



In this problem, you will derive a complete characterization
of the regular polygons in ${\bf R}^2$ that are constructible by 
straightedge and compass, a result that goes back to Gauss.  (He
was so proud of this, in fact, that he had a regular 17-gon, which 
is constructible, carved on his tombstone!)
\itemitem{A)}  Show that the regular $n$-gon is constructible if and
only if $[{\bf Q}(\zeta_n): {\bf Q}]$ is
equal to a power of $2$, where $\zeta_n = e^{2\pi i\over n}$.  (Hint:
$\zeta_n = \cos {2\pi \over n} + i \sin {2\pi\over n}$.)
\itemitem{B)}  Prove that ${\bf Q}(\zeta_n)$ is Galois over ${\bf Q}$.
\itemitem{C)}  By looking at the Galois group
$Gal_{\bf Q}({\bf Q}(\zeta_n))$, deduce that the degree
$[{\bf Q}(\zeta_n): {\bf Q}]$ is equal to the number of integers $m$,
$1 \le m < n$,
that are {\it relatively prime} to $n$.  (Hint:  You can use without
proof the fact we mentioned in class that if
$\phi_k(x)$ is the minimal polynomial of $\zeta_k$ over
${\bf Q})$ (the $k$th cyclotomic polynomial),
then 
$$x^n - 1 = \prod_{d | n} \phi_d(x).$$
Any element of the Galois group $G({\bf Q}(\zeta_n)/{\bf Q})$ must
take $\zeta_n$ to an $n$th root of 1 that is {\it not}
a $d$th root of 1 for any $d < n$; do you see why?)
\itemitem{D)}  Show that if an integer of the form $2^k + 1$ is 
prime, then $k$ must be a power of 2: $k = 2^r$ for some $r \ge 0$.
\itemitem{E)} If the prime factorization of $n$ in ${\bf Z}$ is  
$$n = 2^\ell p_1^{e_1} \cdots p_k^{e_k}\ ,$$ 
where the $p_i$ are odd, then the number
of integers $m$, $1 \le m < n$, relatively prime to $n$ is given by
$$2^{\ell-1} p_1^{e_1-1}(p_1 - 1)\cdots p_k^{e_k-1}(p_k - 1).$$
Deduce from this statement and parts A) and D) that a regular $n$-gon is 
constructible by ruler and compass if and only if $n$ is equal to $2^\ell$ 
times a product of distinct odd primes of the form $p = 2^{2^r} + 1$.  
(Note:  The odd primes here are called {\it Fermat primes}.  Only five of them
are known to exist: $3 = 2 + 1$, $5 = 2^2 + 1$, $17 = 2^4 + 1$, $257 = 2^8 + 1$,
and $65537 = 2^{16} + 1$.  In all other cases where the 
prime factorization of $2^{2^r} + 1$ has been determined, it is a composite
number!)
