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\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}
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\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}
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\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
\[
\begin{tabular}
[c]{ll}
\medskip &amp; $a=r^{5}+r$ for some integer $r,$\\
or &amp; $a=\pm1,\pm6.$
\end{tabular}
\
\]
\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/}.
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