- Hi,

I've read the "new Mersenne Divisors conj." thread, and I propose the following test for Sophie Germain primes > 3 that does not require 2 different cases p = 1 (mod 4) and p = 3 (mod 4).

The test is:

Let p be a prime > 3; p is a Sophie Germain prime <=> 2p+1 divides 4^p-3^p.

Proof:

=> : Suppose q = 2p+1 is prime.

Then (3/q)(q/3) = (-1)^((q-1)/2) = (-1)^p = -1. So (3/q) = -(q/3).

But since p > 3 and q are both primes, q = 2 (mod 3) and (q/3) = -1.

This show that 3 is a square mod q. Let a be such that a^2 = 3 (mod q).

Finally, 4^p-3^p = 2^(2p)-a^(2p) = 2^(q-1)-a^(q-1) = 1-1 = 0 (mod q).

So q = 2p+1 divides 4^p-3^p.

<= : Let q = 2p+1 be a factor of 4^p-3^p. Suppose q be composite.

Let u be the smallest prime factor of q (which implies u^2 <= q and u^2 < q+1).

Then 4^p-3^p = 0 (mod u) and so (3/4)^p = 1 (mod u) (since 4 != 0 (mod u))

Let O be the order of 3/4 (mod u).

O divides p and O divides u-1 => u > p => u^2 > p^2.

But q+1 = 2p+2 > u^2 so 2p+2 > p^2 which is impossible since p > 3.

Mike should appreciate this particular use of our favourite (n+1)^p-n^p numbers ;-)

Best regards,

J-L - --- In primenumbers@yahoogroups.com,

"j_chrtn" <j_chrtn@...> wrote:

> Mike should appreciate this particular

Looks good. But you can speed it up by a factor

> use of our favourite (n+1)^p-n^p numbers ;-)

of 2 by working with Mod(3/4,2*p+1)

JL(p)=Mod(3/4,2*p+1)^p==1;

forprime(p=5,10^7,if(JL(p)!=isprime(2*p+1),print([p,fail])));

David > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

Right. 3/4 appears in the proof.

>

> Looks good. But you can speed it up by a factor

> of 2 by working with Mod(3/4,2*p+1)

>

Do you know if this test is something well known?

I ask this because I've never seen it before.

J-L- --- In primenumbers@yahoogroups.com,

"j_chrtn" <j_chrtn@...> wrote:

> Do you know if this test is something well known?

It's not well known because it's quite unnecessary.

The foolproof "unified Sophie" test that works in

*all* cases is much simpler.

If p is prime, then q = 2*p+1 is prime iff 4^p = 1 mod q.

Proof: Simply use base 2 in Pocklington's theorem and

observe that 2^2-1 is coprime to 2*p+1 for every prime p.

Note that this test detects *every* Sophie pair,

including [2,5] and [3,7]. Here is a sanity check:

Pock(p)=Mod(4,2*p+1)^p==1;

forprime(p=2,10^6,if(Pock(p)!=isprime(2*p+1),print(fail)));

\\ The rest is silence, signifying consent

Entia non sunt multiplicanda praeter necessitatem :-)

http://en.wikisource.org/wiki/The_Myth_of_Occam's_Razor

David [second attempt at a reply; apologies if it appears twice] >--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

Unnecessary but at least correct.

>

>

> It's not well known because it's quite unnecessary.

>

>

This is closed to Henri Lifchitz's test 3^p = 1 (mod q) but your's also handles p=2 and p=3.

> If p is prime, then q = 2*p+1 is prime iff 4^p = 1 mod q.

>

>

"There Is More Than One Way To Do It."

> Entia non sunt multiplicanda praeter necessitatem :-)

>

http://en.wikipedia.org/wiki/There's_more_than_one_way_to_do_it

However, some ways are more efficient than others ;-)

Regards,

J-L