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primes modulo 4

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  • Robert
    No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are [(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then the modulo4 of p is
    Message 1 of 2 , Aug 16, 2007
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      No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are
      [(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then
      the modulo4 of p is either the same as the modulo 2 value or is 1/2 of
      the value.

      It should be possible therefore to determine values k in the power
      series k*4^n+/1, which generate, for all values of n, no factors
      smaller than any given prime value through using a modular sieve
      process which is much more efficient that choosing the equivalent
      primorial or even payam.

      As a result, the potential for long Cunningham Chains base 4 becomes
      apparent.

      So far, the longest I have found (and found after only 3 minutes of
      checking!!) is length 13, namely k= 6703351518. A further 5 hours of
      checking produced a fair selection of length 10 chains.

      Whilst maybe it will be difficult to find one of length 17, it seems
      that the trade off between the modular sieve and working in base 4
      instead of base 2 might work in favour of the sieve, although I have
      no way of proving this.
    • Robert
      ... Hmm, must have been drunk when I wrote this. For modulo read modulo order, or multiplicative order. No base that is square produces a p-1 multiplicative
      Message 2 of 2 , Aug 17, 2007
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        --- In primenumbers@yahoogroups.com, "Robert" <robert_smith44@...> wrote:
        >
        > No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are
        > [(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then
        > the modulo4 of p is either the same as the modulo 2 value or is 1/2 of
        > the value.
        >
        > It should be possible therefore to determine values k in the power
        > series k*4^n+/1, which generate, for all values of n, no factors
        > smaller than any given prime value through using a modular sieve
        > process which is much more efficient that choosing the equivalent
        > primorial or even payam.
        >
        > As a result, the potential for long Cunningham Chains base 4 becomes
        > apparent.
        >
        > So far, the longest I have found (and found after only 3 minutes of
        > checking!!) is length 13, namely k= 6703351518. A further 5 hours of
        > checking produced a fair selection of length 10 chains.
        >
        > Whilst maybe it will be difficult to find one of length 17, it seems
        > that the trade off between the modular sieve and working in base 4
        > instead of base 2 might work in favour of the sieve, although I have
        > no way of proving this.
        >

        Hmm, must have been drunk when I wrote this. For modulo read modulo
        order, or multiplicative order.

        No base that is square produces a p-1 multiplicative order, 4 is the
        smallest square base, it works just as well with bases 9,16,25...
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