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Re: [PrimeNumbers] Re: Prime equivalence theorem

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  • Norman Luhn
    Another argument is: m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q (q-p)=2k
    Message 1 of 6 , Jun 17, 2012
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      Another argument is:

      m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q

      (q-p)=2k


      ________________________________
      Von: Maximilian Hasler <maximilian.hasler@...>
      An: Mark <mark.underwood@...>
      CC: primenumbers@yahoogroups.com
      Gesendet: 8:19 Sonntag, 17.Juni 2012
      Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem


       
      I'm not sure if I understand completely what you meant to say, because you
      use "the same x" for (q-p)/2 (which varies with each pair (p,q))
      and for the m +/- x (which is constant for the whole procedure /
      "theorem", e.g. x=2 in the theorem of this thread).

      But indeed these "theorems" can easily be proved by analysing the possible
      values in a similar way :

      q-p > 2 d => m +/- d are both composite, since strictly comprised between
      the two consecutive primes.

      q-p = 2 d => m +/- d are the two primes p & q.

      Thereafter, there may be remaining cases (for d>1) :

      If d=2,
      q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)

      If d=3,
      q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)
      q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)

      If d=4 or d=5, exceptions are possible.

      If d=6,
      q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)
      q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)
      q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)
      q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)
      q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)

      This shows that the Theorem holds for d=1, 2, 3 and 6.

      I think that for all other d there are infinitely many counter-examples.

      Maximilian

      [Non-text portions of this message have been removed]




      [Non-text portions of this message have been removed]
    • Mark
      Maximilian is correct in that he actually shows why m-k must be p and m+k must be q. It really has to do with admissible 3 - prime constellations of a given
      Message 2 of 6 , Jun 17, 2012
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        Maximilian is correct in that he actually shows why m-k must be p and m+k must be q. It really has to do with admissible 3 - prime constellations of a given length.

        Mark


        --- In primenumbers@yahoogroups.com, Norman Luhn <n.luhn@...> wrote:
        >
        > Another argument is:
        >
        > m-k must be p and m+k must be q,  otherwise  m-k is the consecutive prime to p .. and that is q
        >
        > (q-p)=2k
        >
        >
        > ________________________________
        > Von: Maximilian Hasler <maximilian.hasler@...>
        > An: Mark <mark.underwood@...>
        > CC: primenumbers@yahoogroups.com
        > Gesendet: 8:19 Sonntag, 17.Juni 2012
        > Betreff: Re: [PrimeNumbers] Re: Prime equivalence theorem
        >
        >
        >  
        > I'm not sure if I understand completely what you meant to say, because you
        > use "the same x" for (q-p)/2 (which varies with each pair (p,q))
        > and for the m +/- x (which is constant for the whole procedure /
        > "theorem", e.g. x=2 in the theorem of this thread).
        >
        > But indeed these "theorems" can easily be proved by analysing the possible
        > values in a similar way :
        >
        > q-p > 2 d => m +/- d are both composite, since strictly comprised between
        > the two consecutive primes.
        >
        > q-p = 2 d => m +/- d are the two primes p & q.
        >
        > Thereafter, there may be remaining cases (for d>1) :
        >
        > If d=2,
        > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 2 both are composite (even)
        >
        > If d=3,
        > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 3 both are composite (in 3Z)
        > q-p = 4 => m +/- d = p-1 resp. q+1, both are composite (even)
        >
        > If d=4 or d=5, exceptions are possible.
        >
        > If d=6,
        > q-p = 2 => p=6k-1, q=6k+1 => m +/- d = 6k +/- 6 both are composite (in 6Z)
        > q-p = 4 => p=6k+1, q=6k+5, m=6k+3 => m +/- d both are composite (in 3Z)
        > q-p = 6 => m=p+3=q-3 is even => m +/- d both are composite (even)
        > q-p = 8 => p=6k-1, q=6k+7, m=6k+3 => m +/- d both are composite (in 3Z)
        > q-p = 10 => m=p+5=q-5 is even => m +/- d both are composite (even)
        >
        > This shows that the Theorem holds for d=1, 2, 3 and 6.
        >
        > I think that for all other d there are infinitely many counter-examples.
        >
        > Maximilian
        >
        > [Non-text portions of this message have been removed]
        >
        >
        >
        >
        > [Non-text portions of this message have been removed]
        >
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