,f_k(x$) such that
\[p(x) = \sum_{j=1}^k (f_j(x))^2.\]

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[Ackbach]

Re: Problem Of The Week # 249 - Jan 16, 2017

This was Problem A-2 in the 1999 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct solution, which follows. Honorable mention to Kiwi for a mostly correct solution with only minor holes.

Spoiler

This can be done with $k=2$.

To start with, if $p(x)$ is a real polynomial that is always nonnegative then it must have even degree, and therefore an even number of (real or complex) roots. Also, the coefficient of the highest power of $x$ must be positive and can therefore be written as a square, say $d^2$.

Any real root of $p(x)$ must have even multiplicity (otherwise the polynomial would go negative on one side of the root). So a real root $x=c$ corresponds to a factor of the form $(x-c)^{2m}.$

Complex roots must occur in conjugate pairs of the form $x = a+ib$ and $x=a-ib.$ The corresponding factors of $p(x)$ are $(x-a-ib)(x-a+ib) = (x-a)^2 + b^2.$

Thus $p(x)$ is the product of factors of the form $d^2$, $(x-c)^2$ and $(x-a)^2 + b^2$. But a square times a sum of two squares is again a sum of two squares, and the product of two sums of two squares is also a sum of two squares (because of the identity $(A^2+B^2)(C^2+D^2) = (AC+BD)^2 + (AD-BC)^2$). By repeatedly applying these facts, you see that $p(x)$ must be the sum of two squares (of polynomials).