What is the greatest common divisor of 21001−1 and 21012−1?
The theorem being used here is that
gcd(an−1, am−1)=agcd(m,n)−1
in this case n=1001, m=1012
1001=7⋅11⋅131012=2⋅11⋅23so clearly gcd(1001, 1012)=11211−1=2047 as guest correctly computed
The proof of this theorem can be found online
I think the GCD of ((2^1001) - 1), ((2^1012) - 1) =2^(1012 - 1001) - 1 =2^11 - 1 =2,047