Logic behind a Geometric Construction of regular heptadecagon

Question

I'm reading a Chinese book "Methods of Mathematical Physics" by Wu Chongshi.

During introduction of complex analysis, it explains a Geometric Construction of regular heptadecagon.

Task: to achieve $a=2\sin(\pi/17)$

1. Let $x = e^{i2\pi / 17}$, so $x^0, x^1, \cdots, x^{16}$ are all the roots for equation $x^{17}=1$.

2. Let $s = x^1 + x^9 + x^{9^2} + x^{9^3} + \cdots + x^{9^7}$ and $s'= x^3 + x^{3^3} + x^{3^5} + \cdots + x^{3^{15}}$.

   So $s = x^1 + x^9 + x^{13} + x^{15} + x^{16} + x^8 + x^4 + x^2$, $s'= x^3 + x^{10} + x^5 + x^{11} + x^{14} + x^7 + x^{12} + x^6$.

   Since $s+s'=-1$ , $ss'=-4$, one has $s=\frac{1}{2}(\sqrt{17}-1)$, $s'=-\frac{1}{2}(\sqrt{17}+1)$.

3. Let $s=p+p'$, $s'=q+q'$, where $p = x^1 + x^{13} + x^{16} + x^4$, $p'= x^9 + x^{15} + x^8 + x^2$,$q = x^3 + x^5 + x^{14} + x^{12}$, $q'= x^{10} + x^{11} + x^7 + x^6$.

   Since $pp'=-1$, $qq'=-1$, one has $p=\frac{1}{2}(s+\sqrt{s^2+4})$, $q=\frac{1}{2}(s'+\sqrt{s'^2+4})$.

4. Again, let $r=x^1+x^{16}$, $r'=x^{13}+x^4$, one has $r+r'=p$, $rr'=q$, hence $r=\frac{1}{2}(p+\sqrt{p^2-4q})$.

5. Finally, $a=\sqrt{2-r}$.

Now, with all these steps quite good and smart, my question is, why at the beginning of step 2, $s$ and $s'$ are construct as $$s = x^1 + x^9 + x^{9^2} + x^{9^3} + \cdots + x^{9^7}$$ and $$s'= x^3 + x^{3^3} + x^{3^5} + \cdots + x^{3^{15}}$$?

These are interesting exponential forms, is there anything special behind the form? some reason, logic?

-- update

maybe i explain a little bit more.

From Fermat's small theorem, that $a^{p-1}\equiv 1(\text{mod } p)$ , it's obvious that $$3^{17-1}\equiv 1 (\text{mod } 17)$$ hence $$9^8=(3^2)^8=3^{16}=3^{17-1}\equiv 1 (\text{mod } 17)$$ that's why we could let $$s=\Sigma_i x^{a_i}$$ where $$a_{i+1}=a_i^9$$ similiarly, $$s'=\Sigma_i x^{3a_i}$$

But, when the guy first came out of the idea, how "dare" he do so? How could he be sure that

1. there will have $s+s'=-1$ and $ss'=-4$ so that $s$ and $s'$ are solvable?
2. $s$ and $s'$ could be further broken down to $s=p+p'$, $s'=q+q'$ as above?

Is the guy simply found this idea by luck? Or, there's some math supporting his idea, just like the Fermat's little theorem supports compose $s$ by 8 components? I suspect there's some groundings from number theory...

Answer 1

$\newcommand{\Z}{\mathbb{Z}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\Span}[1]{\langle #1 \rangle}$ Write $\omega = e^{i \frac{2 \pi}{17}}$.

I have only the time to write down the beginning of the argument. Very briefly, the idea is that the extension $\Q(\omega)/\Q$ has degree 16, and can be built via extensions of degree $2$.

Its Galois group $G$ is abelian, isomorphic to the multiplicative group $\Z_{17}^{\star}$, via $\omega \mapsto \omega^{k}$, where $k \in \Z_{17}^{\star}$.

Now $\Z_{17}^{\star}$ is generated by $3$, so $G$ is generated by $\sigma : \omega \mapsto \omega^{3}$. ) So $\Span {\sigma^{2}}$ is a subgroup of $G$ of index two, and corresponds to a field $F$, with $\Q \subseteq F \subseteq \Q(\omega)$ with $\lvert F : \Q \rvert = 2$. This field $F$ contains $$ s = \sum_{\tau \in \Span {\sigma^{2}}} \tau(\omega) $$ (just check that this is left invariant by $\sigma^{2}$, and also note that $\sigma^{2} : \omega \mapsto \omega^{9}$), and then $F$ also contains $\sigma(s) = s'$, as every intermediate field is Galois.

Also note that obviously $\sigma^{2}(s) = s$. Therefore $$ \sigma(s + s') = \sigma(s) + \sigma(\sigma(s)) = \sigma(s) + s = s'+ s, $$ and similarly $\sigma(s s') = s s'$. It follows that $s + s, s s'\in \Q$, and therefore $F$ can be obtained from $\Q$ by adding the roots of the rational polynomial $$ x^{2} - (s + s') x + s s´. $$