Quote:
Originally Posted by Tomazio
How can I prove that if p=3 (mod 4) is a Sophie Germain prime then the Mersenne number 2^p1 is composite?
Thanks in advance.

Because for all such primes p, 2^p1 is divisible by 2*p+1, since 2*p+1 is == 7 mod 8, thus 2 is a
quadratic residue mod 2*p+1, and znorder(Mod(2,2*p+1)) must divide ((2*p+1)1)/2 = p, thus znorder(Mod(2,2*p+1)) is either 1 or p, but cannot be 1 since there is no prime q such that znorder(Mod(2,q)) = 1 (or q must divide 21 = 1, which is impossible), thus znorder(Mod(2,2*p+1)) can only be p, and hence 2*p+1 divides 2^p1, thus 2^p1 is composite.
Similarly, if p == 1 mod 4 is a Sophie Germain prime and p > 5, then the Wagstaff number (2^p+1)/3 is composite.
Here is an exercise for you: prove that if p is a Sophie Germain prime other than 2, 3, and 5, then the dozenal repunit (12^p1)/11 is composite.