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Trurl last won the day on July 18 2022
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Well the shooter was misguided in thinking assignating Trump would change the Republican Party from winning. IMHO. You have to win the Country by becoming a good leader. Taking Trump out of the picture might not even benefit the Democrats. They can’t even decide on the candidate. I saw a YouTube vid of Micho Kaku were he stated he was drafted into the U.S. Army during the Vietnam war and saw a weapons demonstration. He saw the we can blast an enemy and have much more firepower, but this just makes the enemy double down. It is possible to assasinate well known figures, but I don’t think the end result is what the assassin intends. Besides attempting such a thing forfeits the shooter’s life. Trump takes a lot of criticism for stupid things he says, but when he was President there was less wars. He threatened to make NATO pay, but he wouldn’t dismantle it. I find a conundrum with NATO. We join together so that one is attacked we all defend. But instead of securing us we alienate other countries that aren’t members. Does that make other Countries feel they have less influence on what happens in the World? I know that is off topic but I think that should be the issue of the campaign. And assignations will not solve anything.

Need description of Prime# distribution in Riemann hypothesis
Trurl replied to Trurl's topic in Mathematics
On a side note, the instructor in the previous post of Prime distribution made a work of fiction in the form of a short film. Warning this is probably appreciated by nerds. But not telling what it is I thought it was a work of art; something that shows the impact of math. https://www.edwardfrenkel.com/film/ 
Need description of Prime# distribution in Riemann hypothesis
Trurl replied to Trurl's topic in Mathematics
https://youtu.be/d6c6uIyieoo It has been awhile since I ask, but I was wondering how proving that proving that zeroes always do or do not exist on 1/2 would be critical to solving the distribution of Prime numbers. I have to watch this video again to find the time he says this. 15:17 minutes in video Is there a difference in “the distribution of Primes” and “the location of Primes?” By location I mean the value or actual determining if a number is Prime. 
Klapacusis you are right it is trivial. This is a game of trivial pursuit. I don’t know the factoring pair of this 2048 bit semiPrime. But if the number of the equation where it equals zero is correct than the factor is less than that number. And I added the percentage of error which brings us closer to that smaller factor. I know your argument is that it is still a large number. However checking all odd numbers descending from my estimate should be close enough to not require as much computation. 4*10^37 divided by 2. Still a bunch of numbers but more manageable. Still too many numbers but the actual factor should be close (slightly less) than the estimate. And if anyone could add any advice in crunching numbers, particularly in Mathematica I would appreciate the help. There is a function call Monitor. I don’t know how to “keep track” of what is happening behind the scenes when Mathematica is processing. It just says, “Running.” I don’t know if it is going to take hours, days, or months. I understand why someone would not believe my methods. We are not supposed to be able to factor semiPrimes. It is impossible. But is it possible for impossible to exists? I chose the largest RSA number that is open source. I don’t know the factors, but I feel I am getting close. RSA is critical for authentication using digital signatures. It is still one of the most used cryptography schemes on the web. I have recently read that the original Xbox used RSA to secure its hardware. I didn’t know cryptography was used in hardware. But don’t be afraid to factor semiPrimes because it will mean the end of RSA. I think it was flawed from the beginning. Secrets can not be keep if you share them. By encrypting secrets you have just marked them as valuable. That is why misinformation is so prevalent in media. 1977 it made sense to encrypt you secrets. Now it is difficult to determine truth from reports. For instance Trump is found guilty on trial. He says witchhunt ; they say guilty of influencing the election. We hear the verdict, but we weren’t at the trial. We didn’t see all the evidence. And if the County is 50 50 Trump / Biden why wasn’t this reflected in the jury? Politics aside who is right? It all is based on which candidate you believe. But who is trustworthy? I think the true is the same with cryptography. We trust is because we can mathematically test it. If the numbers require too many brute force attempts it is secure. But which crypto systems are trustworthy. If I do go on to break RSA, I hope to break it along with you. If it works, I showed you how to do it. And that you can believe the answer. If I fail and it still is impossible, we have more trust in RSA.

Well it worked here: Clear[x,pnp];pnp=2564855351; eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); Solve[eqn==0&&x>=0,x,Reals]//N (*{{x41350.98025},{x6.387801493*10^11}}*) In[1]:= 41350 is close to 41227 By trivial you mean the division. I did not divide yet. I only used the equations. The magnitude should be close. I still have programming limitations on my own part. But I thought the number seemed reasonable. Close enough for good old brute force. What numbers do you get?

This is the smaller factor of: RSA2048 = 2519590847565789349402718324004839857142928212620403202777713783604366202070 7595556264018525880784406918290641249515082189298559149176184502808489120072 8449926873928072877767359714183472702618963750149718246911650776133798590957 0009733045974880842840179742910064245869181719511874612151517265463228221686 9987549182422433637259085141865462043576798423387184774447920739934236584823 8242811981638150106748104516603773060562016196762561338441436038339044149526 3443219011465754445417842402092461651572335077870774981712577246796292638635 6373289912154831438167899885040445364023527381951378636564391212010397122822 120720357 Verify

I don’t see the point in Russia trying to influence elections. They’d probably want to but so does either side democrat or republican. The idea of collecting data is to know your target audience. Everyone wants an influential voice so the data is valuable. It would take some resources to power such a weapon. And the corrupt message could be drowned out in a sea of messages. No matter what your opinion someone disagrees. I often wonder if the Trump Biden thing is just to piss everybody off. Instead of having announcements that say go vote, they piss off everyone and it is a record turnout. So the morale is that if you don’t know the motivation behind the influence or control you can’t defend against it. Some things like money earned by selling data is common sense, but imho it would take some “power” to manipulate data on that scale. It would be like a president trying to control the entire government let alone the world. I was brainstorming ideas one night and thought what if we had a computer program that spread misinformation. Say it took your online profile and put in different addresses, phone numbers, and metadata. There probably is already software that does this, but I didn’t want to add to garbage already found on the internet, just the online footprint. I understand why China would want to influence us. And I think go ahead, we might learn something. I don’t know crap about China except they are communist and they take American jobs. Is this true. We know better than that. When I was in high school they said Japanese students are better in math and buy American. Our students are just as bright. Besides Japan are our allies. So I guess blaming China or math scores on Japan makes us study harder. But there are misconceptions all over the world. I guess it makes you loyal to your country and respect your leaders. Misdirection.

eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); SetPrecision[NSolve[eqn==0&&x>=0,x, Reals]//N,94] Out[23]= 2519590847565789349402718324004839857142928212620403202777713783604366202070 Out[24]= 7595556264018525880784406918290641249515082189298559149176184502808489120072 Out[25]= 8449926873928072877767359714183472702618963750149718246911650776133798590957 Out[26]= 9733045974880842840179742910064245869181719511874612151517265463228221686 Out[27]= 9987549182422433637259085141865462043576798423387184774447920739934236584823 Out[28]= 8242811981638150106748104516603773060562016196762561338441436038339044149526 Out[29]= 3443219011465754445417842402092461651572335077870774981712577246796292638635 Out[30]= 6373289912154831438167899885040445364023527381951378636564391212010397122822 Out[31]= 120720357 Out[33]= {{x>4.098447549634527496690402228228115660800000000000000000000000000000000000000000000000000000000*10^37},{x>1.975537793137120663017279930682231182622035891422434710723887074852770653001767985540861473587*10^9 Why would a quantum computer render RSA useless?

We choose to factor SemiPrimes. We do this not because it is easy but because it is hard. We choose to post factors not to deceive but because you are smart. We read the writing on the wall because there is math beyond SemiPrimes. We wish there was still rewards for crunching semiPrimes but mathematicians choose to do math not because it makes money but for the quest for knowledge. We choose to break RSA not because it is impossible but because it is not supposed to be possible.

If these numbers hold close, RSA is no more. I realize I have to expand the accuracy to 37 digits, but I got these numbers in 2 seconds. If they hold true so does the Pappy Craylar Conjecture! In[1]:= Clear[x,pnp];pnp=pnp = 2519590847565789349402718324004839857142928212620403202777713783604366202070 7595556264018525880784406918290641249515082189298559149176184502808489120072 8449926873928072877767359714183472702618963750149718246911650776133798590957 0009733045974880842840179742910064245869181719511874612151517265463228221686 9987549182422433637259085141865462043576798423387184774447920739934236584823 8242811981638150106748104516603773060562016196762561338441436038339044149526 3443219011465754445417842402092461651572335077870774981712577246796292638635 6373289912154831438167899885040445364023527381951378636564391212010397122822 120720357 eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); NSolve[eqn==0&&x>=0,x,Reals]//N (*{{x41350.98025},{x6.387801493*10^11}}*) Out[1]= 2519590847565789349402718324004839857142928212620403202777713783604366202070 Out[2]= 7595556264018525880784406918290641249515082189298559149176184502808489120072 Out[3]= 8449926873928072877767359714183472702618963750149718246911650776133798590957 Out[4]= 9733045974880842840179742910064245869181719511874612151517265463228221686 Out[5]= 9987549182422433637259085141865462043576798423387184774447920739934236584823 Out[6]= 8242811981638150106748104516603773060562016196762561338441436038339044149526 Out[7]= 3443219011465754445417842402092461651572335077870774981712577246796292638635 Out[8]= 6373289912154831438167899885040445364023527381951378636564391212010397122822 Out[9]= 120720357 Out[11]= {{x>4.09845*10^37},{x>1.97554*10^94}} 4.09845*10^37 Hopefully is close.

I may have factored a RSA number previously unfactored. That is is the number crunching is right. Clear[x,pnp];pnp=pnp = 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199; eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); NSolve[eqn==0&&x>=0,x,Reals]//N (*{{x41350.98025},{x6.387801493*10^11}}*) 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 {{x>3.83952*10^37},{x>1.67814*10^94}} {{x>3.83952*10^37},{x>1.67814*10^94}} x should be just larger than 3.84*10^37 Test it. I hope it is close. Too bad there is no longer a reward for factoring RSA numbers 🤑

@Ghideon you are right much processing power is needed. I thought if I could get the size needed to test down to the size of a smaller RSA number that has already been factored, I could crunch it. Three hours no results. I had some help with the programming. A tester pointed out that I get 41351 instead of 41227. That is 124 digits off. But I am hoping is close enough to guess. Does anyone here do number crunching? I could use some tips on brute force crunching. Clear[x, pnp]; pnp = 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199; x = 30000000000000000000000000000000000001; While[x <=pnp,If[Divisible[pnp,x], Print[x]]; x+2]; While[x<=pnp,If[Divisible[pnp,x],Print[x]];x+2]; Clear[x,pnp];pnp=2564855351; eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); Solve[eqn==0&&x>=0,x,Reals]//N (*{{x41350.98025},{x6.387801493*10^11}}*) In[1]:= While[x<=pnp,If[Divisible[pnp,x],Print[x]];x+2]; Clear[x,pnp];pnp=pnp = 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199; eqn=((pnp(Sqrt[(x^2*pnp^4+2*pnp*x^5)+x^8])/pnp^4(1(x^2/(2*pnp)))*(pnp^2/x^2))); Solve[eqn==0&&x>=0,x,Reals]//N (*{{x41350.98025},{x6.387801493*10^11}}*) Out[2]= 2211282552952966643528108525502623092761208950247001539441374831912882294140 Out[3]= 2001986512729726569746599085900330031400051170742204560859276357953757185954 Out[4]= 2988389587092292384910067030341246205457845664136645406842143612930176940208 Computational processing required.

We are going to try and factor a RSA number that has yet been factored. No guarantees it will work. My estimate by graphing is that if we start at 4.0 X 10^37 and increase in value using brute force to divide into PNP, we may be rewarded with an answer. I understand it is still many values but at least we have a starting point. I also understand that if we are unsuccessful, we may be aided by finding the error of the factoring equation. Also graphs can be deceiving. If it is wrong starting point, it could be the graph is right, but the view of the graph is wrong. Clear[x,z, pnp]; pnp = 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199 z=pnp/(pnp/2) Show[ Plot[(( (((pnp^2/x) + x^2)) / x) /pnp), {x,5000000000000000000000000000000000000, 200000000000000000000000000000000000000}] ] 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199 2 Clear[x]; Plot[((pnp  (Sqrt[(x^2 * pnp^4 + 2 *pnp*x^5) + x^8])/pnp^4  (1  (x^2/(2*pnp))) *(pnp^2/x^2))), {x,5000000000000000000000000000000000000, 200000000000000000000000000000000000000}] Clear[x]; Plot[(1/((pnp / (((pnp^2/x) + x^2)) / x))), {x,5000000000000000000000000000000000000, 200000000000000000000000000000000000000}]

105951 105950 105909 105910 105911 Ok on an odometer when there are at least 3 matching numbers say 555 or two sets of repeating numbers, say 5500, then we will try to predict when each occur chronologically on the odometer. Sure you could just fill in the numbers with the corresponding numbers. But imagine these numbers were related to Prime distribution. The pattern is harder to see with the odometer as it moves linearly. Imagine you have a regular odometer. It goes zero through nine. So in the singledigit place of the odometer (the start of counting), you note that 3, 5, 7 are Prime. So now you go to the tensdigit. And note that 3+1 or 3+3 or 3+5 or 3+7 are eliminated as Prime. To get another Prime number you would have to add an even number to 3, 5, or 7. Of course adding an even number doesn’t always result in a Prime. This is just a graphical representation of a sieve. But just as repeating numbers in the odometer are hard to predict linearly throughout the revolutions of the odometer, are we not doing the same thing when looking for patterns in Prime numbers? I study Prime numbers because I like finding patterns. Patterns can confuse or look like they could be there, but patterns are what we see in math. I might have stretched the truth when I claimed RSA was in trouble, but that oneway function is why I started all this math in the first place. The odometer sieve would require a lot of calculation. Again it would only show what is going on. It is the same with my graphs. I can only estimate where the semiPrime factor is on a graph. Graphing 128bit numbers and analyzing the graph is a challenge, but necessary to prove as the curve on the graph approaches zero x approaches the smaller Prime factor. That is where I believe if the graph holds true, for larger N’s, it will be superior to brute force. But as the graph becomes larger and more difficult to evaluate so does choosing an N. I have read that if N becomes too large, the enciphering of the message becomes too cumbersome. In a future post I will be sending a graph. But It is important for anyone reading this to share if they had any results with the graph. I have tried to make the graph more useful (less test values) by inverting the equation.

Clear[x, pnp] pnp = 2564855351 Show[ Plot[(( (((pnp^2/x) + x^2)) / x) /pnp), {x, 0, 60000}] Graph this equation. It is the inverse. It should make it easier to find the SemiPrime. It will decrease number of trails, but on small numbers will still be more time than brute force. It will however give you a graphical understanding of the smaller SemiPrime. Remember 2 unknowns should be impossible.