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RE A property of twin primes, only?

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  • Jose Ramón Brox
    ... From: Sebastian Martin Let p q consecutive prime numbers p
    Message 1 of 2 , Jul 5, 2006
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      ----- Original Message -----
      From: "Sebastian Martin" <sebi_sebi@...>


      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.

      ==============================

      Some random (useless?) facts:

      2(z^2 +1) = p^2 + q^2
      2(z+i)(z-i) = (p+q·i)(p-q·i)

      Since z+i <> p+q·i, z+i <> p-q·i and p,q >1,
      z+i and z-i can't be gaussian primes.
      Then z^2+1 is not a prime.

      As 2(z^2+1) is a sum of two squares, it must be a product of factors of 2 and primes of
      the form 4n+1.
      But since we can consider p,q > 2, then p^2 + q^2 = = 2 (mod 4), it only can have one
      factor of 2, so
      z^2+1 is a product of primes of the form 4n+1.

      2(z^2+1) = p^2 + q^2
      2(z^2+1) = = q^2 (mod p)
      2(z^2+1) = = p^2 (mod q)

      2(z^2+1) is a quadratic residue in Zp and in Zq.

      Moreover, if q = p+2d, then

      z^2+1 = = 2d^2 (mod p) and
      z^2+1 = = 2d^2 (mod q)

      2d^2-1 = = z^2 (mod p)
      2d^2 -1 = = z^2 (mod q)

      So, 2d^2 -1 is a quadratic residue in Zp and Zq

      In brief:
      a) z^2+1 is a product of (at least two) primes of the form 4n+1.
      b) 1 = Legendre(2(z^2+1),p) = Legendre(2(z^2+1),q) = Legendre(2d^2-1,p) =
      Legendre(2d^2-1,q)
      c) z^2+1 = = 2d^2 in Zp and in Zq.

      A related question can then be:
      If x is a quadratic residue of p,q primes and 2·x-1 is also a qr of p,q... is it
      necessarily x = 1?

      I think the answer is probably "no", but it is a necessary condition, so if the answer is
      "yes", then the problem is solved.

      Regards. Jose Brox
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