# Constructing regular polygons

Yesterday, as usual, we saw some cool stuff. The gist of it was that for polynomials $P$ that can’t be factored over $\mathbb Q$, if $P(\alpha)=0$ and $P$ is a polynomial of degree $n$, then $[\mathbb Q(\alpha):\mathbb Q]=n$. Here we can replace $\mathbb Q$ with any field $k$, but for constructibility, we only care about $\mathbb Q$.

What about constructing regular polygons? If I want to construct a regular $N$-gon, I need to construct $\cos\left(\frac{2\pi}N\right)$. So for which $N$ can I do this? I’m going to do something a little crazy. I’m going to consider the $\zeta_N=\cos\left(\frac{2\pi}N\right)+i\sin\left(\frac{2\pi}N\right)$. This is a complex number, which you might recognize as an $N$th root of unity. That is $\left(\zeta\right)^N=1$. What I want to know is the smallest polynomial which has $\zeta_N$ as root. I’m not going to justify this, but if you take all the terms $(x-\zeta_N^k)$ where $k$ is relatively prime to $N$, and multiply them together, you get exactly this polynomial.

But the real question is the degree of this polynomial. It has one factor of $x$ for each $k$ relatively prime to $N$. We have a name for that. It’s the Euler totient function $\phi(N)$. I have a prize for anyone who can tell me what the word “totient” means.

It turns out that by working over the complex numbers, I haven’t actually cheated (though I won’t justify this either), and the degree of $[\mathbb Q\left(\cos\left(\frac{2\pi}N\right)\right):\mathbb Q]=[\mathbb Q(\zeta_N):\mathbb Q]=\phi(N)$. Well, we need $\phi(N)$ to be a power of two if we have any chance of constructing it. Here are some facts about $\phi(N)$ you might prove in an introductory number theory course:

• If $p$ is a prime dividing $N$, then $p-1$ divides $\phi(N)$.
• If $n^2$ divides $N$, then $n$ divides $\phi(N)$.
• If $N$ is odd, then $\phi(2N)=\phi(N)$.
• If $N$ is even, then $\phi(2N)=2\phi(N)$.
So the first thing we should do is break up $N$ into its even and odd parts. Write $N=2^k\cdot m$ where $m$ is odd. Repeated applications of the previous rules tells us that $\phi(N)=2^{k-1}\phi(m)$. So $\phi(N)$ is a power of two if and only if $\phi(k)$ is as well. Take some prime $p$ which divides $k$. Then $p-1$ divides $\phi(k)$, so we need $p-1$ to be a power of two. That is, $p$ is a prime which is one more than a power of two.
We have a name for these types of primes. They’re called Fermat primes, and they seem to be exceedingly rare. We only know of 5 (namely 3, 5, 17, 257, and 65537). What’s more, we can only have each of these primes dividing $N$ once (by the second rule).
So we do know of infinitely many constructible $N$-gons. Namely, the square, octagon, 16-gon, 32-gon, etc. But if you want to know how many constructible $N$-gons there were with $N$ odd, you would be asking an extremely difficult question that has been open for nearly 200 years!
Okay. At long last, we’re ready to jump into Galois Theory. Hold on to your socks, because they’re about to get knocked off!

### 5 Responses to Constructing regular polygons

1. Jeffrey Lai says:

hey andy isn’t http://s0.wp.com/latex.php?latex=%5Cphi%28N%29%3D2%5Cphi%28N%29&bg=ffffff&fg=555555&s=0 for N is even supposed to be 2*phi(N/2) on the right side?

2. Jeffrey Lai says:

…which obviously means isn’t for N is even supposed to be
phi(N) = 2*phi(N/2)

3. Louis says:

15 is not a Fermat prime. (Think of what Kelly would say to you!)

4. Louis says:

Also, the first paragraph has some unformatted LaTeX junk in it