Theorem: if p= 1 (mod 4) is prime, then 2*p +1 is also prime iff

2*p +1 divides Mp +2; a 'complimentary' proof to that of Euler's

result for Mersenne Divisors [shown by modifying Lagrange's proof

of the original if p= 3 (mod 4),...].

proof(forward): Suppose that q= 2*p +1 is prime; q= 3 (mod 8) and

2 is a quadratic residue modulo q, and there exists an integer 'n'

such that n^2= 4 (mod q). Thus, 2^p = 2^((q-1)/2) = n^(q-1) = -1

(mod q), and adding 2 to each portion of this equation shows that

(2*p +1) | (Mp +2).

proof(backward): Let 2*p +1 be a composite factor of Mp +2 and let

'q' be the least prime factor; 2^p = 1 (mod q) and the order of 2

divides both p and q-1; so, the conclusion is the same as the orig-

inal Lagrange proof, (2*p +1) +1 > q^2 > p^2 contradicts that p> 2.

I think that proves it. Cheers, Bill

--- On Tue, 1/12/10, Bill Bouris <leavemsg1@...> wrote:

From: Bill Bouris <leavemsg1@...>

Subject: [PrimeNumbers] new Mersenne Divisors conj. ???

To: "pgroup" <primenumbers@yahoogroups.com>

Date: Tuesday, January 12, 2010, 9:10 AM

I think that I found a new Mersenne Divisors conjecture ???; I couldn't

locate it on the web, but that doesn't mean anything.

Let p == 1(mod 4) be prime. 2p +1 is also prime iff 2p +1 divides Mp +2.

p = 5, 29, 41, 53, 89, etc. satisfy the conjecture.

Does anyone like it ??? Bill Mp for p= 5 & 89 happens to be prime.