## 5x^2 - 5x +1 revisited

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• Perhaps someone would enjoy applying some Gaussian concepts to this basic stuff!: Previously, in September, 08 it was conjectured that: x, A(x), B(x), k,
Message 1 of 1 , Nov 28, 2009
Perhaps someone would enjoy applying some
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.

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-