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Simple yet interesting.


Trurl
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N=85

x=5

y=17

All my math problems deal with semiPrimes.

Is there a pattern? Does it only work with semiPrimes? Can you reduce the fraction to N/oversomedenominator? Has it been observed before?

I did not derive it from an equation directly. I am looking for patterns and thought this was one. But I will share if you deem it interesting.

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               (N^2 + x^3)/N = N + (x^3/N)

                              We know from discovery that (x^3/N) = (x^2/y)

                              So, we substitute for y in terms of x

                              The simplest substitution is (y=N/x)

                                             (For another equation we could substitute: (y=(N^2/x)+x^2/N)

 

                              Another equation in terms of y is: (y=(N^2/x + x^2)/N). This can also be substitute into

                                             (x^3/N) = (x^2/y), and graphed.

 

Look at the graphs on the attached pdf. Do you see it now?

GraphicCalcSimple.pdf

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The graph is what’s important.

N=1847*2393=4419871=4.4199871*10^6

For p3 as y approaches 0, x will be 1847 or x.

 

For p5  as x approaches 1847, y=4.4199871*10^6 or N

 

Using graphing in polynomial time to get an answer that algebra can’t solve.

 

 

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Thoughts?

 

Do you believe the math? I’ve shared it with an engineer and he thought it worked but he said he wasn’t a mathematician. I am not a mathematician either. And I am biased. I envisioned it so it makes sense to me. I was hoping for some guidance from these boards. Is it worth publishing? I am not an elite mathematician so message boards and the net are where I publish.

 

So now break large semiPrimes. And if the math holds true, break the DLP next. Primality test by multiplication and seeing if a semiPrime is formed. There are no prizes for factoring anymore. Patterns in the graphs; a normal distribution.

 

The maths range from algebra to calculus. There are similarities to graphing differential equations. So if you believe in my math please let me know. The message boards were built to solve problems like these.

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My maths is weak. That may account for my next observation. I have no idea what you are trying to assert/define/explain/postulate.

An alternative explanation for my lack of understanding is that you have, thus far, done a poor job of presenting your hypothesis.

Until you have addressed that possibility I cannot asses the third option - that you are posting nonsense.

I look forward to your clarification.

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7 minutes ago, Area54 said:

My maths is weak. That may account for my next observation. I have no idea what you are trying to assert/define/explain/postulate.

An alternative explanation for my lack of understanding is that you have, thus far, done a poor job of presenting your hypothesis.

Until you have addressed that possibility I cannot asses the third option - that you are posting nonsense.

I look forward to your clarification.

I'm with you on this, I have no idea what his point is.

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I know I not the best math teacher. What parts do you not understand?

 

I am taking the known semiPrime, PNP, and graphing it along the x axis to find the place where y equals zero. I do this to find x knowing only PNP.

 

Look at the pdf I posted (see previous post attachment). You will see 2 equations, p3 and P5, and the graphs that correspond to them there.

 

 

N=1847*2393=4419871=4.4199871*10^6. (These are N=x*y as an example)

 

 

 

For p3 as y approaches 0, x will be 1847 or x.

 

For p5  as x approaches 1847, y=4.4199871*10^6 or N (here x at 1847 y=N because the equation is structured differently. Also note y in N=x*y is distinct from y on the graph.)

 

 

I know that my equations are confusing. I was trying to show how they were derived algebraically. For now forget the equations and focus on the graph.

 

I recommend using a computer software to graph equations p3 and P5.

 

Trust me this is not a super advanced math problem. Stop treating it as such and go with you graph reading skills. That is why I called it simple yet interesting. Sometimes the simplest solution is best.

 

Let me know if this is a better description. And remember I use N and PNP to mean the same N=PNP.

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PNP is N in this. Originally it was N = p*j. But N is protected in Mathematica so I renamed it pnp. N = p*J = x*y, because I also called p, x and j, y. This is a poor decision because when you graph it ygraph is different from y in the equation. So it created some confusion, but writing this up and coming up with a solution variables get jumbled.

DLP is the discrete logarithmic problem. If this can factor semi-Primes then the DLP is next to go. Remember I am claiming I can factor semi-Primes and thus RSA cryptography would be no more. But it wouldn't matter because other cryptosystems would take its place. I know that is a bold claim. But I am only looking at simple patterns in factoring.

Here is a draft write-up I was working on. I hope it helps and not hinders. Finding math is one challenge. Explaining it is another.

 

Hacking extends to all disciplines and is not limited to technology. The question I ask is can math be hacked? Is there a “math hack” that will solve problems?

The number guessing game is a programming exercise for those learning to program. The user of the program guesses a number and is told if their guess is higher or lower than the desired value. It is very efficient and often takes less than ten guesses to find a number with several digits of length.

Factoring is a problem of cryptography. The most common example is finding the Prime factors of very large numbers. Yes, this is an impossible math problem. But what if we take what is known about the number guessing game and apply it to find the smaller Prime factor?

Here I attempt to demonstrate how a series of guesses can eliminate possible factors.

 

Definitions:

N is the number that is known and the number we will factor. (Or estimate the equation is equal to.)

x is the smaller factor. We will plug this number in for a guess. The guess is found when N calculated is within range of the known N.

y is the larger factor number. We will place it terms of x to build the equations.

 

We start with 3 sample equations:

 

Equation 1:

N = xy

 

Equation 2:

y = Sqrt[(((N^2) / x) – x^2)) / x]

 

Equation 3:

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

** In case you are wondering how these equations are derived, it is simply finding a pattern in y and putting it in terms of x.

** Also note Equation 2 and 3 are equal. They stand alone when we guess. It only takes Equation 2 or Equation 3 to make a guess.

 

We know N and will place a test x into the equation:

N = [Equation 2] * x

Or

N = Sqrt[(((N^2) / x) – x^2)) / x] * x

**Do not simplify further. The rules of simplification results in an imaginary number. Remember we are just guessing there will not be a perfect solution.

Test N = 25, 135,039

Place into equation and use a test x of 5500. So that 5500 is the test x.

25135039 = Sqrt[(((25135039^2) / 5500 – 5500^2) / 5500)] * 5500

 

25135039 != 25131729.1592027

 

The guess on the right side of the equation is larger than the Given N. N does not equal N. So we must decrease the test value x. The value 5500 for x creates to small a number. NGiven > NGuess

 

So we will try a test value of 2500.

N = Sqrt[((N^2) – x^2)) / x] * x

25135039 = Sqrt[(((25135039^2) / 2500 – 2500^2) / 2500)] *2500

25135039 >25134728.17699

This is very close. NGiven is still greater than  NGuess. But it is still within error.

 

We would continue the process until we test an x of approximate value 3581.

 

25135039 = Sqrt[(((25135039^2) / 3581 – 3581^2) / 3581)] *3581

Which equals 7018.440204 * 3581

25135039 is close to 25134125.4942275

 

Then we divide NGiven by x and have

25135039/ 3581 = y = 7019

 

 

Well we tested the value of x = 3581 because we know the answer. I know what you are thinking that is a difference of plus or minus 1000. But N is 8 digits in magnitude.

There is another test: Equation 3. You see, setting y in terms of x is quite easy. There are easily  dozens such equations. Now we test 2500 and 3581 again using Equation 3.

 

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

N = (((N^2/x) + x^2) / N) * x

 

For x = 2500

N = (((N^2/x) + x^2) / N) * x

N = (((25,135,039^2/2500) + 2500^2) / 25,135,039) *2500

25,135,039 is approximated by 25,135,660.6421

 

Is it too close to call again? Did we narrow it done to only 1000 odd numbers we have to test?

 

Place in a test x of 3518:

N = (((25,135,039^2/3581) + 3581^2) / 25,135,039) *3581

25,135,039 is approximated by 25,136,865.98

 

Simple equations that are easy to plug and chug withing a few minutes. The guessing game requires a math program. I use Mathematica to guess and 128-bit Prime numbers. I still do not know if it is simplified enough to prove useful. It is a math hack. Traditional math would want proofs and an equation that gave a perfect solution. But it does make sense that there are patterns in factorization. A number has 2 factors, say x and y. If you increase y then x must decrease. So there are patterns. The challenge is to describe those patterns with a math description.

 

https://github.com/Craylar/Patterns-in-Factorization

 

 

 

 

Equation 3:  y = (((N^2/x) + x^2) / N)

N = (((N^2/x) + x^2) / N) * x, which reduces to

               (N^2 + x^3)/N = N which we know is not true. The error is (x^3/N)

                                             So, (N^2 + x^3)/N = N + (x^3/N)

                                             **

               (N^2 + x^3)/N = N + (x^3/N)

                              We know from discovery that (x^3/N) = (x^2/y) **Will explain further.

                              So we substitute for y in terms of x

                              The simplest substitution is (y=N/x)

                                             (For another equation we could substitute: (y=(N^2/x)+x^2/N)

The goal is to use as simple equation that can be graphed. (That is in terms of the know N in terms of x.) It is irrelevant that we cannot simplify or solve the equations. It is a “math hack.” We use the graph to find x in polynomial time.

                              Another equation in terms of y is: (y=(N^2/x + x^2)/N). This can also be substitute into

                                             (x^3/N) = (x^2/y), and graphed.

 

**

SFNGrap;h20210303.png

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I asked what PNP was and your reply was: 

PNP = N = p * j = x * y.

So based on that I stopped reading any more of your post.  I am not sure about your math knowledge, but your english knowledge leaves a lot to be desired.

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On 2/25/2021 at 2:38 AM, Trurl said:

N=85

x=5

y=17

All my math problems deal with semiPrimes.

Is there a pattern? Does it only work with semiPrimes? Can you reduce the fraction to N/oversomedenominator? Has it been observed before?

No, there isn't because those are just three numbers.

The product of two primes is always a semiprime, because that's the definition of semiprime. Is that what you're on to?

"Some denominator"? Oh, that' clear!!!

Will you just state clearly what you're trying to get at? I don't get it. And I don't seem to be alone in this.

"The graph is what's important" just doesn't cut it.

Make a statement. If you don't make a statement there can't be "any thoughts". Except: "what's this all about?"

Something like "the distance (or the quotient, etc.) between consecutive semiprimes goes like such and such". It can be a conjecture, (a guess, an intuition).

It all sounds to me like mathematical innuendo. And it's very annoying.

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I concur with @Bufofrog and @joigus. Your posts make no sense. If you are unable to post a clear, unambiguous statement of your concept, along the lines joigus suggested, I shall recommend to the moderators that the thread be closed. That would be a shame, if your idea actually is interesting. However, if you are simply trolling, or inept, it would be the perfect solution. I look forward to a positive and understandable reply.

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5 hours ago, Bufofrog said:

I asked what PNP was and your reply was: 

PNP = N = p * j = x * y.

So based on that I stopped reading any more of your post.  I am not sure about your math knowledge, but your english knowledge leaves a lot to be desired.

PNP is the variable name of the semiPrime in question. It is the given. It is usually referred to as N as in N=p*q, where p and q are the Prime number factors. PNP is my name because I cannot call it N because N is a keyword in Mathematica. PNP or N is the factor of 2 Primes and is the given in RSA cryptography. Finding x knowing only N is considered difficult with large Prime factors. The goal of this work is to use a simple mathematical pattern to list N in terms of x, than reveal the value of x in the graph of the equations.

So basically, if you did not understand my write up, you could plug and chug PNP into my equations p3 and p5 and they would both reveal x, the smaller Prime factor at:

for p3 where y (y of the graph) =0

for p5 where y equals N (again y of the graph)

 

i should have have started by explaining it that way, but I wanted to show to some extent, how the equations where simply derived.

 

It will take me some time to address the other questions.

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Here is the summary of my work:

 

The main equation is: (N^2+x^3)/N = N+[(x^2/((N^2/x)+x))*N]

 

All the other math is the simplification of this equation. If you simplify the equation completely it results in zero equals zero. Which is not particularly useful. But we can put N the given semiprime product of x and y, where x and y are the Prime factors. As I have showed in the examples if you “plug and chug” “N” into the equation and graph the equation in its un-simplified form, then my hypothesis says that reading the graph in how I told you to read p3 and p5 should reveal x knowing only N.

 

Test it. Plug in N and x. Or just graph it. You will find it works for any size N. And that graphing is in polynomial time.

 

But if you do not believe me, then please tell me your reasoning. I know the way I wrote my proof to you was hard to understand. Explaining is much more difficult than solving. I tried to word the proof how I thought it through, but if no one understands it, it is not helpful.

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9 minutes ago, Trurl said:

You will find it works for any size N

I do not understand the claims or the proof* but it is interesting, can you show how larger numbers are handled? For instance RSA-100**

1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139

 

14 minutes ago, Trurl said:

And that graphing is in polynomial time

How does the algorithmic complexity compare to other methods for integer factorization?

 

*) Math and primes are not my area of expertise
**) https://en.wikipedia.org/wiki/RSA_Factoring_Challenge

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For a large N use a software like Mathematica and paste in N.

Yes I know math types don’t like that. I apologize. It just needs some properties as an algorithm.

So in the following Mathematica code replace pnp with N and graph and look for a yonthegraph of pnp. Work from the left side.

 

 SFNGrap;h20210303.png

 

 

I don’t know it’s speed compared to other methods. The truth is I don’t know the other methods. I know they rely on series. I took a simpler approach and looked for patterns in the division of semiPrimes.

My question does it work for all factor, not just factors of semiPrimes? Refer to my second post of this thread.

i am going to rewrite the proof so every can understand it. (Not just me.)

i think the confusion is people are looking for a number series and I am just doing algebraic operations.

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11 hours ago, Trurl said:

For a large N use a software like Mathematica and paste in N.

Yes I know math types don’t like that. I apologize. It just needs some properties as an algorithm.

So in the following Mathematica code replace pnp with N and graph and look for a yonthegraph of pnp. Work from the left side.

I meant can you do the example and show the factors of the number I provided? Sorry for being unclear. 

 

11 hours ago, Trurl said:

I don’t know it’s speed compared to other methods. The truth is I don’t know the other methods.

Ok.

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Ok, so I graphed the RSA 100. There is one problem in particular. I just can’t read the axis’ scale. I am researching how to evaluate the graph in Mathematica. I just need to find xongraph where yongraph equals pnp. That would be the range.

I believe you can test for yongraph on a graphing calculator and find points on the graph. I post this as a mathematical challenge to the SFN community.

How do I arrange the scales in mathematic software to find the y value I am looking for?

It takes milliseconds to draw, but I need to be able to evaluate it.

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  • 2 weeks later...

Ok, I hope this makes sense. It is only an idea. I just wanted to post it to see if it makes anyone else see something new in a plain graph.

 

I had trouble finding the values on the graph of x when N is a hundred-digit number. I was certain that I could just zoom in or pan the graph to see the proper value. As I have found, this is difficult to do in practice. So I am back at just approximating x and using test values. But in trying to do this I noticed something else that would help in the search for x knowing only N.

 

So for N = x *y, where x and y are 2 unknown Prime numbers, there exists a pattern. (I hope you can understand the following algebra.)

(x^2/y) = (X^3/N) = ((N/(y^2/x))

So with a x of 5 and a y of 17 and an N of 85 we have:

(25/17) = (125/85) = (85/57.8) = 1.470588

This should be true only of semiPrimes.

 

It is interesting to so values of inequalities in the x axis.

(x^2/y)   <   x   <   y   <   (y^2/x)   <   N   <   (N+(x^3/N))

 

I hypothesized if I transformed the y and x axis for equation p5 I would have a clear plot. Because a value of y would be known to be N. I have tried the function Parametric Plot. I haven’t found software that defines x knowing a known range.

 

So my hypothesis is this:

My equations are based on the value N. If I were to take equation p5 and subtract in by itself, I would have N-N or zero. But what I now have is the equation of p3. So I just translated the graph N in the negative direction. Then from the result, I add (x^2/y) along the x-axis and the y value at this x value will be x.

So for 85=5*17 the y value of equation p3 will equal 5 where the x value equals 1.47.

 

There is still much to be worked out. Please check my logic. And please don’t kill my post. I am being serious. Doing math is one thing. Explaining it is another. But if this works I have just transformed axis using only basic algebra. I think it will be useful to find x knowing only N.

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  • 3 weeks later...

p3 = ((pnp^2 + x^3) / pnp) – ((pnp + (x^2 / (pnp^2 + x)) * pnp))

 

p5 = (pnp^2 + x^3) / pnp + (x^3 / pnp)

where N = pnp for computer computation

 

 

Given equations p3 and p5.

p3 is equal to p5 a distance of N. Since N is the distance subtracted from p5

So,

p3(x_semiPrime)=0
and
p5(x_semiPrime)=N

That is y=0 at an x equal to the smaller semi-Prime at p3. And at p5, y=N.

 

Knowing the above,

p3(x) = p5(x+N)

so p3(N)=p5(N+N)

since these equal the derivative of those should also equal

p3’(x) = p5’(N)

and p3’(N) = p5’(N+N)

 

My goal is to get enough “given” information to form a solvable deferential equation. I find it to be challenging if not outright confusing. I subtracted p5 by another equation with value N to get p3. I thought it translated the x-axis values to the y-axis. However it has shifted the equation a value of N in the x-axis to the left. I believe the above statements are correct. Are they enough to create a solvable differential equation?

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Lock this thread.

I have spent years reading encryption books and tinkering with Prime numbers. It was very valuable in that I could study statistics, game theory, review calculus and work on ideas. It was more than just crunching x’s and pnp’s. That is why I keep it up. I even learned the basics of Mathematica. My end result was to approximate x by graphing a pattern in division. I need to know what x will have a y value of N. I can’t graph it accurately enough with standard programs. I don’t know if there are any computer programs to graph a range that is pnp. I could take guess at the value and solve within 20 guesses, but I really thought the graph was possible.

I am on to other projects. I am really interested on the brilliant, but common sense experiments of the early 1900’s. I am amazed how all the experiments in light and electricity were common place in those scientists’ labs. I have a physics experiment book that is at the undergrade level. I think I’ll follow along and recreate the experiments.

So all my work and I can only approximate x. It’s not wasted time. I learned a lot. But if anyone has some more advanced graphing software plug my numbers in please.

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One more post before I conclude this project. Why did no one recommend Excel for the graph and the list? It would graph and list an N of 100 digits easily. It has a segment of adjustable accuracy. Spread sheets are common to solve graphs and differential equations and it is a tool everyone has access to. It would take more adjustments with Mathematica. Of course if you didn’t believe in my work you wouldn’t mention this simple tool. However if you think it can be estimated with a simple graph you have already plugged it into the spreadsheet without sharing.

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