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Re: Unified test for Sophie Germain primes

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  • 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 1 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 2 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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