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.

Factorizing \(x^{17} - 1 = 0\) yields:

Let \(\zeta = e^{{2\pi i / 17}}\) be a primitive 17th root of unity. Since \(\zeta \ne 1\), we must have:

Since \(3\) is a generator of \(\mathbb{Z}_{17}^*\), the primitive 17th roots of unity can be written in the sequence

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,

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:

To save room, let us calculate the powers of \(3 \pmod{17}\):

Thus

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 \(x_2\)

and solve a quadratic to obtain their values.

Now define \(z_1, z_2\) from \(y_1\) in this fashion:

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

and simply halve \(z_1\).

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