the application of the simple function 5r^2 - 4*C(y)

over relatively small tightly ordered intervals (valid for

any y > some minimum value) in fact generates

valid information about primality is still, I believe,

something of a novelty. The present object of attention,

C(y) = y^2 + 5y + 5, is one such equation and it is a

transform of 5x^2 -5x +1 that was previously examined

here in some detail (PN: Nov 28). When I say that one

equation is a transform of some other equation, I mean

that the total underlying set of primes composing all of

the terms of both of them is identical, except for a finite

number of added factors that are sometimes a by-

product of the transformation process. All of the terms

of y^2 + 5y + 5 and 5x^2 -5x +1 are made solely

of +/- 1 mod 10 primes with the one exception being

that the transform has an added factor of 5 in every

fifth member, and it is no accident that these two

sequences can be represented matrix style as the

y and x axis respectively of the binary quadratic form

5x^2 + 5xy + y^2 for x, y > 0:

----C(y)

A(x) 11, 31, 61, 101, 151....

-----19, 44, 79, 124, 179....

-----29, 59 ,99, 149, 209....

-----41, 76,121,176, 241....

-----55, 95,145,205,275...

----- : : : : :

Conjecture:

y,C(y) ,r,T(r),I ,floor: integers;

Let C(y) = y^2 + 5y + 5;

Let T(r) = 5r^2 - 4*C(y);

Let floor equal the first integer >= sqrt(5/4*C(y))

and I = (y + 2) - floor;

Case 1: If y < 12 then I = 0. C(y) will equal 11,19,

29,41,55/5,71,89,109,155/5 or 181 and all of these

are all prime.

Case 2: If y >= 12, y mod 5 > 0, and floor <= r < y + 2

then C(y) will be prime if no value T(r) exists such that

T(r) is a perfect square, otherwise C(y) will be composite

Case 3: If y >= 12, y mod 5 = 0 and floor <= r < y + 2

then C(y)/5 will be prime if no value T(r) exists such that

T(r) is a perfect square, otherwise C(y) will be composite.

Examples:

_y__C(y)__floor__r__T(r)_y+2__y+3

12__209____13___13__9____14___15

13__239____14___14__24___15___16

14__271____15___15__41___16___17

15__305____16___16__60___17___18

16__341____17___17__81___18___19

By Case 2 above, 239 and 271 are prime,

and 209 and 341 composite. By Case3,

305/5 is prime.

It is not until y = 21 that I(the number of possible r values)

begins to exceed 1. At that point C(21) = 551 and T(21) = 1,

a perfect square, so 551 is composite. At y = 34, I = 3,

and T(33) = 121, so 1331 is composite. At y = 40, I = 4,

and T(38) = 0, so 1805/5 is composite. At y = 177, I = 18,

and T(161) = 729, so 32219 is composite.

If there are several factors in C(y) many

squares are generated by T(r), 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 appears to be closely related to the binomial

coefficients. The simplest of these occurs when all of

the prime factors of C(y) 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)/2;

eg : two factors makes one square, three make

three, four make seven etc

If a perfect square value of T(r) is located for C(y),

two of its factors can be easily be discovered. Let

us examine the composites 209, 341,551,1331,1805,

and 32219 from our examples and have a look at all the

other relevant integers needed to factor them. These

other numbers are just outside the range for r that we

have been using at r = y + 2 and r = y + 3. No

tyranny of small numbers principle applies here, so

these tiny numbers will adequately demonstrate a

factoring procedure that will work for all C(y), no

matter how large. Simply figure the gcd(C(y),w1) and

the gcd(C(y),w2) for each composite below

to find factors. It should be noted that this process

is not identical to the one for A(x) = 5x2 - 5x + 1

described in the earlier posting.

At y = 12, y^2 + 5y + 5 = 209 = 11*19,

r________ : 13,,,| 14,,15

T(r)_____ : 9,,,|

sqrt(T(r)): 3,,,|

w1 = ((14 + 15) *13 + 3)/2 = 19*10;

w2 = w1 - 3 = 11*!7;

At y = 16, y^2 + 5y + 5 = 341 = 11*31,

r________: 17,,|18,,19

T(r)_____: 81,,|

sqrt(T(r)): 9,,|

w1 = ((18 + 19) *17 + 9)/2 = 11*19;

w2 = w1 - 9 = 31*10 ;

At y = 21, y^2 + 5y + 5 = 551 = 19*29,

r________: 21,,| 23,,24

T(r)_____: 1,,|

sqrt(T(r)): 1,,|

w1 = ((23 + 24) *21 + 1)/2 = 19*26;

w2 = w1 - 1 = 29*17;

At y = 34, y^2 + 5y + 5 = 1331 = 11*11*11;

r________: 33,,| 36,,37

T(r)_____: 121,,|

sqrt(T(r)) : 11,,|

w1 = ((36 + 37) *33 + 11)/2 = 121*10;

w2 = w1 - 11 = 11*!09;

At y = 40, y^2 + 5y + 5 = 1805/5 = 19*19,

r________: 38,,| 42,,43

T(r)_____: 0,,|

sqrt(T(r)): 0,,|

w1 = ((42 + 43) *38 + 0)/2 = 19*85;

w2 = w1 - 0 = 19*85;

With a final example, let us look at C(177) , expand the

range of r (to floor <= r < 2y +3) and examine some of

the many closely related ways that 32219 may be

factored that are similar to the one shown above. The

functional equivalence of each method may not be intuitively

obvious, and the hope is that this plethora of data can eventually

lead to some really efficient new way to either prove large

C(y) integers prime or to factor them ( or at least to prove true

the conjectures presented on this page).

At y = 177, y^2 + 5y + 5 = 32219 = 11*29*101,

r________ :161,163,172|179,,180|189,213,228,255,276,327,,|357

sqrt(T(r)):27,,,63,138|177,,182|223,313,362,443,502,637,,|713

w1 = ((179 + 180) *161 + 27)/2 = 29*997;

w2 = w1 - 27 = 11*101*26;

w1 = ((161 + 228) *180 + 67*182)/2 = 11*101*37;

w2 = w1 - 67*182 = 29*997;

w1 = ((161 + 255) *179 + 94*177)/2 = 11*101*41;

w2 = w1 - 94*177 = 29*997;

w1 = (161 + 255) *179 + (161 + 228) *180 -

2*(179 + 180) *161 = 11*101*26;

w1 = ((179 + 180) *163 + 63)/2 = 29*101*10;

w2 = w1 - 63 = 11*2657;

w1 = ((163 + 213) *180 + 50*182)/2 = 11*3490;

w2 = w1 - 50*182 = 29*101*10;

w1 = ((163 + 276) *179 + 113*177)/2 = 11*4481;

w2 = w1 - 113*177 = 29*101*10;

w1 = (163 + 276) *179 + (163 + 213) *180 -

2*(179 + 180) *163 = 11*2657;

w1 = ((179 + 180) *172 + 138)/2 = 29*11*97;

w2 = w1 - 138 = 101*305;

w1 = ((172 + 189) *180 + 17*182)/2 = 101*337;

w2 = w1 - 17*182 = 29*11*97;

w1 = ((172 + 327) *179 + 155*177)/2 = 101*578;

w2 = w1 - 155*177 = 29*11*97;

w1 = (172 + 327) *179 + (172 + 189) *180 -

2*(179 + 180) *172 = 101*305;

Aldrich Stevens