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