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## Re: new Mersenne Divisors conj.

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• ... http://primes.utm.edu/notes/proofs/MerDiv2.html proves that if p = 3 mod 4 and p is prime, then 2p+1 is prime iff 2p+1|2^p-1 Bill claims that if p = 1 mod
Message 1 of 4 , Jan 16, 2010
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Bill Bouris <leavemsg1@...> wrote:

> I think that proves it

http://primes.utm.edu/notes/proofs/MerDiv2.html proves that
if p = 3 mod 4 and p is prime,
then 2p+1 is prime iff 2p+1|2^p-1

Bill claims that
if p = 1 mod 4 and p is prime,
then 2p+1 is prime iff 2p+1|2^p+1

and his facsimile proof looks good to me.

If Chris agrees, than he might like to add
Bill's case to the Euler-Lagrange page, above.

David
• ... Henri has remarked (off list) that he had already given essentially the same proof as Bill, in 1998: http://www.primenumbers.net/Henri/us/NouvTh1us.htm
Message 2 of 4 , Jan 16, 2010
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> http://primes.utm.edu/notes/proofs/MerDiv2.html proves that
> if p = 3 mod 4 and p is prime,
> then 2p+1 is prime iff 2p+1|2^p-1
>
> Bill claims that
> if p = 1 mod 4 and p is prime,
> then 2p+1 is prime iff 2p+1|2^p+1
>
> and his facsimile proof looks good to me.

Henri has remarked (off list) that he had already given
essentially the same proof as Bill, in 1998:

It's always good to see clear minds agreeing.

David
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