In the last section we learned how to bisect an angle. Use this to turn an n-gon into a 2n-gon. For example, the square can be used to build an octagon.
If m is a proper factor of n, an n-gon implies an m-gon. For instance, connect every third point in a 15-gon to get a pentagon. You may want to rescale the pentagon, so its sides have unit length. This is not hard to do. Start at a vertex of the oversized pentagon and move one unit along the side. Draw a line through this point, parallel to the next radial segment. Combine this with the radius incident to the original vertex and find a smaller isosceles triangle, with base 1, and the appropriate apex. Paste 5 of these together to build a pentagon with side length 1.
Now consider an n-gon and an m-gon, where m and n are coprime. Work with central angles, rather than interior angles. Thus the central angle of the n-gon is c/n, where c is a special code for the entire circle. Add the two central angles together to get c(m+n)/mn. Note that m+n is coprime to both m and n, hence coprime to mn. There is some k such that k(m+n) = 1 mod mn. Copy the angle k times over, wrapping around the circle, and find an angle that is equal to c/mn. This is the central angle for an mn-gon.
In summary, an mn-gon is constructible iff an m-gon and an n-gon are constructible, for m and n coprime. It is enough to investigate n-gons where n is a prime power.
We already said central angles could be bisected, and the square is constructible, so all powers of 2 are fair game. This includes a power of 2 times any other constructible shape, such as a 48-gon, which is a triangle enhanced by a factor of 16.
It is more convenient to work with complex numbers, so let's take the plunge. Adjoin i to the rationals to get the field of gaussian rationals Q(i). These points are all constructible - merely the points in the plane with rational coordinates.
Use demoivre's rules to show that points in the complex plane can be multiplied and divided via straightedge and compass. (Multiply the radii and add the angles.) Once again the constructible points form a field.
Let z represent a complex number. Take the square root of the disntance from z to the origin, and cut the angle in half. The square root of z is constructible, as a complex number. Furthermore, if a new point z lies outside the current field, and has been constructed using points in the current field, one of its coordinates was produced by an application of the quadratic formula, and the second coordinate is a linear function of the first. Adjoin the square root of the discriminant, and the point z is brought in. The discriminant is a real number, which is a (degenerate) complex number, so z is produced by adjoining the square root of a certain complex number.
Tweak the earlier proof, and z is constructible iff it lies at the top of a tower of field extensions, each of dimension two over the previous, with Q(i) at the base.
Suppose a p2-gon is constructible. Start at 1,0 on the x axis and work your way around the polygon counterclockwise. The first point you run into is the primitive root of 1, with order p2. This satisfies the polynomial xp2-1 = 0. But this polynomial can be reduced. The irreducible polynomial that contains this root is denoted ζp2(x). See cyclotomic polynomials for more details. It has dimension φ(p2), which is p×(p-1). The extension generated by this point has a dimension that is divisible by p. This is suppose to be part of a tower whose dimension is a power of 2. This is impossible, hence the p2-gon is not constructible. Higher powers of p are also ruled out, as they would imply a p2-gon.
We only need consider the p-gon, where p is prime. Again, look at the irreducible cyclotomic polynomial ζp(x). If has order φ(p) = p-1, which must be a power of 2. Therefore a constructible p-gon forces p to be a fermat prime.
Just because z generates an extension of dimension 2m, that doesn't mean it meets our tower requirements. In this case it does, because a cyclotomic extension is galois, with a cyclic galois group. The chain of subgroups corresponds to a tower of intermediate fields, each a quadratic extension of the previous. So our p-gon is indeed constructible.
In summary, an n-gon is constructible iff n is a power of 2 times a (possibly empty) product of distinct fermat primes.
If you're looking for an intermediate field of dimension 2, complex conjugation is a pretty good guess. The subfield fixed by this automorphism lies on the real line, i.e. the x axis. If r is the fifth root of 1, let s = r+r4. This lies on the real line, yet its value, 2×cos(72), is not rational. Let's see if we can solve for s.
Square s and see what happens.
s2 = r2+2+r8 = r2+r3+2 = 2-r-r4-1 = 1-s
Use the quadratic equation, and your ability to construct square roots, to solve s2+s-1 = 0. Then mark this point on the x axis. Use the segment from the origin to s as a base and draw the perpendicular bisector. This line intersects the unit circle in r and r4. From there you can draw the pentagon.
Gauss (biography) proved the regular n-gon theorem, pretty much as shown above. He then, patiently, developed the procedure for constructing the 17-gon. Talk about a proof of concept!
Computers have since constructed the 257-gon and the 65537-gon.