## Primality and Dr. B's Fab Fibo Formula

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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
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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