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Given a nonnegative polynomial $p(x)$ on $\mathbb{R}$, does there exist some $k$ such that for some polynomials $f_1,\ldots ,f_k$ we have $p(x)=\sum_{i=1}^k(f_i)^2$?

I think yes, because of the nonnegative nature of $p(x)$, but am unable to prove. Any hints. Thanks beforehand.

Rodrigo de Azevedo
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vidyarthi
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  • Every non negative polynomial with real coefficients is the sum of two squares of polynomials with real coefficients. Is that what you're looking for, or $k$ is not necessarily $k=2$? – Watson Jan 17 '17 at 13:00
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    Is your question this: Every nonnegative polynomial is a sum of squares? – lhf Jan 17 '17 at 13:01
  • (For sum of three cubes, see here). – Watson Jan 17 '17 at 13:02
  • Your probably swapped the quantifiers. Is your question this: for every nonnegative polynomial $p(x)$ (on $\Bbb R$), does there exist some $k$ and polynomials $f_1,\ldots ,f_k \in \Bbb R[X]$ such that we have $p(x)=\sum_{i=1}^k(f_i)^2$? – Watson Jan 17 '17 at 13:05
  • My comment is : you first begin with $p$, and then have to find $k$ and $f_1,\dots, f_k$. Your question is: you first begin with $k,f_1,\dots,f_k$, and then have are given $p$. – Watson Jan 17 '17 at 13:07
  • Hint: 1) Show that a non-negative quadratic is either a square or the sum of two squares. 2) Show that a product of two polynomials, both sums of squares, is again a sum of squares. 3) Apply FTA. – Jyrki Lahtonen Jan 17 '17 at 13:08
  • @Watson yes, first begin with $p(x)$ – vidyarthi Jan 17 '17 at 13:08
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    When you say: "... such that for polynomials $f_1, f_2, \ldots$" you assume that you know $f_1,f_2,\ldots$. – Alex Silva Jan 17 '17 at 13:08
  • Closely related: https://math.stackexchange.com/questions/1754076 – Watson Jan 17 '17 at 13:14

1 Answers1

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Every nonnegative real polynomial is a sum of squares.

Let $p$ be a nonnegative real polynomial.

By the Fundamental Theorem of Algebra, $p=af_1^2 \cdots f_n^2 g_1 \cdots g_m$, where $a\in \mathbb R$, $f_i$ has degree $1$ and $g_j$ has degree $2$ and no real roots.

Now if $g$ is a monic real quadratic polynomial with no real roots, then $g(x)=(x+u)^2+v^2$, by completing the square.

An expression of $p$ as a sum of squares now follows from the Brahmagupta–Fibonacci identity.

lhf
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