Consider $n$, $m$ two natural numbers, and $\zeta_{mn}$ the $mn$-th root of unity. How can I use the Bezout identity in order to prove that there exists $k$ an integer such that $\zeta_{mn}^k$ is a power of $\zeta_m$ and $\zeta_{mn}^{k+1}$ is a power of $\zeta_n$?
This is what I considered: I considered $k\in\Bbb Z$ such that $k\in[0,mn-1]$, and went with this: \begin{equation*} \zeta_{mn}^k = \zeta_{mn}^{k(am+bn)} = \zeta_{mn}^{kam}\zeta_{mn}^{kbn} = \zeta_{n}^{ka}\zeta_{m}^{kb} \end{equation*} If we want to get $k$ such that $\zeta_{mn}^k$ is a power of $\zeta_m$, we could consider $ka = n$, in order for $\zeta_{n}^{ka} = 1$ and $\zeta_{mn}^k = (\zeta_{m})^{kb}$. But how can we prove that $\zeta_{mn}^{k+1}$ is a power of $\zeta_n$?
If we use the preceding computations, how could we prove that such $k$ verifies that $(k+1)b = m$? Os is there another way to check the existence of these elements? Thanks in advance!
You want $k$ a multiple of $n$ for $\zeta_{mn}^{k}$ to be a power of $\zeta_m$ and you want $k+1$ a multiple of $m$ for $\zeta_{mn}^{k+1}$ to be a power of $\zeta_{n}.$
This is $$k=nx\\k+1=my$$ for some integers $x,y,k.$ Then subtracting, we see this is equivalent to $$my-nx=1$$ where $k=nx.$
But $my-nx=1$ is essentially Bézout’s identity. It has integer solutions $x,y$ if and only if $\gcd(n,m)=1.$