Dividing an angle into $n$ equal parts My question is simply: for which values of $n$ is it possible to divide any given angle into $n$ equal parts using only a compass and a straight edge? I know that it is possible for $2$ and not possible for $3$, but is it possible for any integers that are not of the form $2^k$?
 A: The only possibility is indeed numbers of the form $2^k$.
We use the famous characterization of constructible regular polygons. The $\frac{360^\circ}{N}$ angle is straight edge and compass constructible if and only if $N$ is of the shape
$$N=2^k p_1\cdots p_s,\tag{1}$$
where the $p_i$ are distinct Fermat primes (possibly none).
This theorem rules out immediately all numbers $N$ not of the shape (1). But it also rules out the numbers of shape (i) where the number $s$ of Fermat primes in the factorization is non-zero.
For the theorem says that if $N$ involves one or more Fermat primes, then the $\frac{360^\circ}{N}$ angle cannot be straight-edge and compass divided into $N$ equal parts. 
A: It is not possible.You know how to construct the square root, so successively you can do with the half of any of 1/2 , 1/4, 1/8,.....(see the formule of sin ($\alpha$/2)). But you can not get 1/3 of the angle which is, as you know, a famous classic problem of impossibility (see the formule of sin ($\alpha$/3)) and not for the integers you are looking for. The simple reason is that you get with compass and a straight edge only arcs of circles and straight lines  so the new points you can get are either of first or second or fourth degree over the field you have previously got. (By the way,construction of $\pi$ would need an infinity of points of the same quadratic kind and this is not other thing that the trascendance of $\pi$). 
