Quartic rationally derived polynomials

Importance: High ✭✭✭
Subject: Number Theory
Recomm. for undergrads: no
Posted by: mdevos
on: May 17th, 2008

Call a polynomial $p \in {\mathbb Q}[x]$ \emph{rationally derived} if all roots of $p$ and the nonzero derivatives of $p$ are rational.

\begin{conjecture} There does not exist a quartic rationally derived polynomial $p \in {\mathbb Q}[x]$ with four distinct roots. \end{conjecture}

Probably anyone who has ever designed simple problems for calculus students has looked for polynomials $p$ with the property that both $p$ and some small derivatives of it are easy to factor. Perhaps inspired by this, Buchholz and MacDougall attempted to classify all univariate polynomials defined over a domain $k$ with the property that they and all their nonzero derivatives have all their roots in $k$. This problem can be split into cases dependent upon the multiplicity of the roots, and Buchholz and MacDougall solved many of the small ones for $k={\mahtbb Q}$. Based on their results and a theorem of Flynn [F], an affirmative solution to the above conjecture would complete this classification problem for $k={\mathbb Q}$.


*[BM] R. Buchholz, and J. MacDougall, \href[When Newton met Diophantus: a study of rational-derived polynomials and their extension to quadratic fields]{http://www.geocities.com/teufel_pi/papers/rdp.pdf}. J. Number Theory 81 (2000), no. 2, 210--233. \MRhref{MR1752251}

[F] E. V. Flynn, On Q-derived polynomials. Proc. Edinb. Math. Soc. (2) 44 (2001), no. 1, 103--110. \MRhref{MR1879212}

* indicates original appearance(s) of problem.


Hyperlink in the bibliography is no longer valid, but the article can be found at: http://web.archive.org/web/20011127182208/http://www.geocities.com/teufel_pi/papers/rdp.pdf

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