On 9/20/2025 at 2:13 PM, Ghideon said:

I know of Euclid's method and its use in cryptography; I got curious since I don't see an obvious connection to trurl's approach ("multiplying by 5"). Can you elaborate on the relation to Euclid's method?

Well I’d explain it to you Ghideon but we both know studiot could explain it better. I only know studiot from their posts, but studiot appears to be some sort of educator. I will say this all my Prime number work is cumulative just like that college final that is worth 70% if you grade. I could explain but no one knows what the heck I am trying to say anyway. That is why it is so important someone made some sense of it. I too am waiting from.a reply from @studiot

Here are two numbers, imagine we don't know which is prime (if any):

X = 7,823

Y = 7,827

Multiply by 5 as you like to do:

X x 5 = 39,115

Y x 5 = 39,135

Exactly what are you going to do with 39,115 and 39,135, that makes it any easier to figure out which of 7,823 and 7,827 (if any) is prime, than just doing the same thing directly with 7,823 and 7,827?

On 9/23/2025 at 10:45 PM, pzkpfw said:

Exactly what are you going to do with 39,115 and 39,135, that makes it any easier to figure out which of 7,823 and 7,827 (if any) is prime, than just doing the same thing directly with 7,823 and 7,827?

Point taken. Even when multiplying by 5 it is hard to see which is a true semiPrime. But in these examples you miss the point of the semiPrime test and that is to efficiently eliminate possibilities. Little lies I think it’s called. Fermat’s Prime test had liars. In these cases I would treat all false positives as Prime. It is not mathematically pretty but may aid in number crunching. Remember the semiPrime rules are still correct we are just having difficulty proving them. The problem is can we factor semiPrimes which is also the key to the mystery of Primes.

Take this equation:

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

Plug in known compost products for pop and a known factor for x. You will get y, the other factor.

You can also test values of max and min x and y.

As x increase y decreases and as x decreases increases.

I did this to give me a realistic range of numbers to work with. Yes it is still huge but I believe the change in size is significant.

I was ask how the rules of creating semiPrimes would be useful so I showed that I would narrow the number range even more by using the semiPrime rules.

People think that I am posting the same Primes number program over and over again. But the truth is it is many separate math problems designed with the intent to crunch large Primes.

I do not always explain it justly, but I thought when I posted in these forums it would resonate.

Can you demonstrate it with the example numbers?

I chose numbers that were not many many digits for a reason.

On 9/26/2025 at 3:52 AM, Trurl said:

Take this equation:

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

Plug in known compost products for pop and a known factor for x. You will get y, the other factor.

How? Please show mathematically step by step what you mean. Use the symbols and derive the result. (Do not use any numerical example)

Here is a temporary explanation. As I review this problem I will try and write a better explanation. I will try and answer your questions. Feel free to ask anything.

IN

Clear {x, y, pnp}

x = 971

y = 2803

pnp = x*y

(((pnp^2/x) + x^2)/pnp)

OUT

Out[44]= {971 Clear, 2803 Clear, 2721713 Clear}

Out[45]= 971

Out[46]= 2803

Out[47]= 2721713

Out[48]= 7857780/2803

In[85]:= N[7857780/2803]

2803.3464145558332`

IN

Clear {x, y, pnp}

x = 945

y = 2803

pnp = x*y

(((pnp^2/x) + x^2)/pnp)

OUT

Out[86]= {945 Clear, 2803 Clear, 2721713 Clear}

Out[87]= 945

Out[88]= 2803

Out[89]= 2648835

7857754/2803

Out[92]= 7857754/2803

In[93]:= N[7857754/2803]

2803.337138779879`

The last example is to show that only the factors work in this equation. Only a y of 2803 will result when multiplied by x.

IN :

Clear {x, y, pnp}

x = 945

y = 2803

pnp = 971*2803

(((pnp^2/x) + x^2)/pnp)

OUT

Out[94]= {945 Clear, 2803 Clear, 2648835 Clear}

Out[95]= 945

Out[96]= 2803

Out[97]= 2721713

Out[98]= 7408565562994/2572018785

In[99]:= N[7408565562994/2572018785]

Out[99]= 2880.45

December 20, 2024

AUTHOR

comment_1281391

Clear[x, y, g, pnp]

pnp = 2211282552952966643528108525502623092761208950247001539441374831\

912882294140\

2001986512729726569746599085900330031400051170742204560859276357953\

757185954\

2988389587092292384910067030341246205457845664136645406842143612930\

176940208\

46391065875914794251435144458199;

x = 1.13056560621865239372901234269585839625544`15.653559774527023*^100

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

g = x*y

N[y]

Out[75]= 1.130565606218652*10^100

Out[76]= 1.955908211597676*10^159

Out[77]= 2.211282552952967*10^259

Out[78]= 1.95591*10^159

(((((pnp^2 / x) + x^2)) / x) / pnp) where y = 0 This is the start equation to find x. That is, numerically or graphically finding x when y is zero. We know pnp and y and use it to graphically get a ball park figure where x is. I left out the y = (((((pnp^2 / x) + x^2)) / x) / pnp) because y here is the y of the Cartesian coordinates and not necessarily the y of the factors. It is confusing but n and p are reserved variables in mathematica.

I believe this is where the x = 1.13056560621865239372901234269585839625544`15.653559774527023*^100 comes from. I have to review my records, but I believe it comes from the graph where y = 0. That is where finding Primes by multiplying by 5 (or other Prime numbers) comes in. x is not precise. Because of the error in the estimate there is an amount of deviation where y equals zero on the coordinate plane. This error should not be exponential. It needs studying on how it effects large numbers.

We then place x into the above equation y = (((pnp^2/x) + x^2)/pnp) and that is where Euclid's method applies.

All of which seems pretty simple for what we are asking it to do. I know it lacks mathematical prettiness. But this is crunching numbers.

I hope this makes some sense. I been busy and I need to write the whole thing out better. But I was just scripting in Mathematica and IN is the information I put in and OUT is the output.

If this works I will show you how everything is derived. It is not complex but there is no sense deriving it until it proves useful.

Well the equations seem to tell the truth. But the question is how I derived x = 1.13056560621865239372901234269585839625544`15.653559774527023*^100 from the graph.

On 10/1/2025 at 4:30 AM, Trurl said:

I will try and answer your questions. Feel free to ask anything.

Can answer the question I asked above?

On 9/30/2025 at 1:22 PM, Ghideon said:

  On 9/26/2025 at 3:52 AM, Trurl said:

Take this equation:

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

Plug in known compost products for pop and a known factor for x. You will get y, the other factor.

How? Please show mathematically step by step what you mean. Use the symbols and derive the result. (Do not use any numerical example)

Note the bold part

23 hours ago, Ghideon said:

(Do not use any numerical example)

On 9/30/2025 at 10:30 PM, Trurl said:

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

This is the equation to graph. We know pnp. When this graphs y equals 1 (or greater; but should be less than 2)

This gives you the factor x from the graph where y on the graph is 1 (or slightly greater due to error). Note this y on the graph is not the factor y! y is near 1 where x on the graph is the factor.

Once we have x we test the magnitude of the factors x and y in this equation:

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

We then test all Prime numbers in this magnitude range by division into pnp. We only need to test the values we think are Prime. These Prime numbers were identified by multiplying by a known Prime say 5.

I wasn’t going to go slightly off topic and explain that the multiply by a known Prime number test is similar to an idea I had about preserving the arms race with semiPrimes.

Say the equations held true and I could factor semiPrimes then you could search for semiPrimes by talking a known Prime number for x and graph numbers where the 1 occurs.

I don’t have all the details but I posted it somewhere in the past. It isn’t just multiplying x and y. It is stepping through numbers until the x Prime number equals 1 with the pnp number we are cycling through and testing if it approaches 1.

It is a test for semiPrimes. And if it holds true RSA remains unbreakable.

The only problem is that pnp divided by estimated pnp does not always equal exactly 1. It has some error. But if the equations hold true we now have enough eyes in the forum to find a solution to the error.

Integer factorisation is defined within the integers (Z) and depends on integer divisibility. A relation that yields non-integers cannot represent or detect factors. Your formula (((pnp^2/ x ) + x^2) / pnp) yields non-integers (except for some trivial case). Hence your equation is algebraically invalid as a method for finding primes or factors.

Edited October 10Oct 10 by Ghideon

On 10/10/2025 at 6:15 PM, Ghideon said:

Integer factorisation is defined within the integers (Z) and depends on integer divisibility. A relation that yields non-integers cannot represent or detect factors. Your formula (((pnp^2/ x ) + x^2) / pnp) yields non-integers (except for some trivial case). Hence your equation is algebraically invalid as a method for finding primes or factors.

Again you missed the point. The equations only give a range where the factors are. There is error. To compensate for this error we must test values in a certain range. x on the graph is still an integer. y on the graph is not. Just because y on the graph is decimal does not change the usefulness of the equation. We are number crunching. The graph allows to narrow the range of the potential factors. I don’t think algebra alone is enough to solve the original equation. The graph gives the range where algebra fails.

Finding 2 factors N= p*q without factoring. I gave p and q (which I call x and y) “conditions” which set them apart from other potential factors. I know an algebraic equation that had no error and returned the exact factor as an integer would be ideal but the fact remains that there is no alternative that I know of.

2 hours ago, Trurl said:

Again you missed the point. The equations only give a range where the factors are. There is error.

What is the mathematical expression* of this error?

3 hours ago, Trurl said:

The graph allows to narrow the range of the potential factors.

Given the mathematical expression for the error, provide an expression for the range.

*) Not a numerical or visual example in a graph

pnpapprox = (((((pnp^2 / x) + x^2)) / x) )

pnp = test semiPrime

pnp/pnp = 1

pnpapprox / pnp is our original equation

Error = pnpapprox / pnp - (1 - (1 - pnpapprox / pnp))

This looks complex but it isn’t. This is to demonstrate how I would handle the error.

On 10/13/2025 at 8:07 PM, Trurl said:

Error = pnpapprox / pnp - (1 - (1 - pnpapprox / pnp))

The expression collapses to zero.

Error = (1 - (1 -(pnpapprox / pnp)))

Good catch. I think this fixes it.

This is tricky. Do you see what I am attempting to do to find where x on graph equals where y = 1?

Adding the error to 1 and finding where y equals the 1+the error.

But this error may or may not help in picking the correct x. More testing is required. I not sure if finding this error narrows the choice but I think it is time for some number crunching either way.

On 10/17/2025 at 3:50 AM, Trurl said:

Error = (1 - (1 -(pnpapprox / pnp)))

The ones cancel out.

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

On 10/13/2025 at 2:07 PM, Trurl said:

pnpapprox = (((((pnp^2 / x) + x^2)) / x) )

On 10/16/2025 at 1:20 PM, Ghideon said:

Error = pnpapprox / pnp - (1 - (1 - pnpapprox / pnp))

The expression collapses to zero.

On 10/25/2025 at 9:38 AM, Ghideon said:

The ones cancel out.

If it cancels out we have found the error. Perhaps if you add the error rather than subtracting it in the y equals 1 equation:

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

Add y + error

Or 1 + error

On 10/6/2025 at 6:13 AM, Trurl said:

I posted it somewhere in the past

I reviewed your earlier thread on primes and this thread is a repetition; the proposals lack a coherent description and the few interpretable parts contradict established mathematics. These issues were already explained, in multiple ways, in the previous discussion but have not been incorporated into your reasoning.
Your replies also continue indicate fundamental confusion regarding factors, integers, and prime numbers; as mentioned in the earlier thread, you may find this resource helpful: Khan Academy – Factors and multiples

Well I completely understand. The solution seems too simple to apply. But I think it is simple enough to determine if it works. That is at least when dealing with small values. It takes more effort to prove. The problem is the same with a different approach. Though the equation is simple it takes a while to derive it. The quote you refer to is me stating if it works then it can be used to find larger Prime numbers. There is more to this problem than pnp/x. It looks similar, but most posts have never been seen before. Some are to document my work or to fix a mistake that I have made along the way. I do not claim to know anything about Prime numbers. I am explaining Prime numbers by use of semiPrimes. I was introduced into this math through cryptography. I read Steven Levy's Crypto. I learned of the RSA one-way-function and ask, "is it truly one-way?" That is where pnp comes from. It is simply the N. N=p*q. This one way function is like having 2 grains of sand and throwing it into the beach. But what if the process to find these 2 grains of sand marked them in some way? What if it made them a different color from the rest of the sand? If you knew the general area they were in you could look for the different colors. I don't really believe in one-way-functions.

The following is a screen shot I took today. It is a know semiPrime. It doesn't prove anything, but may hint the equation is simple enough to be true.

Yes, there are tests for prime numbers. Methods like the trial division test, Sieve of Eratosthenes, and modern algorithms such as the Miller–Rabin test can determine whether a number is prime.

I hope the people who read this thread see the significance of my last post. That is where the known Prime factors plugged into the equation equals 1.

The following photos are feedback from A.I. It puts information more clearly than me.

I just wanted to post a conclusion to this work.

I just wanted to note I use PNP Approximate divided by PNP to find where y on the graph equals 1 to find the corresponding value at x which is the smaller factor.

I could use PNP Approximate minus PNP to find x where y on graph equals zero

Or PNP Approximate equals y where x at y is the smaller factor

I believe it works or else I wouldn’t have posted it. But it is more than if it works it has to be efficient to be useful. I believe AI shows it works with smaller numbers.

The number crunching problem is if you can find where y equals one on the graph for a 600 digit number.

But even if that isn’t efficient new approaches have showed a different approach.