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Old 2022-05-25, 05:33   #3
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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Originally Posted by Tomazio View Post
How can I prove that if p=3 (mod 4) is a Sophie Germain prime then the Mersenne number 2^p-1 is composite?
Thanks in advance.
Because for all such primes p, 2^p-1 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 2-1 = 1, which is impossible), thus znorder(Mod(2,2*p+1)) can only be p, and hence 2*p+1 divides 2^p-1, thus 2^p-1 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^p-1)/11 is composite.
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