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Re: twin primes and 2modp

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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 1 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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