  Trurl

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1. If you are reading this you know that I am looking for established math references to support the Pappy Craylar method.

One observation: a differentiation of the curve between zero and one may have surprising results.

I claimed the graph was an inverted normal distribution. But there is “an irregular logarithmic graph called the Tracy-Widom distribution that explains Prime number distribution.

Again could be nothing but I am searching for anything that would credibility to the PC method.

2. Ok,  so in my last post to Simple Yet Interesting I gave 3 formulas. The math is there. I have no more to argue. It may seem that the math is inconsistent. However if you look at how it evolved the underlying ideas are consistent.

The equations graphed between zero and N (the Semi-Prime) result in the need  to test values between zero and one on that graph.. That is quite the accomplishment. It could be argued for double Prime numbers there are more possible numbers between zero and one. However the idea is the same, only numbers between zero and one need tested. With double Prime numbers there is just an increase in numbers to test. The hypothesis remains true.

My goal now is to use existing knowledge to verify the Pappy Craylar Method (PC method: That is what I call my work.) By existing knowledge it is meant that rules to Primes that are already well established or others work that is already accepted.  For example if you increase one factor it results in a decrease in value of the corresponding  factor. Knowing that there are only 2 factors and they fit together one way allows the factorization of N (the Semi-Prime), knowing only N.

The part about the Semi-Prime is me. I am not saying it is accepted. I am just explaining my idea. Those 2 sentences are why I believe in the PC method.

Any thoughts?

3. Klapaucius, uh um, I mean Ghideon, You act as there is noting of value in this tread. Sure, breaking RSA would be quite a feat. But you must admit there is no other method that does it. There is no guarantee the Pappy Craylar Method works. And I have moved on to other projects. However, we had an active discussion that may lead to something that might find a different approach. You see, The Pappy Craylar Method is based on the fact that if you increase the size of one factor you must decrease the other. And since there is only one way the Semi-Prime factors go together, the equation to find the factors eliminates possibilities. Sure, the number of possibilities isn’t only one and sometimes the possibilities are further or closer to zero. But until a solution exists, the method leads to the ability to guess the factors by trial and error. See my construction of the 2 graphs that intersect at 5. Of course, I already new the answer, but don’t you find the PC method simple yet interesting? In:= pnp = 85 x = 5 Show[{Plot[(x^4/(pnp^2 + x)), {x, 0, 10}], Plot[(pnp - (x*Sqrt[(pnp^3/(pnp* x^2 + x))])), {x, 0, 10}]}] Out= 85 Out= 5 In:= Clear[pnp, x] pnp = 8637 {Plot[(x^4/(pnp^2 + x)), {x, 0, 3}] Plot[(pnp - (x*Sqrt[(pnp^3/(pnp* x^2 + x))])), {x, 0, 3}]} Out= 8637 Does this work for every Semi-Prime? Maybe, maybe not. This is all I have to share. I am moving on to other projects. So, unless some idea comes along, I will not post. Is my last message Simple, yet interesting?
4. You must tell me why RSA is not affected by this thread. We can agree that RSA was based on the Prime factorization problem. That is factoring Semi-Primes into the Prime factors. So you must not trust my algebraic equations that estimate where the factors of semi-Primes occur. That is the basis of my work. I think I have stated what they do. But in your test where N=p*q where do you not believe p cannot be found knowing only N? $$q\sqrt {\frac {\text {pnp}^3} {\text {pnp} q^2 + q}}$$
5. @Ghideon I know you are not going to like my explanation, but q/N is enough to get an estimate of N. Yes I know it is true for all numbers, but using the reciprocal of p and q helps determine where they are positioned on the number line. For semi-Primes my hypothesis is that q/N will reduce to 1/p. Then take 1/p and multiply it by N. You could argue that it is no more useful than recursive division. However, it does give an estimate of where y lies on the number line. And it could further improve estimates of x when determining how far (N minus p) is from (N minus p calculation). I know it seems dumb. Why did we divide in the first place if a simple q/N could solve the solution? And it is a high possibility it doesn’t work. But I am saying the simple solution to the semi-Prime factors problem is helpful. That is: not helpful for a beautiful math equation, but helpful when trying to find a hack to defeat a problem that was never a one-way-function from the start. Clear[pnp, q] pnp = 85 q = 17 Simplify[(q^2 - (pnp*q^2 + q)/ pnp)] Out= 85 Out= 17 -(1/5) ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ \ ___ ___ ___ ___ ___ ___ __ In:= Clear[pnp, q] pnp = 85 q = 19 Simplify[(q^2 - (pnp*q^2 + q)/ pnp)] Out= 85 Out= 19 -(19/85) ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ \ ___ ___ ___ ___ ___ ___ ___ In:= Clear[pnp, q] pnp = 293*3 q = 293 Simplify[(q^2 - (pnp*q^2 + q)/ pnp)] Out= 879 Out= 293 Out= -(1/3) ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ \ ___ ___ ___ ___ ___ ___ ___ _ In:= Clear[pnp, q] pnp = 293*3 q = 291 Simplify[(q^2 - (pnp*q^2 + q)/ pnp)] Out= 879 Out= 291 Out= -(97/293) In:= N[-(97/293)] In:= -0.3310580204778157`* 293*3 Out= -291.
6. Why? Because it is p^2 minus p^2 derived. Same with second equation. It is just how I derived the equations. Does it work? I don’t know. But I am simply comparing patterns in division. Obviously you tried test values. Did it work with your values? i will run more test values. Let me know if you have any more questions or your test values don’t work.
7. Note: These are 3 separate groups of equations. They are separated by lines. $p^3 - \frac {N^2 p^3} {N^2 + p}$ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ _ The above equation should be close to zero, but sometimes the error is closer to 1 $\frac{N p^3}{N^2+p}=\text{fraction}$ because q = N/p $\frac{\text{fraction}}{q}=\frac{\text{fraction}}{\frac{N}{p}}$ $N \sqrt{\frac{\text{fraction}}{\frac{N}{p}}}=p^2$ ___________________________________________________________________ The above equations are the second example. $q = \sqrt {\frac {N q^2 + q} {N}}$ square q $q^2-\frac{ \left(N q^2+q\right)}{N}$ In the case where N = 85 and q = 17, $q^2 - (N*q^2 + q)/ N$ is 1/x or 1/5 or 0.2 $0.2 * 85 = x$ or 5 in this example These are 3 separate groups of equations. They are separated by lines.
8. Here are my equations to review. There are quite possibly hundreds of variations. This is what I meant by isolating x and y in the equations. I mean isolated p and q separately, knowing only N. There are 3 equations. I am just getting used to Latex so I hope this is readable. If necessary I will post corrections. But it is these 3 separate equations I want you to test. p^3 - (p^3*N^2) / (N^2 + p) $p^3 - \frac {N^2 p^3} {N^2 + p}$ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ _ The above equation should be close to zero, but sometimes the error is closer to 1 (p^3 *N) / (N^2 + p) = fraction fraction/ q = fraction / (N /p) Sqrt[fraction / (N /p)] * N = p^2 $\frac{N p^3}{N^2+p}=\text{fraction}$ ___________________________________________________________________ The above equations are the second example. q = Sqrt[(N*q^2 + q)/ N] q^2 - (N*q^2 + q)/ N is approximately 0 In the case where N = 85 and q = 17, q^2 - (N*q^2 + q)/ N is 1/x or 1/5 or 0.2 0.2 * 85 = x or 5 in this example $q^2 - \frac {0\text {approximately}\t ext {is}\left (N q^2 + q \right)} {N}$ Final equations are a proposed method for finding q, knowing only N I have been working with the first equation for awhile. The other 2 are based on the first, but still need tried. There is nothing fancy here. I am just comparing how numbers divide.
9. The new "y isolated" equation does not work. It was based on the "x isolated" equation, but does not work. Does anyone see a way to isolate y knowing only pnp?
10. \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} Yes that is exactly it. But does it work? I think it works, but does it prove useful? I call "N": pnp and "p and q": "x and y" Given N (I call pnp) we estimate x. And in the latest post I am trying to find y, the larger factor. \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} $\frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \t ext {pnp}} {\text {pnp} y^3 + y^2}$
11. (y^3 * pnp^3 - y^2 * pnp + pnp) / (y^3 * pnp + y^2) = pnp^2 y is now isolated as x was. Ready to simplify. No, you uncovered a weakness of the PC method. That is that the error varies between 0 and 1. The first SemiPrime was close to 94. I factored the second to be sure. But by working with the equation for years that it could be a smaller Prime factor. The trouble is without knowing the true error testing numbers less than 94 in the example, narrows the selection but not more useful than division. We need exact results and not an estimate. But does that exist? 3 possible solutions: Isolate y and compare to x Investigate the curve of the graph of the equation between 0 and 1 Find a pattern in the error. An error not spanning such a large range. 3^4/(8637^2+3)=0.000001732 3 < 94 but the number of importance is 0.00001732. It causes 3 to be hidden when starting at 1 and not zero. Questions are good. It is quite possible I am wrong. But I see something or I wouldn’t have put so much effort. From Wikipedia: Semiprimes are highly useful in the area of cryptography and number theory, most notably in public key cryptography, where they are used by RSA and pseudorandom number generators such as Blum Blum Shub. These methods rely on the fact that finding two large primes and multiplying them together (resulting in a semiprime) is computationally simple, whereas finding the original factors appears to be difficult. In the RSA Factoring Challenge, RSA Security offered prizes for the factoring of specific large semiprimes and several prizes were awarded. The original RSA Factoring Challenge was issued in 1991, and was replaced in 2001 by the New RSA Factoring Challenge, which was later withdrawn in 2007. https://en.m.wikipedia.org/wiki/Semiprime
12. Ok. I'll try two examples: 8633, 8637. Both are semi primes. NSolve [( x^4/((8637)^2 + x)) == 1] x≈-92.935 x≈92.935 NSolve [( x^4/((8633)^2 + x)) == 1] x≈-92.914 x≈92.914 @Trurl Is this helpful for someone looking for the prime factors of the numbers 8633 and 8637? If so, how? Yes the numbers are too close to estimate. But I still have faith in the Pappy Craylar method. After all 3 < 92 and 3^4/(8637^2+3)=0.000001732 Which Is close to zero where the error is supposed to be near anyway. Remember when I said we needed to test the logarithmic curve of 3*5; 3*7; 3*11 and so forth to find the error change as 3*Prime number increases? I know your smart and are looking for holes in the PC method. That is what you are supposed to do. But I know you are smart enough to understand it. A fix to to great number of possible estimates is to isolate y as x is isolated. Then we will see a better picture of what is happening when SemiPrimes are factored.
13. y^2 = ((pnp*y^2+y)/pnp) This is one of many isolated y’s. I have much more to share. But I want readers of this to see how simple the equation is. Remember that this thread is called Simple yet Interesting. 😒
14. I will post here how the equations were derived if anyone is interested. My question to you is: Do you agree the Pappy Craylar method is significant to factorization? My second question to you is there a list of all the ciphers that use SemiPrimes? The hash ciphers the message but to have public keys you need a one-way function. But do one-way functions exist? I read on the web the question of whether math is created or discovered. Something that hasn’t been solved is impossible to it’s discovered it is not. Was it your understanding that was discovered? I cannot factor without error. So large SemiPrime increase in range between zero and one. But if I can post here with the derivation of x, maybe a similar equation can be found for y, the larger factor.
15. Trurl,

I like your electric fields idea regarding the applications of the twelve sections of the sphere but since I last posted on that thread I have run into a major roadblock to all the applications I was considering.  The subdivisions do NOT appear to be of equal area as I had figured earlier in the thread, using a spherical calculator.  I made some incorrect assumptions about angles and unfortunately it appears the subdivisions near the center of each diamond are  of greater area then nearer the corners.    So most applications are thwarted until I come up with a way to have each designated subsection span an equal area.

There are still some applications that will work if exactness is not required.  If you quarter a section each of the quarters is of equal area, so if you want to divide the sphere into 48 equal sections for your application the scheme is still good.

One such idea is to for instance mount 48 cameras on a tower or aircraft each pointing in the direction of the center of a quartered section of each diamond.  Then the output of the cameras could be sent in  a one to one way to 48 screens positioned in the same pattern around a viewer.  Thus placing the viewer at the top of the tower or in the craft.

1. Show previous comments  1 more
2. So the 48 cameras are some sort of virtual reality?

Application: What if you divide the Earth into sections to map satellites?

What about video game 3D world positioning?

The changing shape and propagation of the radio wave?

It is hard for me to think of something mathematical that doesn’t have an application no matter how small.

I know my application suggestions are not descriptive and it would take much work to complete them, but I envision a 3D model manipulated by change, shape, and division of spheres. Instead of editing polygons you’d edit spheres. If you tried 3D printing you know the inside of the model is structured like a bridge trusses. Those supports could be your spheres divisions.

3. Nice,i

I do need to learn some CAD and Rendering programs.

There are many applications I think for the general Sperical Rhombic Dodecahedral structure.

And interesting aspect is it filled with symmetries and dual figures. And it is the basis for  a dense packing scheme that puts 12 spheres around a center sphere.  This pattern is a cuboctrahedral pattern that when built out, putting always 12 spheres around each sphere in the same pattern it looks more and more like a cuboctahedron as you build out, and contains 3 intersectinng square planes and 4 intersecting hexagonal planes.

The math is already there, the switch is to think of a cube with the corners cut off to the center of the edges.  The center of each edge is the center of a 1/12th section of the sphere.

I am working on a scheme to each sphere in the scheme.  The scheme fills space and is completely scalable and very symmetric, having direct analogs to the cube, the sphere, the tetrahedron and the cuboctahedron.

I like your ideas.  Please use the system freely. It is a mathematical system so I can not claim any ownership.  What I will claim is the discovery of the way to break the sphere into TAR radians using the four axis of the cube that extend through the corners, assigning a color to each 360 group of great circles and using the intersection of the two colors in each of the diamonds to name each degree sized area.  I recently found out that all the degree sized area are not equal in area, so there is as of yet not a clean mathematical way to use the system.

Where I see some possibility however is in labeling the spheres in the dense packing situation.  There is a relationship between the spheres in the dense packing system and the TAR radian system but I have not discovered it yet.

4. a scheme to label each sphere in the scheme

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