If a regular $n$-gon is constructible i.e. if $\cos (2\pi/n)$ is a constructible number then how to show that $\phi(n)$ is a power of $2$?

Question

If a regular $n$-gon is constructible i.e. if $\cos (2\pi/n)$ is a constructible number then how to show that $\phi(n)$ is a power of $2$ ?

Answer 1

This is equivalent to the primitive $n$th root $\zeta_n=e^{2\pi i/n}$ being constructible, the irreducible polynomial of which has degree $\phi(n)$.

On the other hand, geometric construction can at most solve quadratic equations, i.e., if all given coordinates are in a field $K$, then any point of intersection betwene lines and circles dtermined by these is in a quadratic extension of $K$. Thus geometric constructiility of the $n$-gon means that there are fields $$\Bbb Q=K_0\subseteq K_1 \subseteq K_2\subseteq \ldots\subseteq K_N$$ where $[K_{j+1}:K_j]=2$

Answer 2

Constructible numbers are got through iterated quadratic extensions, so each generates a field of $2$-power order. The cyclotomic field $\Bbb Q(\exp(2\pi i/n))$ has degree $\phi(n)$ and is quadratic over $\Bbb Q(\cos(2\pi/n))$ so that has degree $\phi(n)/2$ (for $n\ge3$). So $\phi(n)/2$ has to be a power of $2$ for the regular $n$-gon to be constructible.