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\title{Vertices of polygons}
\author{David Pierce}
\date{\today}


\begin{comment}

\address{Mathematics Dept\\
Middle East Tech.\ Univ.\\
Ankara 06531, Turkey}

\email{dpierce@metu.edu.tr}
\urladdr{http://www.math.metu.edu.tr/~dpierce/}

\end{comment}


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\newtheorem*{theorem}{Theorem}

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\usepackage{pstricks,pst-plot,pst-math,pst-node,pst-tree}
\usepackage{hfoldsty}  % gives old-style numerals
\usepackage{upgreek}
\newcommand{\mpi}{\uppi}
\newcommand{\mi}{\mathrm{i}}
\newcommand{\me}{\mathrm{e}}
% Standard structures
\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\N}{\stnd{N}}         % natural numbers
\newcommand{\varN}{\omega}        % my usual preference for this
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\newcommand{\Q}{\stnd{Q}}         % rationals
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\begin{document}
  \maketitle
%\tableofcontents
\footnotetext{The inspiration for this article was a conversation with
  Sasha Borovik at
Zencef{}il restaurant in \.Istanbul on a
rainy holiday Monday, October 29, 2007.}

\section{The pentagon}

Vertices of a regular pentagon can be obtained as
the solutions in $\C$ of the equation
\begin{equation}\label{eqn:5}
  x^5-1=0
\end{equation}
(Fig.~\ref{fig:5small}).
\begin{figure}[t]
\begin{center}
  \begin{pspicture}(-2.7,-2.7)(2.7,2.7)
\psline{->}(-2.7,0)(2.7,0)
\psline{->}(0,-2.7)(0,2.7)
\psset{dotsize=3pt 2}
    \parametricplot[plotpoints=5,
plotstyle=dots]{0}{4}
{t 360 mul 5 div cos 2 mul t 360 mul 5 div sin 2 mul}
\uput[ur](2,0){$1$}
\uput[ul](0,2){$\mi$}
\uput[ul](-2,0){$-1$}
\uput[dl](0,-2){$-\mi$}
\psdots[dotstyle=o](0,2)(-2,0)(0,-2)
  \end{pspicture}
\caption{The five 5th roots of 1}\label{fig:5small}
\end{center}
\end{figure}
To find these solutions,
we can use the factorization
\begin{equation*}
x^5-1=(x-1)\cdot(x^4+x^3+x^2+x+1).  
\end{equation*}
So $1$ is a solution of~\eqref{eqn:5}, and the other solutions are
solutions of 
\begin{equation}\label{eqn:x^2}
  x^2+x+1+\frac1x+\frac1{x^2}=0.
\end{equation}
To solve \emph{this,} we can make a substitution, letting
\begin{equation}\label{eqn:x+}
  y=x+\frac1x
\end{equation}
so that
\begin{equation*}
y^2=x^2+2+\frac1{x^2}.
\end{equation*}
Then~\eqref{eqn:x^2} becomes
\begin{equation}\label{eqn:y^2}
  y^2+y-1=0.
\end{equation}
So $y$ is quadratic, and $x$ is quadratic in $y$, and we can find our
solutions.  
The question arises:  Is a similar analysis possible for $x^{17}-1$?
How did Gauss factorize this in the
\emph{Disquisitiones Arithmeticae} \cite{MR0197380}?

\section{The numbers for the pentagon}

Just to see the numbers, let us first continue the analysis
of~\eqref{eqn:5}.  From~\eqref{eqn:y^2} and~\eqref{eqn:x+} we
have\footnote{I prefer 
  writing $\radix 5$ to $\sqrt 5$: the overline or \emph{vinculum} is
  not needed here as a grouping symbol, as it is in $\sqrt{y^2-4}$;
  though even here, one may, as I shall, write $\radix(y^2-4)$.
  Compare $\log x$ with $\log(1+x)$.} 
\begin{align*}
y&=\frac{-1\pm\radix 5}2,&
x^2-yx+1&=0,&
x&=\frac{y\pm\radixp{y^2-4}}2.
\end{align*}
As $y^2-4=-y-3$ by~\eqref{eqn:y^2}, so we can also write
\begin{equation*}
x=\frac{y\pm\radixp{-y-3}}2.  
\end{equation*}
Since $-y-3= -(5\pm\radix 5)/2<0$,
the four solutions to~\eqref{eqn:x^2} are (as in Fig.~\ref{fig:5})  
\begin{align*}
&\frac{-1+\radix 5}4\pm\frac{\mi}2\radix[\frac11]{\frac{5+\radix 5}2},&
&\frac{-1-\radix 5}4\pm\frac{\mi}2\radix[\frac11]{\frac{5-\radix 5}2}.
\end{align*}
\begin{figure}[t]
\begin{center}
  \begin{pspicture}(-3.8,-3.8)(3.8,3.8)
\psline{->}(-3.8,0)(3.8,0)
\psline{->}(0,-3.8)(0,3.8)
\psset{dotsize=3pt 2}
    \parametricplot[plotpoints=5,
plotstyle=dots]{0}{4}
{t 360 mul 5 div cos 3 mul t 360 mul 5 div sin 3 mul}
\uput[ur](3,0){$1$}
\uput[ul](0,3){$\mi$}
\uput[ul](-3,0){$-1$}
\uput[dl](0,-3){$-\mi$}
\psdots[dotstyle=o](0,3)(-3,0)(0,-3)
    \parametricplot[plotpoints=2,
plotstyle=dots, dotstyle=o]{-1}{1}
{t 5 sqrt mul 1 sub 3 mul 4 div 0}
    \parametricplot[plotpoints=2,
plotstyle=dots, dotstyle=o]{-1}{1}
{0 t 5 5 sqrt add 2 mul sqrt mul 3 mul 4 div}
    \parametricplot[plotpoints=2,
plotstyle=dots, dotstyle=o]{-1}{1}
{0 t 5 5 sqrt sub 2 mul sqrt mul 3 mul 4 div}
\uput[r](.93,2.85){$\frac{-1+\radix 5}4+\frac{\mi}2\radix[\frac11]{\frac{5+\radix 5}2}$}
\uput[r](0.93,-2.85){$\frac{-1+\radix 5}4-\frac{\mi}2\radix[\frac11]{\frac{5+\radix 5}2}$}
\uput[u](-2.44,1.77){$\frac{-1-\radix 5}4+\frac{\mi}2\radix[\frac11]{\frac{5-\radix 5}2}$}
\uput[d](-2.44,-1.77){$\frac{-1-\radix 5}4-\frac{\mi}2\radix[\frac11]{\frac{5-\radix 5}2}$}
  \end{pspicture}
\caption{The five 5th roots of 1, evaluated}\label{fig:5}
\end{center}
\end{figure}

\section{Seventeenth roots of unity: first attempt}

Suppose we want to solve
\begin{equation}\label{eqn:17}
  x^{17}-1=0.
\end{equation}
Besides $1$, the solutions are the solutions of
$x^{16}+x^{15}+x^{14}+\dotsb+x^2+x^1+1=0$ and therefore of
\begin{equation}\label{eqn:16}
x^8+x^7+x^6+\dotsb+x^{-6}+x^{-7}+x^{-8}=0.
\end{equation}  
We can attempt here the same substitution~\eqref{eqn:x+} as before.  After
the computations in Table~\ref{table:y}, we obtain
\begin{sidewaystable}
\renewcommand{\arraystretch}{1.6}
  \begin{equation*}
    \begin{array}{r*{2}{@{}r}*{13}{@{\negthickspace\negthickspace}r}*{2}{@{}r}}
y  ={}&&&&&&&&x&&{}+x^{-1}\\
y^2={}&&&&&&&x^2&&{}+2&&{}+\myfract{x}{2}\\
y^3={}&&&&&&x^3&&{}+3x&&{}+\myfrac3x1&&{}+\myfract{x}{3}\\
y^4={}&&&&&x^4&&{}+4x^2&&{}+6&&{}+\myfrac4{x}{2}&&{}+
\myfract{x}{4}\\ 
y^5={}&&&&x^5&&{}+5x^3&&{}+10x&&{}+\myfrac{10}x1&&{}+
\myfrac5{x}{3}&&{}+\myfract{x}{5}\\ 
y^6={}&&&x^6&&{}+6x^4&&{}+15x^2&&{}+20&&{}+\myfrac{15}{x}{2}&&{}+
\myfrac6{x}{4}&&{}+\myfract{x}{6}\\ 
y^7={}&&x^7&&{}+7x^5&&{}+21x^3&&{}+35x&&{}+\myfrac{35}x1&&{}+
\myfrac{21}{x}{3}&&{}+ 
      \myfrac7{x}{5}&&{}+\myfract{x}{7}\\ 
      y^8={}&x^8&&{}+8x^6&&{}+28x^4&&{}+56x^2&&{}+70&&{}+\myfrac{56}x1&&{}+
      \myfrac{28}{x}{4}&&{}+\myfrac8{x}{6}&&{}+
      \myfract{x}{8}  \\\hline
\hline
      y^8={}&x^8&&{}+8x^6&&{}+28x^4&&{}+56x^2&&{}+70&&{}+\myfrac{56}x1&&{}+
      \myfrac{28}{x}{4}&&{}+\myfrac8{x}{6}&&{}+
      \myfract{x}{8}  \\
y^7={}&&x^7&&{}+7x^5&&{}+21x^3&&{}+35x&&{}+\myfrac{35}x1&&{}+
\myfrac{21}{x}{3}&&{}+ 
      \myfrac7{x}{5}&&{}+\myfract{x}{7}\\ 
-7y^6={}&&&-7x^6&&{}-42x^4&&{}-105x^2&&{}-140&&{}-\myfrac{105}{x}{2}&&-
\myfrac{42}{x}{4}&&-\myfrac7{x}{6}\\ 
-6y^5={}&&&&-6x^5&&{}-30x^3&&{}-60x&&{}-\myfrac{60}x1&&
-\myfrac{30}{x}{3}&&{}-\myfrac{6}{x}{5}\\ 
15y^4={}&&&&&15x^4&&{}+60x^2&&{}+90&&{}+\myfrac{60}{x}{2}&&{}+
\myfrac{15}{x}{4}\\ 
10y^3={}&&&&&&10x^3&&{}+30x&&{}+\myfrac{30}x1&&{}+\myfrac{10}{x}{3}\\
-10y^2={}&&&&&&&-10x^2&&{}-20&&{}-\myfrac{10}{x}{2}\\
-4y  ={}&&&&&&&&-4x&&{}-4x^{-1}\\
1  ={}&&&&&&&&&1\\\hline
&x^8&{}+x^7&{}+x^6&{}+x^5&{}+x^4&{}+x^3&{}+x^2&{}+x&{}+1&x^{-1}&{}+x^{-2}&{}+x^{-3}&{}+x^{-4}&{}+x^{-5}&{}+x^{-6}&{}+x^{-7}&{}+x^{-8} 
    \end{array}
  \end{equation*}
\caption{Reduction of $(x^{17}-1)/(x-1)x^8$}\label{table:y}
\end{sidewaystable}
\begin{equation*}
  y^8+y^7-7y^6-6y^5+15y^4+10y^3-10y^2-4y+1=0.
\end{equation*}
It is not clear how to proceed from here.  However, Galois
theory will illuminate the general situation. 

\section{Galois analysis}

Let $n$ be a positive integer, and let $A$ be the set of solutions in
$\C$ of 
\begin{equation*}
  x^n-1=0.
\end{equation*}
Then $A$ is an order-$n$ cyclic subgroup of $\units{\C}$.  Indeed, we
have a homomorphism $x\mapsto \exp(2\mpi\mi x/n)$ from $\Z$ into
$\units{\C}$ with image $A$ and kernel $(n)$.  If
$\zeta$ is a generator of $A$, then $\Q(A)=\Q(\zeta)$.  Let $G$ be the
Galois group (namely, the group of automorphisms) of this field.  Then
we have an isomorphism $x\mapsto\sigma_x$ from $\Zmu[n]$ to $G$, where
\begin{equation*}
  \sigma_x(\zeta)=\zeta^x.
\end{equation*}
In particular, $G$ is cyclic if and only if $n$ has a primitive root:
equivalently, $n$ is $2$, $4$, an odd
prime power, or twice an odd prime power.  
Let us suppose that we are
in the special case where $n$ is a Fermat prime, namely a prime $p$,
where 
\begin{equation*}
p=2^{2^m}+1
\end{equation*}
for some $m$.\footnote{The only way
  $2^k+1$ can be
prime is if $k$ is a power of $2$.  The Fermat number $2^{2^m}+1$ is
apparently \cite{Fermat-numbers} known to be prime when $0\leq m\leq
4$, but composite when $5\leq m\leq32$.  It is also composite for
  certain $m$ as large as $2478782$; but its status is 
  unknown when $m=33$.}   Let $\alpha$ be a generator of
$G$.  Then $\alpha=\sigma_r$ for some primitive root~$r$ of $p$.  The
subgroups of $G$ are just the groups $\gpgen{\alpha^{2^k}}$, where
$k\leq m$.  Moreover,
\begin{equation*}
  \gpgen{\alpha^{2^k}}\included\gpgen{\alpha^{2^{\ell}}}\iff \ell\leq k;
\end{equation*}
so the set of subgroups of $G$ is linearly ordered by inclusion.
Fix a generator~$\zeta$ of $A$: say
\begin{equation*}
  \zeta=\exp\Bigl(\frac{2\mpi\mi}p\Bigr).
\end{equation*}
Suppose $p-1$ is factorized as $k\cdot\ell$.  Then the subgroup
$\gpgen{\alpha^k}$ of $G$ has
order $\ell$.  We define
\begin{equation*}
  \sgsum{\ell}=
  \sum_{\xi\in\gpgen{\alpha^k}}\xi(\zeta)=\sum_{x\in\gpgen{r^k}}\zeta^x,
\end{equation*}
where $\gpgen{r^k}$ is understood as a subgroup of $\Zmu[p]$.  Then
$\sgsum{\ell}$ is the sum of~$\ell$ elements of $A\setminus\{1\}$.  In
particular,
\begin{align*}
\sgsum1&=\zeta,&
\sgsum2&=\zeta+\zeta\inv,&
\sgsum{p-1}&=-1
\end{align*}
(since the coefficient of $x^{p-2}$ in $(x^p-1)/(x-1)$ is $1$).
In general, $\alpha^k$ fixes $\sgsum{\ell}$, but not $\sgsum{2\ell}$
(if indeed $\ell<p-1$).
Therefore the fixed field of $\gpgen{\alpha^k}$ is
precisely $\Q(\sgsum{\ell})$, and we have the Galois correspondence in
Table~\ref{table:Gal-gen}.
\begin{table}[t]
  \begin{equation*}
  \xymatrix{
\gpgen{1}       \ar@{-}[d] & \Q(\zeta)\\
\gpgen{\alpha^{(p-1)/2}}\ar@{-}[d] & \Q(\zeta+\zeta\inv)\ar@{-}[u]_2\\
\gpgen{\alpha^{(p-1)/4}}\ar@{.}[d] & \Q(\sgsum 4)\ar@{-}[u]_2\\
\gpgen{\alpha^2}\ar@{-}[d] & \Q(\sgsum{(p-1)/2})\ar@{.}[u]\\
\gpgen{\alpha  }           & \Q          \ar@{-}[u]_2
}
\end{equation*}
\caption{The Galois correspondence for
  $\Q(\exp(2\mpi\mi/(2^{2^m}+1)))$}\label{table:Gal-gen} 
\end{table}
As $\sgsum 2=\zeta+\zeta\inv$, so
$\sgsum 2\cdot\zeta^{\pm1}=\zeta^{\pm2}+1$,
and therefore $\zeta$ and $\zeta\inv$ are the solutions of
\begin{equation}\label{eqn:x^2-sgsum}
  x^2-\sgsum 2\cdot x+1=0.
\end{equation}
We have seen how to obtain the minimal polynomial of $\sgsum 2$ by
means of the substitution~\eqref{eqn:x+} in
$\sum_{k=0}^{p-1}x^k/x^{(p-1)/2}$.  When $p=5$, this is all we need
do to find the roots of $x^p-1$.  When $p=17$, more work is needed.
As it happens, this work is done in Hardy \&\ Wright
\cite[\S5.8, pp.~57--62]{MR568909}, though not explicitly
with the theoretical framework made possible by Gauss.

\section{Periods}

Let us still suppose $k\cdot\ell=p-1=2^{2^m}$, and now also $\ell<p-1$.
So $\sgsum{\ell}$ is quadratic over $\Q(\sgsum{2\ell})$.  Moreover,
the conjugate of $\sgsum{\ell}$ over this field is
$\sgsum{2\ell}-\sgsum{\ell}$.  This is a sum indexed by the coset of
$\gpgen{\alpha^k}$ in $\gpgen{\alpha^{k/2}}$.
In general, if $s\in\Zmu[p]$, let us define
\begin{equation*}
  \cssum{\ell}s=
  \sum_{x\in\gpgen{r^k}s}\zeta^x=
  \sum_{\xi\in\gpgen{\alpha^k}\sigma_s}\xi(\zeta).   
\end{equation*}
In Gauss's terminology \cite[\S VII, \P343]{MR0197380}, which Hardy
and Wright also use, this is a
\defn{period}.  By the Galois correspondence, we must have
\begin{equation*}
\Q(\cssum{\ell}s)=
  \Q(\sgsum{\ell}).
\end{equation*}
Also, $\cssum{\ell}1=\sgsum{\ell}$,
and the conjugate of $\sgsum{\ell}$ over $\Q(\sgsum{2\ell})$ is some
$\cssum{\ell}s$, where $\cssum{\ell}s+\sgsum{\ell}=\sgsum{2\ell}$.
Then the minimal polynomial of $\sgsum{\ell}$ over $\Q(\sgsum{2\ell})$
takes the form
\begin{equation*}
  x^2-\sgsum{2\ell}x+\sgsum{\ell}\cdot\cssum{\ell}s
\end{equation*}
for some $s$.  The minimal polynomials for other periods are similar.
We shall be able to understand the constant terms of these by means of
one of Gauss's observations.  
To avoid superscripts, let us denote $\zeta^x$ also by $\exp x$.  (One
might write 
$\exp_{\zeta}(x)$ to be precise.)  Then
\begin{equation*}
  \cssum{\ell}s=\sum_{j\in\Zm[\ell]}\exp(r^{jk}\cdot s).
\end{equation*}
This makes sense also when $s=0$, and then we have
\begin{equation*}
  \cssum{\ell}0=\ell.
\end{equation*}
Indeed, the definition of $\cssum{\ell}s$ makes sense whenever
$k\cdot\ell=\phi(n)$ for some $n$ of which $r$ is a primitive root.
The following is \cite[\S VII, \P345]{MR0197380}.

\begin{theorem}[Gauss]
Suppose $n$ has primitive root $r$, and
$k\cdot\ell=\phi(n)$.  Then
\begin{equation}\label{eqn:rnk}
\cssum{\ell}s\cdot\cssum{\ell}t=
  \displaystyle\sum_{x\in\gpgen{r^k}s}\cssum{\ell}{x+t}.
\end{equation}
\end{theorem}

\begin{proof}
Just compute:
\begin{align*}
  \cssum{\ell}s\cdot\cssum{\ell} t
&=\sum_{(i,j)\in(\Zm[\ell])^2}
  \exp(r^{ik}\cdot s+
  r^{jk}\cdot t)\\
&=\sum_{(i,j)\in(\Zm[\ell])^2}
  \exp(r^{jk}\cdot(r^{(i-j)k}\cdot s+t))\\
&=\sum_{(i,j)\in(\Zm[\ell])^2}
  \exp(r^{jk}\cdot(r^{ik}\cdot s+t))
\end{align*}
since $(i,j)\mapsto(i-j,j)$ is a permutation of $(\Zm[\ell])^2$.  Hence
\begin{equation*}
  \cssum{\ell}s\cdot\cssum{\ell} t
=\sum_{i\in\Zm[\ell]}\sum_{j\in\Zm[\ell]}
  \exp(r^{jk}\cdot(r^{ik}\cdot s+t))
=\sum_{i\in\Zm[\ell]}\cssum{\ell}{r^{ik}\cdot s+t},
\end{equation*}
which yields the claim.
\end{proof}

\section{The heptakaidecagon}

Henceforth let $n=p=17$.  We want to find the roots of~\eqref{eqn:17}.
These are the vertices
of a regular heptadecagon, or heptakaidecagon (Fig.~\ref{fig:17}; the
Greek for~17 is
\selectlanguage{polutonikogreek}<eptaka'ideka, 
\selectlanguage{english}\emph{seven and ten}).
\begin{figure}[t]
\begin{center}
  \begin{pspicture}(-5,-5)(5,5)
\psline{->}(-5,0)(5,0)
\psline{->}(0,-5)(0,5)
\psset{dotsize=3pt 2}
    \parametricplot[plotpoints=17,
plotstyle=dots]{0}{16}
{t 360 mul 17 div cos 4 mul t 360 mul 17 div sin 4 mul}
\uput[ur](4,0){$1$}
\uput[ul](0,4){$\mi$}
\uput[ul](-4,0){$-1$}
\uput[dl](0,-4){$-\mi$}
\psdots[dotstyle=o](0,4)(-4,0)(0,-4)
  \end{pspicture}
\caption{The seventeen 17th roots of 1}\label{fig:17}
\end{center}
\end{figure}
Let~$r$ be the primitive root $3$ of $17$.  The
powers of $3$ are in Table~\ref{table:3^}.
\begin{table}[t]
  \begin{equation*}
    \begin{array}{|r||*{8}{r|}|l|}\hline
      x&0&1&2&3&4&5&6&7&(\operatorname{mod}16)\\\hline\hline
3^x&1&3&-8&-7&-4&5&-2&-6&(\operatorname{mod}17)\\\hline
3^{8+x}&-1&-3&8&7&4&-5&2&6&(\operatorname{mod}17)\\\hline
    \end{array}
  \end{equation*}
\caption{Powers of 3 \emph{modulo} 17}\label{table:3^}
\end{table}
The Galois correspondence now is in Table~\ref{table:gal}.
\begin{table}[t]
\begin{equation*}
  \xymatrix{
\gpgen{1}       \ar@{-}[d] & \Q(\zeta)\\
\gpgen{\alpha^8}\ar@{-}[d] & \Q(\sgsum 2)\ar@{-}[u]_2\\
\gpgen{\alpha^4}\ar@{-}[d] & \Q(\sgsum 4)\ar@{-}[u]_2\\
\gpgen{\alpha^2}\ar@{-}[d] & \Q(\sgsum 8)\ar@{-}[u]_2\\
\gpgen{\alpha  }           & \Q          \ar@{-}[u]_2
}
\end{equation*}
\caption{The Galois correspondence for
  $\Q(\exp(2\mpi\mi/17))$}\label{table:gal} 
\end{table}
We aim to express generators of the subfields of $\Q(\zeta)$ by radicals.
Towards this end, we can arrange the powers of $3$ according to the
cosets of subgroups of 
$\Zm[16]$, as in Table~\ref{table:3^-r}.
\begin{table}[t]
  \begin{equation*}
    \begin{array}{|r|||r|r||r|r|||r|r||r|r|||r|}\hline
      x&0&4&2&6&1&5&3&7&(\operatorname{mod}16)\\\hline\hline
3^x&1&-4&-8&-2&3&5&-7&-6&(\operatorname{mod}17)\\\hline
3^{8+x}&-1&4&8&2&-3&-5&7&6&(\operatorname{mod}17)\\\hline
    \end{array}
  \end{equation*}
\caption{Powers of 3 again, in cosets}\label{table:3^-r}
\end{table}
Hence the distinct sums $\cssum{\ell}s$ are the nodes of the tree in
Table~\ref{table:tree}. 
\begin{sidewaystable}
  \begin{center}
\pstree[nodesep=5pt%,treemode=R
]{\TR{$\sgsum{16}$}}
       {\pstree{\TR{$\sgsum8$}}
               {\pstree{\TR{$\sgsum4$}}
                       {\pstree{\TR{$\sgsum2$}}
                               {\TR{$\zeta$}\TR{$\zeta\inv$}}
                        \pstree{\TR{$\cssum24$}}
                               {\TR{$\zeta^4$}\TR{$\zeta^{-4}$}}}
                \pstree{\TR{$\cssum42$}}
                       {\pstree{\TR{$\cssum28$}}
                               {\TR{$\zeta^8$}\TR{$\zeta^{-8}$}}
                        \pstree{\TR{$\cssum22$}}
                               {\TR{$\zeta^2$}\TR{$\zeta^{-2}$}}}}
        \pstree{\TR{$\cssum83$}}
               {\pstree{\TR{$\cssum43$}}
                       {\pstree{\TR{$\cssum23$}}
                               {\TR{$\zeta^3$}\TR{$\zeta^{-3}$}}
                        \pstree{\TR{$\cssum25$}}
                               {\TR{$\zeta^5$}\TR{$\zeta^{-5}$}}}
                \pstree{\TR{$\cssum46$}}
                       {\pstree{\TR{$\cssum27$}}
                               {\TR{$\zeta^7$}\TR{$\zeta^{-7}$}}
                        \pstree{\TR{$\cssum26$}}
                               {\TR{$\zeta^6$}\TR{$\zeta^{-6}$}}}}}
  \end{center}
\caption{Tree of periods}\label{table:tree}
\end{sidewaystable}
In applying Gauss's theorem, it will be useful to note that
 $\cssum2{-s}=\cssum2s$ and to have some other such equations, as in
Table~\ref{table:cosets}.  (In practice, it may be easier just to
 refer to Table~\ref{table:3^-r}.)
\begin{table}[t]
  \begin{equation*}
    \begin{array}{|*{9}{r|}|l|}\hline
 &8& 6&-7& 4&-3& 7& 3& 2&\log(-x)\pmod{16}                      \\\hline
 &0&-2& 1&-4& 5&-1&-5&-6&\log x\pmod{16}                        \\\hline\hline
0&1& 2& 3& 4& 5& 6& 7& 8&x\pmod{17}                             \\\hline
%0&1& 2& 3& 4& 5& 6& 7& 8&y,\text{ where }\cssum2y=\cssum2{\pm x}\\\hline
0&1& 2& 3& 1& 3& 6& 6& 2&y,\text{ where }\cssum4y=\cssum4{\pm x}\\\hline
0&1& 1& 3& 1& 3& 3& 3& 1&y,\text{ where }\cssum8y=\cssum8{\pm x}\\\hline
    \end{array}
  \end{equation*}
\caption{Which periods are which}\label{table:cosets}
\end{table}
We have already seen that $\zeta$ and $\zeta\inv$ are the solutions
of~\eqref{eqn:x^2-sgsum}. 
In particular,
\begin{equation}\label{eqn:r}
  \zeta^{\pm1}=\frac{\sgsum 2\pm\radixp{\sgsum 2{}^2-4}}2.
\end{equation}
So we want now to find $\sgsum2$ by radicals.
From Table~\ref{table:tree},
we have
\begin{equation*}
  \sgsum 2+\cssum 24=\sgsum4.
\end{equation*}
From~\eqref{eqn:rnk} and Tables~\ref{table:3^} and~\ref{table:tree},
we have 
\begin{equation*}
  \sgsum 2\cdot\cssum 24=\sum_{x\in\gpgen{3^8}}\cssum 2{x+4}=
\cssum25+\cssum23=\cssum43.
\end{equation*}
Therefore $\sgsum 2$ and $\cssum24$ are the solutions of
\begin{equation}\label{eqn:R2}
  x^2-\sgsum 4\cdot x+\cssum43=0.
\end{equation}
The two solutions are real, and
$\sgsum2$ is the greater (Fig.~\ref{fig:2}). 
\begin{figure}[t]
  \begin{center}
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 17 div 360 mul cos 2 mul 
 t 17 div 360 mul sin 2 mul}
    \parametricplot[plotpoints=2,
plotstyle=dots,dotsize=3pt 2]{0}{1}
{-1 t exp 17 div 360 mul cos 2 mul 
 -1 t exp 17 div 360 mul sin 2 mul}
\uput[d](0,-2.5){$\sgsum 2$}
  \end{pspicture}
\quad
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 360 mul 17 div cos 2 mul 
 t 360 mul 17 div sin 2 mul}
    \parametricplot[plotpoints=2,
plotstyle=dots,dotsize=3pt 2]{0}{1}
{-1 t exp 4 mul 360 mul 17 div cos 2 mul 
 -1 t exp 4 mul 360 mul 17 div sin 2 mul}
\uput[d](0,-2.5){$\cssum24$}
  \end{pspicture}
\end{center}
\caption{The order-2 subgroup and a coset}\label{fig:2}
\end{figure}
Hence
\begin{equation*}
  \sgsum2=\frac{\sgsum4+\radixp{\sgsum 4{}^2-4\cssum43}}2.
\end{equation*}
To find $\cssum43$, we have
\begin{equation*}
  \cssum43+\cssum46=\cssum83,
\end{equation*}
as well as
\begin{align*}
  \cssum43\cdot\cssum46
=\sum_{x\in\gpgen{3^4}3}\cssum4{x+6}
&=\cssum4{3+6}+\cssum4{-3+6}+\cssum4{5+6}+\cssum4{-5+6}\\
&=\cssum4{-8}+\cssum43+\cssum4{-6}+\cssum41\\
&=\cssum4{8}+\cssum43+\cssum4{6}+\cssum41=-1.
\end{align*}
Hence $\cssum43$ and $\cssum46$ are the solutions of 
\begin{equation}\label{eqn:R43}
  x^2-\cssum83\cdot x-1=0.
\end{equation}
The greater is $\cssum43$ (Fig.~\ref{fig:43}); so
\begin{figure}[t]
  \begin{center}
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 17 div 360 mul cos 2 mul 
 t 17 div 360 mul sin 2 mul}
    \parametricplot[plotpoints=4,
plotstyle=dots,dotsize=3pt 2]{0}{3}
{4 t exp 3 mul 17 div 360 mul cos 2 mul 
 4 t exp 3 mul 17 div 360 mul sin 2 mul}
\uput[d](0,-2.5){$\cssum43$}
  \end{pspicture}
\quad
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 360 mul 17 div cos 2 mul 
 t 360 mul 17 div sin 2 mul}
    \parametricplot[plotpoints=4,
plotstyle=dots,dotsize=3pt 2]{0}{3}
{4 t exp 6 mul 360 mul 17 div cos 2 mul 
 4 t exp 6 mul 360 mul 17 div sin 2 mul}
\uput[d](0,-2.5){$\cssum46$}
  \end{pspicture}
\end{center}
\caption{Two order-4 cosets}\label{fig:43}
\end{figure}
\begin{equation*}
  \cssum43=\frac{\cssum83+\radixp{\cssum83{}^2+4}}2.
\end{equation*}
Similarly,
\begin{align*}
  \sgsum4+\cssum42&=\sgsum8,\\
\sgsum4\cdot\cssum42
&=
\begin{aligned}[t]
\sum_{x\in\gpgen{3^4}}\cssum4{x+2}
&=\cssum43+\cssum41+\cssum46+\cssum4{-2}\\
&=\cssum43+\cssum41+\cssum46+\cssum42=-1,
\end{aligned}
  \end{align*}
so $\sgsum4$ and $\cssum42$ are the solutions of
\begin{equation}\label{eqn:R4}
  x^2-\sgsum8\cdot x-1=0,
\end{equation}
and $\sgsum4$ is the greater (Fig.~\ref{fig:4}), so
\begin{figure}[t]
  \begin{center}
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 17 div 360 mul cos 2 mul 
 t 17 div 360 mul sin 2 mul}
    \parametricplot[plotpoints=4,
plotstyle=dots,dotsize=3pt 2]{0}{3}
{4 t exp 17 div 360 mul cos 2 mul 
 4 t exp 17 div 360 mul sin 2 mul}
\uput[d](0,-2.5){$\sgsum4$}
  \end{pspicture}
\quad
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 360 mul 17 div cos 2 mul 
 t 360 mul 17 div sin 2 mul}
    \parametricplot[plotpoints=4,
plotstyle=dots,dotsize=3pt 2]{0}{3}
{4 t exp 2 mul 360 mul 17 div cos 2 mul 
 4 t exp 2 mul 360 mul 17 div sin 2 mul}
\uput[d](0,-2.5){$\cssum42$}
  \end{pspicture}
\end{center}
\caption{The order-4 subgroup and a coset}\label{fig:4}
\end{figure}
\begin{equation*}
  \sgsum4=\frac{\sgsum8+\radixp{\sgsum8{}^2+4}}2.
\end{equation*}
It remains to find $\sgsum8$ and $\cssum83$.  We have
\begin{equation*}
  \sgsum8+\cssum83=-1,
\end{equation*}
and also
\begin{multline*}
  \sgsum8\cdot\cssum83
=\sum_{x\in\gpgen{3^2}}\cssum8{x+3}\\
=\cssum84+\cssum82+\cssum8{-1}+\cssum87
  +\cssum8{-5}+\cssum8{-6}+\cssum81+\cssum85\\
=\sgsum8+\sgsum8+\sgsum8+\cssum83
  +\cssum83+\cssum83+\sgsum8+\cssum83\\
=4\cdot(\sgsum8+\cssum83)=-4. 
\end{multline*}
Hence $\sgsum8$ and $\cssum83$ are the solutions of
\begin{equation}\label{eqn:R8}
  x^2+x-4=0;
\end{equation}
as $\sgsum8$ is the greater (Fig.~\ref{fig:8}), we have
\begin{figure}[t]
  \begin{center}
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 17 div 360 mul cos 2 mul 
 t 17 div 360 mul sin 2 mul}
    \parametricplot[plotpoints=8,
plotstyle=dots,dotsize=3pt 2]{0}{7}
{2 t exp 17 div 360 mul cos 2 mul 
 2 t exp 17 div 360 mul sin 2 mul}
\uput[d](0,-2.5){$\sgsum8$}
  \end{pspicture}
\quad
  \begin{pspicture}(-2.5,-3)(2.5,2.5)
\psline{->}(-2.5,0)(2.5,0)
\psline{->}(0,-2.5)(0,2.5)
    \parametricplot[plotpoints=17,
plotstyle=dots,dotsize=3pt 2,dotstyle=o]{0}{16}
{t 360 mul 17 div cos 2 mul 
 t 360 mul 17 div sin 2 mul}
    \parametricplot[plotpoints=8,
plotstyle=dots,dotsize=3pt 2]{0}{7}
{2 t exp 3 mul 360 mul 17 div cos 2 mul 
 2 t exp 3 mul 360 mul 17 div sin 2 mul}
\uput[d](0,-2.5){$\cssum83$}
  \end{pspicture}
\end{center}
\caption{The order-8 subgroup and its coset}\label{fig:8}
\end{figure}
  \begin{align*}
    \sgsum8
&=\frac{-1+\rtxvii}2,&
\cssum83
&=\frac{-1-\rtxvii}2.
  \end{align*}
In sum, we have the equations in Table~\ref{table:sum}.
Now we can in principle work backwards to obtain $\zeta$ explicitly by
radicals. 
\begin{table}[t]
\renewcommand{\arraystretch}{2.2}
\begin{equation*}
\begin{array}{|c|c|c|c|}\hline
x&x&f(x)=0&\text{source}\\\hline\hline
  \zeta&\displaystyle\frac{\sgsum 2+\radixp{\sgsum
  2{}^2-4}}2&x^2-\sgsum2\cdot x+1=0&\text{\eqref{eqn:x^2-sgsum}}\\\hline
  \sgsum2&\displaystyle\frac{\sgsum4+\radixp{\sgsum
  4{}^2-4\cssum43}}2& x^2-\sgsum4\cdot
  x+\cssum43=0&\text{\eqref{eqn:R2}}\\\hline   
  \cssum43&\displaystyle\frac{\cssum83+\radixp{\cssum83{}^2+4}}2&
  x^2-\cssum83\cdot x-1=0&\text{\eqref{eqn:R43}}\\\hline
  \sgsum4&\displaystyle\frac{\sgsum8+\radixp{\sgsum8{}^2+4}}2&
  x^2-\sgsum8\cdot x-1=0&\text{\eqref{eqn:R4}}\\\hline 
\cssum83&\displaystyle\frac{-1-\rtxvii}2&
  x^2+x-4=0&\text{\eqref{eqn:R8}}\\\hline 
\sgsum8&\displaystyle\frac{-1+\rtxvii}2&
  x^2+x-4=0&\text{\eqref{eqn:R8}}\\\hline 
\end{array}
\end{equation*}
\caption{The equations for $\exp(2\mpi\mi/17)$}\label{table:sum}
\end{table}

\section{Clean-up}

We may wish to avoid having to use $\cssum43$ and $\cssum83$ in
finding $\zeta$.  To this end, we observe
\begin{align*}
  \sgsum4\cdot\cssum43
=\sum_{x\in\gpgen{3^4}}\cssum4{x+3}
&=\cssum44+\cssum42+\cssum4{-1}+\cssum47\\
&=\sgsum4+\cssum42+\sgsum4+\cssum46;
\end{align*}
but since also $\sgsum4+\cssum42+\cssum43+\cssum46=-1$,
we have
\begin{equation*}
\sgsum4\cdot\cssum43=\sgsum4-\cssum43-1,
\qquad  \cssum43=\frac{\sgsum4-1}{\sgsum4+1}.
\end{equation*}
Also,
\begin{equation*}
  \sgsum4{}^2=\sgsum8\cdot\sgsum4+1,
\qquad
\sgsum8{}^2=4-\sgsum8.
\end{equation*}
Hence, towards an alternative expression for $\sgsum2$, we have
\begin{align*}
\sgsum4{}^2-4\cssum43
&=\sgsum8\cdot\sgsum4+1-4\cdot\frac{\sgsum4-1}{\sgsum4+1}\\
&=\frac{\sgsum8\cdot\sgsum4{}^2+\sgsum8\cdot\sgsum4+\sgsum4+1-4\sgsum4+4}
       {\sgsum4+1} \\
&=\frac{\sgsum8{}^2\cdot\sgsum4+\sgsum8+
	 \sgsum8\cdot\sgsum4-3\sgsum4+5}
       {\sgsum4+1}\\
&=\frac{4\sgsum4-\sgsum8\cdot\sgsum4+\sgsum8+
	 \sgsum8\cdot\sgsum4-3\sgsum4+5}
       {\sgsum4+1} 
=\frac{\sgsum4+\sgsum8+5}
       {\sgsum4+1}.
\end{align*}
Likewise, for $\zeta$ itself, we have
  \begin{align*}
    \sgsum2{}^2-4
=\sgsum4\cdot\sgsum2-\frac{\sgsum4-1}{\sgsum4+1}-4
&=\frac{\sgsum4{}^2\cdot\sgsum2+\sgsum4\cdot\sgsum2-\sgsum4+1-4\sgsum4-4}
	   {\sgsum4+1}\\
&=\frac{\sgsum8\cdot\sgsum4\cdot\sgsum2+\sgsum4\cdot\sgsum2+\sgsum2-5\sgsum4-3}
	   {\sgsum4+1}. 
  \end{align*}
The situation is now as in Table~\ref{table:equations-again}.
\begin{table}[t]
\renewcommand{\arraystretch}{2.5}
\begin{equation*}
\begin{array}{|c|c|c|}\hline
x&x&f(x)=0\\\hline\hline
  \zeta&\displaystyle\frac12\Biggl(\sgsum
  2+\radix[\frac11]
  {\displaystyle\frac{\sgsum8\cdot\sgsum4\cdot\sgsum2+
  \sgsum4\cdot\sgsum2+\sgsum2-5\sgsum4-3}
	   {\sgsum4+1}}\Biggr)&
x^2-\sgsum2\cdot
  x+1=0\\\hline 
  \sgsum2&\displaystyle\frac12\Biggl(\sgsum4+
  \radix[\frac11]{\displaystyle\frac{\sgsum4+\sgsum8+5}{\sgsum4+1}}\Biggr)& 
  x^2-\sgsum4\cdot x+\displaystyle\frac{\sgsum4-1}{\sgsum4+1}=0\\\hline    
  \sgsum4&\displaystyle\frac{\sgsum8+\radixp{8-\sgsum8}}2&
  x^2-\sgsum8\cdot x-1=0\\\hline 
\sgsum8&\displaystyle\frac{-1+\rtxvii}2&
  x^2+x-4=0\\\hline 
\end{array}
\end{equation*}
\caption{The equations for $\exp(2\mpi\mi/17)$,
  again}\label{table:equations-again} 
\end{table}

%\listoftables
%\listoffigures

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\bibliography{../../../TeX/references}

\end{document}

