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

Trurl · September 28, 2013

5 * 17 = 85

Prime Number + Prime Number = Product = N (in RSA)

y = ((85/x) * 85 - x^2)/ x = ((85^2/x) + x^2)/ 85

p = ((85/x) * 85 - x^2)/ x * ( 85^2/((85^2/x) + x^2)) - 85;

sol = NSolve[p0, x]

{{x®-86.893},{x®-7.50438+19.0222 ä},{x®-7.50438-19.0222 ä },{x®16.9017}}

mathematic · September 28, 2013

Your description is too cryptic. The first line y = has an error. Beyond that it needs some explanation, line the meaning of R inside a circle, also what are sol, Nsolve, p0, and a unlaut?

Trurl · October 8, 2013

Yes it is hard to see what I was asking.

I am working on finding what 2 numbers equal a product; In particular what Prime numbers equal a product. 5 times 17 = 85

I am isolating x in an equation. Just basic algebraic substitution.

In particular sqrt( (N * y – x^2) / x) is approximately equal to y. Or in the example 5 * 17 = 85.

So sqrt[(85 *17 – 5^2) / x] = y = 17

As seen in the corrected group of formulas it works for all Prime numbers and possibly all products.

Again this was a pattern in my mind to isolate x so knowing 85 you could find 17.

It is just that simple. But the equations can be confusing fast trying to derive.

There is also the problem of solving the polynomial result. I have isolated x before, but this is the first time that it was simplified enough for Mathematica to solve it.

As you can see the only reasonable value in NSolve is 9.44493. Not 5! However with a degree of variation of the approximation 5 is within range of error of the equations.

9 – x = 4 ; 4 *2 = 8 ; 8 +9 = y where x =5 and y = 17

I am not sure of the above pattern of 5, 9, and 7 but the range of approximated error of the equations allow x and y to be estimated knowing only N and in this case N = 85:

Below is the corrected Mathematica work. Test it for yourself. And keep in mind I am an amateur mathematician.

I believe it works or I wouldn’t have posted it. The only question is if it is useful, because the math world is in need of better methods to solve polynomials.

I ask a friend who hasn’t done math in 20 years, but it takes a while to see what I am attempting to do.

It could be wrong. But I think it points out a new way to estimate 2 Prime numbers from N.

y = sqrt[(((85/x) * 85 - x^2)/ x) ]= ((85^2/x) + x^2)/ 85

p = ((85/x) * 85 - x^2)/ x - (( 85^2/((85^2/x) + x^2) ) ^2);

sol = NSolve[p0, x]

{{x-36.2894},{x27.7376+21.7226 },{x27.7376-21.7226 },{x-10.093+32.5167 },{x-10.093-32.5167 },{x9.44493},{x0.257048+9.23565 },{x0.257048-9.23565 },{x-8.95875}}

This is still an approximation. In the above example 9 solves the equation better than 5. That is where 5 should be the desired answer. N = 85; x = 5; and y = 17 . So there still isn’t a perfect solution just estimated. But take:

y = sqrt[(N * y - x^2) / x]

y = sqrt[(7872197 * 3191 - 2467^2) / 2467]

sqrt[(-x^{2+N sqrt[(-x2+N sqrt[(-x2+1/85 N (7225/x+x2))/x])/x])/x]}

sqrt[10180014]

So sqrt of 10180014 = 3190.613421 which is approx y or 3191.

So y and p equations above hold true. But the problem is it is still an approximation. But Calculus can be used to determine where the graph approaches a correct value.

Trurl · March 26, 2015

I have been working on n = p * q. I can find p knowing only n.

If you do not believe me check my website: <link removed by mod>

But if you decide to Back me at KickStarter it is only $2

If you dont believe it can be done waste your money and be proven wrong.

I dont mean to advertise, but I figured this was the only forum that could prove or disprove my work.

!

Moderator Note

we don't allow appeals - no matter how well intentioned - nor links to website with a commercial aspect. Please explain your ideas here.

pzkpfw · March 26, 2015

Isn't the point of RSA, not that it can't be done, but that with big enough p and q it'll just take too long to be feasible?

What's the largest value you've tested against?

Trurl · March 26, 2015

I am claiming that I can factor n no matter how large because I use an equation and not recursive factoring.

pzkpfw · March 26, 2015

I am claiming that I can factor n no matter how large because I use an equation and not recursive factoring.

Yeah, I figured as much by noting your previous thread: http://www.scienceforums.net/topic/78958-does-this-hold-true-for-all-prime-products/#entry769264

Again, what's the largest value you've tested against?

**imatfaal** · March 26, 2015

I dont understand - your equation y approximates sqrt( (N * y – x^2) / x) gives you no new information that would allow you to attempt to factorise the products of large primes.

1. it is basically a messy way of writing y approximates sqrt(y^2-x) - which clearly shows why it is an approximation and not a very good one

2. You seem to be using the factors as input to your equation - but the very point is that you do not know the factors

3. If you rewrite in terms of N you get N approximates (y^2.x-x^2)/y what use is this?

Trurl · March 30, 2015

The largest is 9 digits.

I am not experienced enough at programming to get a 256 bit number. However the equation will work there just has to factor in the amount of error each side of the equation is between each side.

That is why I need Backers. I have thinking that I could compute the error with the equation since any products can be entered. I also believe that what might work is putting n into the equations and then subtract n/p and q found in each side of the equation. With computed error the equation will find p every time. At least in theory.

That is N = x * y in my equations for n = p * q in standard representation.

BTW I posted a simplified and augmented equation for my Backers.

John Cuthber · March 30, 2015

Please show us how you would factor 14209 using your method.

[it's 13 times 1093, but that's not the point. Please show us your working.]

Trurl · April 1, 2015

Well I can’t show you the complete solution. If you want that you can back me on the KickStarter link for $2. See updates 7 and 8.

But I have shown enough on my website that will show merit. This link:

http://www.constructorscorner.net/Files/PrimeProductSolutionFlyer.pdf

This link will list 3 equations at the top. I set them equal then solved the resulting polynomial by Mathematica. I have simplified these equations for my Backers.

I am not pulling an April Fools joke. Even though I am not releasing my KickStarter information yet, there is still time to Back (for $2). It isn’t about the money it is about the project. I would like to present my findings with a team that has tested them.

I have some good things in my KickStarter updates. This project is only a beginning of many possible projects.

BTW, I am no expert on RSA. I just took the factor problem and found a pattern. I started with n = p * q and broke the one way function of factorization.

I know it sounds unrealistic. It is supposed to be impossible. It would change cryptography as we know it if it worked. The downside is that there could be a man in the middle attacks with digital certificates.

I encourage you to back. I don’t make a fortune. I only get a valuable team member.

John Cuthber · April 1, 2015

"Well I can’t show you the complete solution. "
Then there's nothing here to discuss.

And the failure of your efforts in your previous posting may impact the morality of asking for money.

**Strange** · April 1, 2015

Please show the prime factors of this number: 2260138526203405784941654048610197513508038915719776718321197768109445641817966676608593121306582577250631562886676970448070001811149711863002112487928199487482066070131066586646083327982803560379205391980139946496955261

From here: http://en.wikipedia.org/wiki/RSA_numbers#RSA-220

Edited April 1, 2015 by Strange

**imatfaal** · April 1, 2015

!

Moderator Note

1. Anymore appeals to go to Kickstarter will cause the thread to be locked and you banned. You have been warned

2. If you are not going to go into the details of your algorithm then let us know and we can lock the thread now. This site exists to discuss academic topics not to promote your kickstarter idea.

3. Please start responding to members' questions

Trurl · April 2, 2015


PNP = 605054707

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / 
    PNP^3) == ((PNP^2/x) + x^2)/PNP + 0.335659906865573163514044265, x]

605054707

{{x -> 357636.9295970702230530200 + 
    619609.1770247820759821702 I}, {x -> 
   357636.9295970702230530200 - 
    619609.1770247820759821702 I}, {x -> \
-715273.7465265673198060312}, {x -> -14251.112667573126300008868}, {x \
-> 14251.000000000000000000000}}

The drawback is finding the right error exact to such demanding decimals. I could find it for the range of numbers on a graph, but as n increases so does the need for error between the 2 sides of the polynomial equation.

**imatfaal** · April 2, 2015

PNP = 605054707

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / 
    PNP^3) == ((PNP^2/x) + x^2)/PNP + 0.335659906865573163514044265, x]

605054707

{{x -> 357636.9295970702230530200 + 
    619609.1770247820759821702 I}, {x -> 
   357636.9295970702230530200 - 
    619609.1770247820759821702 I}, {x -> \
-715273.7465265673198060312}, {x -> -14251.112667573126300008868}, {x \
-> 14251.000000000000000000000}}
The drawback is finding the right error exact to such demanding decimals. I could find it for the range of numbers on a graph, but as n increases so does the need for error between the 2 sides of the polynomial equation.

OK - One of the solutions to that particular quintic is one of the roots (or close). But how do you generate that quintic?

If I take a new PNP say 1025876959 - I again get five (complex) roots

x = -1.01706×10^6

x = -18556.6

x = -3.55149×10^-7

x = 18556.5

x = 508532 -880999. i

None of them are within 5000 of either of my primes. You clearly have to customise your polynomial for each PNP!

And I bet dollars to donuts you either have to iterate it multiple times or you have to know the prime factors in advance in order to so customise. And if you are homing in on an answer via iteration then it WILL be slower than the current methods and still far too slow to worry RSA fans. Or if you need the primefactors in advance then you are cheating.

Trurl · April 9, 2015

There is a problem with error. That is the decimal I am adding to the right side of the equation. It should be 0.299… but for a number in the trillions the error is 0.33. I don’t understand why it isn’t working. This is what we need to be working on.

The truth is this equation is too valuable to abandon. Obviously a computer algorithm could test different errors. I just wanted to make the math equation pure. That is having no guesses just a perfect equation that solves for PNP.

I have posted this to my Backers and 2 message boards. I post not so the equations are stolen, but more minds will solve this.

______________________________________

Yes I cheated with the 11 digit example. I know that putting a number into the equation without the error creates imaginary numbers. However the error can be graphed for error as opposed to the difference of the left equation minus the right side of the equation. This should give a relatively easy and good estimate as to the error that the equations fall in.

So far not perfect but I am working with it. On the bottom equation I took the error the known 0.28 and put it in and a real number 941 appears. Place the original PNP and put 941 in for x and subtract the left equation from the right and the error is found.

I know this solution of the error is not mathematically perfect, but remember it started as a pattern and with the assistance of a computer estimating the error should be possible. I know for a 256 bit number this sounds cumbersome. However I would say that graphing y = ((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / PNP^3) - ((PNP^2/x) + x^2)/PNP

PNP = 3163007

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / PNP^3)== ((PNP^2/x)+x^2)/PNP + 0.287159517,x]
3163007
{{x->10766.8 +18673. I},{x->10766.8 -18673. I},{x->-21533.6},{x->-953.082},{x->953.}}

The equation below shows how to solve the error which is 0.287

PNP = 3163007

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / PNP^3)== ((PNP^2/x)+x^2)/PNP + 0.28,x]
3163007
{{x->10767.00553406246` +18672.685351161595` I},{x->10767.00553406246` -18672.685351161595` I},{x->-21533.93266812111`},{x->-941.1247294994954`},{x->941.0463294956866`}}

The error of 0.28 is a starting point. Each side of the NSolve equation. These equations equal the Prime product within this error. I know it isn’t exact yet, but I am working on it.

x = 941

PNP = 3163007

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

941

3163007

8859632789683911270/31644661843413961343

N[8859632789683911270/31644661843413961343,11]

0.27997242737

__________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________

PNP = 85
x = 5




y = sqrt ((PNP*y\[Dash]x^2)/x)

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




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


 

85

5

$RecursionLimit::reclim: Recursion depth of 1024 exceeded. >>

Hold[425 sqrt y \[Dash]]

294/17

N[y, 11]

17.593323835

N[y, 11]

17.593323835


PNP = 85
x = 5
((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / PNP^3) - ((PNP^2/x) + x^2)/PNP

85

5

1470/4913

N[1470/4913, 11]

0.29920618767



PNP = 605054707

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / 
    PNP^3) == ((PNP^2/x) + x^2)/PNP + 
   0.29920618766537756971300630979035212701, x]

605054707

{{x -> 357642.060612364370036337217772965000 + 
    619600.276496119708382454094550208609 I}, {x -> 
   357642.060612364370036337217772965000 - 
    619600.276496119708382454094550208609 I}, {x -> \
-715284.031700385975080353264491082027}, {x -> \
-13455.01084271186475677779496348851204}, {x -> 
   13454.92131836909976445662390864053977}}


PNP = 605054707
x = 13454.9213

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


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


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

605054707

13454.9

44969.3

44969.6

0.299206



PNP = 605054707

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / 
     PNP^3) - .6 == ((PNP^2/x) + x^2)/PNP - 0.299206186842639, x]

605054707

{{x -> 357642. + 619601. I}, {x -> 
   357642. - 619601. I}, {x -> -715284.}, {x -> -13490.7}, {x -> 
   13490.6}}




PNP = 605054707
x = 13490.6

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


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

PNP/x

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

605054707

13490.6

44850.4

44850.7

44850.1

0.300795

44850.5 - 44850.09614

0.40386



44850.7 - 44850.09614

0.60386



PNP = 605054707

NSolve[((PNP^4/x + 2* (PNP^2 * x^2) + x^5) / 
     PNP^3) - .6 == ((PNP^2/x) + x^2)/PNP - 0.4, x]

fiveworlds · April 9, 2015

Isn't the point of RSA, not that it can't be done, but that with big enough p and q it'll just take too long to be feasible?

Preety much. However with RSA

A public key is generated
The file is encrypted with the public key
The public key is shared with third party
The file is decrypted by third party

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

Edited April 9, 2015 by fiveworlds

**imatfaal** · April 9, 2015

Preety much. However with RSA

A public key is generated

The file is encrypted with the public key

The public key is shared with third party

The file is decrypted by third party

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

The above is almost entirely wrong.

Your public keys could only be used to encrypt messages which would then need your private key to decrypt.

1. imatfaal generates a private and public key

2. imatfaal shares public key with fiveworlds

3. fiveworlds uses public key to encrypt message

4. message can be decrypted ONLY with private key that imatfaal has kept secret

fiveworlds · April 9, 2015

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

Edited April 9, 2015 by fiveworlds

**imatfaal** · April 9, 2015

http://en.wikipedia.org/wiki/Public-key_cryptography

fiveworlds · April 9, 2015

I know which is fine on paper but on a computer the rules are slightly different. The enemy can read what you type. Look at windows sales figures

Windows 1.0 sales from its November 1985 launch to April 1987: 500,000 (Computerworld)
Windows sales in 1988 (Windows 2.0 shipped on December 9, 1987): 1 million (InfoWorld)
Windows sales, all versions, 1985 to January 1990: less than 2 million (InfoWorld)
Windows 3.0 sales, first year: 4 million (InfoWorld)
Windows 3.1 sales, first 3 months or so: 3 million (InfoWorld)
Windows 95 sales, first year: 40 million (Network World)
Windows 98 sales, first four days: 530,000 boxed copies through retail channels (New York Times)
Windows 2000 sales, less than a month after launch: 1 million (Microsoft)
Windows ME sales, first three days: 200,000 boxed copies through U.S. retail channels (Network World)
Windows XP sales, first three days: 300,000 boxed copies through U.S. retail channels (Network World)
Windows XP sales, just over two months after launch: 17 million (Microsoft)
Windows Vista sales, one month after launch: 20 million (Microsoft)
Windows 7 sales, first six months: 100 million (Engadget)
Windows 7 sales from October 2009 launch to June 2010: 150 million (Neowin)
Windows 7 sales in less than two years: 450 million (TechCrunch)
Windows 8 sales in a little over two months: 60 million (ZDnet)
Windows 8 sales, first six months: 100 million (Microsoft)

Are you telling me that in 100million similar copies of windows 8 sold 10 people haven't figured how to hack it already?

**imatfaal** · April 9, 2015

There are many exploits against poorly worked RSA implementations - but very very few proven against proper usage. Replicating public keys would not work as public keys are used for encryption only. It might help a semantic insecurity search but any decent implementation will automatically render that impossible with padding.

If third parties have full control of your computer then difficult to hide stuff - but that is akin to saying that the algoritm is faulty cos someone can read the plaintext over your shoulder.

fiveworlds · April 9, 2015

If third parties have full control of your computer then difficult to hide stuff - but that is akin to saying that the algoritm is faulty cos someone can read the plaintext over your shoulder.

Exactly but people will do it and not only that but they can win at least $100000 for figuring it out http://zerodayinitiative.com/Pwn2Own2015Rules.html

**imatfaal** · April 9, 2015

What does any of that have to do with your post which said ...

Preety much. However with RSA

A public key is generated

The file is encrypted with the public key

The public key is shared with third party

The file is decrypted by third party

Which is not how public key encryption works.

Your post also said

...

Preety useless when a skilled hacker could replicate the public keys stored on your computer....

Which would only allow "a skilled hacker" to encrypt messages not decrypt them.

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

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