If $a $ divides $b$, then $(2^a-1) $ divides $(2^b-1)$

Question

Prove that if $a \mid b$, then $(2^a-1) \mid(2^b-1)$.

So, I've said the following:

1. Let $a \mid b$.
2. $ \implies b=am$
3. Assume $(2^a-1) \mid (2^b-1)$
4. $ \implies (2^b-1)=(2^a-1)x$

But I can't figure out what to do with any of this from here forward.

Answer 1

HINT: Sometimes choosing the ‘right’ representation makes things easier to see. Note that the binary (base two) representation of $2^n-1$ is

$$\underbrace{111\ldots 111}_{n\text{ ones}}\;.$$

Suppose that $b=ka$; then

$$\begin{align*} \underbrace{111\ldots111}_{b\text{ ones}}&=\underbrace{\underbrace{111\ldots 111}_{a\text{ ones}}\underbrace{111\ldots 111}_{a\text{ ones}}\ldots\underbrace{111\ldots 111}_{a\text{ ones}}}_{k\text{ blocks}}\\\\ &=\left(\underbrace{111\ldots 111}_{a\text{ ones}}\right)\left(\underbrace{1\underbrace{000\ldots000}_{a-1\text{ zeroes}}1\underbrace{000\ldots000}_{a-1\text{ zeroes}}1\ldots1\underbrace{000\ldots000}_{a-1\text{ zeroes}}1}_{k\text{ ones}}\right)\;. \end{align*}$$

For example, with $a=3$ and $b=12$, we have

$$111111111111=111\cdot1001001001\;.$$

These ideas can be translated into statements about sums of powers of $2$.

Answer 2

When you have a statement you are trying to prove, such as "if $p$, then $q$", you can assume $p$ is true, and then show that $q$ must also be true. This is called a direct proof.

Alternatively, you can assume $q$ is not true, and show that this means $p$ cannot be true. This is called a proof by contradiction.

You are assuming both $p$ and $q$ are true. You should pick one of the two strategies I mentioned above. I would suggest a proof by contradiction in this particular case.

Answer 3

Suppose $a\mid b$ and so $b=am$ for some $m\geq 1$. Then, $$2^b-1=2^{am}-1=(2^a)^m-1^m\\=(2^a-1)\left((2^a)^{m-1}+(2^a)^{m-2}+\cdots+(2^{a})^2+(2^a)+1\right).$$ We see that $(2^a)^{m-1}+(2^a)^{m-2}+\cdots+(2^{a})^2+(2^a)+1\geq 1$ and so $2^b-1=(2^a-1)k$ for some $k\geq 1$. Hence $2^a-1\mid 2^b-1$.

Remember, it's very easy to begin circular reasoning if you're not careful when you assume a result that you're trying to prove. It's much more conventional (especially when first learning proofs) to reach that result by deducing it from the hypotheses in the statement - in this case $a\mid b$.