Formubla-bla-bla to calculate the sum of two Prime numbers just by knowing the product - Page 7

Godzilla · 2016-09-08, 18:43

I have found that.

$ \frac{\sqrt{2178 * (p1*p2-9…9)}}{\sqrt{\frac{p1*p2}{22}}}$

es.

$ \frac{\sqrt{2178 * (30021-9999)}}{\sqrt{\frac{30021}{22}}} =\sim 22 \sqrt{\frac{660726}{10007}} = \sim 22\sqrt{66} = 178,7..$

$22\sqrt{66} = 178,7$ the number 22 is static , and if i catch the divisors of 66 i will found the prime number factor.

Is this possible ?

science_man_88 · 2016-09-08, 19:16

Quote:

Originally Posted by Godzilla

I have found that.

$ \frac{\sqrt{2178 * (p1*p2-9…9)}}{\sqrt{\frac{p1*p2}{22}}}$

es.

$ \frac{\sqrt{2178 * (30021-9999)}}{\sqrt{\frac{30021}{22}}} =\sim 22 \sqrt{\frac{660726}{10007}} = \sim 22\sqrt{66} = 178,7..$

$22\sqrt{66} = 178,7$ the number 22 is static , and if i catch the divisors of 66 i will found the prime number factor.

Is this possible ?

$ \frac{\sqrt{2178 * (p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}}}}{\sqrt{\frac{p1*p2}{22}}}$ might be nice to note how many 9's there are. edit:of course we can always square it and then reduce it as well.

CRGreathouse · 2016-09-08, 19:29

I just want to make sure I understand what you're doing here. You're proposing untested formulas as lower and/or upper bounds on certain undefined classes of semiprimes, using strings of symbols (9...9) which you have not defined.

Godzilla · 2016-09-08, 20:14

Quote:

Originally Posted by CRGreathouse

I just want to make sure I understand what you're doing here. You're proposing untested formulas as lower and/or upper bounds on certain undefined classes of semiprimes, using strings of symbols (9...9) which you have not defined.

I thought maybe you know better the subject which I am discussing and that you can help me to understand

science_man_88 · 2016-09-08, 20:26

Quote:

Originally Posted by science_man_88

$ \frac{\sqrt{2178 * (p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\sqrt{\frac{p1*p2}{22}}}$ might be nice to note how many 9's there are. edit:of course we can always square it and then reduce it as well.

I think my annoyance with it is that you're obfuscating it so well you don't see that it can be simplified:

$ \frac{\sqrt{2178 * (p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\sqrt{\frac{p1*p2}{22}}} = \frac{\sqrt{2178} * \sqrt{(p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\sqrt{\frac{p1*p2}{22}}} = \frac{\sqrt{22}*\sqrt{99} * \sqrt{(p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\sqrt{\frac{p1*p2}{22}}} = \frac{\sqrt{22}*\sqrt{99} * \sqrt{(p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\frac{\sqrt{p1*p2}}{\sqrt{22}}}=\sqrt{22}*\sqrt{99} * \sqrt{(p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}\frac{\sqrt{22}}{\sqrt{p1*p2}}$

above continued:

$= 22*\Bigg(\frac{\sqrt{99} * \sqrt{(p1*p2-\underbrace{9\ldots9}_{\text{\normalsize logint(p1*p2,10)}})}}{\sqrt{p1*p2}}\Bigg)$

retina · 2016-09-08, 20:34

The way I see it the whole purpose of this topic has no value.

You have some number that either a) you don't know how many factors it has so this "formula" is worthless, or b) you already know it has two factors, and subsequently either 1) you already know the factors because you constructed the number yourself so any formula is not needed, or 2) the number was constructed by someone else (probably for RSA) and this "formula" is worthless because it is only a heuristic anyway, and examining the original generator code would already tell the range of values of the two factors it will generate.

Unless some actual mathematics is going to be presented then there really isn't any point in trying to refine anything further.

The REAL maximal sum of factors value for an arbitrary number with two unknown factors is: floor(n/2 + 2) [1]
The REAL minimal sum of factors value for an arbitrary number with two unknown factors is: ceil(sqrt(n) * 2) [1]

If your "formula" generates numbers that do not cover the entire range above then it will always be possible to construct a number n that falls outside your estimates.

[1]At least I think these are correct. So sue me if I'm wrong.

science_man_88 · 2016-09-08, 21:00

Quote:

Originally Posted by retina

The way I see it the whole purpose of this topic has no value.

You have some number that either a) you don't know how many factors it has so this "formula" is worthless, or b) you already know it has two factors, and subsequently either 1) you already know the factors because you constructed the number yourself so any formula is not needed, or 2) the number was constructed by someone else (probably for RSA) and this "formula" is worthless because it is only a heuristic anyway, and examining the original generator code would already tell the range of values of the two factors it will generate.

Unless some actual mathematics is going to be presented then there really isn't any point in trying to refine anything further.

The REAL maximal sum of factors value for an arbitrary number with two unknown factors is: floor(n/2 + 2) [1]
The REAL minimal sum of factors value for an arbitrary number with two unknown factors is: ceil(sqrt(n) * 2) [1]

If your "formula" generates numbers that do not cover the entire range above then it will always be possible to construct a number n that falls outside your estimates.

[1]At least I think these are correct. So sue me if I'm wrong.

I guess my point is that simplified it takes out a factor and maybe make it able to be known what's affecting it's ability to be generalized into potentially smaller range versions. edit: and in your formulae floor and ceil are not needed the two cases represented are the case when n is an even semiprime and when n is a perfect square without them. and technically you can apply them to arbitrary multiplications as if you prove a number is a*b for any a and b the sum of a and b must be in this range.

CRGreathouse · 2016-09-09, 01:40

Quote:

Originally Posted by retina

The REAL maximal sum of factors value for an arbitrary number with two unknown factors is: floor(n/2 + 2) [1]
The REAL minimal sum of factors value for an arbitrary number with two unknown factors is: ceil(sqrt(n) * 2) [1]

If your "formula" generates numbers that do not cover the entire range above then it will always be possible to construct a number n that falls outside your estimates.

[1]At least I think these are correct. So sue me if I'm wrong.

Yes.

Godzilla · 2016-09-13, 19:56

Quote:

Originally Posted by science_man_88

I guess my point is that simplified it takes out a factor and maybe make it able to be known what's affecting it's ability to be generalized into potentially smaller range versions. edit: and in your formulae floor and ceil are not needed the two cases represented are the case when n is an even semiprime and when n is a perfect square without them. and technically you can apply them to arbitrary multiplications as if you prove a number is a*b for any a and b the sum of a and b must be in this range.

I have a similar formula , and it is $\frac{\sqrt{2178 * \frac{(p1*p2 )} {9..9}}}{17,5 or 22} =$ very very near to p1 or p2 (around the minor factor prime number) only if the number are distant example 123456789 * 1234

P.S.

9's are undefined may be two or three or four etc. I will work on it.

Godzilla · 2016-09-13, 21:13

Quote:

Originally Posted by Godzilla

I have a similar formula , and it is $\frac{\sqrt{2178 * \frac{(p1*p2 )} {9..9}}}{17,5 or 22} =$ very very near to p1 or p2 (around the minor factor prime number) only if the number are distant example 123456789 * 1234

P.S.

9's are undefined may be two or three or four etc. I will work on it.

-EDIT-

the numbers nine are defined so :

The first formula $\frac{\sqrt{2178 * (p1*p2 ) - 9..9}}{17,5} =$

The second new formula define how many 9 , about the digits of the result:

example :

997 * 3 = 2991

$\frac{\sqrt{2178 * (2991 - 999)}}{17,5} = 119$ three digits so three nine

$\frac{\sqrt{2178 * \frac{(2991)} {999}}}{17,5 } = 4$ the small factor is near

or

123456789*1234 =152345677626

$\frac{\sqrt{2178 * ( 152345677626 - 99999999999)}}{17,5} = 610142$ six digits so six nine

$\frac{\sqrt{2178 * \frac{(152345677626)} {999999}}}{17,5 } = 1040$ the small factor is near

science_man_88 · 2016-09-13, 21:38

Quote:

Originally Posted by Godzilla

-EDIT-

the numbers nine are defined so :

The first formula $\frac{\sqrt{2178 * (p1*p2 ) - 9..9}}{17,5} =$

The second new formula define how many 9 , about the digits of the result:

example :

997 * 3 = 2991

$\frac{\sqrt{2178 * (2991 - 999)}}{17,5} = 119$ three digits so three nine

$\frac{\sqrt{2178 * \frac{(2991)} {999}}}{17,5 } = 4$ the small factor is near

or

123456789*1234 =152345677626

$\frac{\sqrt{2178 * ( 152345677626 - 99999999999)}}{17,5} = 610142$ six digits so six nine

$\frac{\sqrt{2178 * \frac{(152345677626)} {999999}}}{17,5 } = 1040$ the small factor is near

okay well based on this going back to the huge number in post 2 has a smallest factor near: ...

2016-09-08, 18:43	#67
Godzilla May 2016 242₈ Posts	I have found that. $<br /> <br /> \frac{\sqrt{2178 * (p1p2-9…9)}}{\sqrt{\frac{p1p2}{22}}}$ es. $<br /> <br /> \frac{\sqrt{2178 * (30021-9999)}}{\sqrt{\frac{30021}{22}}} =\sim 22 \sqrt{\frac{660726}{10007}} = \sim 22\sqrt{66} = 178,7..$ $22\sqrt{66} = 178,7$ the number 22 is static , and if i catch the divisors of 66 i will found the prime number factor. Is this possible ?

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2016-09-08, 19:29	#69
CRGreathouse Aug 2006 3×1,993 Posts	I just want to make sure I understand what you're doing here. You're proposing untested formulas as lower and/or upper bounds on certain undefined classes of semiprimes, using strings of symbols (9...9) which you have not defined.

2016-09-08, 20:34	#72
retina Undefined "The unspeakable one" Jun 2006 My evil lair 6226₁₀ Posts	The way I see it the whole purpose of this topic has no value. You have some number that either a) you don't know how many factors it has so this "formula" is worthless, or b) you already know it has two factors, and subsequently either 1) you already know the factors because you constructed the number yourself so any formula is not needed, or 2) the number was constructed by someone else (probably for RSA) and this "formula" is worthless because it is only a heuristic anyway, and examining the original generator code would already tell the range of values of the two factors it will generate. Unless some actual mathematics is going to be presented then there really isn't any point in trying to refine anything further. The REAL maximal sum of factors value for an arbitrary number with two unknown factors is: floor(n/2 + 2) [1] The REAL minimal sum of factors value for an arbitrary number with two unknown factors is: ceil(sqrt(n) * 2) [1] If your "formula" generates numbers that do not cover the entire range above then it will always be possible to construct a number n that falls outside your estimates. [1]At least I think these are correct. So sue me if I'm wrong.