# Trurl

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1. ## Simple yet interesting.

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!

3. ## Simple yet interesting.

(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.
4. ## Simple yet interesting.

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

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.
6. ## Simple yet interesting.

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))
7. ## Simple yet interesting.

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
8. ## Simple yet interesting.

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]
9. ## Simple yet interesting.

Ok. Challenge accepted. It may take me a while, but here is my method: 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 It is that simple. Not an exact result, but a good estimate for ball park figure. I will start on Monday. I can’t use Mathematica because it has an recursion warning. For some odd reason Wolfram alpha will draw it. All I have to do is to draw the 2 graphs on same plot. And it will prove or disprove the ability to factor RSA.
10. ## Simple yet interesting.

Ok this is the steps I will take for this challenge. I want to run this idea by you. I may not have the computer skills to analyze the graph in Mathematica. I will graph this equation as I already did. $frac {x^4} {N^2+x}$ I can't get the latex to work so it is the above (x^4/(N^2+x)) I will take all values on the y-axis where x equals zero. (As before) Then I will take the x value where y equals zero and put those values into this equation: $N=\frac{x\left(\frac{N^2}{x}+x^2\right)}{N}$ And graph. Where the y axis equals N is the semiPrine factor x. Does this make sense. I can’t tell. It is confusing but it may work? It is simply. Can this possibly work?
11. ## Simple yet interesting.

Yes I know you say the graph doesn’t show anything we don’t already know. That is why I showed you the graph. But testing 10^49 to 90^49 you clam is tribal to testing numbers 10^100. You could argue why don’t I just test the bottom half of 10^100. You may be successful or maybe not. But the point is I just did it mathematically in real time. And if I could estimate the error in the expression I could do it in one calculation. But the purpose is not to destroy RSA anyway. The Pappy Craylar conjecture might not destroy RSA. It is a pattern in factorization. It is a significant step in the pattern. If you look at the database numbers on error I posted, you will see the error has potential to find patterns. And if you followed along I think you know the equation isn’t trivial. So I don’t mind if you say I can’t break RSA. But you have to admit it was a good attempt. And you don’t know what other ideas I have. But I appreciate the critic. You tested the problem. Before it was an idea which no one I know could test. Supposedly Primes are impossible. Right or wrong I needed someone to test it. And it may not be wrong just not precise enough. For future explorations It is important to note the statistics of the curve of the graph of those values with a y axis value of zero.
12. ## Simple yet interesting.

You only have to test those odd numbers whose y value is less than one. I know the graph isn’t very descriptive. I need a “real time” plotting software. But as I show by cheating, I arranged the test value to the answer. As you can see y equals zero when x is the answer. I understand you say that numbers below 25% are too small. But they also have y-axis values below zero. So if you are crunching numbers you would start on the right and divide into N until a factor is found. Instead of numbers of 10^10 you are testing 10^50 and most of the time the right most y-axis zero value is the number. If you want a more mathematical explanation you would have to analyze the zero y-axis values with calculus. RSA is still protected because of the computation of the values on the number line. But as long as you can square pnp, RSA is significantly less secure. But don’t believe me. Test the number line. Look for y-axis equals zero. Start on the right and divide into N. RSA can hide in security of large numbers, but the reason the Pappy Craylar Method works is because it works on all Prime numbers. The small numbers test and so should the large. x = plot[ (x^4/(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139^2+x)), {x, 0, 37975227936943673922808872755445627854565536638199 }] https://www.wolframalpha.com/input?i=x+%3D+plot[+(x^4%2F(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139^2%2Bx))%2C+{x%2C+0%2C+37975227936943673922808872755445627854565536638199+}]
13. ## Simple yet interesting.

The plot is the answer. The y axis is expression x^4/(N^2+x). When the expression equals zero the y axis is also zero. So when the y axis is between zero and one then x at that values of the y axis that are less than one is all that needs tested. Start at the right of those x and divide N by x until you find the correct Prime factor x. The graph does in seconds what takes hours. No. We see the answers on the plot because we are testing x in the equation. The graph will show an answer knowing only N (pnp).
14. ## Simple yet interesting.

Here the link is. https://www.wolframalpha.com/input?i=x+%3D+plot%5B+%28x%5E4%2F%281522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139%5E2%2Bx%29%29%2C+%7Bx%2C+0%2C+10%5E51+%7D%5D Here where y = zero x^4/(N^2+x) = y x on the graph where y = 0, is the smaller factor q So it should work with large numbers. I know I know the answers before computation, but where y = 0, x is a possible Prime factor. Loads in seconds. According to Wikipedia “It takes four hours to repeat this factorization using the program Msieve on a 2200 MHz Athlon 64 processor. “ Note. Look at the graph. I not sure about all the numbers.
15. ## Simple yet interesting.

You are correct. I am simply calling the difficulties of factoring N the one-way-function. The difficulty of the Prime number factorization is what makes RSA work. I claim the Pappy Craylar method can make it 2 way. That is the claim of breaking RSA. I believe you call that one way or trap door function. Or N or NP. RSA seems impenetrable because we can’t find a pattern in Prime numbers to test. The PC method says forget for a moment in testing all known Primes. Instead look at how numbers are factored. Prime numbers can hide but factors can’t.
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