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 $ 
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We show by induction that for any $m$ of the form $2^n q$ where $q$ is odd and $n\ge 0$, the highest power of $2$ that divides $3^{m}-1$  is $2^{n+1}$. 
For the base step $n=0$, consider $3^{q}-1$. Since $3^q\equiv 3\pmod{4}$, it follows that $3^q-1$ is of the form $4t+2$, so the highest power of $2$ that divides it is $2^1$.  
Now we do the induction step. Suppose that the result is true for $n=k$. We show the result is true for $n=k+1$. 
Consider $3^{2^{k+1}q}-1$, where $q$ is odd. This is equal to
$$(3^{2^k q}-1)(3^{2^k q}+1).$$
By the induction hypothesis the highest power of $2$ that divides $3^{2^k q}-1$ is $2^{k+1}$. It is clear that the highest power of $2$ that divides  $3^{2^k q}+1$ is $2^1$. So the highest power of $2$ that divides $3^{2^{k+1}q}-1$ is $2^{k+2}$. This completes the induction step. 
