Prove that if $ 2^n $ divides $ 3^m-1 $ then $ 2^{n-2} $ divides $ m $

Question

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 :-)

Answer 1

This statement is a vacuous truth, meaning that the condition $3^m \equiv 0 \mod{2^n}$ is never satisfied.

Answer 2

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 .

Answer 3

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.