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finite essay2

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  • Bill Bouris
    edited from earlier e-mail... no comments? the pair-wise effort has to be continuous to produce more Fer-mat primes. ... should say... has been once removed,
    Message 1 of 1 , Sep 16, 2007
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      edited from earlier e-mail... no comments? the
      'pair-wise effort has to be continuous to produce more
      Fer-mat primes.

      > Group,... know that this is a modest effort to
      > explain why only so few Fermat numbers are 'prime'.
      > Overview:
      > It can be shown that F(1)...F(4) are the only Fermat
      > numbers that can establish and maintain a continual
      > ‘pair-wise’ condition; these Fermat numbers are all
      > ‘prime’ numbers, and F(0) cannot be recognized using
      > the ‘pair-wise’ condition.
      > Demonstration:
      > If a Fermat number is of the form F(x)=
      > 2^(2^(2^x))+1
      > and x is a /whole number/, then a continual ‘pair-
      > wise’ condition can be supported by these modulo
      > relationships:
      > I. If x is a /whole number/, then G(x)=
      > (2^(2^(2^(x+1)))+1) == 2^q (mod (2^x+1)) where q is
      > /zero or an odd natural number/ which reduces to...
      > II. H(x)= (2^(x+1)+1)== (2n+1) (mod (x+1)) where the
      > expression (2n+1) represents the number ‘q’ from the
      > previous equation, and the exponential argument has
      > been neatly preserved.

      should say... has been once removed, not... preserved.

      > First, F(0)= 3 cannot be tested using H(x) since the
      > ‘modulo’ portion of the equation doesn’t make sense;
      > this Fermat number is considered to be similar to
      > that of geometric point at/near infinity.
      > Now, if x=0, then H(0)= 3 == (2n+1) (mod 1) suggests
      > that [n=1] since a unitary modulo implies the ex-
      > pression is ‘to scale’, and 2 == 2n is a valid com-
      > parison without the need for modular reduction;
      > if x=1, then H(1)= 5 == (2n+1) (mod 2) suggests that
      > [n=0].
      > The ‘pair-wise’ condition for both an even and odd
      > ‘x’ has been established, and a change in the value
      > for ‘n’ would indicate a change in the condition.
      > If x=2, then H(2)= 9 == (2n+1) (mod 3) suggests that
      > [n=1], and the ‘pair-wise’ condition for an even ‘x’
      > is maintained.
      > If x=3, then H(3)= 17 == (2n+1) (mod 4) suggests
      > that [n=0], and the ‘pair-wise’ pattern is unchanged
      > for an odd ‘x’.
      > However, if x=4, then H(4)= 33 == (2n+1) (mod 5)
      > suggests that [n=0]. The value of ‘n’ has changed,

      should be [n=1]... 'n' hasn't changed... it will occur
      for the next value of 'x' since the exponential
      argu-ment has been once removed.
      however, G(4) has revealed that 2^q = 62 and 'q' isn't
      an /odd whole number/

      > a loss of continuity is confirmed since F(5) tests
      > as a ‘composite’ number.
      > If x=5, then H(5)= 65 == (2n+1) (mod 6) suggests
      > that [n=2]; the ‘pair-wise’ condition and its con-
      > tinuity for an odd ‘x’ have also been lost as we
      > find that F(6) tests as a ‘composite’ number.
      > Conclusion:
      > If G. H. Hardy and E. M. Wright were correct to in-
      > sist that a ‘pair-wise’ condition is continually
      > linked to the presence of Fermat ‘prime’ numbers,
      > then I have demonstrated that the F(1)...F(4) are
      > the only Fermat numbers that can retain this con-
      > tinuity. F(0) is a small ‘prime’ number also defin-
      > ed by the Fermat number formula, but a ‘pair-wise’
      > condition cannot be recognized.

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