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This question was originally asked in stackoverflow ( 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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up vote 6 down vote accepted

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 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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The answer to your question is yes by the following lemma:

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)

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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