# The Heptadecagon

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 $$c_1$$.

Ben Lynn blynn@cs.stanford.edu 💡