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I came across a homework question asking if this is true or false. After plugging in some numbers, this turned out to be true. May I know a proof or explanation for this? I sort of know this is related to congruence and GCD, but unsure of how to start.

The 'Fe' Man
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  • Special case of the linked Factor Theorem: put $, n = 2^a,\ m=1,\ r = k\ $ to get $,2^a -1\mid (2^a)^k-1\ \ $ – Bill Dubuque Apr 15 '21 at 14:56
  • Here's a previous question on specific powers of 2 with some general answers: https://math.stackexchange.com/questions/2626598/factor-of-a-mersenne-number – Joffan Apr 15 '21 at 15:24

1 Answers1

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By modular arithmetic:

Take $M_a:= 2^a-1$. Then $2^a \equiv 1 \bmod M_a$ and $2^b=(2^a)^k\equiv 1^k\equiv 1 \bmod M_a$

Thus $M_a$ divides $2^b{-}1$.

Joffan
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  • Note this argument can be easily generalized; $a^{\large b}{-}1$ divides $a^{\large bc}{-}1$ – Joffan Apr 15 '21 at 14:49
  • Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Apr 15 '21 at 14:53
  • Hi @BillDubuque - I think the OP might struggle to apply that generalized case to their question - is there a simpler dupe to link? – Joffan Apr 15 '21 at 15:01
  • It's not much of a "struggle" to substitute $,n = 2^a,\ m=1.\ $ The arguments here are already given in the dupe (and hundreds of prior answers). – Bill Dubuque Apr 15 '21 at 15:13
  • I see now you explained it in a comment on the main question, so that helps, good. – Joffan Apr 15 '21 at 15:17