2p + 1 | Mp iff (p-1)/2 is odd
 

Hi,


is it known that


for a prime p, 2^p - 1 cannot have a factor of 2p+1 if



(p-1)/2 is even?

Yes, it is known:

[url]https://en.wikipedia.org/wiki/Sophie_Germain_prime[/url]

[QUOTE]If a Sophie Germain prime p is congruent to 3 (mod 4), then its matching safe prime 2p + 1 will be a divisor of the Mersenne number 2p − 1[/QUOTE]

(p-1)/2 = 0 (mod 2) => p-1 = 0 (mod 4) => p=1 (mod 4)

thanks very much. It'll take me a while to understand your text but don't hold your breath.


(not meaning to be rude).

[QUOTE=ATH;515290]Yes, it is known:[/QUOTE]

I think he asked about the converse though:

If (p−1)/2 is even (i.e., p = 1 mod 4), then 2p+1 cannot be a factor of 2^p−1

[QUOTE=GP2;515302]I think he asked about the converse though:

If (p−1)/2 is even (i.e., p = 1 mod 4), then 2p+1 cannot be a factor of 2^p−1[/QUOTE]

[url=www.primenumbers.net/Henri/us/NouvTh1us.htm]Henri Lifchitz[/url] has the relevant result: If p=1 (mod 4) is prime, q=2p+1 is also prime if and only if q divides 2^p + 1. (Example: p = 5, q = 11 prime <==> q divides 2^5+1 = 33.) Thus ruling said q out as a possible factor of 2^p - 1, obviously.

[QUOTE=GP2;515302]I think he asked about the converse though:

If (p−1)/2 is even (i.e., p = 1 mod 4), then 2p+1 cannot be a factor of 2^p−1[/QUOTE]


Yes I linked to the article saying 2p+1 is only a factor when p=3 (mod 4) and then I showed that his assumption "(p-1)/2 is even" is the same as p=1 (mod 4), so therefore he was correct but it was already known.

[QUOTE]
If a Sophie Germain prime p is congruent to 3 (mod 4), then its matching safe prime 2p + 1 will be a divisor of the Mersenne number 2p − 1

[/QUOTE]
So this is an equivalence as well as an implication