\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin {center} \vskip 1cm{\LARGE\bf On a Sequence of Nonsolvable \\ \vskip .15in Quintic Polynomials} \vskip 1cm \large Jennifer A. Johnstone and Blair K. Spearman\footnote{All correspondence should be directed to this author.}\\ Mathematics and Statistics\\ University of British Columbia Okanagan\\ Kelowna, BC V1V 1V7 \\ Canada\\ \href{mailto:johnstone33@hotmail.com}{\tt johnstone33@hotmail.com}\\ \href{mailto:Blair.Spearman@ubc.ca}{\tt Blair.Spearman@ubc.ca} \end {center} \vskip .2in \begin {abstract} Aleksandrov, Kolmogorov and Lavrent'ev state that $x^{5}+x-a$ is nonsolvable for $a=3,4,5,7,8,9,10,11,\ldots$. In other words, these polynomials have a nonsolvable Galois group. A full explanation of this sequence requires consideration of both reducible and irreducible solvable quintic polynomials of the form $x^{5}+x-a.$ All omissions from this sequence due to solvability are characterized. This requires the determination of the rational points on a genus $3$ curve. \end {abstract} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{proposition}{Proposition} \theoremstyle{remark} \newtheorem{remark}{Remark} \newcommand{\thmref}[1]{Theorem~\ref{#1}} \newcommand{\secref}[1]{\S\ref{#1}} \newcommand{\lemref}[1]{Lemma~\ref{#1}} \section{Introduction} \noindent Let $f(x)$ be a polynomial with rational coefficients. The polynomial $f(x)$ is solvable if its Galois group is solvable. Equivalently, the zeroes of $f(x)$ can be expressed in radical form. Dummit \cite{Dmt} provides a description of this process for quintic polynomials. Aleksandrov, Kolmogorov and Lavrent'ev \cite{AKL} state that \textquotedblleft the equation $x^{5}+x-a=0,$ where $a$ is a positive whole number, in most cases cannot be solved by radicals. For example, it is not solvable in radicals for $a=3,4,5,7,8,9,10,11,\ldots$ \textquotedblright. The purpose of this paper is to give a complete explanation of this sequence for all integers. Our main Theorem is \begin{theorem} \label{thm:main} Let $f(x)=x^{5}+x-a$ where $a$ is an integer. Then $f(x)$ is not solvable by radicals unless \[ \begin{tabular} [c]{ll} \medskip & $a=r^{5}+r$ for some integer $r,$\\ or & $a=\pm1,\pm6.$ \end{tabular} \ \] \end{theorem} \noindent In Section \ref{sec:prelim} we give some preliminary results we shall need for the proof of our theorem, which is given in Section \ref{sec:proof}. \section{Preliminary Results} \label{sec:prelim} \noindent If the polynomial $f(x)=x^{5}+x-a$ is reducible over $\mathbb{Q}$ then $f(x)$ is solvable by radicals since the factors have degree at most four. In this section we begin by enumerating the values of the integer $a$ for which $f(x)$ is reducible over $\mathbb{Q}$. The first class of reducibility is the easiest and is described in the following lemma, where the proof is obvious and thus omitted. \begin{lemma} \label{lem:first} Let $f(x)=x^{5}+x-a$ where $a$ is an integer. If $f(x)$ has a rational zero, say $r$, then $r$ is an integer and $a=r^{5}+r.$ \end{lemma} \noindent For the second class of reducible polynomials Rabinowitz \cite{Rab} provides us with the following result: \begin{proposition} \label{prop:second} The only integers $a$ for which $x^{5}+x-a$ factors into the product of an irreducible quadratic and an irreducible cubic are $a=\pm1$ and $a=\pm6.$ \end{proposition} \noindent Now we need a characterization of solvable irreducible quintic trinomials $x^{5}+x-a$. This is given by a Theorem of Dummit \cite{Dmt}, specialized to our polynomials. \begin{proposition} \label{prop:char} The irreducible quintic $x^{5}+x-a\in\mathbb{Q}[x]$ is solvable by radicals if and only if the resolvent sextic polynomial \begin{equation}\label{sextic} t^{6}+8t^{5}+40t^{4}+160t^{3}+400t^{2}+(512-3125a^{4})t+256-9375a^{4} \tag{1}% \end{equation} has a rational zero $t$. \end{proposition} To finish this section we give a determination of the set of rational points on a genus 3 curve. This result will be required in the proof of our theorem. \begin{lemma} \label{lem:norat} The affine curve $y^{2}=(20x^{4}-1)(5x^{4}+1)$ has no rational points. \end{lemma} \begin{proof} The computational method of proof used here is described in depth, with examples, by Bremner and Tzanakis \cite{BRT}. We shall work in the number field $K=\mathbb{Q}(\theta)$ where $\theta$ is a zero of $z^{4}+3z^{2}+1.$ The number field $K$ is a bicyclic quartic field, specifically $K=\mathbb{Q}\left( \sqrt{-1},\sqrt{5}\right) .$ The maximal order $O_{K}$ of $K$ is $\mathbb{Z}[\theta]$, the class number of $O_{K}$ is 1, and the fundamental unit of $O_{K}$ is $\varepsilon=1+\theta^{2}.$ Let $(x,y)=\left( X/Z,Y/Z^{4}\right) $ where $X,Y,Z\in\mathbb{Z}$ and $\gcd(X,Z)=1$ with $Z\neq0.$ This gives us the following equation: \begin{equation}\label{affineCurve} Y^{2}=(20X^{4}-Z^{4})(5X^{4}+Z^{4}).\tag{2} \end{equation} If $X$ and $Z$ are both odd then (\ref{affineCurve}) reduces to $Y^{2}\equiv 2$ (mod $4$), which is impossible. If $X$ is even and $Z$ is odd then (\ref{affineCurve}) reduces to $Y^{2}\equiv 3$ (mod $4$), which is also impossible. Therefore, we may assume throughout the remainder of this proof that $X$ is odd and $Z$ is even. We also observe that $X \neq 0.$ Factoring (\ref{affineCurve}) over $K$ gives \[ F_{1}F_{2}=25Y^{2}. \] where \begin{equation}\label{F1F2} \begin{tabular} [c]{l}% $\medskip F_{1}=(10X^{2}+(2\theta^{2}+3)Z^{2})(5X^{2}+(\theta^{3} +4\theta)Z^{2}),$\\ $F_{2}=(10X^{2}-(2\theta^{2}+3)Z^{2})(5X^{2}-(\theta^{3}+4\theta)Z^{2}).$ \end{tabular} \ \ \tag{3} \end{equation} Now we have the following factorization of ideals in $O_{K}$% \[ \left\langle 2\right\rangle =\wp_{2}^{2}\text{, \ \ \ \ }\left\langle 5\right\rangle =\wp_{51}^{2}\wp_{52}^{2}. \] We also have the identities% \[ \begin{tabular} [c]{l} $\medskip(-10X^{2}+(2\theta^{3}+2\theta^{2}+8\theta+3)Z^{2})F_{1} +(10X^{2}+(2\theta^{3}+2\theta^{2}+8\theta+3)Z^{2})F_{2}$\\ $=10Z^{6}(\theta^{3}-4\theta^{2}+4\theta-6),$ \end{tabular} \ \ \] and \[ \begin{tabular} [c]{l} $\medskip((10\theta^{3}+10\theta^{2}+40\theta+15)X^{2}-(5\theta^{3} +10\theta)Z^{2})F_{1}$\\ $\medskip+((10\theta^{3}+10\theta^{2}+40\theta+15)X^{2}+(5\theta^{3} +10\theta)Z^{2})F_{2}$\\ $=500X^{6}(2\theta^{3}+2\theta^{2}+8\theta+3).$ \end{tabular} \ \ \] Then since $X$ and $Z$ are relatively prime and the norms of the elements $(\theta^{3}-4\theta^{2}+4\theta-6)$ and $(2\theta^{3}+2\theta^{2}+8\theta+3)$ from $K$ to $\mathbb{Q}$ are both equal to $5^{4}$ we see that the gcd ideal of $F_{1}$ and $F_{2}$ involves only prime ideals dividing $2$ and $5.$ Recalling the observation stated just after equation (\ref{affineCurve}) that $X$ is odd and $Z$ is even, we see from equation (\ref{F1F2}) that $\wp_{2}^{4}\parallel F_{1} F_{2},$ $\wp_{2}^{2}\parallel F_{1}$ and $\wp_{2}^{2}\parallel F_{2}.$ Furthermore, if $5\nmid Z$ then as ideals, $\left\langle 2\theta ^{2}+3\right\rangle = \left\langle \theta^{3}+4\theta\right\rangle = \wp_{51} \wp_{52},$ so that (\ref{F1F2}) implies $\wp_{51}^{4}\wp_{52}^{4}\parallel F_{1}F_{2},$ $\wp_{51}^{2}\wp_{52}^{2}\parallel F_{1},$ and $\wp_{51}^{2}\wp_{52} ^{2}\parallel F_{2}.$ Finally, if $5\mid Z$ then $5\nmid X$ so that (\ref{F1F2}) gives $\wp_{51}^{8}\wp_{52}^{8}\parallel F_{1}F_{2},$ $\wp_{51}^{4}\wp_{52} ^{4}\parallel F_{1}$ and $\wp_{51}^{4}\wp_{52}^{4}\parallel F_{2}.$ $\ $We conclude that modulo squares the gcd ideal of $F_{1}$ and $F_{2}$ is $\left\langle 1 \right\rangle $. $\ $We now deduce equations \begin{equation}\label{deduction} F_{1}=gU^{2}\text{ \ \ and \ \ }F_{2}=gV^{2},\tag{4} \end{equation} with $5Y=gUV$ and $g=(-1)^{i_{0}}\varepsilon^{i_{1}},$ $i_{0}, i_{1} = 0, 1.$ Using the first of these two equations, a $K$-rational solution $(X,Z,U)$ to this equation would yield a $K$-rational point $(x,y)$ on the elliptic curve \begin{equation}\label{ellCurve} y^{2}g=x(10x+2\theta^{2}+3)(5x+\theta^{3}+4\theta).\tag{5} \end{equation} This elliptic curve arises by multiplying the first equation in (\ref{deduction}) by $X^{2}/Z^{6}$ and defining $y=XU/Z^{3}$ and $x=X^{2}/Z^{2}.$ We wish to determine the Mordell-Weil group of each of these elliptic curves over the number field $K.$ To do this we appeal to Magma and the routine {\tt PseudoMordellWeilGroup}. This routine uses a \textit{2-isogeny descent} when available, returning \textit{true} in the output when the rank is determined. Bruin \cite{Bru} gives detailed information on this routine. For all four choices of $g$ in (\ref{ellCurve}), using {\tt PseudoMordellWeilGroup,} we found that the group of $K$-rational points is isomorphic to $\mathbb{Z}/2\mathbb{Z+Z}/2\mathbb{Z}.$ Thus the only $K$-rational points on (\ref{ellCurve}) are the obvious 2-torsion points $(0,0),$ $\left( -\dfrac{2\theta ^{2}+3}{10},0\right) $ and $\left( -\dfrac{\theta^{3}+4\theta}{5},0\right) .$ If these points were to yield integer pairs $(X,Z)$ with $X,Z\neq0,$ satisfying (\ref{affineCurve}), then one of the nonzero $x$ coordinates would have to be equal to the square of a nonzero rational number, which is impossible as they are not even in $\mathbb{Q}$. Hence there are no rational points on the curve $y^{2}=(20x^{4}-1)(5x^{4}+1)$ except the two points at infinity. Equally well, for the last part of this proof, the Magma command {\tt IsLocallySolvable }can be used to show that the curves $F_{1}=gU^{2}$, given in (\ref{deduction}), have no finite rational points. \end{proof} \section{Proof of Theorem} \label{sec:proof} \begin{proof} Suppose that $f(x)=x^{5}+x-a$ is solvable by radicals. If $f(x)$ is reducible then either $f(x)$ has a rational root or $f(x)$ factors into the product of and irreducible quadratic and an irreducible cubic polynomial over $\mathbb{Q}$. The case of reducible quintics was covered by Lemma \ref{lem:first} and Proposition \ref{prop:second}, in Section \ref{sec:prelim} . This gives us the exceptional values of $a$ which are stated in our theorem. Now we search for integral values of $a$ where the polynomials $x^{5}+x-a$ are solvable and irreducible. By Proposition \ref{prop:char}, this leads to a consideration of the rational solutions $(t,a)$ satisfying (\ref{sextic}). The curve (\ref{sextic}) is birationally equivalent to the genus 3 curve% \begin{equation}\label{genus3} y^{2}=(3125x^{4}-4)(3125x^{4}+16), \tag{6} \end{equation} using the transformations \begin{equation}\label{transformations} \begin{tabular} [c]{l} $\bigskip x=\dfrac{a}{t+2},$ \ \ $y=\dfrac{t^{2}+6t-16}{t+3},$\\ $t=\dfrac{3125x^{4}+y}{2},$ \ \ $a=2x+\dfrac{3125x^{5}+xy}{2},$ \end{tabular} \tag{7} \end{equation} and after scaling (\ref{genus3}) becomes \[ y^{2}=(20x^{4}-1)(5x^{4}+1). \] As shown in Lemma \ref{lem:norat}, there are no rational points on this curve. We conclude from the birational transformations given in (\ref{transformations}) that there are no finite rational points $(t,a)$ satisfying (\ref{sextic}) so that the determination of the integral values of $a$ is complete. \end{proof} \begin{remark} \textrm{In the proof of Theorem \ref{thm:main} we actually showed that there were no rational values of} $a$ \textrm{such that} $f(x)=x^{5}+x-a$ \textrm{is solvable and irreducible. An analog of Theorem \ref{thm:main} for sextic trinomials} $x^{6}+x-a$ \textrm{would be an interesting calculation since for these sextics there exist rational values of }$a$\textrm{\ for which }% $x^{6}+x-a$ \textrm{is irreducible and solvable over} $\mathbb{Q}.$ \textrm{The sextic trinomial} $x^{6}+x+\dfrac{41}{8}$ \textrm{is irreducible and solvable with Galois group} $6T13$ \textrm{in Maple which is isomorphic to} $C_{3}^{2}\mathbb{o}D_{4}$ \textrm{\cite{Coh}.} \end{remark} \section{Acknowledgements} Both authors received funding from the Natural Sciences and Engineering Research Council of Canada and wish to thank them for their support. \begin{thebibliography}{9} \bibitem {AKL}A. D. Aleksandrov, A. N.\ Kolmogorov and M. A. Lavrent'ev, \textit{Mathematics: Its Content, Methods and Meaning}, MIT Press, 1963. \bibitem {WIE}W. Bosma, J. Cannon, and C. Playoust, The Magma algebra system I: the user language, \textit{J. Symbolic Comput.} \textbf{24} (1997), 235--265. \bibitem {BRT}A. Bremner and N. Tzanakis, On squares in Lucas sequences, \textit{J. Number Theory} \textbf{124} (2007), 511--520. \bibitem {Bru}N. Bruin, Some ternary diophantine equations of signature $(n,n,2)$, in W. Bosma and J. Cannon, eds., \textit{Discovering Mathematics with Magma: Reducing the Abstract to Concrete}, Springer, 2006, pp.\ 63--91. \bibitem {Coh}H. Cohen, \textit{A Course in Computational Algebraic Number Theory}, Springer, 2000. \bibitem {Dmt}D. S. Dummit, Solving solvable quintics, \textit{Math. Comput.} \textbf{57} (1991), 387--401. \bibitem {Rab}S. Rabinowitz, The factorization of $x^{5}\pm x+n$, \textit{Math. Mag.} \textbf{61} (1988), 191--193. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 12E05; Secondary 14G05. \noindent {\it Keywords}: Nonsolvable polynomial, Galois group. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received November 11 2008; revised version received February 13 2009. Published in {\it Journal of Integer Sequences}, February 15 2009. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}