Let $a$ and $b$ be integers $\ge 1$. prove that $(2^a-1) | (2^{ab}-1)$.

Asked Mar 07 '19 at 20:16

Active Mar 07 '19 at 20:36

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Let $a$ and $b$ be integers $\ge 1$. prove the following:

$(2^a-1) | (2^{ab}-1)$

My attempt:

$2^{ab}-1=(2^a)^b-1$

$= (2^a-1)((2^a)^{b-1}+(2^a)^{b-2}+...+2^a+1)$

Since $(2^a)^{b-1}+(2^a)^{b-2}+...+2^a+1\in \Bbb{Z}$, then

$(2^{ab}-1)\equiv 0 \mod (2^a-1)$.

Is that true, please ?

edited Jun 12 '20 at 10:38

Community

asked Mar 07 '19 at 20:16

Dima

See also this duplicate. – Dietrich Burde Mar 07 '19 at 20:20
Am I missing something? Your second equals sign makes no sense. – The Count Mar 07 '19 at 20:20
I'm not convinced this is an exact duplicate, as it asks for a very special case, more friendly to newer students of the material... – The Count Mar 07 '19 at 20:22
@DietrichBurde Thank you. – Dima Mar 07 '19 at 20:25
@TheCount sorry, I edited it. – Dima Mar 07 '19 at 20:25
1

@Dima, looks good. The only complaint might be how rigorous it is supposed to be, but it is perfectly clear to me. – The Count Mar 07 '19 at 20:27
@TheCount Yes, you are right. The second duplicate is better, because it is for $x=2$ more specific, i.e., $2^a-1$ divides $2^d-1$ for $a\mid d$. Now take $d=ab$ here. It is very useful to compare the own proofs with the duplicate ones, I think. – Dietrich Burde Mar 07 '19 at 20:28
@TheCount Thank you so much. – Dima Mar 07 '19 at 20:28

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