I predict that 2^2147483647-1 is prime

But how can I prove it fast?

[MrRepunit]

You mean the double Mersenne number M(M(31))=2^(2^31-1)-1. But it has 4 known factors, the smallest one is 295257526626031
See also here: http://mathworld.wolfram.com/DoubleMersenneNumber.html

[cheesehead]

Originally Posted by micmicky

But how can I prove it fast?

You could start by studying here:

http://primes.utm.edu/prove/

[davieddy]

Originally Posted by cheesehead

You could start by studying here:

http://primes.utm.edu/prove/

Leave some jokes for me.

[science_man_88]

Originally Posted by MrRepunit

You mean the double Mersenne number M(M(31))=2^(2^31-1)-1. But it has 4 known factors, the smallest one is 295257526626031
See also here: http://mathworld.wolfram.com/DoubleMersenneNumber.html

could have meant 2^2 * 147483647-1 it's 1 mod 6 but again it's not prime.

[LaurV]

Originally Posted by MrRepunit

You mean the double Mersenne number M(M(31))=2^(2^31-1)-1. But it has 4 known factors, the smallest one is 295257526626031
See also here: http://mathworld.wolfram.com/DoubleMersenneNumber.html

no, he meant the triple mersenne number M(M(M(5))). :P:P:P