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Re: [PrimeNumbers] More twin primes

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  • Jens Kruse Andersen
    ... (p+2)^p % p = 2^p % p, for all numbers. Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp. Fermat s little theorem says all primes
    Message 1 of 3 , Oct 4, 2004
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      David Litchfield wrote:

      > I've just noticed that if p and p+2 are prime then
      >
      > (p+2)^p % p = 2
      > p^(p+2) % (p+2) = p
      >
      > If the first condition is met then p is prime and if the second condition is
      > met then p+2 is prime.
      >
      > I've tested this for twin primes up to 101/103; but of course it needs to be
      > proved. Any pointers on going about proving this?

      (p+2)^p % p = 2^p % p, for all numbers.

      Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp.
      Fermat's little theorem says all primes are 2-prp.
      However some 2-prp are composites called pseudoprimes, specifically 2-psp in
      this case.
      There are infinitely many 2-psp: 341, 561, 645, 1105, 1387, ...
      All Fermat numbers 2^(2^n)+1 are 2-prp.
      All 2-psp below 10^13 have been computed by Richard Pinch:
      http://www.chalcedon.demon.co.uk/carpsp.html

      Setting q = p+2 makes the second condition:
      (q-2)^q % q = q-2

      q satisfying this are called (q-2)-prp, and are exactly the same numbers as
      2-prp.

      --
      Jens Kruse Andersen
    • David Litchfield
      Thanks, Jens! ... (p+2)^p % p = 2^p % p, for all numbers. Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp. Fermat s little theorem
      Message 2 of 3 , Oct 4, 2004
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        Thanks, Jens!

        Jens Kruse Andersen <jens.k.a@...> wrote:David Litchfield wrote:

        > I've just noticed that if p and p+2 are prime then
        >
        > (p+2)^p % p = 2
        > p^(p+2) % (p+2) = p
        >
        > If the first condition is met then p is prime and if the second condition is
        > met then p+2 is prime.
        >
        > I've tested this for twin primes up to 101/103; but of course it needs to be
        > proved. Any pointers on going about proving this?

        (p+2)^p % p = 2^p % p, for all numbers.

        Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp.
        Fermat's little theorem says all primes are 2-prp.
        However some 2-prp are composites called pseudoprimes, specifically 2-psp in
        this case.
        There are infinitely many 2-psp: 341, 561, 645, 1105, 1387, ...
        All Fermat numbers 2^(2^n)+1 are 2-prp.
        All 2-psp below 10^13 have been computed by Richard Pinch:
        http://www.chalcedon.demon.co.uk/carpsp.html

        Setting q = p+2 makes the second condition:
        (q-2)^q % q = q-2

        q satisfying this are called (q-2)-prp, and are exactly the same numbers as
        2-prp.

        --
        Jens Kruse Andersen



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