Question

Suppose that p is prime. For a regular p-gon to be Euclidean constructible, then the roots of r- 1 must be constructible. The roots of zp-4z?-2+ . . . +z+1 together with 1-1 form a regular p-gon. They would need to be constructible. Since z-1 +···+ z + 1 is irreducible, that means the degree of a root of this is p-1. Using this prove that p 22 + 1 for some nonnegative integer n

Answer 1

$\phi(p)=p-1$

If a regular p-gon is (Euclidean) constructible then $[\mathbb{Q}(p):\mathbb{Q}]=2^n$

Thus, we must have $\phi(p)=p-1=2^n$ so that

p=2n + 1

We claim that now n but itself be a power of 2. If not we have

where s is odd

Then $(2^s+1)|(2^n+1)$ which is impossible as is prime

p=2n + 1

Thus, we must have $n=2^k$ itself and so is of the form

p=2n + 1

$p=2^{2^k}+1$