#### How do you prove that constructing a heptagon is impossible?

Theorem
Constructing a regular heptagon using compass and straight edge is impossible.

Proof
Please note that this proof assumes the knowledge that an construction that can be shown to be algebraically equivalent to a cubic containing rational coefficients and having irrational roots is impossible, proved in the Impossible Constructions document.

The proof for the construction of a heptagon (7-gon) is a quite difficult to follow and makes use of complex numbers.

We begin by recognising that, by using complex numbers, the seventh root of unity yields seven solutions. That is, z7 = 1 ⇒ z7 − 1 = 0. By considering the angle between the real number axis and the first root we shall produce the required angle, 360/7 degrees.

So we proceed by writing z7 − 1 = (z − 1)(z6 + z5 + z4 + z3 + z2 + z + 1) = 0. As z ≠ 1 does not provide the given angle, the root must be a solution of z6 + z5 + z4 + z3 + z2 + z + 1 = 0.

Dividing by z³ gives z³ + z² + z + 1/z + 1/z² + 1/z³ = 0.

This can be shown to be equivalent to (z + 1/z)³ + (z + 1/z)³ − 2(z + 1/z)² − 1 = 0.

Let x = z + 1/z, so x³ + x² − 2x − 1 = 0. To show that the construction is not possible all we need demonstrate is that no rational roots exist.

Assume that x = a/b, where a and b have no common factors, so
a3/b3 + a2/b2 − 2a/b − 1 = 0, leading to a3 + a2b − 2ab2b3 = 0.

By writing a3 = b(b2 + 2aba2) and b3 = a(a2 + ab − 2b2), we see that a3 is a multiple of b and b3 is a multiple of a. As a and b have no common factors, each must be ±1. But clearly x = ±1 does not satisfy the cubic equation, so we conclude that the construction of a regular heptagon by the use of compass and straightedge is impossible.