Prove that a positive polynomial function can be written as the squares of two polynomial functions

Solution 1:

Consider roots of $f(x)$, as $f(x)\geq0,\forall x\in\mathbb{R}$, so $f(x)$ can be rewritten as following: $$f(x)=a^2(x-a_1)^2\cdots(x-a_k)^2[(x-\alpha_1)(x-\bar{\alpha_1})]\cdots[(x-\alpha_l)(x-\bar{\alpha_l})]$$ Where $a,a_1,\cdots,a_k\in\mathbb{R},\alpha_1,\cdots,\alpha_l\in\mathbb{C}$.

Denote $g(x)=a(x-a_1)\cdots(x-a_k),h(x)=(x-\alpha_1)\cdots(x-\alpha_l)=h_1(x)+ih_2(x)$, then \begin{align*} f(x)&=g^2(x) \, h(x) \, \bar{h}(x)\\ &=g^2(x) \, [h_1(x)+ih_2(x)] \, [h_1(x)-ih_2(x)]\\ &=(g(x)h_1(x))^2+(g(x)h_2(x))^2 \end{align*}

Solution 2:

Survey article by Bruce Reznick called Some Concrete Aspects of Hilbert's 17th Problem, includes your case in the paragraph on Before 1900:

Solution 3:

Assume that the leading coefficient of $f$ is $1$. If $f(x)=x^k$ for some $k$, then $k$ is even and the result follows. By completing the square we see that the result holds if the degree of $f$ is less than or equal to $2$. If $f$ has a real root, we may assume this root is at 0 by replacing $x$ with $x-a$. We can see that this must be a multiple root by the fact that $f(x)\geq 0$ for all $x$, so by factoring out $x^2$ we reduce the problem to a polynomial of lower degree. We may thus assume that $\deg(f)>2$, the result holds for polynomials smaller degree, and that $f$ has a nonzero complex root. The nonzero roots of $f$ are $z_1,z_1',z_2,z_2',\ldots,z_n,z_n'$ where $z_i$ is the complex conjugate of $z_i'$ for all $i$. Thus there is a quadratic polynomial $x^2+bx+c=(x-z_1)(x-z_1')$ and a polynomial $p(x)$ of degree two less than the degree of $f$ such that $$f(x)=(x^2+bx+c)p(x)$$

By induction, $p(x)=A(x)^2+B(x)^2$ for some polynomials $A(x)$ and $B(x)$. Thus $$f(x)=(x^2+bx+c)(A(x)^2+B(x)^2)$$ We also have that $$f(x)=((x+\frac{b}{2})^2+c-\frac{b^2}{4})(A(x)^2+B(x)^2)$$ so $$f(x)=\left((x+\frac{b}{2})A(x)+\sqrt{c-\frac{b^2}{4}}B(x)\right)^2+\left((x+\frac{b}{2})B(x)-\sqrt{c-\frac{b^2}{4}}A(x)\right)^2$$ by the Brahmagupta–Fibonacci identity, so the result follows by induction provided $c\geq b^2/4$. However, if this were not the case then $x^2+bx+c$ would have a real root, so we are done.