Simple yet interesting.

[Ghideon]

On 10/17/2022 at 4:41 AM, Trurl said:

Plot 2 graphs. One has already been plotted.

 

y=x^4/(N+x)

 

and where it crosses y=(((N^2/x)+(x)/N*x) – N

 

 

The second equation is not a valid mathematical expression.

[Trurl]

Yes, it does not look good for the Pappy Craylar Conjecture. But let's hope it has potential like the Pittsburgh Steelers. The Steelers has offensive weapons, but can't produce offense. The PCC cannot be solved by solving for x only knowing N. But if you place it into a plot it may prove useful.

I am still working on the challenge. I want a usable process that will find x fast and accurately. This is what I have. But I can't get the loop to work yet. It is a Hail Mary for the PCC.

 

clear [i, pnp]

 

pnp= 2564855351

 

 

i=3; while[ ((((pnp^2/i + i^2) / pnp * i) – (i^3/pnp)) << pnp, Print; i=i+2]

 

[Ghideon]

17 minutes ago, Trurl said:

I can't get the loop to work yet

You could try to have matching parentheses? There seems to be unbalanced "(" and ")" in the statement:

18 minutes ago, Trurl said:

i=3; while[ ((((pnp^2/i + i^2) / pnp * i) – (i^3/pnp)) << pnp, Print; i=i+2]

 

 

[Trurl]

I will be taking a break after a weekend crunching numbers. I don’t know if it is the Pappy Craylar Conjecture’s fault. I thought finding lines that intersect would be easier. Go figure.

 

I graphed xthefactor = x, in the first graph.

https://www.wolframalpha.com/input?i2d=true&i=plot\(40)Cbrt[Divide[\(40)Power[x%2C3]+Power[2564855351%2C2]\(41)%2C\(40)+Power[2564855351%2C2]%2Bx\(41)]%2BDivide[Power[x%2C4]%2CPower[2564855351%2C2]+%2Bx]]\(44)+Sqrt[\(40)Divide[\(40)Power[x%2C2]*+Power[2564855351%2C2]\(41)%2C2564855351%2Bx]\(41)+]+%2B+\(40)Divide[Power[x%2C2]%2CPower[2564855351%2C2]%2Bx]+\(41)+from+0+to+50000\(41)

The second graph is N=N. Where x should be the factor at N for both plots. Here the PCC looks promising. I just don’t know of a way for the computer to give me the intercepts and the scale of the graph so I can read it.

 

https://www.wolframalpha.com/input?i=plot((x*Sqrt[2564855351^3%2F(256485531*x^2%2Bx))]+%2B+x^4%2F(2564855351^2%2Bx))%2C+((2564855351^2%2Fx%2Bx^2)%2F2564855351*x-(x^3%2F2564855351))+from+0+to+50000

 

[Ghideon]

5 hours ago, Trurl said:

I will be taking a break after a weekend crunching numbers. I don’t know if it is the Pappy Craylar Conjecture’s fault. I thought finding lines that intersect would be easier. Go figure.

 

Maybe it is a good thing that you begin to find out for our self that your ideas regarding factorisation of semiprimes does not work?

 

[Trurl]

You are right I cannot factor a large Prime. But the fault is mine and not the Pappy Craylar Conjecture.

 

I don’t know of any math software that will let me find x on the graph while y equals N.

 

I believe the PCC is finding the factors, but for me it is impossible to find an answer on the scale of the graph. Do you know of any graphing programs. I could also fix the loop, but I don’t know the advantage of just looping the equation versus looping division.

 

You could guess at the position of x because the PCC equation would tell you if you were higher or lower than the correct value.

 

I found the error. So an N of 85 will occur where x equals 5, exactly with no error. But when you try and solve for x in the equation the square roots stop the equation from being solved.

 

But until I can say f[x} = N, where y = N. I cannot solve the graph without typing in the correct scale. And the scale is hard to find with several hundred digits N.

 

So RSA remains safe for now. I have put a lot of work into this problem. Good thing this isn’t my thesis or I’d fail. But I leave you with one more graph. That is 85 = 5*17:

 

 

https://www.wolframalpha.com/input?i=plot((x*Sqrt[85^3%2F(85*x^2%2Bx))]+%2B+x^4%2F(85^2%2Bx))%2C+((85^2%2Fx%2Bx^2)%2F85*x-(x^3%2F85))+from+0+to+20

One more. I don't think it is the correct factors, but it was a true test of the PCC method.

 

https://www.wolframalpha.com/input?i=((x*Sqrt[2564855351^3%2F(256485531*x^2%2Bx)]+%2B+x^4%2F(2564855351^2%2Bx))%3D%3D+((2564855351^2%2Fx%2Bx^2)%2F2564855351*x%2B(x^3%2F2564855351))

[Ghideon]

9 hours ago, Trurl said:

You are right I cannot factor a large Prime.

As expected your methods and ideas unfortunately failed the challenge. The ideas does not work at all, but it is good that you tested so that you see it for yourself.

 

9 hours ago, Trurl said:

So RSA remains safe for now.

That is good, those acknowledgements may be the first step. Maybe you begin to see where the real issue is? 

 

9 hours ago, Trurl said:

Do you know of any graphing programs.

Yes, as part of my profession I need to know several different programs, frameworks and programming languages. But the software is not relevant to your issues, do you begin to see why?

If you objectively analyse your issues and struggling and its connection to how RSA and similar encryption actually work I think the root cause of your problems will be obvious. I have already tried to explain (several times) but you may learn more from discovering it for yourself. 

 

9 hours ago, Trurl said:

But I leave you with one more graph. That is 85 = 5*17:

Sorry, I'm not interested in yet another repetition in those numbers. 

 

9 hours ago, Trurl said:

One more. I don't think it is the correct factors, but it was a true test of the PCC method.

I have no clue what that is supposed to mean. (Note: 256485531 is not a prime number and also not a semi prime)

 

[Trurl]

17 hours ago, Ghideon said:

As expected your methods and ideas unfortunately failed the challenge. The ideas does not work at all, but it is good that you tested so that you see it for yourself.

Yes, most attempts at finding patterns in Primes fail. I gave it a solid effort.

The one thing that bothers me is that I only needed to know where y on the graph equals N.

If you plug in N and plug in the already known x it equals N. That is why I never gave up. I thought it would be easy to analyze the graph. I conclude here.

This is why I keep going.

https://www.wolframalpha.com/input?i=((41227*Sqrt[2564855351^3%2F(2564855351*41227^2%2B41227)]+%2B+41227^4%2F(2564855351^2%2B41227))

 

41227 is the only x that will produce N. As seen. I just wish there was a way to solve the equation for x for real numbers.

[Ghideon]

 

18 hours ago, Trurl said:

I conclude here.

That is a reasonable conclusion when a conjecture, an idea or similar is a failure.

 

(Note, you are probably wasting time posting the links, all of them have errors or needs additional efforts.)

[Trurl]

 

 

This is what the links that don't work look like. Remember the musing about the NSA cryptographer and the graphic artist?😜

[Ghideon]

The explanation is in clear text in your screen shot. Below it is highlighted for you. 

So the links does not work; I do not know if they goes to a page displaying anything meaningful.

 

(Note: I have pointed out this several times in the past)

On 5/16/2021 at 10:02 PM, Ghideon said:

One trouble is that the first one has unbalanced parentheses

 

On 10/28/2022 at 11:14 PM, Ghideon said:

try to have matching parentheses

 

Note 2:Some links may give other errors such as this. 

[Trurl]

(x*Sqrt[2564855351^3/(2564855351*x^2+x)] + x^4/(2564855351^2+x)) - 2564855351), (x from 10000 to 52000)

 

 

_________________________________________________________________

 

plot( 2564855351-( Sqrt[(((((x^2 * 2564855351^4 +2* 2564855351*x^5) +x^8)/ 2564855351^4) – (1-x^2/ (2* 2564855351))) * ( 2564855351^2/x^2))] ),(from x= 10000 to 50000))

 

 

 

Paste these in Wolfram's Alpha computation bar. If it times out you need Pro.

 

This was my final attempt. That is why I stopped posting. The graph of these equations are in different windows. Alpha would not give me enough space to combine them. But x should be the factor where (if) they intersect between 1000 and 50000. I am working on a better presentation and file format. Alpha correct the parenthesis, so I didn't mess with it. I didn't want to break it.

[Trurl]

[Trurl]

 

 

 

Ok, so the work is over. But I have one question which is Did anyone find it useful? I know 2564855351 is easy to factor with computers. I take the values from 1 to zero from the right to left and test. A 10^9 is reduced to 10^4. I know it doesn’t seem useful. But I haven’t worked with hundreds of digits. The precision of the numbers close to zero (a pnp==pnp) is untested. My programming skills are terrible. But I see what the theme is: “If it does factor semiPrimes it is more than just theory. You should be able to produce the factors.” But it leads me back to the question is it useful? When I thought it up I thought it was gold dust. I cannot factor very well with it. But I am a terrible programmer. I have read about the discovery of determining how many Prime numbers there are under a given number. The thought is that it would lead to a pattern. It never did. However, it led to patterns of series that produced large Prime numbers. I’m not saying the Pappy Craylar Conjecture finds a pattern in Prime numbers, but it is a simple method of predicting factors, approximated (where the graph is between zero and one.

 

Well, it is on to other projects for me. I posted so much because I believed in my hypothesis. I leave you with my corrected equations screenshot. But if anyone does find a use for the PCC: Post it here!

[Trurl]

There is a way to eliminate the possibilities of 4590 in 2564855351.

 

There is enough variation between values to find which value is the answer. And also to find y.

 

[Trurl]

Where x * y = pop

When y = Sqrt[pnp^3/(pnp*x^2+x)]

So that, x* Sqrt[pnp^3/(pnp*x^2+x)] = pnp

 

pnp – pnp = 0

 

 

Let x equal any Prime number. 5 for instance.

Graph x = 5 at x on the graph at that instance = pop

So where 5* Sqrt[pnp^3/(pnp*5^2+5)] – pnp = 0, then pnp is a semiPrime and we can plug it into the equation and find y (the larger Prime number). 

 

And if we continue to graph over all real numbers we will find every Prime number in existence.

 

This is my programming challenge to you.  Remember the math does not have to work to complete the challenge. The challenge is to see how fast you can prove it wrong or correct.

[Trurl]

3^rd Challenge: Graphical Representation:

 

I have claimed that a logarithmic spiral could be drawn to show a pattern in Prime numbers. It is easy to claim but hard to draw. Here I challenge you to draw a logarithmic spiral that explains the last post’s graph of PNP and where it equals zero.

 

But what is the previous post finding a semiprime where the graph equals zero, there was a modified sine wave that explained the graph? So, you have numbers at PNP at zero with x and y (the Prime factors), with the sine curve osculating above and below zero. And the sine curve at the same time is experiencing resonance. Growing larger in magnitude as the value of PNP increases. Could not a logarithmic spiral explain this? And couldn’t a differential equation explain this resonance like a differential equation explains resonance on a spring?

 

And if you put the graph in 3D, does it make a helix? And could it be useful to define changes in helix such as those that describe DNA. Obviously, that is wishful thinking but new ideas bread new ideas.

[Trurl]

p=2564855351 

For[i = 3, i <= p, i+2,
If[(Sqrt[p^3/(p*i^2+i)] - p) < 0.5,
x = i,
Print[i],
i=p+1
]
]
While[x<=  p, x+2

If[ Divisible[p,x],
Print[x],
p = 1];
]

This was an attempt to target the range of divisors into the semi-Prime. I know I am doing division in the second loop, but the range is smaller, where the test equation of the first loop equals approximately zero.

I know you can argue the usefulness. However, I still argue that at large semi-Primes with this algorithm, the number of test values is significantly less.

And a shout-out to Chat gpt. The reason I am a terrible programmer is syntax; too many languages; too many unknown classes. Now that I have uploaded my algorithm, gpt will render all public key encryption invalid 😜--- only kidding of course. No one believes my work anyway. That was until gpt …

[Trurl]

One of my resolutions this year is to stop trying to enhance RSA cryptography. I have been at it as early as 2006. In my first attempts, I stated that a logarithmic spiral could define a pattern in Prime numbers. I still believe this. And it is inherently feasible because logarithms already explain many patterns.

 

But my ideas are the same in all my posts: x * y = pnp.

 

If I had just left the equation at: x * (pnp / x) = pnp ; it wouldn’t have been helpful. I had to find a way to put y terms of x and pnp that was useful and equaled more than 1 =1. The enhanced equation still may be difficult to solve, but with only x as the unknown I could substitute x along the graph and when that value minus pnp approximately equaled zero, the value of x that caused that zero is the smaller Prime factor.

 

Well despite the fact if you believe me or not Or if this is even useful (I believe it is useful), you could set up an equation similar to this isolating x and testing it along a graph for any pattern. It may be a challenge to put y in terms of x and pnp that only solves the desired pattern, but it would be a shot to analyze the series in ways not done before.

 

So, as I break or fulfill my resolution, I ask even though it would still be challenging to break RSA with this method, is the method itself useful in trying to find patterns where this is no pattern?

 

 

x* y = pnp

 

x * (Sqrt[ (pnp^3/(pnp * x^2 + x) + (x^4/(pnp^2 + x) )] ) = pnp

 

x * (Sqrt[ (pnp^3/(pnp * x^2 + x) + (x^4/(pnp^2 + x) )] ) – pnp = 0

 

Let pnp = 2564855351

 

When x = 41227, the (x*y) part of equation equals 2564855365.5

 

Not bad for government work.

 

[Trurl]

@Sensei

 

@Ghideon

 

I had help on this program. A good Samaritan put the code into ChatGPT and it supped-up the code.

 

I still don’t know if it will run faster with larger Primes compared to standard recursive division. I think it gives a picture of what is going on with the factoring. x will only occur around zero on the graph.

 

Again, I don’t know how to measure any efficiency of the program. Obviously with small Prime products (pnp), it will not show any improvement with factoring the semiprime.

 

Again, I am breaking a resolution not to enhance RSA. But I wanted to share some final thoughts on this project I have spent great effort on. I think my next project will be Microsoft Kinect or 3D Printing. A friend bought me a book on Nichola Tesla for Christmas. It has his messages. However, the messages are encoded in puzzles and cryptograms. I thought I could write a simple Mathematica program and decode them. If the program works, I will share every detail. Here we go again. LoL. At least it’s not RSA. Even video games would be a good break from RSA.

 

I know what you are thinking why would I give such information away. That is if you really believe in my program. I know such information is valuable, but if no one attempts to find large Primes with it I don’t know if the program even works. That is why I shared everything in the message board. I have shown my case. Please prove it right by finding large Primes.

 

 

 

p = 2564855351;
x = 3;
Monitor[While[x <= p, If[(Sqrt[p^3/(p*x^2 + x)] - p) < 0.5, Print[x];
   Break[];];
  x += 2;];
 If[x <= p, While[x <= p, If[Divisible[p, x], Print[x];
     Break[];];
    x += 2;];], x]

 

[Ghideon]

On 1/8/2024 at 2:19 AM, Trurl said:

Please prove it right by finding large Primes.

You can test using the numbers I provided:

https://www.scienceforums.net/topic/124453-simple-yet-interesting/?do=findComment&comment=1219386

On 1/8/2024 at 2:19 AM, Trurl said:

I know such information is valuable, but if no one attempts to find large Primes with it I don’t know if the program even works.

Finding large primes with your approach unfortunately does not work; mathematical reasons why has been presented already.

[Trurl]

Quote

Finding large primes with your approach unfortunately does not work; mathematical reasons why has been presented already.

 

@Ghideon

 

And anyone interested

 

 

Yes. I am still dividing and finding a square root in my calculations. How could it be faster than the original recursive division?

 

Probably not faster, since when you start at 3 the unknown x is closer to zero than pnp anyway.

 

But the advantage is you don’t need to test those values in my equation that don’t occur next to zero. This zero is unique to pnp. It also occurs in sequential order. So, if you subtract the left side of the equation by pnp and it is, say 5, you know that x is smaller than the tested x at that value.

 

I know it has many faults, but it does provide a unique graph that may contain some useful analysis.

 

I think it will prove more useful as pnp increases in value. But it is difficult for me to test. I will try to factor the largest semi-Prime posted earlier. But that is why I posted here in the first place for help testing.

 

I realize my algorithm isn’t faster than plain division, but that isn’t what it was intended to do. I need outside eyes because I created the algorithm and am biased, but finding semi-Primes is not the only part. Instead, it is to exploit a pattern in factorization.

[Ghideon]

Maybe it helps to view the lack of patterns in more dimensions?
I've generated a surface plot of the function x×y for x and y ranging from 3 to 100. This plot represents a smooth, continuous surface, as expected from the multiplication of real numbers.

Within this plot, every prime and semi-prime number in the given range is represented. Notice that the surface is uniformly smooth. There are no distinct features, patterns, or anomalies that visually distinguish primes or semi-primes from other numbers: 

Let's modify the surface plot; the function is based on @Trurl's program code: 

This plot is also smooth. Again there are no distinct features, patterns, or anomalies that visually distinguish primes or semi-primes from other numbers.

Using another algebraic function will not help; there are no patterns. Plotting a larger area does not help either; the surface is smooth for any numbers. 

 

20 hours ago, Trurl said:

I will try to factor the largest semi-Prime posted earlier.

please report your progress so far.

[Trurl]

There are no  anomalies. When you graph the  equation, the x of the graph (you just graphed) equals the smaller Prime factor where y of the graph is or around zero.

 

There are no tricks. That is why I called it Simple Yet Interesting.

 

 

The code I posted was generated by ChatGPT. It simplified the equation. You should try it in  Mathematica. It is not meant to be efficient but the first function should break (and print) near zero and estimate the smaller factor x.

 

I have already posted on this.

 

 

 

Let pnp = 2564855351

 

When x = 41227, the (x*y) part of equation equals 2564855365.5

https://www.scienceforums.net/topic/124453-simple-yet-interesting/page/7/#elControls_1258420_menu

 

Hope this explains what I was attempting to do. I need to post  another, larger example.

 

If the program from ChatGPT works you should be able to modify the testing steps ( from every odd number to x+=22 or larger)

 

The one thing that needs checking is how GPT simplified the equation.

 

That is why I said test for lager Primes. Because 2 known Primes multiplied together will equal zero in the equation if they form a SemiPrime.

 

[Ghideon]

11 hours ago, Trurl said:

The code I posted was generated by ChatGPT. It simplified the equation. You should try it in  Mathematica. It is not meant to be efficient but the first function should break (and print) near zero and estimate the smaller factor x.

You may have misunderstood the output from the LLM*. Your claims does not seem to match the code you posed above.

 

11 hours ago, Trurl said:

If the program from ChatGPT works you should be able to modify the testing steps ( from every odd number to x+=22 or larger)

 

The one thing that needs checking is how GPT simplified the equation.

What are your results from testing your program? 

 

12 hours ago, Trurl said:

There are no  anomalies

Ok, you missed the point. 

 

*) LLM=Large Language Model