Get ready Part 30

Since $p(x) \ge 0$ for all real $x$ , $p(x) \in \mathbb{R}[x]$ . Thus the complex roots of $p(x)$ come in complex conjugate pairs. Moreover, any real root of $p(x)$ must have even multiplicity. Hence we can find complex numbers $z_1,z_2,...,z_n$ such that $p(x) \; =\; A\prod_{j=1}^n(x-z_j)(x-\overline{z}_j)$ where $A > 0$ . Thus $p(x) = |q(x)|^2$ , where $q(x) \; = \; \sqrt{A}\prod_{j=1}^n(x-z_j)$ Thus, if we define $f_1 = \mathfrak{Re}\,q$ and $f_2 = \mathfrak{Im}\,q$ , then $f_1(x),f_2(x) \in \mathbb{R}[x]$ and $p(x) \equiv f_1(x)^2 + f_2(x)^2$ .

We prove this by induction. Note that this polynomial will have to be of even degree. We will show that if it is true for a polynomial of degree $2n$ , it will be true for a polynomial of degree $2n+2$ . Base case $n = 0$ is true. If $k$ is a non negative constant. Then $k = c^2$ , for some c.

Let $p(x)$ be a polynomial of degree $2n+2$ . If $p(x)$ is non negative for all real $x$ , then $p(x) = f(x) + c^2$ , where $f(x)$ has root $r$ of multiplicity at least 2. Then $p(x) = (x - r)^2h(x) + c^2$ . Since $(x - r)^2 \geq\ 0$ for all real $x$ , and $f(x) \geq\ 0$ for all real $x$ , $h(x) \geq\ 0$ for all real $x$ . By the induction hypothesis, $h(x) = f_1\left(x\right)^2+f_2\left(x\right)^2+..+f_m\left(x\right)^2$ ( $h(x)$ is of degree $2n$ ). Then $p(x) = \left(\left(x-r\right)f_1\left(x\right)\right)^2+\left(\left(x-r\right)f_2\left(x\right)\right)^2+..+\left(\left(x-r\right)f_m\left(x\right)\right)^2+c^2$ . Thus $p(x)$ has been expressed as a sum of the squares of a finite number of polynomials. Hence the conjecture is true.