I was trying to prove this equation I did these but I don't know what to do next?!
$2^a-1\equiv 2^{a\ \bmod\ b}-1 \ \bmod\ (2^b -1)$
My solution:
$$ a = cb+r$$ $$ 2^{cb+r} \equiv 2^r (\bmod \ 2^b -1) $$ $$ 2^{cb} \equiv 1 (\bmod \ 2^b -1) $$ (in this line I should prove that $gcd(2^b-1,2^r)=1$
and I dont know what to do next! any suggestions?
From: $ 2^{cb} \equiv 1 (mod \ 2^b -1) $
${\left(2^b\right)}^c \equiv 1 (mod \ 2^b -1)$
$1^c \equiv 1 (mod \ 2^b -1)$
$1 \equiv 1 (mod \ 2^b -1)$