Gaussian concepts to this basic stuff!:

Previously, in September, '08 it was

conjectured that:

x, A(x), B(x), k, T(k) : integers;

Let A(x) = 5x^2 - 5x +1;

Let B(x) = 25x^2 - 40x + 16;

Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

If x > 3 and k > x , then A(x) will

be prime if no value T(k) exists such

that 1 < T(k) < B(x) is a square,

otherwise A(x) will be composite.

The second part of this conjecture

was soon proved to be true, but

the first part was only shown to be

a necessary condition for primality in

this situation and it remains inconclusive as to

whether it is also a sufficient condition.

Since then a bit more information has

come to light about the composites of

5x^2 -5x + 1, though not about its

primes. It can be empirically observed

that if the notion of the 'k' values, hitherto

consistantly used to denote integers generating

odd squares under the right conditions

(when factors are present) , is expanded

to include all of the numbers k -1/2

across the same interval, that these 'half'

values of k turn out to also indicate

compositeness when they generate (even)

square values of T(k). These squares

invariably come paired with odd squares

on the other side of the divide that is at

a point approximately 1/4 of the way

from the lower limit of the interval where

T(k) applies. All squares come in pairs

with the sole exception of when the A(x)

value is itself a square that is also equal

to a value of T(k).Such pairs can be in

odd-even, even-odd and odd-odd

combinations but never even-even.

If there are several factors in A(x) many

squares are generated, and their exact

total is always the same for composites with

the same numbers of distinct and redundant

prime factors.Each case has its own formula

and the simplest of these occurs when all of

the prime factors of A(x) are distinct . If we

let 'f ' denote this number of prime factors

then the formula for the number of squares

that will be present in the simple case is 2^f - 2;

eg : two factors makes two squares, three make

six, four make fourteen etc. Half of these will

occur in in the first quadrant of the interval, and

half later. The first square is always paired with

the last, the second with the second to last and

so forth.

To illustrate:

At x = 65, 5x^2 - 5x + 1 = 20801 = 11*31*61,

and the relevant interval has a total of 63 terms

beginning at k = 65.5 and ending at 91.5. The

significant values are as follows:

k : 65.5 66 68 / 79.5 86 89

2k -1 : 130 131 135 / 158 171 177

sqrt(T(k)) : 36 51 89 / 204 251 271

Let us call the 2k-1 values above R1,R2,R3,R4,

R5,R6 and the sqrt(T(k)) values r1,r2,r3,r4,r5,r6.

Actually, only R1,r1,R2,r2,R3,r3 have to be found

before the other values can be easily calculated from

them, and A(x) factored as well. Note that

R1 + R6 = r1 + r6, R2 + R5 = r2 + r5, and

R3 + R4 = r3 + r4. This relationship means that

if say R1 is known, then its paired number R6

can be found by means of a fast binary search

pattern. Then it is an easy calculation to find

two factors of A(x) because

gcd((x - 1)(R1 + R6) + R1, A(x)) gives

one and gcd((x - 1)(R1 + R6) + R6, A(x)) another :

Specifically for 20801, 64 * (130 + 177) + 130 =

11*31*58 and 64 * (130 + 177) + 177 = 61*325.

This process can also be done for the other paired

values of 2k - 1 so that 64 * (131 + 171) + 131 =

11*61*129 and 64 * (131 + 171) + 171 = 31*629,

and 64 * (135 + 158) + 135 = 11*1717 and 64 *

(135 + 158) + 158 = 31*61*101.

Locating even one square by trying out all of the

possibilities of 5*(2*k -1)^2 - 4*A(x) for an

interval approximately x/4 in length can be difficult

for large values of x. There are fortunately ways to

eliminate most of these by exploiting the fact that

valid sqrt(T(k)) solutions occur on other quadratic

equations whose middle three terms can be expressed

as ...A(x), A(x) + 10*k, A(x) + 2*10*k + 10...

For k = 79.5 in our example this would work out to

20801, 21596, 22401....41616 = 204^2. The terms

of these equations with solutions are composed

exclusively of primes +/- 1 mod 10 and also squares

of any sort, whereas all of the other possible

equations of the same form will have some terms

with odd powers of +/-3 mod 10 primes that

can thus be eliminated with a sieving scheme.

I have developed a crude Pascal demo program

that will show that the idea works well for

A(x) < 1e19 and anyone interested can have

a copy upon request. It is hard to gauge at this point

just how large a number with non-trivial factors

could be really cracked in this manner. Perhaps

with top-notch equipment, a sophisticated sparse-

matrix mathematic,and advanced software array

forms (plus a better programmer than myself)

the technique could be competitive as a brute-

force algorithm.

Aldrich Stevens