Is it true that $p\mid [\mathbb Q(\cos{\frac{\pi}{p^2}}):\mathbb Q]$? Is it true that $$p\mid [\mathbb Q(\cos{\frac{\pi}{p^2}}):\mathbb Q],\forall p\in \mathbb P,$$ where $[K:\mathbb Q]$ is the degree of a field extension ?
 A: To quote my previous answer on the Galois machinery behind all of this:

Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then
$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,
$\quad \rm(x-\beta)$ all coprime, $\rm \beta\in G\alpha$, implies $\rm f(x):=\prod\limits_{\beta\in G\alpha}(x-\beta)\mid m$ in $\rm L[x]$,
$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,
$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.
Therefore the zeros of $\rm\alpha$'s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.

Now, $\alpha=2\cos(2\pi\frac{n}{m})=\zeta^n+\zeta^{-n}$ where $\zeta=e^{2\pi i/m}$ is a primitive $m$th root of unity (we assume $n/m$ is reduced). As ${\rm Gal}({\bf Q}(\zeta)/{\bf Q})\cong({\bf Z}/m{\bf Z})^\times=U(m)$, the Galois conjugates of $\alpha$ are $\zeta^\sigma+\zeta^{-\sigma}$ as $[\sigma]$ ranges over $U(m)$. For each $[\sigma]\in U(m)$, $[\sigma]\alpha=[-\sigma]\alpha$ by symmetry. Finally, $\cos$ is injective on the interval $[0,\pi]$ so $\{\cos(2\pi r/m):0\le r\le m/2,[r]\in U(m)\}$ is a complete set of conjugates without any repetitions. As $[K(\alpha):K]=\deg{\rm minpoly}_{\alpha,K}$, we therefore have
$$[{\bf Q}(\cos2\pi\frac{n}{m}):{\bf Q}]=\begin{cases}\varphi(m)/2 & m>2 \\ 1 & m=2 \end{cases} $$
since $m/2\in{\bf N}$ and $(m/2,m)=1$ iff $m=2$. Your case corresponds to $n=1$, $m=2p^2$.
More information: Galois theory and Cyclotomic fields.
