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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.

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.

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.

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.

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.

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.

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.

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

That is a good question. You could test them in real time. I mean dividing them into: pnp = 2211282552952966643528108525502623092761208950247001539441374831\ 912882294140\ 2001986512729726569746599085900330031400051170742204560859276357953\ 757185954\ 2988389587092292384910067030341246205457845664136645406842143612930\ 176940208\ 46391065875914794251435144458199; Should be close to: 1.130565606218652*10^100Though it might be worth the effort to store the numbers that were multiplied by 5 to make it easier and have a list of Primes. Sensei volunteered to be lead programmer.🤪

Here is my code to find the range. It is a script I wrote in Mathematica. This is how I find the designated range that I want to multiply the odd numbers by 5. It is linked to above where it is derived. 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

Sensei, thanks for the descriptive post. It will take me some time to respond. Remember I am filled with abstract thoughts. Here I will share very simply my thought process. Simply stated there is no a, b, or c. That is the definition of composite. That is just to describe composites. Only if you didn’t know they weren’t composite would you multiply a arbitrary Prime (5) d stands alone just as numbers increase new Primes appear. So d times Prime equals a semiPrime. So if I know the range of numbers I wanted which I confirm they are Prime by creating a semiPrime. I am avoiding the database by selecting a designated range I believe the Prime factor is. Then I take my arbitrary well known Prime of 5 and multiply it by all odd numbers in my designated range. The only numbers that are semiPrime end in 5. Some numbers may be composite but I don’t care I have just decreased my sample size. Only database needed is to store the numbers that when multiplied by 5 end in 5.

Right on studiot! I didn’t know that is what is was but I am using that method to narrow the number of factoring by division tests. Out[75]= 1.130565606218652*10^100 Out[76]= 1.955908211597676*10^159 Out[77]= 2.211282552952967*10^259 Out[75] is my estimate of the smaller semiPrime factor Out[76] is the larger semiPrime factor HCF is then used because as out75 increases out76 decreases and vise versa. Out[77] is the estimate of N or the semiPrime factor we wish to factor The Prime numbers between the HCF are those we need to divide into N. That is were multiplying by 5 comes in. Also I thought if the 6x+-1 could find Primes within a set area I could find the exact Prime and not just an estimate. I thought I could use the sieve to magnify the values to the N semiPrime I wished to find.

Well man not believing in God would not mean he couldn’t make him believe. God just gave him a choice. You are right punishment would result in man acknowledging God.

Are there any methods to factor large semiPrimes? I showed you how I would approach the Prime factorization problem. If it works or not. I am concerned with the method. I multiply all odd numbers by five so as studiot pointed out you only have to test the numbers that end in 5. I would not bother to factor the ones that end in 5. Instead I would take the original Prime numbers and divide into the semiPrime factor that is too large to factor. This may be hard to understand but I think even if you don’t believe my method to find the large semiPrime you might agree that multiplying by 5 eliminates impossible work. I think you will be able to follow my process because all the steps are there. There is much going on here and I am not the best teacher.

Well if you know methods to find semiPrimes please share. I will show you a use to find the Prime number that multiplying by a known Prime number would aid in finding the product of the semiPrime. I need to find all Prime numbers where the smaller Prime factor is between 1.0 * 10^90 and 1.0 * 10^130 So I would multiply every odd number by the arbitrary Prime 5. Find new semi semiPrimes and divide those original Prime number values into: Clear[x, y, g, pnp] pnp = 2211282552952966643528108525502623092761208950247001539441374831\ 912882294140\ 2001986512729726569746599085900330031400051170742204560859276357953\ 757185954\ 2988389587092292384910067030341246205457845664136645406842143612930\ 176940208\ 46391065875914794251435144458199;

Well there is a difference between not making sense and saying it is not useful. One of the reason to use this method is having no knowledge of a factor being Prime or semiPrime. You say you’d never use it on a composite, but what if the composite was 32 digits and ended in 99? (I’m still working to make better examples. I know this one needs clarification.) You say it is redundant and makes the given number larger. I argue that as little as 3 times the size may increase it but yields a number easier to factor; sometimes. But what if I didn’t have a sieve and I wanted to find Primes between two larger numbers. Finding a Prime I didn’t know would be hard. So I would take every odd number between those numbers and multiply by 5. (It’s not division and it is only multiplying one time each number.) Would factoring these new numbers and proving they are semiPrime be harder then recursive division of these same numbers? That is the question. I know there are other more advanced algorithms, but now I am concentrated on this one. It may or not be useful. But I simply test would be to take your entire table you are sieving and multiply each number in question by 3 and sieve again. It produces slightly larger numbers you would factor in the future of the sieve anyway. By this I mean you will sieve these large numbers later as the sieve progresses anyway. Why not skip the smaller numbers? But does a standard sieve make it possible to eliminate Prime values without a continuous sieve? By this I mean starting at a given values not a continuous line from start to zero. Can your sieve start at numbers in the millions without a known starting point? I could be wrong of course but the forum is to test ideas. But as you guys pointed out a semiPrime test would give my test reason. But as I have stated in the past factoring semiPrimes is the key to finding a pattern in Primes.

What I said makes perfect sense: Prime times composite equal composite. Prime time semiPrime equals composite. Prime times Prime equals semiPrime. And if you can factor that semiPrime within reasonable, you have just solved the pattern of Prime numbers. The purpose of the multiplication is to increase the number of factors and know what the Prime factors of the newly created semiPrime. That is all. It is less than 1 minute observation. I don’t know that you are trying to understand what I am saying or just trying to destroy my idea. Which is ok because I welcome the challenge, but the above is what I claimed from the 1st post.

You are correct. But I am did not lie. I said unknown number if Prime, times known Prime would create 2 more factors because the number is now a semiPrime. Which doesn’t prove it is useful. But I will leave you with this thought. Do you agree or disagree that the ability to factor semiPrimes or even identifying them would result in the solution of all Primes numbers? That is why I take unknown Prime times Prime to form semiPrimes as a Primality test. That is why I mention this method that seems redundant and foolish but contains a hidden problem. I am not trained in number theory. It would take me days just to learn the notation to some of these sieves. It could be my redundant multiplication makes the numbers larger. But what if I state the redundancy of this method is useful if you can find within doubt that a semiPrime is formed; By multiplying the unknown number by a known Prime.

You are correct a composite number made lager isn’t always necessary. But if that number is so large that you can’t factor it, multiplying it by a Prime number may change the perspective and create more factors. But I am concerned with Prime numbers. A Prime times a Prime is a semiPrime. If I suspect that a number is Prime and it is so large I can’t factor it maybe I can determine if its product is a semiPrime. Yes it can lead to larger numbers but it may be the only information you can get about the number. My method is not rocket science. It is a simple observation. I was not high when I thought of it. I may have increased the magnitude of the number (relatively small) but I have also increased the number of factors. But the method would prove more useful if there was a reliable method to factor semiPrimes. And my question to you is why sieve Primes and not keep track of the semiPrimes at the same time? But I welcome your challenge. It is only an idea until someone tests it.

It will take me some time to write this up but I will test one of your numbers with the calculator on my phone. Say we don’t know if 4829995653 is Prime, semiPrime, or composite. Then we multiply it by a known Prime. I usually start with 5. 5* 4829995653 =24149978265 This product is now either semiPrime or composite. But trial and error we find 24149978265 / 3 =8,049,992,755 So the product was semiPrime it is composite, so unknown number times known Prime is composite so the unknown number is composite. ——— Very simple. May or not prove useful. But if you are dealing with a sieve, the sieve is practically doing the same simple process. But the only difference is this is jumping to steps that would occur in the future of the sieve. This is just a tool. It doesn’t always simplify the process. Like you said Prime times Prime gives a semiPrime that isn’t always verified easily. And if the unknown number is already a semiPrime the product still could be difficult to factor. I don’t know how useful it is, but what if you were completely your sieve in both the present and future numbers.

WIREDThe Kryptos Key Is Going Up for SaleJim Sanborn is auctioning off the elusive solution to K4, the outdoor sculpture that sits at CIA headquarters. But isn’t the value decreased every time someone new learns the secret? Besides isn’t it cheating and defying the original purpose?

No I am just biased because it is my method. But as KJW said even Fermat’s prime test has liars. Prime *Prime *Prime should always equal a composite but even * 5 can be hard to find factors. I know it sounds stupid in application: multiple by five and see what you get. I don’t always approach math scientifically. Sometimes intuitively. The Prime by prime equals semiPrime I don’t think is a complex thought. That is to be not tried before. It is just a fast observation. I will have to research these algorithms. I have never used them. I have tried to avoid Primes. I tried to study circuits and amateur radio. Something more tangible. But here I am again attempting to do data science. But you are right that I don’t always use the scientific method. To me finding Primes in a series is misleading. Greater minds than us have tried and failed. My odometer hit 116811 then 116816 then 116818. Suppose I was looking to find 2 pairs of matching numbers. These would be easy to see but difficult if I tried to find an equation that described when this occurs in sequence from zero to 116828. I mention this because I think prime numbers share the same elusive behavior. Mathematicians look at them and see patterns that aren’t there or see chaos.

Remember I am not factoring but testing for primality. So again I multiply 156,423,343 by 5 and get 782,116,715 which according to my rules is semiPrime. So 156,423,343 is Prime. I didn’t cheat and thought you may have given me a counter example. But if were to try and factor I would start on the right side of the digits. For 782,116,715 I would start on the right with what factors would go into 15. (3 & 5 ) 7-3 =4 __—////// But I’m not sure dividing right to left will factor, but as far as I know the number is prime. If the given number was semiPrime to start multiplying it by 5 should produce a composite number. I would have to write it up. I don’t have time tonight. I have made posts about prime testing and factoring semiPrimes. However those problems are not the path I was steering towards on this thread.

Well they are in no order. I multiplied by 5 and could factor the result easily (not at all). So far a second test I tied implying 9,991,991 by 23 and again tried to factor the product with no success. I repeated the multiplication of 9,991,991 by 7 and could not factor it either. I chose the Prime multiple randomly. I was on my phone so I wanted to keep it small. My logic is this: Number in question if it is Prime multiplied by known Prime results in Prime times Prime Or semi Prime (which I won’t be able to factor) And this means both numbers that were multiplied are Prime —-—————————————————————— But if the number in question is not a Prime number the multiplication with the known Prime will result in a composite number that should be easier to factor (and it helps simplify the knowledge gained by the sieve) ______-_ So basically it is just multiplying by a Prime and factoring the product to work around division of N by every smaller number. I don’t know how useful it is or that it works all the time. Or why Guass didn’t mention it because he didn’t have a computer and was doing arithmetic by hand. But as Ghideon mentioned that semiPrimes are even more difficult to factor. But there is no factoring of semiPrimes (other than difficult in that the formed semiPrimes have no other factors) . If this work and proves useful (which it hasn’t yet), this “math hack” could speed up many of number crunching. To me the main weakness is the difficulty of factoring the semiPrime like Ghideon said, but the sieve information should help. And the factoring that results in a composite number should be relatively easy.

Well I’m not saying a semi prime is easier to factor. I’m saying the number in question is not Prime, multiplying it by a known Prime will not result in a semi prime and thus easier to factor. If both numbers are Prime there will be a difficult semiPrime to factor. You could check by trying the process again with a different Prime number multiple. I test it with 9,991,991 It is difficult to factor. I multiplied by 5, 23, and 7 First look I’d say it was Prime because it ends in 5 and the 49,959,955 is difficult to factor. 9,991,991*36=359,711,676 not prime. That is just a random Prime number. That shows nothing. Just an illustration of 9,991,991 wasn’t Prime it might be easier to factor the product of a prime and non Prime. I need some time to look at this further. But a shoutout to @studiot can you see what I was attempting to do?