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I know they were calculated by Gauss, but how? Is there a method for calculating them?

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Google gives: with an explicit construction – mrf Feb 29 '12 at 22:47
See if the following helps: – Dave L. Renfro Feb 29 '12 at 22:48
See section 2 of Gauss' Method of Constructing a Regular Heptadecagon by Yutaka Nishiyama. – Américo Tavares Feb 29 '12 at 23:12
up vote 3 down vote accepted

If I remember correctly(from what I read), Gauss proved that

We can factor a rational multiple of $x^{17} - 1$ as $(x-1)(P(x)^2 + aP(x) + b)$

Where $P(x)$ is an $8^{th}$ degree polynomial with rational coefficients.

This $P(x)$ could in-turn be represented as a quadratic $Q(x)^2 + cQ(x) + d$, where this $Q(x)$ is an $4^{th}$ degree polynomial.

That $Q(x)$ itself was a quadratic of a quadratic!

The actual method of geometrically constructing the polygon came a few years later.

In more modern terms, basically (again, if I remember the terms correctly), the splitting field of $x^{17}-1$ lies in a tower of quadratic extensions, starting from $\mathbb{Q}$.

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Let $\omega = e^{2 \pi i/17}$. Since $3$ is a primitive root mod $17$, i.e. a generator of the multiplicative group of nonzero integers mod $17$, write $R_j = \omega^{3^j}$ for $j = 0, 1, \ldots, 15$. These and $1$ are the $17$'th roots of unity. For $2^j \le i < 2^{j+1}$ let $x_i = \sum_{k \equiv i \mod 2^j} R_k$. Then for $2^j \le i < 2^j + 2^{j-1}$, $x_i + x_{i+2^{j-1}}$ and $x_i x_{i+2^{j-1}}$ can be expressed in terms of the previous $x_k$'s, which allows $x_i$ and $x_{i+2^{j-1}}$ to be obtained by solving a quadratic equation: if $x + y = c$ and $xy = d \ne 0$, then $x$ and $y$ are the roots of $z^2 - c z + d$. Thus:

$$x_1 = \sum_{j=0}^{15} R_j = -1$$

$x_2 = R_0 + R_2 + \ldots + R_{14}$ and $x_3 = R_1 + R_3 + \ldots R_{15}$ satisfy $x_2 + x_3 = x_1 = -1$ and $x_2 x_3 = 4 x_1 = -4$.

$x_4 = R_0 + R_4 + R_8 + R_{12}$ and $x_6 = R_2 + R_6 + R_{10} + R_{14}$ satisfy $x_4 + x_6 = x_2$, $x_4 x_6 = -1$

$x_5 = R_5 + R_9 + R_{13} + R_1$ and $x_7 = R_7 + R_{11} + R_{15} + R_3$ satisfy $x_5 + x_7 = x_3$, $x_5 x_7 = -1$

$x_8 = R_0 + R_8$ and $x_{12} = R_4 + R_{12}$ satisfy $x_8 + x_{12} = x_4$, $x_8 x_{12} = x_5$

$x_{16} = R_0 = \omega$ and $x_{24} = R_8$ satisfy $x_{16} + x_{24} = x_8$, $x_{16} x_{24} = 1$

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