Let $a$ and $b$ be positive integers. If $b=ak$ for some $k \gt 0$ then $2^b − 1 = (2^a)^k − 1 = (2^a − 1)m$ for some $m$.

Question

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.

Answer 1

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 $2^b{-}1$.