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twin primes conjecture: A property of twin primes, only?

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  • Kermit Rose
    Re: Discussion between Sebastian Martin Ruiz and Mike Oakes ... sqrt([ (23^2 + 25^2)/2 -1) = sqrt(529 + 625)/2 - 1) = sqrt(1154/2 -1) =sqrt(577 - 1) =
    Message 1 of 1 , Jul 4, 2006
      Re: Discussion between

      Sebastian Martin Ruiz
      and
      Mike Oakes




      > Let p q consecutive prime numbers p<q.
      > Let z=sqrt[(p^2+q^2)/2-1]
      >


      > Conjecture: p&q ares twin primes IF AND ONLY IF z is
      > integer.



      sqrt([ (23^2 + 25^2)/2 -1)

      = sqrt(529 + 625)/2 - 1)
      = sqrt(1154/2 -1)
      =sqrt(577 - 1)
      = sqrt(576)
      = 24

      Is this a counter example to the if part?

      The only if part can be easily proven,
      for it follows from the fact that of two twin primes,
      p,q with p < q, that

      p is of the form 6 D -1 and q is of the form 6 D + 1.

      [ Except for the twin primes 3 and 5 ]

      Let D be a positive integer that is not any of the four
      forms

      6 m n - m - n
      6 m n - m + n
      6 m n + m - n
      6 m n + m + n

      with m > 0, n > 0.

      Then both 6D -1 and 6 D +1 are twin primes,

      and all sets of twin primes are of this form.

      D = 1 is permitted because 1 is not any of the four forms.

      4 = 6 - 1 - 1 is the smallest positive integer of any of these forms.


      So D= 1 yields the twin primes 5 and 7.
      D = 2 yields the twin primes 11 and 13.
      D = 3 yields the twin primes 17 and 19.

      Because 4 = 6 - 1 -1,

      23 and 25 are not twin primes.



      Let p and q be twin primes,
      p > 3.

      Set p = 6 D -1
      q = 6 D + 1

      [ p^2 + q^2 ] = [ 36 D^2 - 12 D + 1 ] + [ 36 D^2 + 12 D + 1 ]
      = 72 D^2 + 2


      [ p^2 + q^2 ] /2 = 36 D^2 + 1


      sqrt[(p^2+q^2)/2-1] = 6 D = p + 1
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