Exercise 10.3.18

Proof. A regular $n$-gon can be constructed by origami if and only if ${\zeta }_n=e^{2\mathit{\pi i}∕n}$ is an origami number (see Exercise 10.1.2, where the figures constructible by straightedge and compass are a fortiori constructible by origami).

The splitting field of ${\zeta }_n$ over $\mathbb{Q}$ is $\mathbb{Q}({\zeta }_n)$. By Theorem 10.3.6, ${\zeta }_n$ is an origami number if and only if $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=2^a{3}^b$ for some integers $a,b\ge 0$, and the minimal polynomial of ${\zeta }_n$ over $\mathbb{Q}$ is ${\Phi }_n(x)$, so $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\Phi }_n)=\varphi (n)$, so

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First suppose that $n=2^u{3}^v{p}_1\cdots p_s,u,v\ge 0$, where $p_1,\dots ,p_s$ are distinct Pierpont numbers.

Write $p_k=2^{u_k}{3}^{v_k}+1,u_k,v_k\in \mathbb{N}$, for $k=1,\dots ,s$. As $2^u,3^v,p_1,\dots ,p_s$ are relatively prime, if $u\ge 1,v\ge 1$

$$\begin{array}{llll}\hfill \varphi (n)& =\varphi (2^u)\varphi (3^v)\varphi (p_1)\cdots \varphi (p_s)& \hfill & \\ \hfill & =(2^u-2^{u-1})(3^v-3^{v-1})(p_1-1)\cdots (p_s-1)& \hfill & \\ \hfill & =2^{u-1}(2\times 3^{v-1})(2^{u_1}{3}^{v_1})\cdots (2^{u_s}{3}^{v_s})& \hfill & \\ \hfill & =2^a{3}^b,a,b\in \mathbb{N}.& \hfill & \end{array}$$

If $u=0$, $\varphi (2^u)=1$, and if $v=0$, $\varphi (3^v)=1$. In all cases,

$$\varphi (n)=2^a{3}^b,a,b\in \mathbb{N}.$$

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Conversely, suppose that $\varphi (n)=2^a{3}^b,a,b\ge 0$, and let the factorization of $n$ be $n=q_1^{a_1}\cdots q_s^{a_s}$, where $q_1,\dots ,q_s$ are distinct primes and the exponents $a_1,\dots ,a_s$ are all $\ge 1$. Then $$\varphi (n)=2^a{3}^b=q_1^{a_1-1}(q_1-1)\cdots q_s^{a_s-1}(q_s-1).$$

If $a_i>1$, then $q_i\mid 2^a{3}^b$, so $q_i=2$, or $q_i=3$, and $q_i^{a_i}$ is a power of $2$ or $3$.

If $a_i=1$, then $q_i-1\mid 2^a{3}^b$, therefore $q_i-1=2^{u_i}{3}^{v_i}$, and so $q_i$ is a Piermont prime number, and $q_i^{a_i}=q_i=2^{u_i}{3}^{v_i}+1$.

So $n=2^a{3}^b{p}_1\cdots p_s$ where $a,b\ge 0$ and $p_1,\dots ,p_s$ are distinct Pierpont primes.

Conclusion: a regular $n$-gon can be constructed by origami if and only if $n=2^a{3}^b{p}_1\cdots p_s$ where $a,b\ge 0$ and $p_1,\dots ,p_s$ are distinct Pierpont primes.