Let $p$ be an odd prime, and let $n_1,\dots,n_{p-2}, m$ be even integers such that $n_1 < n_2 < \dots < n_{p-2}$ and \begin{equation} 2m > \sum_{i=1}^{p-2} n_i^2. \end{equation} Consider the polynomial \begin{equation} g(x)=(x^2 + m)(x - n_1) \dots (x - n_{p-2}). \end{equation} From Rolle's Theorem, we know that for each $i=1,2,\dots,p-3$, there exists $x_i \in (n_i,n_{i+1})$ such that $g'(x_i)=0$. So $g'(x)$ has at least $p-3$ distinct real zeroes. My question is: can $g'(x)$ have more than $p-3$ distinct real zeroes?
I do not know the answer, but for sure the constraints on the parameters are relevant here. For example the polynomial $g(x)=(x^2+1)(x-4)(x-2)(x+2)$ has derivative $g'(x)=5x^4-16x^3-9x^2+24x-4=(x-1)(5x^3-11x^2-20x+4)$ which has four distinct real roots, as you can check on WolframAlpha.
NOTE This strange polynomial arises in the construction given by R. Brauer of a polynomial $f(x) \in \mathbb{Q}[x]$ of degree $p$ whose Galois group over $\mathbb{Q}$ is isomorphic to the symmetric group $\mathcal{S}_p$: see Jacobson, Basic Algebra I, $\S 4.10$. The question I asked is clearly irrelevant for the construction, but has intrigued me, since I could not answer it in the negative nor I could find some counterexample.
from Hot Weekly Questions - Mathematics Stack Exchange
Maurizio Barbato
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