+121 To find the greatest common divisor (GCD) of \(2^{1001}-1\) and \(2^2-1\), we can use the difference of squares factorization.
Notice that \(2^2 - 1\) is a difference of squares: \(2^2 - 1 = (2+1)(2-1) = 3\).
Now, let's use the difference of squares formula for \(2^{1001}-1\):
\[2^{1001} - 1 = (2^{500} + 1)(2^{500} - 1).\]
Again, \(2^{500} - 1\) is another difference of squares: \(2^{500} - 1 = (2^{250} + 1)(2^{250} - 1)\).
This pattern continues, and we eventually reach \(2^2 - 1 = 3\).
So, we have:
\[2^{1001} - 1 = (2^{500} + 1)(2^{500} - 1) = (2^{250} + 1)(2^{250} - 1) = \dots = 3^8.\]
Now, we can see that the greatest common divisor of \(2^{1001}-1\) and \(2^2-1\) is simply \(3\).
Therefore, the greatest common divisor of \(2^{1001}-1\) and \(2^2-1\) is \(\boxed{3}\).
Partial factorization of: 2^(1001) - 1
23×89×127×911×6007×8191×724153×112901153×23140471537×158822951431×6120360210855167691724912383945435257223721665116428926703276245201747560818857289386060189570105107203570707721973897692732984143624435639152229768723838829356129391428019897940054854494968142136307334184139636418088074652883916814682909620534381239
(10 prime factors, 1 composite factor)
2^2 - 1 ==3
Prime factorization of:
3 ==1 x 3
Since there no common factor between 2^(1001) - 1 and 2^2 - 1, therefore:
The GCD of [2^(1001) - 1, 2^2 - 1]==1
