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Formular for Factorization

As far, I've found the following three (in a way, two) formulars to determine n for a factor (x-n) , as far considered for numbers with the structure

"nteger base of the approximated root" ^2 + integer base of the approximated root + rest (= number to be factorized)

 

I write this as

prox ^2 + prox + rest = Z

 

An analytic formular to get (integer) factors is

(with "prox" = "x")

 

(x-n)* ( n - 2* x + 1 + rest)= Z

 

we get the equivalence

nx - n^2 + (2* x + 1 + rest)*x + (2* x + 1 + rest)*n = Z

2* x + 1 + rest is given by the structure above, x is given, Z of course.

So we can calculate n by some well known formulars for quadratic equations. 

(x+ (2* x + 1 + rest))*n - n^2 = Z - (2* x + 1 + rest)*x

 and hence 0 = n^2 - (x+ (2* x + 1 + rest))*n - Z - (2* x + 1 + rest)*x

with an^2 + bn + c = 0

we have b = (x+ (2* x + 1 + rest))

and c = Z - (2* x + 1 + rest)*x

 

Whether the formular fits or not, depends on the role of the rest in the numbeer's structure.

For some numbers, n fits. If not,

there is the square_difference formular

(3*prox – n)^2 – (2*prox)^2 = Z

with (5prox - n) * (prox-n) = Z

and n easily to derive by Z - (2prox)^2

 

The square-difference form for

(x-n)* ( n - 2* x + 1 + rest)= Z

is (so far)

(3*prox + 1 + Rest)/2)^2 - ((2n + prox + 1 + Rest)/2)^2 = Z

 

Examples will follow.

 

 

 

 

 

 

 

 

 

ZifferZahlZitat 29.09.2017 0 1114
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