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Extended conjecture

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  • Kermit Rose
    twin prim conjecture extended Posted by: Sebastian Martin sebi_sebi@yahoo.com sebi_sebi Date: Thu Jul 6, 2006 4:32 pm (PDT) Hello I have extended the
    Message 1 of 1 , Jul 8, 2006
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      twin prim conjecture extended
      Posted by: "Sebastian Martin" sebi_sebi@... sebi_sebi
      Date: Thu Jul 6, 2006 4:32 pm (PDT)

      Hello



      I have extended the conjecture:

      p, q consecutive primes p<q

      are equivalents:

      1) p q are twin primes

      2) z=sqrt[(p^2+q^2)/2-1] is integer

      3) w=sqrt[p*q+1] is integer

      considering 2) ---> 3) where q and p are consecutive primes.

      Suppose

      z^2 = [p^2 + q^2 ] /2 - 1

      (q-p)^2 = p^2 + q^2 - 2 p q

      (q-p)^2 /2 = [ p^2 + q^2 ] /2 - p q

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

      (q-p)^2 /2 - 1 = z^2 - pq

      (q-p)^2 / 2 - 1 - 1 = z^2 - pq - 1

      (q - p)^2 / 2- 2 = z^2 - [ pq + 1 ]


      (q-p)^2 / 2 = z^2 - [pq + 1 ] + 2

      (q - p)^2 = 2 ( z^2 - [ pq + 1 ] ) + 4


      That is probably what you did to get your extension of the conjecture.

      We still need to see why z^2 = pq + 1,

      where z^2 = [ p^2 + q^2 ]/2 - 1,

      and

      we know only that q is the next prime after p.

      As you have already shown,

      If we know that z^2 = pq + 1, then we can prove, q = p + 2.


      (q - p)^2 = 2 ( z^2 - [ pq + 1 ] ) + 4


      That is, you have shown that

      if q is the next prime after p, and

      z^2 = [ p^2 + q^2]/2 - 1 = pq + 1

      then q = p + 2.


      This is not quite your original conjecture, but is now
      cast in a provable form.


      Kermit < kermit@... >















      Kermit < kermit@... >
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