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Let $P\in\mathbb{R}[X]$ be a positive polynomial.

I want to show that there exists $A,B\in\mathbb{R}[X]$ so that $P=A^2+B^2$

$\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)$ where $a$ is the leading coefficient, $a_k$ are the real rooks, $b_k$ are the complex roots.

What can i do now ? I know that the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ should come in handy, but I can't figure out :

1) How to show that the product real roots is a square

2) How to use the above identity for the complex roots.

If I can solve those two points, then I can solve the problem.

Hippalectryon
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1 Answers1

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Your identity shows that the product of terms, each of which is a sum of two squares, is a sum of two squares. So we just have to write your polynomial as a product of terms, each of which is a sum of two squares of real polynomials.

  1. If $P(x) \ge 0$ for all real $x$, and $\alpha$ is a real root of $P$, then it must have even multiplicity (otherwise $P(x)$ would change sign as you move from $< \alpha$ to $> \alpha$). $(x-\alpha)^2 = (x-\alpha)^2 + 0^2$.
  2. If $\beta = s + i t$ and $\overline{\beta} = s - i t$ are a complex-conjugate pair, $(x - \beta)(x-\overline{\beta}) = (x - s - it)(x - s + it) = (x-s)^2 + t^2$ is a sum of two squares.
Robert Israel
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