# Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?

This question was originally asked in stackoverflow (http://math.stackexchange.com/questions/103941/every-positive-polynomial-is-above-a-completely-q-factorized-positive-polynomial) but as it has remained without further feedback for a week I migrate it here.

Let $P$ be a unitary polynomial with rational coefficients in one variable $x$, such that $P(x) \gt 0$ for all $x \in \mathbb R$. Then $P$ is of even degree, say $2d$. Is it true that there always exist $d$ rational numbers $q_1,q_2, \ldots ,q_d$ such that

$$(*) \ \ \ P(x) \geq \bigg( \prod_{k=1}^{d} (x-q_k)^2\bigg)$$

for all $x\in \mathbb R$ ?

I can show that the answer is YES when $d=1$ or $d=2$.

When $d=1$, P has a canonical form $(X-a)^2+b$ with $b>0$, so we may take $q_1=a$ ($a$ will be rational since the coefficients of $P$ are) and we are done.

Now assume $d=2$. Then $P$ has a global minimum $\mu_1>0$, attained at one (or several) value $\eta_1$. Then $Q=\frac{P-\mu_1}{(X-\eta_1)^2}$ is a unitary polynomial of degree $2$ in $X$ and is nonnegative everywhere, so we can write $Q=\mu_2 + (X-\eta_2)^2$ with $\mu_2 \geq 0$. If we write $P$ explicitly as $P=X^4+a_3X^3+a_2X^2+a_1X+a_0$, where $a_0,a_1,a_2$ and $a_3$ are rational, then we have

$$Q=X^2 + (a_3 + 2\eta_1)X + (a_2 + (2a_3\eta_1 + 3\eta_1^2))$$

So that $\eta_2=-\frac{a_3 + 2\eta_1}{2}$. We deduce the identities $$Q=\mu_2+(X+\frac{a_3 + 2\eta_1}{2})^2$$

$$P=\mu_1+(X-\eta_1)^2Q=\mu_1+(X-\eta_1)^2\bigg( \mu_2+(X+\frac{a_3 + 2\eta_1}{2})^2\bigg)$$

Now $$\Omega=\Bigg\lbrace r \in {\mathbb R} \Bigg| \forall x\in {\mathbb R}, \ P(x) \gt \frac{\mu_1}{2}+(x-r)^2\bigg( \mu_2+(x+\frac{a_3 + 2r}{2})^2\bigg) \Bigg\rbrace$$

is an open set in $\mathbb R$. It is nonempty, since by construction it contains $\eta_1$. So it will always contain a rational number $q$. Then, we may take $q_1=q$ and $q_2=-\frac{a_3 + 2q}{2}$ and (*) holds.

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The function has no real roots, so all its roots are complex numbers, in conjugate pairs. Thus we can factor it into terms of the form $((x-a_i)^2+b_i)$. First consider the product of all the $(x-a_i)^2$. This will be strictly smaller than the original polynomial. The difference will have degree $2d-2$. Our challenge is to replace each $a_i$ with a nearby rational number $d_i$ without making the difference negative anywhere. Since we can choose $d_i$ arbitrarily close to $a_i$, we can make the difference arbitrarily small, so our only concern is its rate of growth. This will be determined by the coefficient of $x^{2d-1}$, which will be $\sum 2(a_i-d_i)$. Since $\sum a_i$ is a rational, we can choose very nearby rationals satisfying $\sum d_i=\sum a_i$, and these will satisfy your inequality.

More formally, the coefficients of $P-\prod (x-d_i)^2$ are continuous functions of $d_i$, and where the $x^{2d-1}$ coefficient is $0$, the minimum value of $(P-\prod(x-d_i)^2)/(1+x^{2d-2})$ is a continuous function of the coefficients. (Adding or subtracting $\epsilon x^k$ changes the result by no more than $\epsilon \max |x^k/(1+x^{2d-2})|.$) So there must be some open ball in the hyperplane where $\sum a_i=\sum d_i$ where it is still positive. Choose a rational point in that open ball.

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I already thought of your idea, but unfortunately it does not work. The problem is that your construction will yield a case where () holds for "almost all" x (i.e. when $x$ is large enough), but not all x, and mathematicians do mind this ... Also, note that () does not hold when $P$ has an irrational root : e.g. if $P=(x-\sqrt{2})^2$, there is no $q_1$ such that $P \geq (x-q_1)^2$ for all $x$. –  Ewan Delanoy Feb 3 '12 at 19:13
That polynomial does not satisfy the axiom $P(x)>0$. Getting it to hold for small $x$ is easy - if you don't change the roots very much, you don't change the polynomial very much, and since there must be a buffer of positive size between $P$ and the polynomial with roots, which will be enough for all small $x$. –  Will Sawin Feb 3 '12 at 19:29
Of course, $P(x)=(x-\sqrt{2})^2$ does not satisfy my hypotheses, but it illustrates my point : you can find a $q_1$ such that $|x-\sqrt{2}| \geq |x-q_1|$ for all small $x$, and another $q_1$ for large $x$. But one cannot find a $q_1$ that works for all $x$ at once. –  Ewan Delanoy Feb 3 '12 at 21:19
You're misinterpreting "small" and "large" here. In that case, since $P(x)$ is not positive, there is no construction that works for small $x$ because small $x$ means $x$ near $\sqrt{2}$. Similarly, there is no construction that works for large $x$ because large $x$ means large $|x|$. –  Will Sawin Feb 3 '12 at 23:39
When you have a bit of breathing room, my two constructions do not interfere. The small-$X$ construction is a ball around the irrational points, the large-$x$ is a hyperplane through them. The ball and the hyperplane must have nontrivial rational intersection. –  Will Sawin Feb 3 '12 at 23:41

This question looks as if the classical "catalecticant" (see http://en.wikipedia.org/wiki/Catalecticant) would be relevant. It is an invariant of binary forms of degree 2n which vanishes if and only if the form is a sum of only n powers. (I think n=1 will always do but cannot remember.) For example, a quadratic is a square iff its discriminant is 0; a quartic is a sum of 2 squares iff its catalecticant (one of the two invariants classically denoted I and J) vanishes.

I seem to remember that Cassels wrote a paper on this, but cannot find it.

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Let $f$ be a polynomial with rational coefficients which is (strictly) positive on the real line and has degree at least $4$. Then there is a nonnegative quadratic polynomial $p$ with rational coefficients such that $f-p$ is nonnegative on the real line and has a multiple rational root.

In my Diplomarbeit (written in German)

http://www.math.uni-konstanz.de/~schweigh/publications/diploma.thesis.pdf

from 1999, I proved this lemma. More precisely I proved ("Satz 2.27" in the Diplomarbeit):

Let $f\in\mathbb R[X]$ be a polynomial of degree $>0$ such that $f(x)>0$ for all $x\in\mathbb R$. Denote by $a$ the smallest global minimizer of $f$. Then there is $\varepsilon>0\in\mathbb R$ such that for all $t\in\mathbb R$ satisfying $a-\varepsilon< t < a$ $$p_t := f(t)+f'(t)(X-t)+\frac{(f'(t))^2}{4f(t)}(X-t)^2\in\mathbb R[X]$$ is a polynomial of degree 2 such that $p_t \leq f$ on $\mathbb R$.

Note that $p_t$ is a parabola with vanishing discriminant and that $f-p_t$ has a multiple root at $t$ with even multiplicity. If you choose $t$ rational, then $p_t$ has rational coefficients.

In fact, in my Diplomarbeit you find a code which implements in the computer algebra system REDUCE (version 3.6) to find such a $t$. Thus you can compute the $q_k$ in your question using Sturm sequences.

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