Prove that if $p\equiv 5\pmod{8}$ then $\zeta_p$ not constructible

Question

Prove that if $p\equiv 5\pmod{8}$, $p>5$ then $\zeta_p$ not constructible

How to do this? There is a theorem in my book that says that the regular $n$-gon is constructible iff $n=2^k\cdot n_0$ where $n_0$ is the product of distinct Fermat primes, but I don't know how to apply it here since we are talking about an infinitude of primes.

Answer 1

Suppose that $p > 5$ is constructible and $p \equiv 5 \pmod{8}$. Since $p$ must be a Fermat prime, we have $p=2^{2^n}+1$ for some $n \geq 2$. But $$8=2^3 \mid 2^{2^n} \implies p \equiv 1 \pmod{8},$$ a contradiction.

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The same proof would show that if $q \equiv 3,5,7 \pmod{8}$ with $3, 5 \nmid q$, then $\zeta_q$ is not constructible.

Answer 2

$\zeta_p$ has degree $\phi(p)=p-1 \equiv 4 \bmod 8$.

Now, $p-1 > 4$ because $p>5$. If $p-1$ were a power of $2$, then $p-1 \equiv 0 \bmod 8$.

Therefore, $p-1$ cannot be a power of $2$ and so $\zeta_p$ cannot be constructible.