Can RSA be a 2 Lane street?

[Trurl]

All right, here I have a concept. It can be proven right or wrong within 3 minutes. If you liked my SSA triangle then you will love this.

 

http://www.constructorscorner.net/ideas_and_gadgets/math/math_hunch/hunch_00001/hunches_section0007/RSA2Lane.html

 

Remember I am a student, not a scientist. It is a concept not a theorem. The point is not to call me stupid. It is to give guidance. I posted here for instruction. I will not claim it works. Remember student not scientist.

[DrRocket]

All right, here I have a concept. It can be proven right or wrong within 3 minutes. If you liked my SSA triangle then you will love this.

 

http://www.construct...7/RSA2Lane.html

 

Remember I am a student, not a scientist. It is a concept not a theorem. The point is not to call me stupid. It is to give guidance. I posted here for instruction. I will not claim it works. Remember student not scientist.

 

You seem to have overlooked the fact that in the ring of rational functions with rational coefficients [math] xy=1[/math] and that there is no solution to the equation stated for [math] x[/math] which if [math]x \ne 0 [/math] simplifies to

 

[math] 85^2 +x^3 = 85^2 \\ \rightarrow x=0 [/math]

[DrRocket]

All right, here I have a concept. It can be proven right or wrong within 3 minutes. If you liked my SSA triangle then you will love this.

 

http://www.construct...7/RSA2Lane.html

 

Remember I am a student, not a scientist. It is a concept not a theorem. The point is not to call me stupid. It is to give guidance. I posted here for instruction. I will not claim it works. Remember student not scientist.

 

You seem to have overlooked the fact that in the ring of rational functions with rational coefficients [math] xy=1[/math] and that there is no solution to the equation stated for [math] x[/math] which if [math]x \ne 0 [/math] simplifies to

 

[math] 85^2 +x^3 = 85^2 \\ \rightarrow x=0 [/math]

[Trurl]

I know this equation is too good to be true. I don’t mean to sound dumb, but where is that error occurring?

 

I am setting xy = xy.

 

1 = 0

But I set xy*x = y

 

And test to see if the polynomial = 17 which is 85/5.

 

There may be some confusing in the last step after I check if y = the polynomial.

 

I then set y = 85/5 which it is. I do not attempt to solve for x to test because of the 6th degree polynomial.

 

I put some effort into this and when I say polynomial = y checked, I thought that it was too close to be a random number.

 

Do you agree when you put 5 into x you get a y of 17?

 

You are probably right, there is probably a flaw. And if it does work the polynomial is too complex. I just thought it stood out and was worth some attention.

 

But your feedback helps. That is how you tell if you got something or not.

[Trurl]

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

 

sol = NSolve[p  0]

 

-5+52200625/(52200625/x+14450 x2+x5)

 

{{x10.7235 +26.6243 },{x10.7235 -26.6243 },{x-22.8216+10.9987 },{x-22.8216-10.9987 },{x19.0002},{x5.19606}}

 

 

Does anyone see a pattern now?

 

My question is how to analyze the data by programming. NSolve gave decimal numbers but only 2 are true. Really only 5 is true. I think 19 is close enough to 17 that 85/19 = 4.7 .

 

If you think about it something that doesn’t have a pattern with symmetry and proportions as a shape like a parabola has, can only be described by higher order polynomial equation, because it allows for variation. The trouble is solving those polynomials.

 

I want more analysis. I need the answer for NSolve in a variable so I can do calculations with this. I can do calculations in Mathematica I just can’t program. Is there any way to program such intangible data? Help would be much appreciated.

[phillip1882]

so, here you can use newtons method, or even a standard guess and check in a program to get a better and better approximations of the solution, assuming one exists.

85^4/((85^4/x)+ 2*85^2*x^2 +x^5) - 5 =0

here would be the guess and check program, in python.

def guessCheck(value):
  a =1.0; b = value/2.0
  mid = (a+b)/2.0
  aprox = value**4/(value**4 /mid +2*value**2 *mid**2 +mid**5) -5.0
  while aprox > 0.00001 or aprox < -0.00001:
 	if aprox > 0:
		a = mid
 	else:
		b = mid
 	mid = (a+b)/2.0
 	aprox = value**4/((value**4 /mid) +2*value**2 *mid**2 +mid**5) -5.0
  return mid
print guessCheck(85)

i haven't read your paper, so you may need to correct the above formula but hopefully you get the general gist.

to execute the above code, you can either download a copy of python for free at www.python.org

or you can go to the site www.ideone.com to have an online compiler, that requires no download.

edit: the above isnt looking right in the the post, not sure why, but the if and else should be 3 spaces i front of the while, with the a = and b= being indented anther 3 spaces, , and the mod and aprox on the same indent as the if else.

Edited April 21, 2012 by phillip1882