Constructible polygon

A constructible polygon is a regular $n$-gon which can be constructed using a straight edge and compass. For instance equilateral triangles and regular pentagons are well-known to be constructible, however, regular heptagons are not constructible.

Criterion for constructability

Using the characterization of constructible numbers one can show that a regular $n$-gon is constructible iff $n$ is the product of a power of $2$ and distinct fermat primes. For instance, a $17=2^4+1$-gon, a $257=2^8+1$-gon, a $65537 = 2^{16}+1$-gon and a $673720360=2^3\cdot 5\cdot257\cdot 65537$-gon are all constructible, but a $13$-gon isn't.


First notice that constructing a regular $n$-gon is equivalent to constructing the $n^{th}$ roots of unity in the complex plane, starting with the points $0$ and $1$ (because the $n^{th}$ roots of unity form a regular $n$-gon).

Now consider a primitive $n^{th}$ root of unity $\zeta_n = e^{\frac{2\pi i}{n}}$. So now the $n^{th}$ roots of unity are $1,\zeta_n,\zeta_n^2,\ldots,\zeta_n^{n-1}$. As the set of constructible numbers is a field, these numbers will be constructible iff $\zeta_n$ is constructible. Hence a regular $n$-gon is constructible iff $\zeta_n$ is a constructible number.

Now by the characterization of constructible numbers, $\zeta_n$ will be constructible iff there is a chain of field extensions $\mathbb Q = K_0\subseteq K_1\subseteq \cdots\subseteq K_m = \mathbb Q(\zeta_n)$ such that each extension $K_i\subseteq K_{i+1}$ is quadratic (i.e. $[K_{i+1}:K_i]=2$). We claim that this happens iff $\phi(n)$ (where $\phi(n)$ is Euler's totient function) is a power of $2$.

First we note that $\mathbb Q(\zeta_n)$ is the $n^{th}$ cyclotomic field, and hence $[\mathbb Q(\zeta_n):\mathbb Q]=\phi(n)$.

Now assume that such a chain of field extensions exists. Then by the tower law:

\[\phi(n) = [\mathbb Q(\zeta_n):\mathbb Q] = [K_m:K_{m-1}][K_{m-1}:K_{m-2}}\cdots[K_1:K_0] = (2)(2)\cdots(2) = 2^m,\] (Error compiling LaTeX. ! Extra }, or forgotten $.)

as desired.

Now assume that $\phi(n) = 2^m$, for some $m$. $\mathbb Q(\zeta_n)$ is the splitting field of $n^{th}$ cyclotomic polynomial, $\Phi_n(x)$ and hence $\mathbb Q(\zeta_n)/\mathbb Q$ is a Galios extension. Now the order of the Galios group $G = \Gal(\mathbb Q(\zeta_n)/\mathbb Q)$ (Error compiling LaTeX. ! Undefined control sequence.) is just $[\mathbb Q(\zeta_n):\mathbb Q]=2^m$. Thus $G$ is a 2-group, and hence there must exist a chain of subgroups \[G = G_m>G_{m-1}>\cdots>G_0=1\] with $[G_{i+1}:G_i]=2$ for all $i$. Now by the fundamental theorem of Galios theory, if $K_i$ is the fixed field of $G_i$ then \[\mathbb Q = K_m\subseteq K_{m-1}\subseteq \cdots\subseteq K_0=\mathbb Q(\zeta_n)\] and $[K_i:K_{i+1}] = [G_{i+1}:G_i] = 2$. Therefore $\zeta_n$ is indeed constructible.

We have now shown that a regular $n$-gon is constructible iff $\phi(n)$ is a power of $2$. It only remains to show that the integers which satisfy this are precisely the integers in the given form.

Let the prime factorization of $n$ be $n = 2^ap_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ (where $p_1,p_2,\ldots,p_k$ are distinct odd primes, $e_i\ge 1$ for all $i$ and $a\ge 0$). Then by the formula for $\phi(n)$ we have \[\phi(n) = 2^{a-1}p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1-1)(p_2-1)\cdots(p_n-1)\] (or $\phi(n) = p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1-1)(p_2-1)\cdots(p_n-1)$ if $a=0$).

Now assume that $\phi(n) = 2^m$. Now a power of $2$ cannot be divisible by an odd prime, so we cannot have $e_i>1$ for any $i$ (otherwise we would have $p_i|2^m$), so $e_1=e_2=\cdots=e_n=1$. Also, any divisor of a power of $2$ must also be a power of $2$, so $p_i-1$ is a power of $2$ for each $i$, from which is easily follows that each $p_i$ is a fermat prime. Hence $n$ is in the desired form.

Conversely assume that $n = 2^ap_1p_2\cdots p_k$, were $p_1,p_2,\cdots,p_k$ are distinct fermat primes, say $p_i = 2^{2^{s_i}}+1$. Then \[\phi(n) = 2^{a-1}(p_1-1)(p_2-1)\cdots(p_n-1) = 2^{a-1}2^{2^{s_1}}2^{2^{s_2}}\cdots2^{2^{s_k}}\] or $\phi(n) = (p_1-1)(p_2-1)\cdots(p_n-1)=2^{2^{s_1}}2^{2^{s_2}}\cdots2^{2^{s_k}}$ if $a=0$. In either case this is a power of $2$, as desired. This completes the proof.

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