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# Help me find the error of these RSA equations in Mathematica

Posted 9 years ago
 This is the post I posted yesterday. It is relevant to Mathematica because it cannot be solved without it. Review the Mathematica code and help me find how to determine the error of the 2 equations! _______________________________________________________________________________- 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] `
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Posted 9 years ago
 We are group of mathematicians. Although I am only an amateur, does no one see that this is a polynomial representation of factoring?This is factoring semi-prime numbers. And an error of 0.17 is the difference of finding an 11 digit semi Prime n where n = p * q?Search the community for my name: bobby joe snyder and see the discussion How do I find x in a Polynomial when x is ApproximateWe are talking the defeat of RSA. All using Mathematica.
Posted 9 years ago
 simplified equation 