Gauss' Java Trip
A Mathematical Travel Report

The 65537-gon

On this page I will not  list all roots of the 65537-gon, rendered by LaTeX, like I did with the pentagon, the 17-gon and the 257-gon. I will not even present the complete output of the programs Hermes and Kompressor in raw format. However, both can be found on the -pages, the first as PDF of about 350 pages and the second as about 23400 lines of ASCII text. On this page I give only exerpts from the said documents and some annotations.

Let's start with the raw data printed by the program Hermes as LaTeX code. They start: 

% Running with arguments: 65537 -tex
\[p_{0,0} = -1,000000000000\]
\[p_{1,0} = \tfrac{p_{0,0} + \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2} = \frac{-1+\sqrt{65537}}{2}\]
\[p_{1,1} = \tfrac{p_{0,0} - \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2} = \frac{-1-\sqrt{65537}}{2}\]
\[p_{2,0} = \tfrac{p_{1,0} - \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\]
\[p_{2,2} = \tfrac{p_{1,0} + \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\]
\[p_{2,1} = \tfrac{p_{1,1} - \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\]
\[p_{2,3} = \tfrac{p_{1,1} + \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\]
\[p_{3,0} = \tfrac{p_{2,0} - \sqrt{p_{2,0}^2 - 4(992p_{2,0}+1024p_{2,2}+1040p_{1,1})}}{2}\]
\[p_{3,4} = \tfrac{p_{2,0} + \sqrt{p_{2,0}^2 - 4(992p_{2,0}+1024p_{2,2}+1040p_{1,1})}}{2}\]
\[p_{3,2} = \tfrac{p_{2,2} + \sqrt{p_{2,2}^2 - 4(1024p_{2,0}+992p_{2,2}+1040p_{1,1})}}{2}\]
\[p_{3,6} = \tfrac{p_{2,2} - \sqrt{p_{2,2}^2 - 4(1024p_{2,0}+992p_{2,2}+1040p_{1,1})}}{2}\]
\[p_{3,1} = \tfrac{p_{2,1} + \sqrt{p_{2,1}^2 - 4(1040p_{1,0}+992p_{2,1}+1024p_{2,3})}}{2}\]
\[p_{3,5} = \tfrac{p_{2,1} - \sqrt{p_{2,1}^2 - 4(1040p_{1,0}+992p_{2,1}+1024p_{2,3})}}{2}\]
\[p_{3,3} = \tfrac{p_{2,3} - \sqrt{p_{2,3}^2 - 4(1040p_{1,0}+1024p_{2,1}+992p_{2,3})}}{2}\]
\[p_{3,7} = \tfrac{p_{2,3} + \sqrt{p_{2,3}^2 - 4(1040p_{1,0}+1024p_{2,1}+992p_{2,3})}}{2}\]

{\footnotesize
\[p_{4,0} = \frac{1}{2}p_{3,0} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,0}^2 - 4(284p_{3,0}+256p_{3,4}+272p_{3,2} \\
&+256p_{3,6}+237p_{3,1}+269p_{3,5}+237p_{2,3})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,8} = \frac{1}{2}p_{3,0} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,0}^2 - 4(284p_{3,0}+256p_{3,4}+272p_{3,2} \\
&+256p_{3,6}+237p_{3,1}+269p_{3,5}+237p_{2,3})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,4} = \frac{1}{2}p_{3,4} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,4}^2 - 4(256p_{3,0}+284p_{3,4}+256p_{3,2} \\
&+272p_{3,6}+269p_{3,1}+237p_{3,5}+237p_{2,3})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,12} = \frac{1}{2}p_{3,4} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,4}^2 - 4(256p_{3,0}+284p_{3,4}+256p_{3,2} \\
&+272p_{3,6}+269p_{3,1}+237p_{3,5}+237p_{2,3})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,2} = \frac{1}{2}p_{3,2} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,2}^2 - 4(256p_{3,0}+272p_{3,4}+284p_{3,2} \\
&+256p_{3,6}+237p_{2,1}+237p_{3,3}+269p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,10} = \frac{1}{2}p_{3,2} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,2}^2 - 4(256p_{3,0}+272p_{3,4}+284p_{3,2} \\
&+256p_{3,6}+237p_{2,1}+237p_{3,3}+269p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,6} = \frac{1}{2}p_{3,6} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,6}^2 - 4(272p_{3,0}+256p_{3,4}+256p_{3,2} \\
&+284p_{3,6}+237p_{2,1}+269p_{3,3}+237p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,14} = \frac{1}{2}p_{3,6} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,6}^2 - 4(272p_{3,0}+256p_{3,4}+256p_{3,2} \\
&+284p_{3,6}+237p_{2,1}+269p_{3,3}+237p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,1} = \frac{1}{2}p_{3,1} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,1}^2 - 4(237p_{2,0}+237p_{3,2}+269p_{3,6} \\
&+284p_{3,1}+256p_{3,5}+272p_{3,3}+256p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,9} = \frac{1}{2}p_{3,1} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,1}^2 - 4(237p_{2,0}+237p_{3,2}+269p_{3,6} \\
&+284p_{3,1}+256p_{3,5}+272p_{3,3}+256p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,5} = \frac{1}{2}p_{3,5} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,5}^2 - 4(237p_{2,0}+269p_{3,2}+237p_{3,6} \\
&+256p_{3,1}+284p_{3,5}+256p_{3,3}+272p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,13} = \frac{1}{2}p_{3,5} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,5}^2 - 4(237p_{2,0}+269p_{3,2}+237p_{3,6} \\
&+256p_{3,1}+284p_{3,5}+256p_{3,3}+272p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,3} = \frac{1}{2}p_{3,3} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,3}^2 - 4(269p_{3,0}+237p_{3,4}+237p_{2,2} \\
&+256p_{3,1}+272p_{3,5}+284p_{3,3}+256p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,11} = \frac{1}{2}p_{3,3} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,3}^2 - 4(269p_{3,0}+237p_{3,4}+237p_{2,2} \\
&+256p_{3,1}+272p_{3,5}+284p_{3,3}+256p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,7} = \frac{1}{2}p_{3,7} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,7}^2 - 4(237p_{3,0}+269p_{3,4}+237p_{2,2} \\
&+272p_{3,1}+256p_{3,5}+256p_{3,3}+284p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{4,15} = \frac{1}{2}p_{3,7} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{3,7}^2 - 4(237p_{3,0}+269p_{3,4}+237p_{2,2} \\
&+272p_{3,1}+256p_{3,5}+256p_{3,3}+284p_{3,7})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,0} = \frac{1}{2}p_{4,0} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,0}^2 - 4(80p_{4,0}+68p_{4,8}+57p_{4,4}+65p_{4,12} \\
&+60p_{4,2}+64p_{4,10}+61p_{3,6}+62p_{4,1}+64p_{4,9}+60p_{4,5} \\
&+70p_{4,13}+64p_{4,3}+58p_{4,11}+60p_{4,7}+70p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,16} = \frac{1}{2}p_{4,0} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,0}^2 - 4(80p_{4,0}+68p_{4,8}+57p_{4,4}+65p_{4,12} \\
&+60p_{4,2}+64p_{4,10}+61p_{3,6}+62p_{4,1}+64p_{4,9}+60p_{4,5} \\
&+70p_{4,13}+64p_{4,3}+58p_{4,11}+60p_{4,7}+70p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,8} = \frac{1}{2}p_{4,8} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,8}^2 - 4(68p_{4,0}+80p_{4,8}+65p_{4,4}+57p_{4,12} \\
&+64p_{4,2}+60p_{4,10}+61p_{3,6}+64p_{4,1}+62p_{4,9}+70p_{4,5} \\
&+60p_{4,13}+58p_{4,3}+64p_{4,11}+70p_{4,7}+60p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,24} = \frac{1}{2}p_{4,8} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,8}^2 - 4(68p_{4,0}+80p_{4,8}+65p_{4,4}+57p_{4,12} \\
&+64p_{4,2}+60p_{4,10}+61p_{3,6}+64p_{4,1}+62p_{4,9}+70p_{4,5} \\
&+60p_{4,13}+58p_{4,3}+64p_{4,11}+70p_{4,7}+60p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,4} = \frac{1}{2}p_{4,4} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,4}^2 - 4(65p_{4,0}+57p_{4,8}+80p_{4,4}+68p_{4,12} \\
&+61p_{3,2}+60p_{4,6}+64p_{4,14}+70p_{4,1}+60p_{4,9}+62p_{4,5} \\
&+64p_{4,13}+70p_{4,3}+60p_{4,11}+64p_{4,7}+58p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,20} = \frac{1}{2}p_{4,4} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,4}^2 - 4(65p_{4,0}+57p_{4,8}+80p_{4,4}+68p_{4,12} \\
&+61p_{3,2}+60p_{4,6}+64p_{4,14}+70p_{4,1}+60p_{4,9}+62p_{4,5} \\
&+64p_{4,13}+70p_{4,3}+60p_{4,11}+64p_{4,7}+58p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,12} = \frac{1}{2}p_{4,12} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,12}^2 - 4(57p_{4,0}+65p_{4,8}+68p_{4,4}+80p_{4,12} \\
&+61p_{3,2}+64p_{4,6}+60p_{4,14}+60p_{4,1}+70p_{4,9}+64p_{4,5} \\
&+62p_{4,13}+60p_{4,3}+70p_{4,11}+58p_{4,7}+64p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,28} = \frac{1}{2}p_{4,12} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,12}^2 - 4(57p_{4,0}+65p_{4,8}+68p_{4,4}+80p_{4,12} \\
&+61p_{3,2}+64p_{4,6}+60p_{4,14}+60p_{4,1}+70p_{4,9}+64p_{4,5} \\
&+62p_{4,13}+60p_{4,3}+70p_{4,11}+58p_{4,7}+64p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,2} = \frac{1}{2}p_{4,2} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,2}^2 - 4(61p_{3,0}+60p_{4,4}+64p_{4,12}+80p_{4,2} \\
&+68p_{4,10}+57p_{4,6}+65p_{4,14}+70p_{4,1}+60p_{4,9} \\
&+64p_{4,5}+58p_{4,13}+62p_{4,3}+64p_{4,11}+60p_{4,7} \\
&+70p_{4,15})
\end{aligned}
}\]
}%footnotesize
{\footnotesize
\[p_{5,18} = \frac{1}{2}p_{4,2} - \frac{1}{2}\sqrt{
\begin{aligned}
& p_{4,2}^2 - 4(61p_{3,0}+60p_{4,4}+64p_{4,12}+80p_{4,2} \\
&+68p_{4,10}+57p_{4,6}+65p_{4,14}+70p_{4,1}+60p_{4,9} \\
&+64p_{4,5}+58p_{4,13}+62p_{4,3}+64p_{4,11}+60p_{4,7} \\
&+70p_{4,15})
\end{aligned}
}\]
}%footnotesize

and they end with: 

\[p_{13,0} = \tfrac{p_{12,0} + \sqrt{p_{12,0}^2 - 4(p_{12,1}+p_{12,1265}+p_{12,4003}+p_{12,1899})}}{2}\]
\centerline{\footnotesize 5 unreferenced roots were skipped}
{\footnotesize
\[p_{13,3072} = \frac{1}{2}p_{12,3072} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{12,3072}^2 - 4(p_{12,3073}+p_{12,241}+p_{12,2979}+p_{12,875})
\end{aligned}
}\]
}%footnotesize
\centerline{\footnotesize 1203 unreferenced roots were skipped}
{\footnotesize
\[p_{13,2980} = \frac{1}{2}p_{12,2980} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{12,2980}^2 - 4(p_{12,2981}+p_{12,149}+p_{12,2887}+p_{12,783})
\end{aligned}
}\]
}%footnotesize
\centerline{\footnotesize 4 unreferenced roots were skipped}
{\footnotesize
\[p_{13,8100} = \frac{1}{2}p_{12,4004} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{12,4004}^2 - 4(p_{12,4005}+p_{12,1173}+p_{12,3911}+p_{12,1807})
\end{aligned}
}\]
}%footnotesize
\centerline{\footnotesize 643 unreferenced roots were skipped}
{\footnotesize
\[p_{13,6236} = \frac{1}{2}p_{12,2140} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{12,2140}^2 - 4(p_{12,3405}+p_{12,2141}+p_{12,4039}+p_{12,2047})
\end{aligned}
}\]
}%footnotesize
\centerline{\footnotesize 2 unreferenced roots were skipped}
{\footnotesize
\[p_{13,3164} = \frac{1}{2}p_{12,3164} + \frac{1}{2}\sqrt{
\begin{aligned}
& p_{12,3164}^2 - 4(p_{12,333}+p_{12,3165}+p_{12,967}+p_{12,3071})
\end{aligned}
}\]
}%footnotesize
\centerline{\footnotesize 39 unreferenced roots were skipped}
\[p_{14,0} = \tfrac{p_{13,0} + \sqrt{p_{13,0}^2 - 4(p_{13,8100}+p_{13,3164})}}{2}\]
\centerline{\footnotesize 11 unreferenced roots were skipped}
\[p_{14,3072} = \tfrac{p_{13,3072} + \sqrt{p_{13,3072}^2 - 4(p_{13,2980}+p_{13,6236})}}{2}\]
\centerline{\footnotesize 5 unreferenced roots were skipped}
\[p_{15,0} = \tfrac{p_{14,0} - \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\]
\begin{verbatim}
% 1/2 * p_{15,0} = NaN
% cos(2*pi/65537): 0.9999999954042476
% Taking reference values!
\end{verbatim}
\centerline{\footnotesize 1 unreferenced roots were skipped}
\[p_{15,0} = \tfrac{p_{14,0} + \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\]
\begin{verbatim}
% 1/2 * p_{15,0} = 0.9999999954039087
% Time used: 33.214807469sec
Used: 2105; Skipped: 8009; Roots: 1365686447
\end{verbatim}

The LaTeX output of Hermes has been compressed by the program Kompressor, which also performed a line-wrapping. The compression took place by discarding the square roots, which are not used by roots of a higher level. The omissions are marked by skipped....

However, the program Hermes itself does not compute all roots. Starting at level 13 a lot of roots are omitted, if they are not used by higher levels. So level 13 stops at $p_{13,3164}$ (instead of $p_{13,8191}$) and level 14 stops at $p_{14,3072}$ (instead of $p_{14,16382}$). In level 15 the value of $\cos(2\pi/65537)$ can immediately be computed on basis of $p_{15,0},$ the 32766 additional roots of this level can safely be skipped. The limits had been identified during the first test runs of the program, and then hard-wired into the program code, to reduce run-time. It is checked by the program Kompressor, whether there are roots missing, that are needed for computations in higher levels. Obviously, this is not the case.

About 2100 square root expressions remained after this procedure. That's almost identical with the about 2000 roots, published by [Brakke 2011]. Why there is a dicrepancy at all, is discussed in chap.11 under the given link.

The line-wrapping by Kompressor is performed using the LaTeX statement aligned, which splits a long line under the root sign into several shorter lines. In addition a smaller font has been applied for the long expressions (\footnotesize).

The output of the program Kompressor has been rendered by LaTeX. Only a few lines of the output have been edited by hand: the simplification of the roots $p_{1,0}$ and $p_{1,1},$ and the comment lines with arguments and CPU-time. However, because of the amount of data, errors may have slipped through undetected. If you find one, don't hesitate to drop me a note.

It follows a once again shortened extract of the output, renderd by LaTeX. Because MathJax does not work outside of math environments the \footnotesize statements had to be discarded:

\[p_{0,0} = -1,000000000000\] \[p_{1,0} = \tfrac{p_{0,0} + \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2} = \frac{-1+\sqrt{65537}}{2}\] \[p_{1,1} = \tfrac{p_{0,0} - \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2} = \frac{-1-\sqrt{65537}}{2}\] \[p_{2,0} = \tfrac{p_{1,0} - \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\] \[p_{2,2} = \tfrac{p_{1,0} + \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\] \[p_{2,1} = \tfrac{p_{1,1} - \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\] \[p_{2,3} = \tfrac{p_{1,1} + \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\] \[p_{3,0} = \tfrac{p_{2,0} - \sqrt{p_{2,0}^2 - 4(992p_{2,0}+1024p_{2,2}+1040p_{1,1})}}{2}\] \[p_{3,4} = \tfrac{p_{2,0} + \sqrt{p_{2,0}^2 - 4(992p_{2,0}+1024p_{2,2}+1040p_{1,1})}}{2}\] \[p_{3,2} = \tfrac{p_{2,2} + \sqrt{p_{2,2}^2 - 4(1024p_{2,0}+992p_{2,2}+1040p_{1,1})}}{2}\] \[p_{3,6} = \tfrac{p_{2,2} - \sqrt{p_{2,2}^2 - 4(1024p_{2,0}+992p_{2,2}+1040p_{1,1})}}{2}\] \[p_{3,1} = \tfrac{p_{2,1} + \sqrt{p_{2,1}^2 - 4(1040p_{1,0}+992p_{2,1}+1024p_{2,3})}}{2}\] \[p_{3,5} = \tfrac{p_{2,1} - \sqrt{p_{2,1}^2 - 4(1040p_{1,0}+992p_{2,1}+1024p_{2,3})}}{2}\] \[p_{3,3} = \tfrac{p_{2,3} - \sqrt{p_{2,3}^2 - 4(1040p_{1,0}+1024p_{2,1}+992p_{2,3})}}{2}\] \[p_{3,7} = \tfrac{p_{2,3} + \sqrt{p_{2,3}^2 - 4(1040p_{1,0}+1024p_{2,1}+992p_{2,3})}}{2}\] \[p_{4,0} = \frac{1}{2}p_{3,0} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{3,0}^2 - 4(284p_{3,0}+256p_{3,4}+272p_{3,2} \\ &+256p_{3,6}+237p_{3,1}+269p_{3,5}+237p_{2,3}) \end{aligned} }\] \[p_{4,8} = \frac{1}{2}p_{3,0} - \frac{1}{2}\sqrt{ \begin{aligned} & p_{3,0}^2 - 4(284p_{3,0}+256p_{3,4}+272p_{3,2} \\ &+256p_{3,6}+237p_{3,1}+269p_{3,5}+237p_{2,3}) \end{aligned} }\] \[p_{4,4} = \frac{1}{2}p_{3,4} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{3,4}^2 - 4(256p_{3,0}+284p_{3,4}+256p_{3,2} \\ &+272p_{3,6}+269p_{3,1}+237p_{3,5}+237p_{2,3}) \end{aligned} }\] \[p_{4,12} = \frac{1}{2}p_{3,4} - \frac{1}{2}\sqrt{ \begin{aligned} & p_{3,4}^2 - 4(256p_{3,0}+284p_{3,4}+256p_{3,2} \\ &+272p_{3,6}+269p_{3,1}+237p_{3,5}+237p_{2,3}) \end{aligned} }\]

At the end of the output you find the following roots:

\[p_{13,2980} = \frac{1}{2}p_{12,2980} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{12,2980}^2 - 4(p_{12,2981}+p_{12,149}+p_{12,2887}+p_{12,783}) \end{aligned} }\] 
 4 unreferenced roots were skipped
\[p_{13,8100} = \frac{1}{2}p_{12,4004} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{12,4004}^2 - 4(p_{12,4005}+p_{12,1173}+p_{12,3911}+p_{12,1807}) \end{aligned} }\] 
 643 unreferenced roots were skipped
\[p_{13,6236} = \frac{1}{2}p_{12,2140} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{12,2140}^2 - 4(p_{12,3405}+p_{12,2141}+p_{12,4039}+p_{12,2047}) \end{aligned} }\] 
 2 unreferenced roots were skipped
\[p_{13,3164} = \frac{1}{2}p_{12,3164} + \frac{1}{2}\sqrt{ \begin{aligned} & p_{12,3164}^2 - 4(p_{12,333}+p_{12,3165}+p_{12,967}+p_{12,3071}) \end{aligned} }\] 
 39 unreferenced roots were skipped
\[p_{14,0} = \tfrac{p_{13,0} + \sqrt{p_{13,0}^2 - 4(p_{13,8100}+p_{13,3164})}}{2}\] 
 11 unreferenced roots were skipped
\[p_{14,3072} = \tfrac{p_{13,3072} + \sqrt{p_{13,3072}^2 - 4(p_{13,2980}+p_{13,6236})}}{2}\] 
 5 unreferenced roots were skipped
\[p_{15,0} = \tfrac{p_{14,0} - \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\] 
% 1/2 * p_{15,0} = NaN
% cos(2*pi/65537): 0.9999999954042476
% Taking reference values!
1 unreferenced roots were skipped
\[p_{15,0} = \tfrac{p_{14,0} + \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\] 
	% 1/2 * p_{15,0} = 0.9999999954039087
% Time used: 33.214807469sec
Used: 2105; Skipped: 8009; Roots: 1365686447

Problems with the Computation

The occurrence of the value NaN of $p_{15,0}$ deserves some explanation. For that purpose, have a look at the (raw) output of the program Hermes, when started with the arguments -tex -ref. It starts 

% Running with arguments: 65537 -tex -ref
\[p_{0,0} = -1,000000000000\]
p_{0,0} = -1,000000000000
 Ref = -1,000000000002 Diff: 1,977529E-12
\[p_{1,0} = \tfrac{p_{0,0} + \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2}\]
p_{1,0} = 127,500976558775
 Ref = 127,500976558774 Diff: 1,094236E-12
\[p_{1,1} = \tfrac{p_{0,0} - \sqrt{p_{0,0}^2 - 4(16384p_{0,0})}}{2}\]
p_{1,1} = -128,500976558775
 Ref = -128,500976558776 Diff: 8,242296E-13
\[p_{2,0} = \tfrac{p_{1,0} - \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\]
p_{2,0} = -26,582920578356
 Ref = -26,582920578357 Diff: 8,846257E-13
\[p_{2,2} = \tfrac{p_{1,0} + \sqrt{p_{1,0}^2 - 4(4096p_{0,0})}}{2}\]
p_{2,2} = 154,083897137131
 Ref = 154,083897137130 Diff: 7,673862E-13
\[p_{2,1} = \tfrac{p_{1,1} - \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\]
p_{2,1} = -154,937451202068
 Ref = -154,937451202068 Diff: 1,421085E-13
\[p_{2,3} = \tfrac{p_{1,1} + \sqrt{p_{1,1}^2 - 4(4096p_{0,0})}}{2}\]
p_{2,3} = 26,436474643293
 Ref = 26,436474643293 Diff: 8,917311E-13

and it ends with 

\[p_{14,0} = \tfrac{p_{13,0} + \sqrt{p_{13,0}^2 - 4(p_{13,8100}+p_{13,3164})}}{2}\]
p_{14,0} = 3,999291821205
 Ref = 3,999397646582 Diff: 1,058254E-04
\[p_{14,8192} = \tfrac{p_{13,0} - \sqrt{p_{13,0}^2 - 4(p_{13,8100}+p_{13,3164})}}{2}\]
p_{14,8192} = 3,847868109895

....

p_{14,3072} = 3,998797152204
 Ref = 3,998795293170 Diff: 1,859034E-06
\[p_{14,11264} = \tfrac{p_{13,3072} - \sqrt{p_{13,3072}^2 - 4(p_{13,2980}+p_{13,6236})}}{2}\]
p_{14,11264} = 3,695520656103
 Ref = 3,695522954334 Diff: 2,298230E-06
\[p_{14,7168} = \tfrac{p_{13,7168} - \sqrt{p_{13,7168}^2 - 4(p_{13,7076}+p_{13,2140})}}{2}\]
p_{14,7168} = 0,000095869075
 Ref = 0,000095870532 Diff: 1,456434E-09
\[p_{14,15360} = \tfrac{p_{13,7168} + \sqrt{p_{13,7168}^2 - 4(p_{13,7076}+p_{13,2140})}}{2}\]
p_{14,15360} = 3,980739581638
 Ref = 3,980739201293 Diff: 3,803448E-07
\[p_{14,512} = \tfrac{p_{13,512} - \sqrt{p_{13,512}^2 - 4(p_{13,420}+p_{13,3676})}}{2}\]
p_{14,512} = -0,859198008421
 Ref = -0,858636561910 Diff: 5,614465E-04
\[p_{14,8704} = \tfrac{p_{13,512} + \sqrt{p_{13,512}^2 - 4(p_{13,420}+p_{13,3676})}}{2}\]
p_{14,8704} = -0,801971489347
 Ref = -0,802476068394 Diff: 5,045790E-04
\[p_{15,0} = \tfrac{p_{14,0} - \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\]
p_{15,0} = NaN
 Ref = 1,999999990808 Diff: NAN
\[p_{15,16384} = \tfrac{p_{14,0} + \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\]
p_{15,16384} = NaN
 Ref = 1,999397655774 Diff: NAN
% 1/2 * p_{15,0} = NaN
% cos(2*pi/65537): 0.9999999954042476
% Taking reference values!
\[p_{15,0} = \tfrac{p_{14,0} + \sqrt{p_{14,0}^2 - 4(p_{14,3072})}}{2}\]
p_{15,0} = 1,999999990808
 Ref = 1,999999990808 Diff: 6,776801E-13
% 1/2 * p_{15,0} = 0.9999999954039087
% Time used: 33.214807469sec

It turns out that run-time and memory requirements don't cause any problems, but so do rounding errors. You see that the difference Diff between the square root expression and the refernce value (Ref =), computed from the $cos,$ is of magnitude $10^{-13}$ at the beginning, but approaches $10^{-4}$ at the end of the computations. That leads to the effect, that just the very last periods, which should have the value of $2\cos(2\pi/65537),$ contain a square root of a negative value, and thus yield NaN. Computing the value of $p_{14,0}^2 - 4p_{14,3072}$ from the square root expressions, one gets \[p_{14,0}^2 - 4p_{14,3072} = 3,999291821205^2 - 4\cdot 3,998797152204 = -0,000831017002794312347975\] using the reference values from the cosines, gives a positive value: \[p_{14,0}^2 - 4p_{14,3072} = 3,999397646582^2 - 4\cdot 3,998795293170 = 0,000000362805640176282724\] However, scanning the whole output, generated with the -ref option, it turns out that the cumulated rounding errors never exceed the magnitude of $10^{-4},$ so one can be pretty sure that the square root expressions themselves are correct. Hence, we compute the value of $p_{15,0}$ again, employing the procedure refactor (see the Java code), on basis of the reference values of $p_{14,0}$ and $p_{14,3072},$ and get a result, differing only in the 12th decimal place. Of course, one could have implemented a soft correction, stepping back to the reference value, whenever the difference exceeds, say $10^{-8}.$ But to me it appears more honest, to run the program uninfluenced for as long as possible.

Finally, let's discuss the total number of square roots, printed by the program Kompressor at the end of the output Roots: 1365686447. This is the number of roots needed for a complete representation of $\cos(2\pi/65537)$ by square roots, replacing all the nested $p_{m,n}$ successively by their root expressions, like we did it in case of the 17-gon. You need more than 1.3 billion roots. Spending more intelligence into the implementation of the procedure simplify (see Java code), one can expect, as already said discussing the 257-gon, to reduce this number. But I dare to make the prediction, that nobody will ever see the square root representation of $\cos(2\pi/65537)$ in closed form. Actually, there exists such a closed square root expression, thanks to the work of the Australian mathematician Joan Taylor (see here for details). However, to quote Joan Taylor herself, [...] an expression of about 2000 pages is difficult to regard as able to be seen !.

For all those, having missed the subtle hint at the beginning of this page, concerning the complete  listing of the roots of the 65537-gon, we repeat it here: