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User:Bogart B. Strauss

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This is my approach to the elementary factorization of integers:

Given an integer with a, b, c, v, and r equal to the positive integers and zero.

Since we can find factors of 2 and 3 with a glance and the sign of the integer is of no importance for factorization, they are considered trivial and thus we are left with which combines two cases (r-even and r-odd). If this set of numbers equals R, then R*R is an element of R.

is the sequence .....sequence A007310 of the OEIS.

Considering that this can be formulated with a single variable, let's define a new sequence z.

with x equal to the positive integers or zero.

Using a sieving method (if you are curious, email me), we can then derive something to equate with.

with y equal to the positive integers or zero.

Thanks to Fermat realizing that a difference of squares with integer bases turns into a product of integers, we can then add that methodology to what we already have.

where n and m are positive integers and m can equal zero with .

Considering that the factors will be found in divisor pairs, we can state that .....with the set of all f being equal to the set of all z.

Notable relations:

where k is a positive integer or zero

Things start getting strange when we look at the resulting patterns from n and m as they relate to . There is a clear split into 6 cases (3 for r-even, 3 for r-odd) and a monolith of a difference of squares formula:

can be split in three: , , and .

6v+(-1)^r=18p+(6q+1)*(-1)^r=( ((1/4)*(18d+18+9*(-1)^r+(6*(-1)^q-1-q*(1+(-1)^q))*(-1)^(d+r)))^2 - ((1/2)*(3y+3k+(3y-3k)*(-1)^r))^2 )*(-1)^r where p and d are positive integers or zero while q={0,1,2}. I wouldn't read too much into this formula without figuring a desired q and r first.

I could not derive any more obvious patterns from this point and had to mess around with, well, anything. That anything is 3 times the number to be factored as another difference of squares.

where g is a positive integer not divisible by 3 and h is an integer not divisible by 3 with .

Notable resulting relations:

From that last formula, we realize that either or has only one possible factor of two or "critical 2". The reason is that is always an odd number and thus it is always a sum of an odd number and an even number.

If we list the first 15 numbers in the sequence , we can identify a pattern of critical 2's. Every unique divisor pair has a unique combination of h,m,n,g.

1*1____h=1, m=0, n=1, g=2

1*5____h=-1, m=2, n=3, g=4

1*7____h=-2, m=3, n=4, g=5

1*11___h=-4, m=5, n=6, g=7

1*13___h=-5, m=6, n=7, g=8

1*17___h=-7, m=8, n=9, g=10

1*19___h=-8, m=9, n=10, g=11

1*23___h=-10, m=11, n=12, g=13

1*25___h=-11, m=12, n=13, g=14

5*5____h=5, m=0, n=5, g=10

1*29___h=-13, m=14, n=15, g=16

1*31___h=-14, m=15, n=16, g=17

1*35___h=-16, m=17, n=18, g=19

5*7____h=4, m=1, n=6, g=11

1*37___h=-17, m=18, n=19, g=20

1*41___h=-19, m=20, n=21, g=22

1*43___h=-20, m=21, n=22, g=23

The pattern of critical 2's is repeating. Also notice that the critical 2's are unique to the number to be factored and thus relates each divisor pair of the number. This splits things up into 8 different cases (four for r-even and four for r-odd).

I can no longer reasonably subject anyone to a formula as we now have a total of 48 unique cases to deal with. Here's an example of just one case with B equaling positive integers and zero:

(from the three-split) (from the four-split) (from the forty eight split)

m is an element of

n is an element of

h is an element of

g is an element of

Recall that z is defined earlier. You can see here that the critical 2 is g for all numbers equal to .

So . While .

That's as far as I've gotten with the elementary factoring of integers and I encourage anyone to try and dig deeper, I certainly am. Patterns exist, we merely have yet to find them. But here it seems that there will always be a "next" pattern.

I will be transferring more summaries of my work here soon. The next piece will be an N choose K type of formula for figuring out the number of divisor pairs with a given prime factorization. This one will be a while as I am trying for a better notation for the extension of Sigma summation I have to use.

This is no place for philosophical statements or meandering through things I may perhaps misunderstand. For the things I may be mistaken about, I have yet to understand the proofs otherwise. I don't believe, I think. It does not escape me that this is a very precious responsibility.