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A property of prime twins, only?

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  • Sebastian Martin
    Hello All: you can see this conjecture at Primepuzzles.net: Problems & Puzzles: Conjectures Conjecture 44. A property of prime twins, only? Sebastián Martín
    Message 1 of 3 , Jul 1 6:05 AM
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      Hello All:

      you can see this conjecture at Primepuzzles.net:

      Problems & Puzzles: Conjectures

      Conjecture 44. A property of prime twins, only?

      Sebastián Martín Ruiz sends the following conjecture
      from his own invention:

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

      Conjecture: p&q are prime twins iff z is integer.

      Question: Prove it or show it false.

      Sincerely

      Sebastian Martin Ruiz

      http://perso.wanadoo.es/smaranda









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    • Ronny Edler
      ... Since p and q are twin primes q=p+2 So z becomes sqrt( (p^2 + (p+2)^2)/2 -1) = sqrt( (p^2 + p^2 + 4p +4)/2 -1) = sqrt( (2p^2+4p+4)/2 -1 ) = sqrt(p^2+2p+2 -
      Message 2 of 3 , Jul 1 7:33 AM
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        > Let p, q be consecutive prime numbers, p<q.
        > Let z=sqrt[(p^2+q^2)/2-1]
        >
        > Conjecture: p&q are prime twins iff z is integer.

        Since p and q are twin primes q=p+2

        So z becomes
        sqrt( (p^2 + (p+2)^2)/2 -1) =
        sqrt( (p^2 + p^2 + 4p +4)/2 -1) =
        sqrt( (2p^2+4p+4)/2 -1 ) =
        sqrt(p^2+2p+2 - 1) =
        sqrt(p^2+2p+1) =
        sqrt( (p+1)^2 ) =
        p+1


        Now let p and q be any (non-twin!) consecutive primes with 2 < p < q
        Now there exist an n>1 such that q=p+2*n

        Doing the same as above leads to:

        z = sqrt((p^2 + p^2+4np+ 4n^2 )/2 -1)
        = sqrt(p^2+2np+2n^2-1)

        Its roots are -n +/- sqrt( (-n)^2 - 2n^2+1 ) = -n +/- sqrt( n^2 - 2n^2+1 )
        = -n +/- sqrt( -n^2+1 )

        which has solutions in integers only for n=0,1

        So the conjecture is indeed true.

        Ciao,
        Ronny
      • David Cleaver
        ... Let p = a and q = a+2. Let a = any integer. Then: (a^2 + (a+2)^2)/2 - 1 = (a^2 + a^2 + 4a + 4)/2 - 1 = (2a^2 + 4a + 4)/2 - 1 = (a^2 + 2a + 2) - 1 = a^2 +
        Message 3 of 3 , Jul 1 8:21 AM
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          > Let p, q be consecutive prime numbers, p<q.
          > Let z=sqrt[(p^2+q^2)/2-1]
          >
          > Conjecture: p&q are prime twins iff z is integer.
          >
          > Question: Prove it or show it false.

          Let p = a and q = a+2.
          Let a = any integer. Then:
          (a^2 + (a+2)^2)/2 - 1 =
          (a^2 + a^2 + 4a + 4)/2 - 1 =
          (2a^2 + 4a + 4)/2 - 1 =
          (a^2 + 2a + 2) - 1 =
          a^2 + 2a + 1 =
          (a+1)*(a+1) =
          (a+1)^2

          So, the the sqrt[(a+1)^2] = (a+1) an integer.
          There is no requirement for (a, a+2) to be prime. They can be any two
          integers that differ by 2.

          The equation does hold for twin primes and none of the other primes, but
          it also holds for non-prime integers as well. So, the equation, by
          itself, tells us nothing about the character of the numbers.

          Also, its not a very helpful test since you start with the statement:
          > Let p, q be consecutive prime numbers, p<q.
          At this point, since you already know that p and q are prime numbers,
          all you have to do is take their difference to see if they are twins.

          However, if you remove the constraint of knowing beforehand that they
          are primes, then the test itself doesn't reveal anything about the
          prime/non-prime character of the numbers. It will just tell you if the
          numbers differ by 2.

          -David C.
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