## Speculations concerning the primes and composites of y^2 + 5y + 5

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• The idea that for a great many quadratic equations the application of the simple function 5r^2 - 4*C(y) over relatively small tightly ordered intervals (valid
Message 1 of 1 , Jan 7, 2010
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The idea that for a great many quadratic equations
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

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
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