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Show that if $a|b \ \ then\ \ 2^a-1|2^b-1$
I've seen this assertion in the proof of another problem but the author hadn't given any reason:
[Original problem containing this assertion][1]
[1]: Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$ .
Seemingly simple,I have no idea how to progress in solving this problem!!

Community
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Hamid Reza Ebrahimi
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    For any $x,y\in\mathbb R, n\in\mathbb Z^+$ we have $x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\cdots+y^{n-1}\right)$. Let $x=2^a$, $y=1$, $n=\frac{b}{a}$. – user236182 Apr 07 '16 at 14:09
  • @DietrichBurde Unfortunately , when typing the question the related questions are not fully listed,and the Math search engine is not perfect! – Hamid Reza Ebrahimi Apr 07 '16 at 14:14
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    @HamidRezaEbrahimi No problem. There are enough answers here, and for the original problem. – Dietrich Burde Apr 07 '16 at 14:15

4 Answers4

2

Rewrite $b=ac$ with c an integer. Then $2^b-1=2^{ac}-1=(2^a)^c-1$.

Now you use the identity $x^n-1=(x-1)(1+...+x^{n-1})$ with $x=2^a$ and $n=c$ and here you go!

pmichel31415
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2

If $a \mid b$, then $b = ka$. We want to find $2^b - 1 \pmod{2^a - 1}$. But this is $2^{ka} - 1 \equiv (2^a)^k - 1 \equiv 1^k - 1 \equiv 0 \pmod{2^a - 1}$. So $2^a - 1 \mid 2^b - 1$.

Note that this applies to any integer $n$, not just 2.

shardulc
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If $a|b$ then $b = m.a$ where $m$ is an integer.

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

Thus $2^a - 1 | 2^b - 1$.

crbah
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$a|b \Rightarrow b=ka, k \in \mathbb Z$ $$2^b-1=2^{ka}-1=(2^a)^k-1=(2^a-1)((2^a)^{k-1}+(2^a)^{k-2}+...+1)$$

Roman83
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