# A polynomial $p(x) \in \Bbb R_{2n-1}[x]$, $p(0) = 0$, $p(x) \geq 0 \ \forall x \geq 0$, can be written as $p(x) = xq_1(x)^2 + q_2(x)^2$

Let $$p(x) \in \Bbb R_{2n-1}[x]$$ be a polynomial such that $$p(0)= 0$$ and $$p(x) \geq 0 \ \forall x \geq 0$$. Then there exists $$q_1, q_2 \in \Bbb R_{n-1}[x]$$ such that $$p(x) = xq_1(x)^2 + q_2(x)^2$$.

Here, $$\Bbb R_{k}[x]$$ is the set of polynomials with real coefficients of degree less than or equal to $$k$$.

Note that we always have $$p(x) = xp_o(x^2) + p_e(x^2)$$ for some $$p_o, p_e \in \Bbb R_{n-1}[x]$$ by separating the terms of even and odd degree of $$p$$.

It is related to Prove that a positive polynomial function can be written as the squares of two polynomial functions but does not seem to follow from it.

This is stated without proof and reference at page 77 from The Classical Moment Problem, Oliver and Boyd Ltd, 1965, by Akhiezer.

The key to the solution is the identity $$(a^2x+b^2)(c^2x+d^2) = (ad-bc)^2x + (acx+bd)^2. \tag1$$ You can put polynomials for $$a,b,c,d$$.

Let $$\mathcal{P} = \Big\{ xq_1(x)^2+q_2(x)^2 \Big\}$$; we need to prove $$p(x)\in\mathcal{P}$$.

Notice that all complete squares and all nonnegative constants are elements of $$\mathcal{P}$$. The identity $$(1)$$ shows that $$\mathcal{P}$$ is closed for multiplication.

Consider the prime factorization of $$p(x)$$: $$p(x) = A \cdot \prod(x+a_i)^{\alpha_i} \prod (x^2+c_ix+d_i).$$

From $$x\to\infty$$ we find that the leading coefficient $$A$$ is nonnegative, so $$A\in\mathcal{P}$$.

All prime factors $$x+a_i$$ with $$a\ge0$$ are in $$\mathcal{P}$$.

In every irreducible factor $$x^2+cx+d=(x-\sqrt{d})^2+(2\sqrt{d}-c)x$$ we have $$2\sqrt{d}-c>0$$, so they are in $$\mathcal{P}$$.

Finally, the positive roots of $$p(x)$$ must have even multiplicities, so every factor $$(x+a_i)^{\alpha_i}$$ with $$a_i<0$$ is a complete square.

Hence $$p(x)$$, being the product of some elements of $$\mathcal{P}$$, is an element of $$\mathcal{P}$$.

• Nice and simple $(+1).$ Dec 10 '18 at 9:48