Loading ...
Sorry, an error occurred while loading the content.

18177Re: [PrimeNumbers] A property of prime twins, only?

Expand Messages
  • Ronny Edler
    Jul 1, 2006
      > 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
    • Show all 3 messages in this topic