Prove that if $ 2^n $ divides $ 3^m-1 $ then $ 2^{n-2} $ divides $ m $ I got a difficult problem. It's kind of difficult to prove. Can you do it? 
Let $ m,n\geq 3 $ be two positive integers. Prove that if $ 2^n $ divides $ 3^m -1$ then $ 2^{n-2} $ divides $ m $ 
Thanks :-)

Because $n \geq 3$ we get $8 \mid 3^m-1$ and so $m$ must be even .
Let $m=2^l \cdot k$ with $k$ odd . Now use the difference of squares repeatedly to get :
$$3^m-1=(3^k-1)(3^k+1)(3^{2k}+1)\cdot \ldots \cdot (3^{2^{k-1} \cdot l}+1)$$
Each term of the form $3^s+1$ with $s$ even has the power of $2$ in their prime factorization exactly $1$ because: $$3^s+1 \equiv 1+1\equiv 2 \pmod{8}$$
Also $k$ is odd so :
$$3^k+1 \equiv 3+1 \equiv 4  \pmod{8}$$ has two factors of $2$ .
Finally the term $3^k-1 \equiv 3-1 \equiv 2 \pmod{8}$ has one factor of $2$ .
This means that $3^m-1$ has $1+2+l-1=l+2$ two's in his prime factorization .
But $2^n \mid 3^m-1$ so $n \leq l+2$ and then $l \geq n-2$ .
This means that $2^{n-2} \mid m$ as wanted .
