Construction of Regular Heptagon by Compass and Straightedge Construction is Impossible

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Theorem

There is no compass and straightedge construction for a regular heptagon.


Proof

Construction of a regular heptagon is the equivalent of constructing the point $\tuple {\cos \dfrac {2 \pi} 7, \sin \dfrac {2 \pi} 7}$ from the points $\tuple {0, 0}$ and $\tuple {1, 0}$

Let $\epsilon = \map \exp {\dfrac {2 \pi} 7}$.

Then $\epsilon$ is a root of $x^7 - 1$.

We have:

\(\displaystyle x^7 - 1\) \(=\) \(\displaystyle \paren {x - 1} \paren {x^6 + x^5 + x^4 + x^3 + x^2 + x + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^6 + \epsilon^5 + \epsilon^4 + \epsilon^3 + \epsilon^2 + \epsilon + 1\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^3 + \epsilon^2 + \epsilon + 1 + \epsilon^{-1} + \epsilon^{-2} + \epsilon^{-3}\) \(=\) \(\displaystyle 0\)

But we have:

\(\displaystyle \epsilon\) \(=\) \(\displaystyle \cos \dfrac {2 \pi} 7 + i \sin \dfrac {2 \pi} 7\)
\(\displaystyle \epsilon^{-1}\) \(=\) \(\displaystyle \cos \dfrac {2 \pi} 7 - i \sin \dfrac {2 \pi} 7\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon + \epsilon^{-1}\) \(=\) \(\displaystyle 2 c\) where $c = \cos \dfrac {2 \pi} 7$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^2 + \epsilon^{-2} + 2\) \(=\) \(\displaystyle 4 c^2\) squaring
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^3 + \epsilon^{-3} + 3 \paren {\epsilon + \epsilon^{-1} }\) \(=\) \(\displaystyle 8 c^3\) cubing
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^2 + \epsilon^{-2}\) \(=\) \(\displaystyle 4 c^2 - 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \epsilon^3 + \epsilon^{-3}\) \(=\) \(\displaystyle 8 c^3 - 6 c\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {8 c^3 - 6 c} + \paren {4 c^2 - 2} + 2 c - 1\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 8 c^3 + 4 c^2 - 4 c - 1\) \(=\) \(\displaystyle 0\)

Thus $2 c$ is a root of the polynomial $x^3 + x^2 - 2 x - 1$

But from Irreducible Polynomial: $x^3 + x^2 - 2 x - 1$ in Rationals:

$x^3 + x^2 - 2 x - 1$ is irreducible over $\Q$.


Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos \dfrac {2 \pi} 7$ is not an element of any extension of $\Q$ of degree $2^m$.

The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2.

$\blacksquare$


Sources