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Re: [PrimeNumbers] more equivalence theorem

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  • Norman Luhn
    Hello Sebastian,  I think it is trivial. m=(p+q)/2 z1=m-k and z2=m+k are prime. m-k+m+k=2m=p+q. So m-k =p and m+k=q , your primes to calculate m. Norman
    Message 1 of 2 , Jun 16, 2012
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      Hello Sebastian, 
      I think it is trivial.

      m=(p+q)/2

      z1=m-k and z2=m+k are prime.

      m-k+m+k=2m=p+q. So m-k =p and m+k=q , your primes to calculate m.

      Norman







      ________________________________
      Von: Sebastian Martin Ruiz <s_m_ruiz@...>
      An: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
      Gesendet: 20:19 Samstag, 16.Juni 2012
      Betreff: [PrimeNumbers] more equivalence theorem


       
      Hello all:

      Also we have:

      Theorem:

      Let p and q consecutive prime numbers >=5

      Let m=(p+q)/2

      m+6 is prime if and only if m-6 is prime.

      Sincerely

      Sebastian Martin Ruiz 

      then for k=2 , 3 , and 6
       m+k is prime if and only if m-k is prime.

      I think is possible to found more k....

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