determine the minimum value of $n$ in a way for every odd $ m$, $2^{2012}$ divides $m^n-1$
I think I have to use LTE for the case where $u_2(m^2-1)=3$ but unfortunately, I have already learned LTE lemma and I'm not good at using it.
can you solve it not using LTE?
please share your ideas in comments and post an answer even if your solution isn't complete. thanks
We want the smallest $n$ such that $m^n\equiv 1\pmod{2^{2012}}$. for all odd $m$. This number is known as $\lambda(2^{2012})$, where $\lambda$ is the Carmichael function. Since the modular base is a power of $2$, and greater than $4$, we have $\lambda(2^{2012}) = \phi(2^{2012})/2 = 2^{2010}$