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Instantly Factorize Any Product Of Two Small Or Large Twin Prime Numbers.


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There is a paper at vixra (dot) org 

Number theory, citation  2103.0181

Instantly Factorize Any Product Of
Two Small Or Large Twin Prime Numbers.

Simple Method -
72 is the constant integer used in the process to find repeated addition in the
series.
First Step –
Repeated Addition Series.
Following the steps ask your colleague to add 72 and 36 as show below.
 72 * 1 = 72 + 36 = 108
 72 * 2 = 144 + 36 = 180
 72 * 3 = 216 + 36 = 252
 72 * 4 = 288 + 36 = 324 ......... 'Last Sum Of Series'


Counting can be done as many times like 72 *5 , 72 * 6, 72 * 7 ........ and
one time adding 36 for each series.
Series can go up to infinity.
Last sum of series is 324.


Second Step -
Finding ‘r ‘ Total Sum Of Series -
Ask your colleague to add all the sums together with number 35 to get total
sum of series ' r ' as shown below.
108+ 180 + 252 + 324 + 35 = 899 ...... 35 is the constant to be
added at last in total sum of series each time you calculate this series.
Here we get r = 899
Now get this two information from your colleague .
1) Last sum of series ie 324.
2) Total sum of series ie 899 .
You should know this to calculate the formula.
Therefore,
Ask your colleague to show the last sum of series i.e 324 and the total sum of
series i.e 899.
Note - Total sum of series is also a product of some two twin prime numbers
or prime number and composite number or may be of two composite
numbers.
So 899 is the product of p*q = 899 which we don’t know yet and we are going
to factorize it to know p & q using formula explained below.


Third step -
Now, ask your colleague that, 'Can they immediately guess what is the
multiple factors of given total sum of series is, without factorizing ?'
Answer for your colleague must be 'no', since no one can easily guess or
reverse the p * q = n if the ''n ' is any large integer.
But wait, using my new researched method you can factor in few minutes, no
matter what large integer 'n' is.
So without showing your colleague, calculate the process explained below.
Calculation Process – Finding ‘s’.
There are two method to find ' s ’.
1) First method -
Notice the above 'repeated 72 series', those bold highlighted integers 72 * 1,
72 * 2, 72 * 3, 72 * 4 ........
Series of Integers in line i.e 1,2, 3, 4.....
Find the last integer i.e 4
Substitute 4 with 0 of 0.83 (constant).
We get 4.83
Therefore, s = 4.83.
Each time you calculate to find ' s’ always find the last integer in the line as
explained above.
2) Second Method -
You know that last sum of series is 324. ( you got the information from
your colleague)
Taking 324 / 72 = 4.5
Get the left hand side integer before the decimal point i.e 4.
Substitute 4 with 0 of 0.83 (constant).
As per second method, we get s = 4.83
Next,
Apply the ‘r’ and ‘s’ in the below formula.
r / s = m
m/ 6 = n
Where,
‘r’ is the total sum of series.
‘s’ in this case is the substitution of 4 with 0 of 0.83 constant to get as 4.83.
6 is the constant in the formula.
We got r = 899, s = 4.83
Finding ' m ' -
r / s = m
899 / 4.83 = 186.12836....
Notice integer on the left hand side before the decimal point i.e 186
So, consider only those integers as ' m ' and ignore integers on the right
hand side of decimal point.
Therefore, m = 186
Finding ' n ' -
m / 6 = n
186 / 6 = 31
n = 31 ....... is the answer.
Check it dividing 899 by 31.
899/ 31 = 29
So the factors of 899 is 31, 29.  

Immediately show the answer to your colleague.

One can surprise their colleague with this method.

This method works even for factorizing any larger product of multiplied twin primes. One can try checking.

 

 

 

twin prime.pdf

Edited by olvin dsouza
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from the paper:

Quote

This can be any two twin prime numbers such a that is either 2 less or 2 more than another prime numbers.

5, 7 11, 13

Or any numbers that is either 2 less or 2 more than another composite numbers or prime numbers.

23, 25 65, 67

Note - this method works only for any multiplied products ‘n’ of any above example of numbers,

Trying to understand: The method works on equations of the form
[math]n(n+2)=m[/math] , [math]n \in N, m \in N [/math]
The integer [math]m[/math] is known and the method will find the unknown integer [math]n[/math] ?

Edited by Ghideon
math notations
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I'm not sure i have understood this.

Does this only work with primes where the two factors differ by two- which is interesting but not much use (I think there's an easier way)?

 



If not, please show us how it works with 33,033,660,080,507

(which isn't a product of very big primes, just fairly big.)

Edited by John Cuthber
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Posted (edited)
On 4/11/2021 at 2:46 PM, John Cuthber said:

I'm not sure i have understood this.

Does this only work with primes where the two factors differ by two- which is interesting but not much use (I think there's an easier way)?

 



If not, please show us how it works with 33,033,660,080,507

(which isn't a product of very big primes, just fairly big.)

Prime numbers differ by two is called twin prime numbers.

This method works only if you get 'n' by following repeated 72  additon series.

If you give me any n this method doesn't work.

Any how, i have found a new method where,
If you multiply any of the below following p & q and give me n.
 
Twin prime × Twin prime = n

Prime number × Composite number = n ( not divisible by 2 or 3, having a gap of 2).

Composite number × Composite number = n ( not divisible by 2 or 3, having a gap of 2).

Then i can factorise it in few minutes. Even if the n is 1000+ digit large.
Very soon i will publish paper on this another method.

 

 

Edited by olvin dsouza
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2 hours ago, olvin dsouza said:

This method works only if you get 'n' by following repeated 72  additon series.

What makes 72 special? Can you show how the method works in the general case [math]n(n+2)=m[/math] where [math] n \in N, m \in N [/math]? 

 

 

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Given a product of two primes n, and n+2 the product is going to be very close to (n+1)^2

So you can take the nearest integer to the square root of the product, and the factors will be  that +/- 1

 

Eg.
 imagine I give you the product 19043

the square root of that is 137.996...

Very close to 138

And the integers 1 away from it are 137 and  139.

And those are the factors of 19043.

 

 

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On 4/15/2021 at 11:11 PM, John Cuthber said:

Given a product of two primes n, and n+2 the product is going to be very close to (n+1)^2

So you can take the nearest integer to the square root of the product, and the factors will be  that +/- 1

 

Eg.
 imagine I give you the product 19043

the square root of that is 137.996...

Very close to 138

And the integers 1 away from it are 137 and  139.

And those are the factors of 19043.

 

 

I agree, 

What if, 

n = 1661521

And condition is- use only simple calculator.

Try to find sqrt of n.

By the time you calculate or by guessing.

Using my explained method and formula. I can get the two factor of n in less then 1 minute.

No matter how big is  n in digits, calculation time is less than 1 minute.

 

 

On 4/15/2021 at 10:34 PM, Ghideon said:

What makes 72 special? Can you show how the method works in the general case n(n+2)=m  where nN,mN

 

 

If you give me n by simply multiplying p×q 

Then the only way i can find p & q  is by sqrt method. It will take some time or lot of time for me to factor using sqrt method by simple calculator.

 

Addition 72 series is the only method to get the 'last sum of series'.

If i get the 'last sum of series'. - It is the magic number and input it in 0.83 constant. I can factor any small or even larger  1000 + digits n in less then 1 min. (As explained in my method formula)

Take this theorm as alternate to sqrt method.

Which can provide fast result using simple calculator, but with conditions such as following addition 73 series.

 

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2 hours ago, olvin dsouza said:

If you give me n by simply multiplying p×q 

Then the only way i can find p & q  is by sqrt method. It will take some time or lot of time for me to factor using sqrt method by simple calculator.

 

Addition 72 series is the only method to get the 'last sum of series'.

If i get the 'last sum of series'. - It is the magic number and input it in 0.83 constant. I can factor any small or even larger  1000 + digits n in less then 1 min. (As explained in my method formula)

Take this theorm as alternate to sqrt method.

Which can provide fast result using simple calculator, but with conditions such as following addition 73 series.

Thanks for your reply. Can you elaborate on the steps to use to solve a general case; a case constructed in such a way that your method works? I give you the number [math]m[/math]. I know [math]m[/math] is the product of two integers [math]n[/math] and [math]n+2[/math] so your method should be applicable. Please show the steps required to calculate factors [math]n[/math] and [math]n+2[/math] using your method.

In other words; in your posts so far I see numerical examples, I would like to see the formulas and steps expressed for any number [math]m[/math] where [math]m[/math] is the product of two integers [math]n[/math] and [math]n+2[/math] as per your statements about the limitations of your method.

 

Edited by Ghideon
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3 hours ago, olvin dsouza said:

What if, 

n = 1661521

Then it's not of the form N (N+2)

 

How long would it take you if the number was a thousand digits?

What about a million?

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