# Constructible polygon

A constructible polygon is a regular -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 -gon is constructible iff is the product of a power of and distinct fermat primes. For instance, a -gon, a -gon, a -gon and a -gon are all constructible, but a -gon isn't.

## Proof

First notice that constructing a regular -gon is equivalent to constructing the roots of unity in the complex plane, starting with the points and (because the roots of unity form a regular -gon).

Now consider a primitive root of unity . So now the roots of unity are . As the set of constructible numbers is a field, these numbers will be constructible iff is constructible. Hence a regular -gon is constructible iff is a constructible number.

Now by the characterization of constructible numbers, will be constructible iff there is a chain of field extensions such that each extension is quadratic (i.e. ). We claim that this happens iff (where is Euler's totient function) is a power of .

First we note that is the cyclotomic field, and hence .

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

Now assume that , for some . is the splitting field of cyclotomic polynomial, and hence is a Galois extension. Now the order of the Galois group is just . Thus is a 2-group, and hence there must exist a chain of subgroups with for all . Now by the fundamental theorem of Galois theory, if is the fixed field of then and . Therefore is indeed constructible.

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

Let the prime factorization of be (where are distinct odd primes, for all and ). Then by the formula for we have (or if ).

Now assume that . Now a power of cannot be divisible by an odd prime, so we cannot have for any (otherwise we would have ), so . Also, any divisor of a power of must also be a power of , so is a power of for each , from which is easily follows that each is a fermat prime. Hence is in the desired form.

Conversely assume that , were are distinct fermat primes, say . Then or if . In either case this is a power of , as desired. This completes the proof.