# Simple yet interesting.

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On 9/23/2022 at 3:00 PM, Ghideon said:

It says x is semiprime.

Should say x is smaller factor of SemiPrime pnp.

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On 9/26/2022 at 11:54 PM, Trurl said:

Should say x is smaller factor of SemiPrime pnp.

I still do not see any practical or useful connection to RSA. Assuming realistic RSA keys* and your definition** $(pnp=x*y, x<y)$ then $0.25 < \frac{ x^{4} }{ pnp^{2}+x } < 1$. That gives no new information about the prime factors x and y, in RSA only x*y is known.

*) Using two large prime numbers of similar length.
**) It's hard to understand the picture, some guesswork required.

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Pnp is known. Which happens to be x*y.

The question is what x with known pnp will be close to or moving away from zero.

The given pnp will only be true (within error) when x puts the equation near zero.

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3 hours ago, Trurl said:

Pnp is known. Which happens to be x*y.

Yes, I know. That's why it is in my calculation above.

3 hours ago, Trurl said:

The question is what x with known pnp will be close to or moving away from zero.

The given pnp will only be true (within error) when x puts the equation near zero.

That does not make sense. Can you express it mathematically? I've already provided you with a general calculation, see the lover (0.25) and upper limits (1)).

You can get the expression arbitrarily close to zero by using any sufficiently large numbers where x<<y. That means that the closer you get to zero in your expressions the more wrong you are if finding RSA related primes is you goal. That insight should be trivial if you are familiar with RSA?

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That is fair constructive criticism. It is a bold claim to make to break RSA. If the Pappy Craylar conjecture was true RSA and any ciphers that relied on factoring, Prime numbers, or logarithms would have to have larger keys.

Yes I know that as pnp gets larger the number of values less than zero, increase. I don’t see this as a flaw. It is just how the numbers fall. If pnp was a hundred digit number it is still possible 5 could still be a factor.

So does it apply to the pnp value of RSA? It isn’t perfect, but I am not aware of any other tool to take its place. But it is useful in RSA factoring because there are 4 times less numbers to test. Maybe even less number are possible if a pattern is found. You would test lager odd numbers first. But as you test values there are hints to how large the test value should be. For example with a pnp of 85, 3 would be between 0 and 1 but 85/3 isn’t a whole number. If I was crunching numbers, I would test the higher numbers first.

But the second thing to note is what happens with larger numbers. Will the distance between large numbers become smaller or will there be more zeros with more numbers to test? That is why I post this thread. I think I have a simple enough pattern that may challenge RSA.

I post this picture to show the less than zero values of the Prime number 3. Yes it approaches zero as pnp increases. But you forget as x approaches zero the corresponding y value approaches pnp.

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Posted (edited)
9 hours ago, Trurl said:

It is a bold claim to make to break RSA.

Bold and so far unsupported.

9 hours ago, Trurl said:

Yes I know that as pnp gets larger the number of values less than zero, increase.

This is confusing ,what is less than zero? The expressions you posted returns positive numbers*. I'm still referring to your expression $\frac{ x^{4} }{ pnp^{2}+x }(pnp=x*y, x<y)$

9 hours ago, Trurl said:

If pnp was a hundred digit number it is still possible 5 could still be a factor.

How? RSA is about the product of 2 large primes as far as I know.

9 hours ago, Trurl said:

But it is useful in RSA factoring because there are 4 times less numbers to test.

Compared to what method?

9 hours ago, Trurl said:

I post this picture to show the less than zero values of the Prime number 3. Yes it approaches zero as pnp increases. But you forget as x approaches zero the corresponding y value approaches pnp.

Again you post an image so it can't be quoted. It's confusing; I don't see any negative numbers in the picture, and I do not see x in any expression and there are some errors in the statements.

Would you mind commenting on the fact that 0.25 is the lower limit of your expression $\frac{ x^{4} }{ pnp^{2}+x }(pnp=x*y, x<y)$?

Clarification & details: Prime numbers x and y, realistically used in encryption, would be large and of similar size for instance n bits (typically n>512). The smallest such number x would be the smallest prime larger than 10...02 and the largest number y would be the largest prime less than 11...12. So the two primes can differ by most 2n-1. Minimum difference y-x is 2 if x,y are prime twins. This results in the your expression having limits: $0.25 < \frac{ x^{4} }{ pnp^{2}+x } < 1$ where $(pnp=x*y, x<y, x>0, y>0)$ , x and y are primes and requiring same number of bits for binary representation. This provides the mathematics needed to show that your approach, as far as it is possible to understand it, is incorrect. As long as you guess numbers that results in your expression being close to zero (<0.25) we are guaranteed that you are guesses are wrong in RSA.

*) The mainstream definition is that prime numbers are positive so x>1 and y>1

Edited by Ghideon
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18 hours ago, Ghideon said:

This is confusing ,what is less than zero? The expressions you posted returns positive numbers*. I'm still referring to your expression x4pnp2+x(pnp=xy,x<y)

For values where the expression (the main equation) will equal a number between 0 and 1. As pnp gets larger the x value will become smaller. For example if an x=5 is less then 1 when put into the expression for a pnp test value of 85 and we keep the pnp=85 but try x=3 the expression is still between 0 and 1.

That is what I show in the spreadsheet. That smallness is the error (distance from zero of the expression) that occurs. Sure we can have a million digit SemiPrime with a factor of 3. But as pnp increases the expression becomes close to zero which in turn y becomes larger. But y cannot become larger than pnp.

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19 hours ago, Ghideon said:

Compared to what method?

Division

19 hours ago, Ghideon said:

As long as you guess numbers that results in your expression being close to zero (<0.25) we are guaranteed that you are guesses are wrong in

Is 0.25 your lower limit because y approaches pnp?

To me it seems like we are solving something different. 3 in my equation would be smaller and smaller as the SemiPrime increases. But 3 has infinity large SemiPrimes. That is why in a status update that if you could factor SemiPrimes you could test for Primality. But the trouble is all the 3s then the 5s and the 7s all lead to infinity SemiPrimes. But you would only need one number to factor and graph. So if 3*Prime# is between 0 and 1 then they are Prime.

However the problem is that the expression often has many values between zero and one. That is why it isn’t perfect. But that doesn’t mean it isn’t useful.

19 hours ago, Ghideon said:

How? RSA is about the product of 2 large primes as far as I know.

RSA uses large Primes but SemiPrimes use all Primes. But remember I’m a graphic artist not a cryptographer. I have done some reading. Basically the code is open source. My method is not mathematically complex. I am simply isolating x as compared to pnp. That expression between zero and one is just pnp minus pnp estimate.

That is why with 100 digit numbers x of 3 is very small and falls into the test area but if x of higher values don’t test to be the correct factor 3 is possible. Download the complete spreadsheet and test for yourself.

I’m am glad you are challenging my work. That is why I put it here. I very likely could be wrong. But look at the stumbling block our ancestors gave us. They want a pattern of Primes but it’s impossible. But once you start down the rabbit hole you can’t stop. But it means everything to computer science.

While I was in school I did busy work with a cryptographer. I really didn’t appreciate his job. But it was pretty serious stuff. He had to get a polygraph and testing for weeks. I think I’ll stick with open source crypto.

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Posted (edited)

17 hours ago, Trurl said:

Is 0.25 your lower limit because y approaches pnp?

No, 0.25 is the lover limit because of assumptions regarding primes used in RSA. Everything is in the mathematics above* where I compared your expression to typical numbers used in RSA. Feel free to ask for clarifications or for more details if needed.

17 hours ago, Trurl said:

To me it seems like we are solving something different.

(I have no clue what you try to solve or do with your musings on prime numbers, your explanations does not make sense. I adress the posted claims regarding RSA.)

Edited by Ghideon
clarification
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2 hours ago, Ghideon said:

(I have no clue what you try to solve or do with your musings on prime numbers, your explanations does not make sense. I adress the posted claims regarding RSA.)

I will explain better. I keep talking about Prime numbers because I focused on the N=p*q part of RSA. It is where my work began. I focused on p and q. If p is so large that would determine the size of q.

In RSA we are given N and it is a challenge to factor, thus a one way function. This is my focus. I figured let’s put N in terms of p alone. So I wrote an equation that would use N and compare (subtract N -N calculated) N in terms of N and p.

And it worked. Except there was always a slight error when subtracting N - Ncalculated. An error between zero and one.

So eventually I solved the error or at least found how it was calculated. This error is why the expression is between zero and one. That means the expression to be useful in breaking RSA there is a need to test all values between zero and one. (I don’t know how complex that is with hundred digit numbers.)

That is how many p’s (you know p as x) there are when the main equation (or expression) results in a value between zero and one. (Again N-Ncalculated approaches zero.)

I believe factoring N is the key to find a pattern in Prime numbers. Because a pattern in factoring correlates with a measurable pattern that can be graphed.

Picture this: If p is small q is large. If p is large q is small. I made an equation that would approximate N with an unknown p. And since N is the know we will match unknown p’s until the value makes the known and unchanging N true. Because of the error the N=N is between zero and one.

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13 hours ago, Trurl said:

I will explain better.

Sorry, the provided explanation is not better.

13 hours ago, Trurl said:

thus a one way function.

If it is a one-way function then you are guaranteed to fail to find the factors of a given large integer as used in RSA. You seem to use an approach that require that it is not a one-way function.

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10 hours ago, Ghideon said:

If it is a one-way function then you are guaranteed to fail to find the factors of a given large integer as used in RSA. You seem to use an approach that require that it is not a one-way function.

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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(Bold by me)

5 hours ago, Trurl said:

I claim the Pappy Craylar method can make it 2 way.

This is clearer than before but still a claim and not yet an explanation (or evidence or mathematical proof). And as far as I can tell (also see the mathematics I provided) you do the opposite; making harder or impossible. I must admit that I'm not interested enough to try to provide an explanation or attempt at a mathematical proof.

5 hours ago, Trurl said:

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.

Sorry, it does not make sense.

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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.

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9 hours ago, Trurl said:

Here where y = zero

x^4/(N^2+x) = y

x on the graph where y = 0, is the smaller factor q

Sorry, the description does not seem to match the link. I see you use RSA-100 but not what the other means.

9 hours ago, Trurl said:

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.

Sorry, I have no clue what this means. The opposite seems true; y=0 means x is not a possible prime factor.

9 hours ago, Trurl said:

But it does not give any information about factors. Throwing a dice would be even quicker and give the same amount of information (zero).

9 hours ago, Trurl said:

Look at the graph.

I still have no clue what you expect someone to see that has a connection to actually finding factors of a semi prime. But it supports my note above about the lover limit of 0.25 of large semi primes related to RSA.
(We can see this since RSA-100 factors are known by adjusting the plotted x-values)

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17 hours ago, Ghideon said:

But it does not give any information about factors. Throwing a dice would be even quicker and give the same amount of information (zero

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.

17 hours ago, Ghideon said:

I still have no clue what you expect someone to see that has a connection to actually finding factors of a semi prime. But it supports my note above about the lover limit of 0.25 of large semi primes related to RSA.
(We can see this since RSA-100 factors are known by adjusting the plotted x-values)

17 hours ago, Ghideon said:

We can see this since RSA-100 factors are known by adjusting the plotted x-values)

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).

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At least this comment seem illustrate your belief

4 hours ago, Trurl said:

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.

How many numbers do you have to test?  (answer: too many when the semi prime is of realistic length)
How long does the test take? (answer: very long when the semi prime is of realistic length)
Have you reduced the amount of numbers that needs to be tested.  (answer: no)

4 hours ago, Trurl said:

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).

You just need to check my description of how 0.25 is a limit and how I am correct on this. And a quick look at the graph and/or a binary conversion of the smallest of the factors of RSA-100 supports my statement. Feel free to ask for clarifications.

I don't understand your claims at all. Obviously the graph tells us that 0 < x < 1051  which is already trivial given that the semi prime is an RSA number. Your graph gives no new information and is therefore not useful. What you say is already trivially known without your graph.

The central part of the problem is just swept away without any explanation:

4 hours ago, Trurl said:

all that needs tested.

It is just that the numbers of factor are huge and no improvement is presented which means that RSA (and similar methods ) is unaffected.

Edited by Ghideon
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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 }]

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Your response does not seem to adress my questions and also the mathematics I presented.

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7 hours ago, Ghideon said:

Your response does not seem to adress my questions and also the mathematics I presented

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.

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9 hours ago, Trurl said:

But testing 10^49 to 90^49 you clam is tribal to testing numbers 10^100.

If I said that pleas provide a quote. It seems incorrect.

9 hours ago, Trurl said:

You could argue why don’t I just test the bottom half of 10^100.

Why would anyone argue that? That wouldn't be very clever if RSA is involved.

9 hours ago, Trurl said:

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.

Makes no sense, sorry.

9 hours ago, Trurl said:

But the purpose is not to destroy RSA anyway.

On 3/3/2021 at 7:15 PM, Trurl said:

I am claiming I can factor semi-Primes and thus RSA cryptography would be no more.

9 hours ago, Trurl said:

The Pappy Craylar conjecture might not destroy RSA.

If "might" is replaced with "does, with 100% certainty," it will be closer to the truth.

Quote

It is a pattern in factorization.

From the descriptions it seems more a kind of untestable personal belief or a wish, not a real pattern that someone can use on factorization

9 hours ago, Trurl said:

If you look at the database numbers on error I posted, you will see the error has potential to find patterns.

There was no pattern when I looked, it sounds apophenia.

9 hours ago, Trurl said:

And if you followed along I think you know the equation isn’t trivial.

I have followed along for six pages. I have seen no non-trivial equations so far.

9 hours ago, Trurl said:

So I don’t mind if you say I can’t break RSA.

Thats cool, disagreement can be healthy in discussions. And in this case I see no reason not to hold on to my position.

9 hours ago, Trurl said:

But you have to admit it was a good attempt.

No, my opinion is quite the opposite.

9 hours ago, Trurl said:

you don’t know what other ideas I have.

True. Any reason to care?

9 hours ago, Trurl said:

But I appreciate the critic. You tested the problem. Before it was an idea which no one I know could test.

Thanks. Note that I did not perform a formal test. Things claimed lacks substance; tests can't be defined. I compared some mathematical expressions in this thread to the mathematics of RSA.

Edited by Ghideon
clarifications
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Here is a test for your ideas. I have generated some semi primes*.  Please apply your ideas and post the factors. By your ideas I mean your graphs, equations, expressions, "patterns".

16 bit semiprime n: 34189
32 bit semiprime n: 2564855351
64 bit semiprime n: 14830573937595324521
96 bit semiprime n: 51710300225695220586621873149
127 bit semiprime n: 105885478296634626184079475557631302167
184 bit semiprime n: 12767878405113739778678228941014070709490602057822917487
256 bit semiprime n: 65987296772226902159803529660127173701120758847708653228034824460492065707557
512 bit semiprime n: 7836720177069608003500755381905379696113734820415437059107302880557005627301137450218819438830220854264769471529917213850983558125480442066927341587395863
1023 bit semiprime n: 69490761693879024515322409543501909771269878484507282046636369964444778350330138534492998629992038016396256793090732559749217992926715622188685818640312898075316196657706595786953437471020980242957623038475766495858035713914761703256554473705391771835499591933107282105938025772818934223909581971045391352957
2048 bit semiprime n: 24526390543922276163761960363928196029766688656050108715445599450389739269747070746319365337941271616416750589041896232434918183132656902143747734873329125429339367399911892083925190236294913521057711207069839256771767342182218357681756228812990287212506112441822553026785861136879985887637208094501026429985550942440215340084813276521593663885143089395413706069476749023299175391323954138329879311921588226542287367889696999663985429219142007825586735068016851507637293578802967253836923946950044656545745427656518620040242839141101217310921847170940220268514592632894815641661819532833727871690527012340299696283369


*) using python and cryptodome.

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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?

Edited by Trurl
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Thanks for the description. What are the factors of the semiprimes I posted?

Note 1: When you post the factors you find with your method, can you please aslo supply the number of x-values you tested?
Note 2: I will not take part in the test, but I may comment on the results.

5 hours ago, Trurl said:

It is simply. Can this possibly work?

I have given my opinion (several times).

Edited by Ghideon
clarification
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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.

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