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Primality and Dr. B's Fab Fibo Formula

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  • aldrich617
    Yes! We have more and yet similar conjectures concerning consistant patterns of p s and c s, found this time on quadratic sequences derived from the quartic
    Message 1 of 1 , Dec 30, 2008
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      Yes! We have more and yet similar conjectures concerning
      consistant patterns of p's and c's, found this time on quadratic
      sequences derived from the quartic forms of the complicated
      expression shown next.

      On 3/30/2006 Dr David Broadhurst wrote:

      " Let C be any positive integer all of whose prime divisors
      are congruent to 1 module 10. Let B be the smallest positive
      even integer coprime to C such that B^2 + C^2 = 5a^2
      for some positive integer a.
      Then for every pair of integers (k,x), f(k,a,B,C,x) =
      (5*(2*x -1)^4 + 5*(2*a*fibonacci(6*k + 1) + (B - a)*
      fibonacci(6*k)) * (2*x -1)^2 + C^2)/16 is a positive integer
      all of whose prime divisors are congruent to 1 modulo 10.

      The Fibonacci series, extended in both directions, generates
      all of the polynomials for each value of C."

      Fortuitously, it appears that all of the equations represented
      by Dr. Broadhurst's formula, their number infinite in three
      directions, can be manipulated in a rational and consistent
      manner to produce quadratic equations whose prime and
      composite terms may be distinguished by simply counting the
      number of squares occurring in in a relatively small
      calculated interval. The method used to accomplish this
      is very similar to the one explained in the recent posting on
      this website on 12/22/8. Although all assertions herein remain
      to be proven, it seems a good bet that they are true, as the
      first 15000 iterations of twenty randomly chosen polynomials
      from Dr. Broadhurst's list yielded no counterexamples. Below
      are two typical examples of fourth degree equations on the list,
      together with a description of how they may be changed into
      quadratics with the desired characteristics suitable for simple
      theorems on primality. Perhaps it is not a stretch to believe
      that more general conjectures along these lines applying to the
      entire list are possible and may one day be proven.

      [A]

      Starting with F(x) = 5x^4 - 10x^3 + 100x^2 - 95x + 31,
      Let a1 = F(1) = 31 and a3 = F(2 ) = 241, and interpolate
      a2 = (a1 +a3 -10)/2 = 131 into the sequence 31, 131, 241....
      A quadratic equation for these values that begins with 31
      at x = 0 is A(x) = 5x^2 + 95x + 31. The second difference
      between any three of its consecutive terms is always 10,
      and the values of the original fourth degree form are all
      embedded within it. Let q = 1 + 2(a2 -a1)/10 = 21 and
      let q(x) = q + 2x, so q(0) = q = 21.

      For some particular value y then, there will exist
      q(y) = q + 2y,and 5( q(y) + 6p)^2 - 4( A(y + 2p) ) =
      (r + 10p)^2 where for all p >= 0 , r is an integer. At
      y = 13 these relations hold (ex : at p = 5, 5( 47 + 30)^2
      - 4(4861) = (51 + 50)^2 ) so we will rewrite A(x) at a
      new starting point as A(x) = 5x2 + 215x + 1891, x >= 0,
      q = q(0) = 45.
      For some particular value m then there will exist
      5( q(0) + 2m + 6p)^2 - 4( A(2p) ) = (s + 10p)^2 . where
      for all p>= 0 , s is an integer. At m = 20 these relations hold
      (ex: at p = 4, 5( 45 + 40 +24)^2 - 4(3931) = (169 + 40)^2 )

      x, A(x), q, q(x), z, T(z),s : integers;
      T(z) = 5z^2 - 4(A(x));
      new A(x) = 5x2 + 215x + 1891 ;
      q = 45,
      q(x) = q + 2x,
      z odd integer;

      19) For all x >= 0, A(x) will be prime if exactly
      one z value exists in the interval
      q(x) <= z <= q(0) + 40 + 3x
      such that T(z) is a square of an integer.
      Otherwise A(x) will be composite.

      (*note: only the square values
      of T(z), and the first value of T(z) will appear in
      the examples. All other T(z) will be designated as '-'.)

      Example: if x = 2, then A(x) = 2341, 49 <= z <= 91
      and the values of T(z) are:
      2641--------------------179^2
      One of these is a square, therefore 2341 is prime.

      Example: if x = 3, then A(x) = 2581, 51<= z <= 94
      and the values of T(z) are:
      2681 61^2---91^2-------139^2--------
      Three of these are squares, therefore 2581 is composite.

      [B]

      Starting with F(x) = 5x^4 - 10x^3 + 350x^2 - 345x + 401,
      Let a1 = F(1) = 401 and a3 = F(2 ) = 1111, and interpolate
      a2 = (a1 +a3 -10)/2 = 751 into the sequence 401, 751, 1111....
      A quadratic equation for these values that begins with 401
      at x = 0 is A(x) = 5x^2 + 345x + 401. The second difference
      between any three of its consecutive terms is always 10,
      and the values of the original fourth degree form are all
      embedded within it. Let q = 1 + 2(a2 -a1)/10 = 71 and
      let q(x) = q + 2x, so q(0) = q = 71.

      For some particular value y then, there will exist q(y) = q + 2y,
      and 5( q(y) + 6p)^2 - 4( A(y + 2p) ) = (r + 10p)^2 where
      for all p>= 0 , r is an integer. At y = 42 these relations hold
      (ex : at p = 5, 5( 155 + 30)^2 - 4(31861) = (159 + 50)^2 )
      so we will rewrite A(x) at a new starting point as
      A(x) = 5x2 + 755x + 22951, x >= 0, q = q(0) = 153.
      For some particular value m then, there will exist
      5( q(0) + 2m + 6p)^2 - 4( A(2p) ) = (s + 10p)^2 . where
      for all p>= 0 , s is an integer. At m = 74 these relations hold
      (ex: at p = 4, 5( 153 + 148 +24)^2 - 4(29311) = (601 + 40)^2 )

      x, A(x), q, q(x), z, T(z),s : integers;
      T(z) = 5z^2 - 4(A(x));
      A(x) = 5x2 + 755x + 22951 ;
      q = 153,
      q(x) = q + 2x,
      z odd integer;

      20) For all x >= 0, A(x) will be prime if exactly
      one z value exists in the interval
      q(x) <= z <= q(0) + 148 + 3x
      such that T(z) is a square of an integer.
      Otherwise A(x) will be composite.

      (*note: only the square values
      of T(z), and the first value of T(z) will appear in
      the examples. All other T(z) will be designated as '-'.

      Example: if x = 3, then A(x) = 25261, 159 <= z <= 310
      and the values of T(z) are:
      25361 169^2----------------------------------------
      ----------------------------------
      One of these is a square, therefore 25261 is prime.

      Example: if x = 4, then A(x) = 26051, 161<= z <= 313
      and the values of T(z) are:
      25401-------229^2---------------------------------
      ---------------------------------621^2
      Two of these are squares, therefore 26051 is composite.

      Well, my clear-seeming little mental picture has proven
      to be quite tricky to explain, and that together with my
      very small set of experimental data has me feeling that I
      may be really off into the vast azure on a wing and a
      prayer with this idea. Still I have a bit of confidence that
      this approach to primality testing is valid for all of the
      infinities of equations represented by Dr Broadhurst's formula,
      and I have heard that there exists a Euclidean procedure that
      can calculate the factors of a composite from the discovery of
      squares in this manner. Whether all of this posseses an
      immanent 'usefulness' is another matter.

      Aldrich Stevens
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