In 1796, a teenage Gauss proved that a regular 17-gon can be constructed using a straight-edge and compass by showing that a primitive 17th root of unity can be found by solving a succession of quadratic equations over the rationals.

Let $\zeta =e^{2\pi i/17}$ be a primitive 17th root of unity. Then

$$\zeta +...+{\zeta }^{16}=-1$$

(by considering the equation $x^{17}-1=0$; the sum of the roots is zero, and the other root is $1$).

Pick a generator of ${\mathbb{Z}}_{17}^{*}$. Let us choose $3$. Then the 17th roots of unity can be written in the sequence

$${\zeta }^{3^0},{\zeta }^{3^1},...,{\zeta }^{3^{15}}$$

Define $x_1$ to be the sum of every second member of the sequence, and $x_2$ to be the sum of the other members, that is,

$$x_1={\zeta }^{3^0}+{\zeta }^{3^2}+...+{\zeta }^{3^{14}}$$ $$x_2={\zeta }^{3^1}+{\zeta }^{3^3}+...+{\zeta }^{3^{15}}$$

Then $x_1+x_2=-1$. By construction, $x_1$ and $x_2$ are Gaussian periods which means it is easy to compute $x_1{x}_2=-4$ (or use brute force(!)), thus $x_1,x_2$ are roots of a quadratic equation with integer coefficients, namely $(-1\pm \sqrt{17})/2$. The solution $x_1$ is the positive one since only two terms in its sum point to the left on the complex plane.

Next define $y_1,y_2$ from the elements used to construct $x_1$ in a similar way:

$$y_1={\zeta }^{3^0}+{\zeta }^{3^4}+{\zeta }^{3^8}+{\zeta }^{3^{12}}$$ $$y_2={\zeta }^{3^2}+{\zeta }^{3^6}+{\zeta }^{3^{10}}+{\zeta }^{3^{14}}$$

To save room, let us calculate the powers of $3(mod17)$:

$$1,3,9,10,13,5,15,11,16,14,8,7,4,12,2,6$$

Thus

$$y_1=\zeta +{\zeta }^{13}+{\zeta }^{16}+{\zeta }^4$$ $$y_2={\zeta }^9+{\zeta }^{15}+{\zeta }^8+{\zeta }^2$$

Then $y_1+y_2=x_1$. It turns out $y_1{y}_2=-1$, thus $y_1,y_2$ are roots of a quadratic equation with coefficients involving the integers and $x_1$.

Similarly we can define $y_3,y_4$ from $x_2$

$$y_3={\zeta }^3+{\zeta }^5+{\zeta }^{14}+{\zeta }^{12}$$ $$y_4={\zeta }^{10}+{\zeta }^{11}+{\zeta }^7+{\zeta }^6$$

and solve a quadratic to obtain their values.

Now define $z_1,z_2$ from $y_1$ in this fashion:

$$z_1=\zeta +{\zeta }^{16}$$ $$z_2={\zeta }^{13}+{\zeta }^4$$

We have $z_1+z_2=y_1$ and $z_1{z}_2=y_3$, so $z_1,z_2$ can be found from a quadratic whose coefficients we know. Lastly we either note that both the sum and product of $\zeta$ and ${\zeta }^{16}$ are known so they can be found from a quadratic, or use the fact that

$$\zeta +{\zeta }^{16}=2\cos (2\pi /17)$$

and simply halve $z_1$.

We can generalize this procedure to find expressions for any root of unity.

Using the above, we can give an elementary method for finding $\cos (2\pi /17)$ that seems to work magically. If we don't mention generators the solution appears mysterious.

Let $c_m=\cos (2\pi m/17)$. By considering the sums of the roots of unity we have $2(c_1+...+c_8)=-1$.

Set

$$a=c_1{c}_4,b=c_3{c}_5,c=c_2{c}_8,d=c_6{c}_7.$$

By basic trigonometric identities we have

$$2a=c_3+c_5,2b=c_2+c_8,2c=c_6+c_7,2d=c_1+c_4.$$

(These correspond to the $y_i$ s above.) Thus $a+b+c+d=-1/4$. Also,

$$ac=(c_3+c_5)(c_6+c_7)/4=(c_1+...+c_8)/4=-1/16.$$

Similarly $bd=-1/16$. We also find $16ab=-1+4a+4b$, along with similar equations for $bc,cd,da$. Define

$$a+c=2e,b+d=2f$$

(Naturally, $e,f$ correspond to $x_1,x_2$ above.) Then

$$e+f=-1/8,4ef=ab+bc+cd+ad=-1/4$$

so we can solve a quadratic equation to find $e,f$. Once we have them, we can solve a quadratic equation to find $a,c$, and another to find $b,d$. With these values we can solve for $c_1$.

[I found this version in a solution that also describes a practical straight-edge-and-compass construction.]