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Re: [PrimeNumbers] Extension of Dirchlet Theorem

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  • Lawrence Hon
    I dont think any exist besides a few trivial cases. First, p must be congruent to either 1 or 2 mod 3 because if it was congruent to 0 it would be a multiple
    Message 1 of 2 , Nov 1, 2003
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      I dont think any exist besides a few trivial cases.

      First, p must be congruent to either 1 or 2 mod 3 because if it was
      congruent to 0 it would be a multiple of 3.

      so we have two cases:

      case 1:
      p = 1 mod 3
      2p-1 = 1 mod 3
      4p-1 = 3 mod 3 = 0, so 4p-1 is a multiple of 3 (not prime)

      case 2:
      p = 2 mod 3
      2p-1 = 3 mod 3 = 0 so 2p -1 is a multiple of 3 (not prime)
      4p-1 = 5 mod 3 = 2

      thus, this never works unless we take 3 to be that multiple.

      so this implies in case 1, p = 1 (not a solution)
      or in case 2, 2p - 1 = 3, p = 2
      so p = 2, 2p-1 = 3, 4p-1 = 7 so (2,3,7) is a solution

      lastly, p could be = 0 mod 3, so p = 3, 2p-1 = 5, 4p-1 = 11
      so (3,5,11) is also a solution.

      Thus there are only two solutions and your search is futile :)

      Lawrence




      On Sat, 1 Nov 2003, Adam wrote:

      > RE: Adam Karasek post for pythagorean triples with a and c primes and
      > b a product of at almost four primes, I seek primes p with 2p-1 and
      > 4p-1 simlutaneously prime (to reformulate, we could write p=2n+1, and
      > then seek 2n+1,4n+1,8n+1 simultaneously prime).  I am having poor
      > luck finding such p values, contrary to the belief in the extension
      > of Dirichlet's theorem.  For instance, I generate a random 12 digit
      > prime, search the through the next 5000 primes, and find about 200-
      > 300 that have 2p-1 prime, but only 0 of those having 4p-1 prime.
      >
      > Am I not being patient enough or...?
      >
      > Adam
      >
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