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Re: [PrimeNumbers] twin primes and 2modp

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  • Phil Carmody
    ... That s equivalent to: for every prime p 3, p-2 has at least one prime factor and thus is always satisfied. Phil () ASCII ribbon campaign ()
    Message 1 of 3 , May 1, 2006
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      --- Robert <rw.smith@...> wrote:
      > "Every prime p, greater than 3, is 2mod(at least one prime less than p)"

      That's equivalent to:
      for every prime p>3, p-2 has at least one prime factor
      and thus is always satisfied.

      Phil


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    • Robert
      ... than p) ... Phil, as always, to the rescue. Good man! Think I ll stick to prime searching rather than mathematics. I am not yet at class 101 stage, and it
      Message 2 of 3 , May 1, 2006
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        --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
        >
        > --- Robert <rw.smith@...> wrote:
        > > "Every prime p, greater than 3, is 2mod(at least one prime less
        than p)"
        >
        > That's equivalent to:
        > for every prime p>3, p-2 has at least one prime factor
        > and thus is always satisfied.
        >
        > Phil

        Phil, as always, to the rescue. Good man!

        Think I'll stick to prime searching rather than mathematics. I am not
        yet at class 101 stage, and it shows. Still, managed to waste 3 hours
        even if non-productively.

        For what its worth, the jumping champions (refer to top message) were
        as below.

        Regards

        Robert

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