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Re: Unified test for Sophie Germain primes

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  • j_chrtn
    ... Right. 3/4 appears in the proof. Do you know if this test is something well known? I ask this because I ve never seen it before. J-L
    Message 1 of 5 , Jan 19, 2010
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      > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > Looks good. But you can speed it up by a factor
      > of 2 by working with Mod(3/4,2*p+1)
      >

      Right. 3/4 appears in the proof.

      Do you know if this test is something well known?
      I ask this because I've never seen it before.

      J-L
    • djbroadhurst
      ... 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 =
      Message 2 of 5 , Jan 20, 2010
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        --- 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]
      • j_chrtn
        ... Unnecessary but at least correct. ... This is closed to Henri Lifchitz s test 3^p = 1 (mod q) but your s also handles p=2 and p=3. ... There Is More Than
        Message 3 of 5 , Jan 20, 2010
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          >--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
          >
          >
          > It's not well known because it's quite unnecessary.
          >
          Unnecessary but at least correct.

          >
          > If p is prime, then q = 2*p+1 is prime iff 4^p = 1 mod q.
          >
          This is closed to Henri Lifchitz's test 3^p = 1 (mod q) but your's also handles p=2 and p=3.

          >
          > Entia non sunt multiplicanda praeter necessitatem :-)
          >

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