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

Re: [PrimeNumbers] A property of twin primes, only?

Expand Messages
  • Sebastian Martin
    I dont spoke about you. There are other e-mails in this list that said this result is trivial. You don t understand me correctly. Sorry my english is very
    Message 1 of 2 , Jul 4, 2006
    View Source
    • 0 Attachment
      I dont spoke about you. There are other e-mails in
      this list that said this result is trivial.
      You don't understand me correctly. Sorry my english is
      very little.

      Your e-mail is very intersting. I put it by a correct
      example.

      I am sorry that I have been understood badly.
      --- mikeoakes2@... escribió:

      > Sebastian
      >
      > I didn't say it was trivial.
      >
      > I said
      > "p&q are twin primes => z is integer"
      > is trivial.
      >
      > Isn't it??
      >
      > Mike Oakes
      >
      >
      > -----Original Message-----
      > From: Sebastian Martin <sebi_sebi@...>
      > To: lista de primos <primenumbers@yahoogroups.com>
      > Sent: Tue, 4 Jul 2006 17:03:44 +0200 (CEST)
      > Subject: [PrimeNumbers] A property of twin primes,
      > only?
      >
      >
      > 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.
      >
      > This is not a trivial result.
      >
      > Sincerely
      > Sebastian Martin Ruiz
      >
      > See you below:
      >
      > The "only if" is trivial:
      > If p and q are twin primes, then q = p+2, so z =
      > (p+1).
      >
      > As for the "if":
      > The conjecture is true for p = 2, so assume p >= 3.
      >
      > If we write
      > q = p+2*d
      > then
      > z^2 = (p+d)^2 + (d^2-1)
      >
      > If (p+d)^2 + (d^2-1) < (p+d+1)^2, then z^2 is not a
      > perfect square.
      > This condition is:
      > q < p + 2*(sqrt(2*p+3)+1)
      > which is easily verified to hold for p <= 10^9.
      > It is believed (but has never been proved) to hold
      > for
      > all p >= 3.
      >
      > -Mike Oakes
      >
      >
      > ______________________________________________
      > LLama Gratis a cualquier PC del Mundo.
      > Llamadas a fijos y móviles desde 1 céntimo por
      > minuto.
      > http://es.voice.yahoo.com
      >
      >
      >




      ______________________________________________
      LLama Gratis a cualquier PC del Mundo.
      Llamadas a fijos y móviles desde 1 céntimo por minuto.
      http://es.voice.yahoo.com
    Your message has been successfully submitted and would be delivered to recipients shortly.