Can all positive real polynomials be expressed as a sum of polynomial squares?

Question

For $P(x)=y$, where $P(x)$ is some real polynomial strictly above the $x$-axis, can it be expressed as the sum of polynomial squares?

Answer 1

If $P(x)\in\mathbb{R}[x]$ is such that $P(x)\geqslant 0$ for $x\in\mathbb{R}$, then $P(x)$ cannot have odd-degree roots in $\mathbb{R}$ (otherwise it would change its sign in a neighborhood of such a root). Hence each of its roots is either even-degree real one, or comes in a pair with its complex-conjugate. In other words, $$P(x)=\prod_{k=1}^n\big((x-a_k)^2+b_k^2\big)$$ for some real $a_k$ and $b_k$ for $1\leqslant k\leqslant n$. But $$(A^2+B^2)(C^2+D^2)=(AC+BD)^2+(AD-BC)^2,$$ hence (by induction on $n$) $P(x)=A^2(x)+B^2(x)$ for some polynomials $A(x)$ and $B(x)$.

Answer 2

If "sum of linear squares" means $$ P(x) = \sum_{j=1}^n (a_jx+b_j)^2, \tag1$$ then the answer is no. For example $x^4+1$ is not of that form. Indeed, for anything of the form $(1)$ we have $$ \lim_{x \to \infty}\frac{P(x)}{x^4+1} = 0 $$

Answer 3

Yes. We induct on the degree. If $P$ is constant we are done. Otherwise we must have the degree of $P$ be even and the limit of $P$ is infinity in both $\pm \infty$ dirrections. Hence there exists an $a \in \mathbb R$ such that $P$ obtains it's minimum value at $a$. But then $(x - a)^2$ divides $P(x) - P(a)$. Thus we can write

$$P(x) = (x-a)^2 \frac{P(x) - P(a)}{(x-a)^2} + \sqrt{P(a)}^2$$ and by inductive hypothesis $\frac{P(x) - P(a)}{(x-a)^2}$ can be written as the sum of polynomial squares so we are done.