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Simple Factorizing Function & *new* Function with primes as result

So well...

to summarize the current results in a language for everybody:

I found two ways to design the structuring of a number in respect to squares.

The first is to find squares and to show their specific relation to the factorized number.

This is the function the dev c++ programm has been written for.

Numbbers (Z), within this context, integers, can be analyzed using an approximatd highest square with integer base:The rest is partitioned in the remaining multiple of the approximated root and a further integer rest that is to small to be a multiple of the root factor

(1) Z = x^2 + v*z + res

The programm shwos, that a partitiong of numbers in this way shows that square-differences can represent Z (and its factors), in the following way:

(2) (x +v)^2 - v^2 = (x +2v)*v

The equivalence is trivial and well known as third binomial formular. I started my investigation because I wanted to get a better understanding which numbers build the squares that help factorizing numbers.

Such squares have the structure: factor (x) plus multiple of the factor beyond the root (v) suqred minus multiple squares. with the third binomial formular, we have the factors (factor x plus multiple v for two times) multiplied with (multiple).

a lot of factorizable numbers have additionaly a rest of the factor x for some times, whose number are to low for the multiple v. They don't build factors by the third binomial formular, but of course the structure - square - multiple*factor - rest - shows nevertheless in which way they can divide the number Z, as it shows the factor x in all three parts.

I found a second way to use the structure (1) Z = x^2 + v*z + res differently. It came out to explain how an approximated root of a number can be used to factorize the number

The difference to (1) is, that the second part of th formular was standardized, not optimized. I created a series of factors of a number in this way:

(3) Z =(a-n)^2 - 1*(a-n) + res1 and v = 1

Z =(a-n-1)^2 - 3*(a-n-1) + res2 and= 3

Z = (a -n-2)^3 - 5(a-n-1) + res3 and 5=v

I started with (a-n) as high as possible to get a positive rest res. the multiple "v" grows by convention in the way 1+2n. The root-base is minored by 1 with every step. The res grows, and as I could determine, in the way

n^2 + n + res1.

Hence, every number can be partitioned into

Z = (a-n)^2 + v*n^2 + res.

If a number is not prime, res can be divided by x,

Hence, the number has the form

(a-n)^2 + v*n + a new factor f *(a-n)

Investigating the third structure (3) Z =(a-n)^2 - 1*(a-n) + res1 and v = 1

it became obvious, that the rest (res) developes by n^2 + n +- res of formular (3).[To sketch the proof, Gauss series or squares lies in between the squares:

x^2 +vx + res - (x-1)^2 - vx2 -res2 = x^2 - (x-1)^2 +vx + res -res2 - vx2

n^2 + n +- res (res the number for res for v=1 or v=2

n^2 simply means, that 2 majores the given rest with +2 more than it is maximized before

continues soon

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08.09.2017 (13 days ago)

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