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Show that there are no $4$-variable polynomials $p(a_1,a_2,a_3,a_4)$ such that the quartic $x^4+a_1x^3+a_2x^2+a_3x+a_4$ has $4$ real roots if and only if $p\ge0$. A rather natural way to attempt this problem is to write the polynomial as a product of $2$ quadratic polynomials and then to expand, by the system I obtain gets out of hand quite fast and I'm not sure how to continue down this path. Thanks in advance for your help !

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    I (and probably anyone else reading this) know what you want to ask. But to be pedantic and nit-picky, that's not what your post says. When you say "Given the quartic $x^4+a_1x^3+a_2x^2+a_3x+a_4$ with real coefficients", whether or not that quartic has real roots is established. So either the constant polynomial $p = 1$, or the constant polynomial $p = -1$ will suffice. If, instead, you reverse it, and say "Show that there is no 4-variable polynomial $p(a_1, a_2, a_3, a_4)$ such that $p\geq 0$ iff the real quartic $x^4+a_1x^3+a_2x^2+a_3x+a_4$ has real roots." that resolves this issue. – Arthur Jul 08 '20 at 08:20
  • You are right, it has been edited –  Jul 08 '20 at 08:32
  • I don't understand the question. According to the way the question is written, $p$ is a polynomial in $a_1$, $a_2$, $a_3$, and $a_4$. Even if I treat $p(a_1,a_2,a_3,a_4)$ as a polynomial in $x$, then $p(0,0,0,0)\geq 0$ for all $x$ with real roots $x=0$ (having multiplicity $4$). So, there exists at least one polynomial that works (and in fact, infinitely many polynomials). If you mean that the $4$ roots need to be distinct, then the question is trivial. Let $\alpha<\beta<\gamma<\delta$ be the roots. Then, when evaluated at $\dfrac{\alpha+\beta}{2}$, the polynomial becomes negative. – Batominovski Jul 10 '20 at 10:08
  • $p(a_1,a_2,a_3,a_4)$ is not $x^4+a_1x^3+a_2x^2+a_3x+a_4$. It is simply a polynomial in $a_1$, $a_2$, $a_3$ and $a_4$. –  Jul 10 '20 at 12:17
  • @MikeDruther I’m confused. In your question, you state that we are dealing with the quartic $x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4$, so how is it that $p(a_1,a_2,a_3, a_4)\neq{x_4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4}$ ? We aren’t dealing with a polynomial of unknown degree, are we? – Radial Arm Saw Jul 10 '20 at 13:53
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    Alright, I will give you an example if you don't understand the question. Consider the polynomial $P(x)=x^2+a_1x+a_2$. Let $p(a_1,a_2)=a_1^2-4a_2$ (discriminant). Then $P$ has $2$ real roots if and only if $p\ge0$. Essentially the question asks to show that there is no multivariate polynomial $p(a_1,a_2,a_3,a_4)$ such that the polynomial $P(x)=x^4+a_1x^3+a_2x^2+a_3x+a_4$ has $4$ real roots if and only if $p\ge0$ –  Jul 10 '20 at 14:38

1 Answers1

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Suppose that there is such a polynomial $p(a_1,a_2,a_3,a_4)$.

Then, we see that $p(a_1,a_2,a_3,a_4)$ has to satisfy the followings :

$$\small p(0,a_2,0,a_4)\begin{cases}\ge 0&\text{if ($a_2\le 0$ and $a_2^2-4a_4=0$) or $(a_2\lt 0,a_4\ge 0$ and $a_2^2-4a_4\gt 0)$}\\\\\lt 0&\text{otherwise}\end{cases}$$

Now, let us see $p(0,a_2,0,a_4)$ as a polynomial on $a_2$. Let $m$ be the degree of the polynomial and let $f(a_4)$ be the coefficient of $a_2^m$ where $m$ is a non-negative integer.

  • Case 1 : $m=0$
    It follows from $p(0,-2,0,1)\ge 0$ that $p(0,2,0,1)\ge 0$ which contradicts the fact that $p(0,2,0,1)\lt 0$.

  • Case 2 : $m\gt 0$
    Suppose that $m$ is odd. Let us consider the case when $a_4=-1$. It follows from $\displaystyle\lim_{a_2\to +\infty}p(0,a_2,0,-1)=-\infty$ that $f(-1)\lt 0$. Since it follows that $\displaystyle\lim_{a_2\to -\infty}p(0,a_2,0,-1)=+\infty$, there exists a constant $k_1\lt 0$ such that $p(0,k_1,0,-1)\ge 0$ which contradicts the fact that $p(0,a_2,0,-1)\lt 0$ for any $a_2$. So, we see that $m$ has to be even.

    Next, let us consider the case when $a_4=1$. It follows from $\displaystyle\lim_{a_2\to +\infty}p(0,a_2,0,1)=-\infty$ that $f(1)\lt 0$. Since it follows that $\displaystyle\lim_{a_2\to -\infty}p(0,a_2,0,1)=-\infty$, there exists a constant $k_2\lt -2$ such that $p(0,k_2,0,1)\lt 0$ which contradicts the fact that $p(0,a_2,0,1)\ge 0$ for any $a_2\lt -2$.

From the two cases above, we see that there is no such $p(a_1,a_2,a_3,a_4)$.

mathlove
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