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Fermat primes

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  • Bill Bouris
    Group, Q: Why are the number of them... finite? A: They are bounded by two modulo conditions, namely... G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1)) H(x):
    Message 1 of 2 , Sep 25, 2007
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      Group,

      Q: Why are the number of them... finite?

      A: They are bounded by two 'modulo' conditions,
      namely...

      G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1))
      H(x): 2^(x+1) +1 == (2n+1) mod (x+1)

      G. H. Hardy and E. M. Wright predicted the latter
      condition and would have labeled the Fermat primes as
      'pair-wise' when an F(odd x)/F(even x) gave the result
      of [n=1]/[n=0], respectively. I noticed that this
      'modulo' equation explains why they thought that the
      Fermat primes were consecutively linked.

      I conjecture that if two consecutive Fermat numbers
      are 'pair-wise' by H(x) and also 'congruent' by G(x),
      then F(odd x)/F(even x) are necessarily prime!

      It's an idea that would undoubtedly be hard to prove,
      but never-the-less, H(x) can be easily calculated, and
      G(x) can be verified using the 'Try GMP!' evaluator on
      the googled 'GNU Bignum Library' website.

      That's why F(1)/F(2) and F(3)/F(4) are the only Fermat
      primes in the sequence; the G(x) and H(x) 'modulo'
      clocks are wound together so tightly.

      Bill



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    • leavemsg1
      ... Fermat or Fantasy ????? Let F(x)= 2^(2^x)+1 represent a Fermat number such that `x is a /whole number/. The first few... F(0) = 3, F(1) = 5, F(2) = 17,
      Message 2 of 2 , Oct 14, 2007
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        --- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...>
        wrote:
        >
        > Group,
        >
        > Q: Why are the number of them... finite?
        >
        > A: They are bounded by two 'modulo' conditions,
        > namely...
        >
        > G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1))
        > H(x): 2^(x+1) +1 == (2n+1) mod (x+1)
        >
        > G. H. Hardy and E. M. Wright predicted the latter
        > condition and would have labeled the Fermat primes as
        > 'pair-wise' when an F(odd x)/F(even x) gave the result
        > of [n=1]/[n=0], respectively. I noticed that this
        > 'modulo' equation explains why they thought that the
        > Fermat primes were consecutively linked.
        >
        > I conjecture that if two consecutive Fermat numbers
        > are 'pair-wise' by H(x) and also 'congruent' by G(x),
        > then F(odd x)/F(even x) are necessarily prime!
        >
        > It's an idea that would undoubtedly be hard to prove,
        > but never-the-less, H(x) can be easily calculated, and
        > G(x) can be verified using the 'Try GMP!' evaluator on
        > the googled 'GNU Bignum Library' website.
        >
        > That's why F(1)/F(2) and F(3)/F(4) are the only Fermat
        > primes in the sequence; the G(x) and H(x) 'modulo'
        > clocks are wound together so tightly.
        >
        > Bill
        >
        > more info...

        Fermat or Fantasy ?????

        Let F(x)= 2^(2^x)+1 represent a Fermat number such that `x' is
        a /whole number/.

        The first few... F(0) = 3, F(1) = 5, F(2) = 17, F(3) = 257, F(4) =
        65537... are all prime. F(5) has a factor of 641.

        Here's evidence that "No more Fermat primes can exist!".

        I have characterized the notion of `pair-wise'ness that G. H. Hardy
        and E. M. Wright have conjectured about `F(odd x)/F(even x)' compan-
        ions.

        Fermat numbers are `pair-wise' iff... the condition H(x): 2^(x+1)+1 ==
        (2n+1) (mod (x+1)) implies that [n=1]/[n=0], respectively and in suc-
        cession.

        Fermat numbers are `prime' iff... the `pair-wise' compan-ions both
        satisfy the congruency G(x): F(x+1) == 2^q (mod (2^x +1)); the `pair-
        wise' companions must imply an integral solution for `q'.

        F(0) can't be evaluated under G(x), so it's considered to be similar
        to a geometric point at/near infinity.

        If F(1)= 5 and F(2)= 17, then we have...
        F(1)=> H(0): 3 == (2n+1) (mod 1) implies [n=1], and...
        F(2)=> H(1): 5 == (2n+1) (mod 2) implies [n=0].
        F(1) & F(2) are `pair-wise', but are they `prime'???

        G(0): 5==(1 = 2^0)(mod 2) and G(1): 17==(2 = 2^1)(mod 3); F(1) & F(2)
        are considered `prime' because the congruency, G(x), has the integral
        solutions 0 & 1, respectively.
        Note: determining the primality of a Fermat number is possible using G
        (x) because of its unusual `double-exponential' form.

        If F(3)= 257 and F(4)= 65537, then we have...
        F(3)=> H(2): 9 == (2n+1) (mod 3) implies [n=1], and...
        F(4)=> H(3): 17 == (2n+1) (mod 4) implies [n=0].
        F(3) & F(4) are `pair-wise', but are they `prime'???

        G(2): 257==(2 = 2^1)(mod 5) and G(3): 65537==(8 = 2^3)(mod 9); F(3) &
        F(4) are `prime' again because of the congruency G(x) has integral
        solutions,... namely 1 & 3.

        Someone may eventually notice the following:
        [The `pair-wise' Fermat candidates] have the form...
        If F(Mp)=> H(Mp-1): 2^(Mp)+1 == (2n+1) mod(Mp) and F(Mp+1)=> H(Mp): 2^
        (Mp+1)+1 == (2n+1) (mod(Mp+1)) both imply [n=1], [n=0], respectively,
        then the F(odd x)/F(even x) companions are `pair-wise', but are
        they `prime'???
        where Mp is the usual Mersenne prime.

        They are `prime' iff both are congruent to 2^q, for some q.
        F(Mp):G(Mp-1): 2^(2^(Mp))+1 == 2^q (mod(2^(Mp-1)+1)) and
        F(2^y):G(2^y-1): 2^(2^(2^y))+1 == 2^q (mod(2^(2^y-1)+1))
        using the `Try GMP!' feature on the GNU Bignum Library web-site to
        evaluate the modulo equation, G(x), when it becomes too large.
        Again, Mp is the usual Mersenne prime, and 2^y is the power of two
        that is just one larger than a Mp.

        Look at the `2^q' values in following table...

        For those `x' values that passed the `pair-wise' condition:

        (x+1) | (x) | (2^q)== (after applying the G(x) congruency)

        1 0 1 == 2^0
        2 1 2 == 2^1
        They are `pair-wise' and `prime'!

        3 2 2 == 2^1
        4 3 8 == 2^3
        They are `pair-wise' and `prime'!
        ...

        7 6 62 <> 2^q
        8 7 17 and so on...
        They are `pair-wise' but `composite'(not congruent)
        ...

        31 30 257
        32 31 17
        They are `pair-wise' but `composite'...
        ...

        127 126 85070591730234615865843651857942052862 <> 2^q
        128 127 17
        They are `pair-wise' but `composite'...
        ...

        8191 8190 (odd), however large it is...
        8192 8191 17
        They are `pair-wise' but `composite'...
        ...

        131071 131072 (evaluator couldn't compute modulo)
        131072 131073 (same)
        ...

        Can someone compute 2^(2^(131071+1))+1 == 2^q (mod ((2^131071)+1))...
        to further end the mystery ???

        I believe that another Fermat prime will not be found.

        Bill
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