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 💡