Math-1849 · mathematics · Analysis

Math-1849

Research Problems in Function Theory — Problem 4.27

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Statement

Definitions and mathematical statement

Let $n\geq1$ and let $p\in\mathbb R[x]$ have degree $n$ and $n$ distinct rational roots. Is there always a nonzero real number $t$ for which $p(x)-t$ also has $n$ distinct rational roots? All roots in both fibers are required to be rational and simple; one additional rational point in a fiber does not suffice.

The question has the following equivalent arithmetic normalization. Write $p(x)=c q(x)$, where $c\ne0$ is its leading coefficient and \[ q(x)=\prod_{j=1}^{n}(x-a_j),\qquad a_1,\ldots,a_n\in\mathbb Q\text{ distinct}. \] Must there be $s\in\mathbb Q\setminus\{0\}$ and distinct $b_1,\ldots,b_n\in\mathbb Q$ such that \[ \prod_{j=1}^{n}(x-b_j)=q(x)-s? \] Any rational root of $q-s$ forces $s$ itself to be rational, so this normalization loses no possibility. The desired shift of the original polynomial is $t=cs$.

Equivalently, the two rational root multisets should have the same elementary symmetric functions of orders $1,\ldots,n-1$, but different products. Their sets are automatically disjoint because the two polynomial levels are distinct. The known low-degree cases motivate the unrestricted question; the remaining general target is to establish this property in every degree $n\geq4$, or exhibit a polynomial for which no such nonzero shift exists.

Short English statement

Does every polynomial with distinct rational roots have another nonzero level at which all its roots are again distinct and rational?

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