My ultimate goal is showing that $\mathbb{Q}(\sqrt 2) \subset \mathbb{Q}(\zeta_n) \iff n = 8 \cdot m$ for $ m \in \mathbb{N}$. I have showed that for $n= 8 \cdot m$ the result holds. There is only one direction left to show.
I was wondering if there is an elementary proof showing that $\sqrt 2$ is no element of $\mathbb{Q}(\zeta_n)$ if $n = l \cdot m +k $ for $m,l \in \mathbb{N}$ and $ k \in {1,2,3,4,5,6,7}$. Since that would solve the problem immediately. I know it is proven here(When is $\sqrt{2}$ in $\mathbb{Q}(\zeta_n)$?) however, the proves relies heavily on the fact that $ \mathbb{Q}(\zeta_n) \cap \mathbb{Q}(\zeta_m) = \mathbb{Q}(\zeta_{gcd(m,n)})$, which the proofs use Galois theory which has yet to be developed in class. My idea was using the identities with $sin$ and $cos$ for $e^{ix}$, yet I hadn't had much success. A different strategy, has been showing that $\mathbb{Q}(\zeta_8)$ is a vectorspace of dimension $4$ over $\mathbb{Q}$ and than showing that the intersection for any $m$ is exactly the gcd. However, I have not made any progress using $sin$ and $cos$ identities. Is there a better strategy utilizing the fact that we can work with $n=8$ in proving this identity $ \mathbb{Q}(\zeta_n) \cap \mathbb{Q}(\zeta_m) = \mathbb{Q}(\zeta_{gcd(m,n)})$ or a better strategy in solving the entire problem.
