Does this hold true for all Prime products? / RSA isn' the encryption...

[Strange]

did you read what i said?

 

Preety useless when a skilled hacker could replicate the public keys stored on your computer/keep a log file of. Essentially you might use unusual encryption algorithms however your operating system is still bog standard and can be easily exploited

 

The whole point about a public key is that is public. So if someone steals it from your PC, no problem.

 

If you are really concerned, then keep your private keys on a non-networked computer and use that to do all encryption. Then copy the encrypted message to your networked PC in order to send it.

[Enthalpy]

[i remove my answer, since Strange already told it]

Edited May 1, 2015 by Enthalpy

[Trurl]

PNP = 7723 * 7919

 

x=7723

 

61158437

 

7723

 

x * (((PNP^2 / x) + x^2 ) / PNP - (x^2/PNP)) == PNP

 

True

 

input: Still Useless???

 

output: Maybe Not!

[Keen]

 

The whole point about a public key is that is public. So if someone steals it from your PC, no problem.

 

If you are really concerned, then keep your private keys on a non-networked computer and use that to do all encryption. Then copy the encrypted message to your networked PC in order to send it.

Or you could encrypt the private key by some symetric encryption algorithm like AES or blowfish and decrypt it in your RAM only when needed. That's how for example PGP works if I'm not mistaken.

[Endy0816]

PNP = 7723 * 7919

 

x=7723

 

61158437

 

7723

 

x * (((PNP^2 / x) + x^2 ) / PNP - (x^2/PNP)) == PNP

 

Your terms cancel out, any two numbers will work here.

 

ie.

 

777 * (((999^2 / 777) + 777^2 ) / 999 - (777^2/999)) = 999

[Trurl]

Ok, consider the last equation a brain fart proof of the error and proof of the algebra of the equation.

 

But can’t some math genius here help me find these equation equal; as this post shows. I have found the symbolic form of the error. The error was why my equations were declared useless here. Now I have it and accurate to 10 +++ digits.

 

I am not claiming to have a perfect solution. I just need input on what approach I should take to solve this polynomial perfectly.

 

This is it. The end of my work. But was nothing discovered here? Even if we just had an equation to test. It would not be significant, but it may be used for other applications.

In[6]:= PNP = 605054707

NSolve[((PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3) == ((PNP^2/x) + x^2)/
    PNP + x^2/PNP, x]

Out[6]= 605054707

Out[7]= {}




In[18]:= PNP = 605054707
x = 14251

((PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3)

((PNP^2/x) + x^2)/PNP + x^2/PNP

Out[18]= 605054707

Out[19]= 14251



Out[20]= 3249406777843210000/76532854417993

In[24]:= N[3249406777843210000/76532854417993, 10]

Out[24]= 42457.67132



In[23]:= N[1802625351/42457, 10]

Out[23]= 42457.67131

[Strange]

 

Ok, consider the last equation a brain fart proof of the error and proof of the algebra of the equation.

 

But can’t some math genius here help me find these equation equal; as this post shows. I have found the symbolic form of the error. The error was why my equations were declared useless here. Now I have it and accurate to 10 +++ digits.

 

I am not claiming to have a perfect solution. I just need input on what approach I should take to solve this polynomial perfectly.

 

This is it. The end of my work. But was nothing discovered here? Even if we just had an equation to test. It would not be significant, but it may be used for other applications.

In[6]:= PNP = 605054707

NSolve[((PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3) == ((PNP^2/x) + x^2)/
    PNP + x^2/PNP, x]

Out[6]= 605054707

Out[7]= {}




In[18]:= PNP = 605054707
x = 14251

((PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3)

((PNP^2/x) + x^2)/PNP + x^2/PNP

Out[18]= 605054707

Out[19]= 14251



Out[20]= 3249406777843210000/76532854417993

In[24]:= N[3249406777843210000/76532854417993, 10]

Out[24]= 42457.67132



In[23]:= N[1802625351/42457, 10]

Out[23]= 42457.67131

 

Can you explain what you are doing here? I can't follow it at all. You seem to be taking a number (PNP=605054707) and one of its divisors (x=14251) and then taking a long winded route to get an approximation to PNP/x. What is the point?

[Trurl]

The goal is for the first equation to find N knowing only p. (N=NPN and p = x here.) I do not know how to program Mathematica to find solutions 42457.67132 is approximately 42457.67131. You can’t just say < 0.00002 and have Mathematica evaluate the answer.

 

I believe this would be simple for someone who programs numbers. I am just not at that level yet. If y = y is true for only the Prime Products then the equation can find one unique and correct x.

 

I printed the decimal values of the given answer to show the equation is approaching 0. However it isn’t as useful if “other” products also equal each other in the y = y equation.

 

The second equation was just to show knowing x you can solve for y. obviously you don’t need these 2 equations to find N/x = y. The top is the empty set. I believe because Mathematica isn’t looking for a difference of 0.00001

 

What I am asking is if there a way to write a program that will see a difference of 0.00001? There is a problem if more than just the Prime Products approach zero. But for the time being how do I say close enough to 0 or y approximately equals y?

 

If you look in this posts previous post, you will see that my equations were called useless because finding the error was too great. Well know I have found the error symbolically in y/N (also known as x^2/N). But I want to know has finding the error helped make this anymore possible?

 

Too me I designed these equations with a design in mind. I wondered if a simple polynomial could attack RSA cryptography. My idea is simple to me but I cannot program it. I need an expert’s advice as to tell if this would improve the solution.

 

Simply put: Can I solve for x in this equation knowing only PNP? Remember the second equations are just to show the closeness of the values knowing x.

[Strange]

The goal is for the first equation to find N knowing only p.

 

What are N and p, and what is the relationship between them?

 

Is N a number and p one of it's prime factors?

 

In your example, you know both N and p, so what are you trying to find?

 

 

(N=NPN and p = x here.)

 

Having multiples names (x and p, for example) for the same thing is very confusing.

[Trurl]

What are N and p, and what is the relationship between them?

 

Is N a number and p one of its prime factors?

 

In your example, you know both N and p, so what are you trying to find?

 

The N = p * q is the integer factoring problem. It is used in cryptography often to form a one way function. The difficulty of factoring a number with the products being 2 Prime numbers is difficult with extremely large numbers. There is no equation that will solve the products in polynomial time. I want to find p which here I called x without knowing q which I called y. If I could do this the factoring problem would be no more. Yes, I am aware that this doesn’t seem solvable by 2 equations. I have heard and know this. But I had an intuitive yet simple approach. I originally started by saying that a pattern in Prime numbers could be explained by a logarithmic spiral. However this work is related but relies on patterns I derived by geometry, algebra, and intuition. I just need to know if there is something significant here. I think the last equations I posted show a significance. It doesn’t mean an answer is concrete, but it may be a different yet simple approach to the problem.

 

N = p * j or PNP = x * y

 

I have more work, but I want you to look at this before there is even more confusion.

 

I know it is a one way function. But the numbers are close. If I had assistance programming this thing maybe we would have an answer.

 

I know that there is much confusion here. I am building this in real time. Just concentrate on the correctness of the last equation. I will answer any question to the best of my ability.

 

 

 

 

 

Here is some videos. You will probably laugh at me, but it might answer any questions. There are inaccuracies due to my presentation ability and a few blunders in the equations. I have over 6 equations where y is in terms of x. I will share more, but I am not a professional so I am expected to be called naive. But don't let my ignorance stand in the fact that the last equations I posted to SFN may be something. I don't mind the feedback. I seriously believe in my equations.

[Strange]

The N = p * q is the integer factoring problem. It is used in cryptography often to form a one way function. The difficulty of factoring a number with the products being 2 Prime numbers is difficult with extremely large numbers.

 

I thought that might be what you were doing. But you start off with one of the factors known (x=14251) so all you have to do is N/x. Which makes all your work pointless.

 

Can you use your formula to find the (approximate) prime factors of 141620173 ?

 

 

Here is some videos.

 

 

Sorry, I can't watch videos. And I wouldn't if I could as they are a ridiculous form of communication.

[Trurl]

141620173

 

 

 

 

Without cheating I say: 655

 

I debated how much I should show here. I don't want to deceive anyone or not get credit for my work. But then I realized I could inspire someone or get help working towards a solution. But first I need to make sure it works. I will share the equation here. My other equations were correct, but when finding y it could not differentiate the factors being decimal or integers. So I had an elaborate equation that set y equal to y but it was always true. I have dozens of equations, but I think my original equation posted here shows a difference where y = y is unique. I could be wrong, but I keep trying because my ideas seem so intriguing to me. So take a look at this code and tell me what you think.

In[1]:= PNP = 85


NSolve[(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP) == 
  Sqrt[((PNP * (PNP/x) - x^2))/x] + 1/x, x]

Out[1]= 85

Out[2]= {{x -> -18.4935 + 11.9812 I}, {x -> -18.4935 - 
    11.9812 I}, {x -> 7.54812 + 13.0737 I}, {x -> 
   7.54812 - 13.0737 I}, {x -> 5.44949}}

In[3]:= PNP = 8141620173


NSolve[(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP) == 
  Sqrt[((PNP * (PNP/x) - x^2))/x] + 1/x, x]

Out[3]= 8141620173

Out[4]= {{x -> -3.87566*10^6 + 2.50985*10^6 I}, {x -> -3.87566*10^6 - 
    2.50985*10^6 I}, {x -> 1.55455*10^6 + 2.69255*10^6 I}, {x -> 
   1.55455*10^6 - 2.69255*10^6 I}, {x -> 2534.62}}

In[5]:= PNP = 141620173


NSolve[(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP) == 
  Sqrt[((PNP * (PNP/x) - x^2))/x] + 1/x, x]

Out[5]= 141620173

Out[6]= {{x -> -260189. + 168496. I}, {x -> -260189. - 
    168496. I}, {x -> 104363. + 180762. I}, {x -> 
   104363. - 180762. I}, {x -> 656.727}}

[Strange]

Without cheating I say: 655

 

Very obviously not. Just by inspection: the original number does not end in 0 or 5, therefore 655 cannot be a factor.

 

And 655 is not a prime number.

Edit: and the two nearest primes are not factors of the original number either.

 

So your system doesn't work.

 

 

 

So take a look at this code and tell me what you think.

 

Utterly incomprehensible, as always, without some explanation.

Edited January 26, 2016 by Strange

[Trurl]

Back to the drawing board. I will state my problem more cleary. I have had it find x knowing only N in previous tries. But this time it is wrong and I am not sure while because when you do plug in x it does seem the left and right side of the equation works. What is the error here? If discoverable?

Let PNP = N ; x = p and y = q where N = p * q where N is the product of 2 Prime numbers p and q.

x is the smaller Prime number. The 2 equations equal y to solve a polynomial equation where N is given 

Here is my best equation so far. I think that y = y (each side of equation equals y). x is unknown so solving the polynomal
knowing N may work if the 2 seperate equations for y are unique to Prime factors and not true for all integers.

So you see it is simple algebra except for the part of solving the polynomial. I have other equations that eliminate the square
and I believe this is why Mathematica is not solving the polynomal correctly. Also Mathematica can not distinquish from 0 and 0.0007.

There error of the equations because alone they are off by an error of (x^2 / x) and (x^2/ (2*PNP) respectively.

As the error becomes closer there may be more numbers y = y is true. However the error is a problem and is why the solution
would have to be programmed into a algorithm that can determine the value of x. I realize this is not a perfect algebraic solution.
However I think that as N increases in value the accuracy also increases.


Shown below is the main equation. As you can see Mathematica is not finding the proper x even though each equation, within
reasonable error, is y = y.

This is where I need help. I am not trained as a mathematician. I do not know how to solve this polynomial without the help of
Mathematica. But do you agree that these equations equal?

I know it seems impossible to break the one way function. And this proves its strength. But I still feel there is potential here.

The equations following show different values of the y part of the equation knowing both N and x. (Yes I realize the equation is
not useful already knowing both PNP and x. However it is just to demonstrate the values and patterns of the equations.







PNP = 85


NSolve[(PNP^4/x + 2*(PNP^2*x^2) + x^5)/
    PNP^3 - (2* 
     x^2 / PNP) == (Sqrt[((PNP * (PNP/x) - x^2))/x] + (x^2/ (2*PNP))), x]

85

{{x -> -17.7913 + 13.1603 I}, {x -> -17.7913 - 13.1603 I}}

In[7]:= 

PNP = 85
x = 5

(Sqrt[((PNP * (PNP/x) - x^2))/x] + (x^2/ (2*PNP)))

(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP)

Out[7]= 85

Out[8]= 5

Out[9]= 5/34 + 2 Sqrt[71]

In[11]:= N[5/34 + 2 Sqrt[71]]

Out[11]= 16.9994




Out[6]= 83546/4913

In[12]:= N[83546/4913]

Out[12]= 17.0051


The above is where x = 5. However x =3 would also be within reasonable error. 7 and 11 also may be close. However there is still
a significance of the Prime factors which should increase in distance as PNP (or N) increases in value.

I could show more values, but it may complicate the presentation of the problem.



In[23]:= 
PNP = 6911 * 4129
x = 4129

(Sqrt[((PNP * (PNP/x) - x^2))/x] + (x^2/ (2*PNP)))

(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP)

Out[23]= 28535519

4129



Out[25]= 4129/13822 + 4 Sqrt[2984862]

In[27]:= N[4129/13822 + 4 Sqrt[2984862], 20]

Out[27]= 6910.9999935435190680








Out[21]= 2281201114658882/330082636031

In[22]:= N[2281201114658882/330082636031, 20]

Out[22]= 6911.0000516496147904


As seen above y = y. However it takes a computer to see the difference. Next we will test x = 4127.




PNP = 6911 * 4129
x = 4127

(Sqrt[((PNP * (PNP/x) - x^2))/x] + (x^2/ (2*PNP)))

(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP)




Out[33]= 28535519

Out[34]= 4127

Out[35]= 17032129/57071038 + Sqrt[814205553002978]/4127

In[37]:= N[17032129/57071038 + Sqrt[814205553002978]/4127]

Out[37]= 6914.35




Out[31]= 663045156038911226701180291010/95894079886244829190262593

In[32]:= N[663045156038911226701180291010/95894079886244829190262593, 20]

6914.34921556632278818568561229321882459748`20.








So there is a computable difference between x = 4129 and x = 4123, in that
y = y is becomming untrue as the value of the number increases of decreases
from the value of the factor 4129.


In[43]:= 

PNP = 6911 * 4129
x = 4123

(Sqrt[((PNP * (PNP/x) - x^2))/x] + (x^2/ (2*PNP)))

(PNP^4/x + 2*(PNP^2*x^2) + x^5)/PNP^3 - (2* x^2 / PNP)

Out[43]= 28535519

Out[44]= 4123

Out[45]= 16999129/57071038 + Sqrt[814205757190494]/4123

In[47]:= N[16999129/57071038 + Sqrt[814205757190494]/4123]

Out[47]= 6921.06




Out[41]= 663045156010247586320631832010/95801136750905604737449157

In[42]:= N[663045156010247586320631832010/95801136750905604737449157, 20]

Out[42]= 6921.0572911493124224

[Trurl]

More equations. More evidence of patterns. Can such a polynomial be solved and can it be done by hand?

Here are more examples. That is values being tested.

There is my signiture PNP = 85 and x = 5.

As you can see y = y again. And there is more distrinction
between possible Prime products at different values.

Here I square both sides and y no longer equals y until we take the
square root of the answers.

Again y = y with distinction in the Prime factors, but it would take a 
computer algorithm to find x. I cannot use the NSolve command in
Mathematica and get accurate results.

How do I go about solving such a polynomial?






In[82]:= PNP = 85
x = 5

((PNP^4/x + (2* (PNP^2 * x^2) + x^5))/ PNP^3)



(PNP * PNP/x - x^2)/ x



Out[82]= 85

Out[83]= 5

Out[84]= 86436/4913

Out[85]= 284

In[81]:= N[86436/4913]

In[86]:= 17.593323834724202`

Sqrt (284)

Out[86]= 17.5933

284 Sqrt









Below I try NSolve. Even though the equations are algebr

In[89]:= PNP = 85
NSolve[(((PNP^4/x + (2* (PNP^2 * x^2) + x^5))/ PNP^3))^2 == (PNP * PNP/x - 
     x^2)/ x, x]



Out[89]= 85

Out[90]= {}



Below I squared both sides to elminate the square root. Did not help, but I do have equations where I have
eliminated it.

Again how to I go about simplifying and solving these polynomial equations by hand?





In[102]:= 

PNP = 85
x = 5

(((PNP^4/x + (2* (PNP^2 * x^2) + x^5))/ PNP^3))^2



(PNP * PNP/x - x^2)/ x

Out[102]= 85

Out[103]= 5

Out[104]= 7471182096/24137569

In[106]:= N[7471182096/24137569, 20]

Out[106]= 309.52504355347466847

In[107]:= Sqrt (309.52504355347466847220612813162750565312`20.)

Out[107]= 309.52504355347466847 Sqrt

284



I like the square root here. It is showing that increments in y 
signify small changes in x. It also seems to help with small 
value changes in value of x. For example x =3 value significantly 
deferes from x = 5. This is more evident in larger numbers.






In[117]:= 

PNP = 85
x = 3

(((PNP^4/x + (2* (PNP^2 * x^2) + x^5))/ PNP^3))^2



(PNP * PNP/x - x^2)/ x

Out[117]= 85

Out[118]= 3

Out[119]= 2765866292982016/3394345640625

In[121]:= N[2765866292982016/3394345640625]

Out[121]= 814.845



These may not seem spetacular. But there are important 
relationships here. Look at the numbers and the results of testing
knowing both PNP and x. It is not simple plug and chug.

The question is if the polynomial can be solved.

[imatfaal]

1. This is a fool's errand there is a wealth of beautiful mathematics on prime numbers - your time would be much better spent following the work of some of the greats of the past than floundering around here. This study might in turn give you a firm base to move forwards

 

2. If you really want help here (and that will be an explanation of what you are doing wrong / how you are mis-thinking) I think you need to state your questions in a more accessible format - the mixture of pseudocode and formulae is pretty impenetrable.

 

3. You might want to look at the complexity level of your calculations - ie how many steps / much computer time you would take to find out a prime - and then compare it to some of the now fairly rapid prime factorisation methods. It is possible that your system (which will not work) would be slower than current algorithms even if it did work.

 

The fastest deterministic algorithm works by multiplying together blocks of integers modulo your test number and if that set of integers contains a factor then the greatest common divisor of the product and the test number will be greater than one. the non-deterministic algorithms (ie using vast arrays of random numbers which are processed alongside the test number) are in fact quicker but do need more computation space due to the need for random number feed (not seeded pseudo random) - it is beyond me to say what these actually do