Getting factors of Mersenne numbers. Why does $2^p-1|2^b-1$ where $p|b$ and $p$ is prime and $b\in Z$?

Asked Nov 14 '19 at 21:38

Active Nov 14 '19 at 21:51

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Why does $2^p-1|2^b-1$ where $p|b$ and $p$ is prime and $b\in Z$?

I'm working with Mersenne numbers and factorization of numbers to a large power minus 1 and the solution to the problems are to find a prime that divides $b$ then $2^p-1$ will divide $2^b-1$

edited Nov 14 '19 at 21:51

asked Nov 14 '19 at 21:38

user8714896

1

If $b=pc$ then $2^b-1=(2^p)^c-1=(2^p-1+1)^c-1=\sum_{n=\mathbf{1}}^{c}\binom{c}{n}(2^p-1)^n$ – conditionalMethod Nov 14 '19 at 21:48
that's kind of cool actually – user8714896 Nov 14 '19 at 21:49
@user8714896 The first coimment boils down to $, a\mid (a!+!1)^n-1,,$ which can be proved in many ways besides the Binomial Theorem, e.g. Factor Theorem, Taylor's Theorem, modular arithmetic (congruences), induction, etc. – Bill Dubuque Nov 14 '19 at 21:56

Getting factors of Mersenne numbers. Why does $2^p-1|2^b-1$ where $p|b$ and $p$ is prime and $b\in Z$?

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