New One Way Function

• 12-21-2013, 10:53 PM
trurl_
3 equations that should have some relationships; 3 equations of this thread so far.
In:= soln = Solve[y == Sqrt[(n*y - x^2)/x], x]
soln //. {n -> 85, y -> 16.85229955} // N
Out= {{x -> 1/2 (-y^2 - Sqrt[4 n y + y^4])}, {x ->
1/2 (-y^2 + Sqrt[4 n y + y^4])}}
Out= {{x -> -288.957}, {x -> 4.95729}}

Sqrt[(85*(85/x) – x^2)/ x] = x

Abs(tan(85 / (2*Pi))) * x + x = 85/(2*Pi) tan in radians of course.

((85^4/x + 2*(85^2 * x^2) + x^5) / 85^3 ) * x – 85 = 0

These 3 equations each approximate x and y when 85 = x * y.

In my program I tested each value from 1 to 85 to see which numbers worked for x.

There is some error, but it x should approximately equal 5.

N = p *q or N = x * y.

I cannot solve the polynomial to find x from only knowing N. But when you test y proves the equation works.

The trouble is it is not so much different from factoring. No speed increase. But it is a legitimate equation. I am researching new ways to solve polynomials.

Any more questions just ask or see my site: www.constructorscorner.net

There is a link on the Homepage ?
• 12-27-2013, 09:11 PM
trurl_
5.02984
Equation – Error = Equation + Error
17.5933238347242 – 0.6 = 16.85229954635272 + 0.15

I cannot due the calculation of this complex polynomial by hand. That takes much higher math. But I can set the 2 equations equal to y and plug them into Wolfram alpha. With the error the answer is in range.
I am researching methods to better solve polynomials with no concept yet to start with. But you have to admit that though the pattern of Primes is yet to be solved, there is something here. The multiplication is what makes the RSA algorithm vulnerable, but the logarithmic spiral is a main way to visualize this multiplication.
I will address the logarithmic spiral later for know notice Wolfram Alpha came up with 5.02984 as a real solution. This is a design not just plugging and jugging. Let me know what you think. If it interests anyone I will further explain the logarithmic spiral.

http://www.wolframalpha.com/input/?i...+x%5D+%2B+0.15
• 01-05-2014, 11:46 PM
trurl_
Trurl's Challenge
I have attempted to encrypt a file using RSA. Let me know if this is the public key. Attempt to use my equations to reverse the encryption process.

http://www.constructorscorner.net/id...Challenge.html
• 01-09-2014, 02:26 AM
BryanFreeman
I don't understand that. That's pretty complex. Bouncing over my head.
• 01-21-2014, 07:51 AM
Selena
Looks good. But i leaved using c from last 4 years. :)
• 02-21-2014, 09:57 PM
trurl_
A more descriptive picture.
Sqrt[(85*(85/x) – x^2)/ x] = y

Abs(tan(85 / (2*Pi))) * x + x = 85/(2*Pi) tan in radians of course.

((85^4/x + 2*(85^2 * x^2) + x^5) / 85^3 ) * x – 85 = 0

________ These three equations solve show how two Prime numbers x and y equal 85.

This is important because in asymmetric cryptography such as RSA, the public key is the product of 2 unknown Prime numbers. If we can set a relationship to find x or y when given N, which in this case is 85, the one way function that RSA is based on is mathematically defeated!!!

_______ So if we set equation 1 equal to equation 3 the polynomial produced, if solvable, will solve for x.

Equation used to find 5 knowing 85:

((85^4/x + 2*(85^2 * x^2) + x^5) / 85^3 ) – 0.6 = Sqrt[(85*(85/x) – x^2)/ x] + 0.15

= approximately 5.02 when solved by Wolfram Alpha!

I believe this will defeat RSA. However no one believes me or just doesn’t care. To me it is interesting, but when I ask people about it no response. If it worked I think it would go viral. What do you think?

This is the same post to another community. Either people don’t believe me and my math or they just don’t care about math. I thought that knowing N and solving to find one of its products x could be found knowing only N. (There is no reason that the equations cannot be manipulated to solve for y.)

I’m posting this problem everywhere I can. I believe it has merit. On another message board a math guru showed that x and y (knowing them both) complete the equation. However for every value I have tested, within error, I have found the answer to be in acceptable range of the Prime number we are solving for knowing only 85 (N).

This is as simple as I can make the equations in a page format. I used Wolfram Alpha to solve the polynomial. I have had good results.

There is an error in the approximation within a certain range.

You may not believe me but there is merit to this math write-up.
• 04-15-2014, 11:07 AM
trurl_
http://community.wolfram.com/groups/..._auth=P7T3UwOb

I have been writing about the logarithmic spiral and Prime numbers for quite some time. I don't want people to think that is all I do.

Here at this link I made a comment on an observation I made about this mathematician's "snow flakes". It applies to 3DBuzz in that it is a way to do 3D modeling. I think it is call "subdivisional surfaces", but I'm unsure the correct term. I know that when you model you want nice "round" geometry. You don't want sharp of flate geometry when creating a human figure for example.

This is why I say use an ellipse. It is made up of straight lines and can be modified to fit any form. If you follow the link and see the pictures you will instantly understand.

As for the logarithmic spiral, I have one more geometry to show that should prove the logarithmic spiral has potential to map Prime numbers. But I will post that once I write it up. Until then look at this mathematician's project and see what I had to comment on his work. It is a break from Prime numbers.
• 05-09-2014, 01:10 AM
trurl_
A simple trigonometric definition:

s = R\[Theta]

A vector where x + y = N

But this isn't useful in this form so we solve with a triangle that is similar that we do know some values.

Attachment 74200

"x" is the value we are solving for. We know N - ((y/2\[Pi]). The obtuse angle between the 2 "x" radiuses is (y/2\[Pi]). This is if we had
taken the angle of y and rotated it around the cirlce with radius "x". So "x" * "y" equals N. And as in the trigonometric definition s = R\[Theta].

But we know that the smaller, similar triangle n = (N/x) / 2\[Pi]) by inspection. (Where N is the given product of 2 Prime numbers
and the angle wraps around the circle is divided by 2\[Pi]. And we know that y = (N/x).

So by vector division u + v = the longest chord or N. So we form the following equation:

|tan(y/(2\[Pi]))| * x + x = (y/(2\[Pi]))

substitute (N/x) for y

|tan((N/x)/(2\[Pi]))| * x + x = ((N/x)/(2\[Pi]))

multiply both sides by x

|tan(N//(2\[Pi]))| * x + x^2 = (N/(2\[Pi]))

example where N = 85

|tan(85//(2\[Pi]))| * x + x^2 = (85/(2\[Pi]))
• 05-11-2014, 09:34 PM
trurl_
Just by logic not proven:

((85/[Pi]) / (2/[Pi]) = x*(remainder of (y/(2*[Pi])) where N= 85
• 05-17-2014, 06:12 PM
trurl_
A little relationship
I know what you are thinking after the last equation: "You call yourself Trurl the Constructor and you throw a junk equation into an already hard to follow problem?"

Well I will show you a relationship between Prime numbers that doesn't take Mathematica to calculate some high degree polyonmial. But you may say, "There are infinitely many of such relationships. What makes this one significant?"

Nothing. An equation is only as significant as you make it. I am just saying we are taught that Prime numbers are impossible to work with and I say they are not.

Where x = 571 and y = 1381

1381 / 2\[Pi] = 219.7929764

2\[Pi] * (remainder(1381/2\[Pi]) * 2\[Pi]) * 571 = 13208.65186
rewritten: 2\[Pi] * ((1381/2\[Pi]-whole number part) * 2\[Pi]) * 571 = 13208.65186

13208.65186 / (1381/2\[Pi]) = 60.09

13208.65186 / 60.09 = 219.7929764

219 = 219

So judge for yourself if this is significant. This is very confusing so I hope I wrote this down right. But as I once invented a machine that made all things "N'. Numbers. I am in the business of constructing numbers.

If the last equation didn't work logically. But sometimes you plug numbers into a calculator along the lines of your equation and it gives you a "feel" for what is happening with those numbers.

I have other relations, but it would be too confusing to share them at the same time.