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 $3 \pmod{17}$:

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

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.

## A Magic Solution

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,

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

Similarly $b d = -1/16$. We also find $16 a b = -1 + 4a + 4b$, along with similar equations for $b c, c d, d a$. Define

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

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

$e+f = -1/8, 4e f = a b + b c + c d + a d = -1/4$

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 $c_1$.