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RE reciprocal consecutive primes

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  • Jose Ramón Brox
    ... From: Jan van Oort p^2 mn which is true iff ( m a ) ( condition i ) ...
    Message 1 of 3 , Nov 3, 2005
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      ----- Original Message -----
      From: "Jan van Oort" <glorifier@...>

      p^2 > mn

      which is true iff

      ( m < p AND n < p ) OR ( m = p + a AND n = p - b AND b > a ) ( condition i )

      ----------------------------------------------------
      Jose:

      That's not true.

      Consider b = a-1, then we have

      m*n = (p+a)(p-b) = (p+a)(p-a+1) = p^2 +p-a(a+1)

      If a(a+1) > p, then m*n < p^2, with b < a, m=p+a, n = p-b. You can get similar results
      with different b < a.

      ----------------------------------------------
      Jan:

      now condition ( i ) only holds iff Bertrand's postulate holds
      ------------------------------------------------------------
      Jose:

      Can you explain why? I can't see it at the moment.


      Jose Brox
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