<s>[INST] Heptadecagon Derivation I am currently very interested in the derivation of the constructability of the 17-gon by Carl Friedrich Gauß.
Has someone got an easy explanation for the solution of 
$$x^{17} - 1=0?$$
That was the equation he solved with which he showed
\begin{align}\cos \frac{360^\circ}{17}&=\frac{1}{16}\left( -1 + \sqrt{17} + \sqrt{ 2\left(17 -\sqrt{17} \right)}+ 2 \sqrt{ 17 + 3 \sqrt{17} - \sqrt{2 \left(17- \sqrt{17} \right)} - 2 \sqrt{2 \left(17+ \sqrt{17} \right)} } \right) \\&\approx 0.93247222940435580457311589182156.\end{align}
Can someone briefly explain his derivation, please? 
 [/INST] This is an elementary proof.
Let $\varphi=\frac\pi{17}$,
$$S=-\sum_{n=1}^8(-1)^n\cos(n\varphi)$$
Multiplication by $2\cos(\varphi/2)$ gives:
\begin{align}
2S\cos\left(\frac\varphi 2\right)
&=-\sum_{n=1}^8(-1)^n\left(\cos\left(\frac{2n-1}2\varphi\right)-\cos\left(\frac{2n+1}2\varphi\right)\right)\\
&=\cos\left(\frac 12\varphi\right)-\cos\left(\frac{17}2\varphi\right)\\
&=\cos\left(\frac\varphi2\right)
\end{align}
so that $S=\frac 12$.
Now let
\begin{align}
X&=\cos(3\varphi)+\cos(5\varphi)-\cos(6\varphi)+\cos(7\varphi)\\
Y&=-\cos(\varphi)+\cos(2\varphi)+\cos(4\varphi)+\cos(8\varphi)
\end{align}
so that $X-Y=\frac 12$.
Moreover, $XY=4S=2$, hence $XY=1$ which gives
\begin{align}
&X=\frac{\sqrt{17}+1}4&&Y=\frac{\sqrt{17}-1}4
\end{align}
Now let
\begin{align}
z&=\cos(3\varphi)+\cos(5\varphi)\\
x&=\cos(6\varphi)-\cos(7\varphi)
\end{align}
so that $X=z-w$.
Then $2zx=S=\frac 12$, so that we obtain
\begin{align}
z&=\frac{1+\sqrt{17}+\sqrt{34+2\sqrt{17}}}8\\
x&=\frac{-1-\sqrt{17}+\sqrt{34+2\sqrt{17}}}8
\end{align}
Similarly, $y=\cos(\varphi)-\cos(4\varphi)$ and $v=\cos(2\varphi)+\cos(8\varphi)$ satisfy $Y=v-y$ and $yv=\frac 14$, thus giving
\begin{align}
y&=\frac{1-\sqrt{17}+\sqrt{34-2\sqrt{17}}}8\\
v&=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}}8
\end{align}
Finally $\cos(2\varphi)+\cos(8\varphi)=v$ and $\cos(2\varphi)\cos(8\varphi)=\frac x2$ from which we get
$$\cos(2\varphi)=\frac 1{16}\left(-1+\sqrt{17}+\sqrt{34-2\sqrt{17}+2\sqrt{17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}}}\right)$$
</s>