20190207, 15:31  #23  
Aug 2006
3·5^{2}·79 Posts 
Quote:
I think that it would help your intuition to learn other ways of creating the primes. For example, just as the sieve of Eratosthenes corresponds to the use of the binary quadratic form \(xy\), the sieve of Atkin and Bernstein corresponds to the use of the forms \(4x^2+y^2\), \(3x^2+y^2\), and \(3x^2y^2\). Do you think even Eratosthenes wrote down the even numbers? You realize that all highperformance sieves today store 30 values to a byte (since \(\varphi(30)=8\))... disregarding multiples of 2 and 3 is an optimization between a thousand and two thousand years old, hardly a grand simplification to which to lay claim. 

20190207, 17:21  #24  
"William"
May 2003
New Haven
2358_{10} Posts 
Quote:
Suppose you we looking for primes among the numbers 2^(n^2)1. The first few candidates are 2^11 = 1 2^41 = 15 = 3*5 2^91 = 511 = 7*73 2^161 = 65535 = 3*5*17*257 2^251 = 33554431 = 31*601*1801 These first few candidates will be sieved out when the sieve reaches 31. Reasoning in parallel to you twin prime argument, there should either be an infinite number of primes, or there should be a last prime that sieves away all the remaining candidates. But neither is true. There are no primes in this list because 2^(n^2)1 is always divisible by 2^n1. But there also cannot be a last sieving prime. For a prime number p, divisors of 2^(p^2)1 are all of the form 2kp+1, so the smallest possible divisor is 2p+1. For any "stopping point" in the sieve of Eratosthenes there will primes bigger than the stopping point, and hence candidates 2^(p^2)1 that have all divisors larger than the stopping point. 

20190207, 17:29  #25 
Aug 2006
1725_{16} Posts 

20190208, 09:32  #26 
Feb 2019
100_{2} Posts 
Proof of the twin prime conjecture
The proof of the twin prime conjecture must show all the steps, even the one you think are obvious
Pdf file of twin prime conjecture https://drive.google.com/open?id=1Kf...6Z56Y1630AaUwq 
20190208, 17:16  #27 
Feb 2019
7×13 Posts 
I don't see how you can formulate an entirely different problem and begin to draw an analogy to my twin prime proof. In fact, come to think of it, I can dispense with all my talk about if the twin prime conjecture is false, it will take one and only one special prime to eliminate all the 6n1 and 6n+1 pairs at some point in its elimination process. This has been the main statement that a lot of you have been hanging on to. I still believe in this statement but it doesn't even need to be in the proof because the mere fact that every prime, 5 or greater, will perpetually keep hopping over uneliminated 6n1 and 6n+1 pairs during its turn on the number line with all multiples of 2 and 3 removed is enough to prove the twin prime conjecture is true. Every prime having this property means that these pairs can never run out. No one, as yet, has come up with a way that these pairs can run out apart from one prime doing the job if this was possible. All you keep doing is formulating new problems and trying to use analogical arguments. That doesn't work for me. Stick to this twin prime problem at hand and explain if there is any other way for all these pairs to be eliminated at some point if the twin prime conjecture is false. None of you can because there is no other way i.e. if the twin prime conjecture is false. A fraction of infinity is a lesser infinity but still infinity. With my way that fraction of infinity is 1. This means that this prime takes out all the infinity number of pairs and not say a third or half or a quarter of infinity. Of course this latter part of my discussion is based on if the twin prime conjecture is false. I don't need to read some difficult philosophical books about infinity by the 'Greats' to come up with my own common sense concept of infinity.

20190208, 17:28  #28  
If I May
"Chris Halsall"
Sep 2002
Barbados
2×4,657 Posts 
Quote:
But if you haven't read (and, importantly, understood) and applied the prior art to your claims and methods, you are doomed to failure. This is how this process works. 

20190208, 17:37  #29 
Jun 2003
37×127 Posts 
There's your problem right there...
Last fiddled with by axn on 20190208 at 17:40 Reason: show the "there" there 
20190208, 17:44  #30 
Aug 2006
3·5^{2}·79 Posts 
wblipp presented his analogy because you haven't understood the criticism everyone has leveled at your proof, so it seemed prudent to present something simpler to follow. I thought it did a great job presenting the core of the problem with your proof. If you can't understand the problem with your proof, as explained repeatedly, and you can't understand how wblipp's post applies to your proof, then all I can do is repeat my admonition to rewrite your proof with utmost rigor.

20190208, 18:36  #31 
"Ben"
Feb 2007
3·1,097 Posts 
You like to talk about number lines so I'll approach things that way.
In your argument you are fixated on the 6n1 and 6n+1 aspect of twin primes, where you consider only numbers with factors of 2 and 3 removed. This is another way of saying that every twin prime pair must be +1 and 1 mod 6. But there is nothing special about 6, or about writing down only the numbers with factors of 2 and 3 removed. Instead, write down the number line with all factors of 2, 3, and 5 removed. Whether the twin prime conjecture is true or false on your number line with 2 and 3 removed, it will also be true or false on this version of the number line as well, right? In this number line, twin primes greater than 5 can be (30n+11, 30n+13) or (30n+17, 30n+19), or (30n+29, 30n+31). However in this number line the prime 7 doesn't "perpetually keep hopping over uneliminated pairs". For example it hops over 101 and 103, but then eliminates the candidate pair 119,121 by eliminating 119. If I understand your argument correctly, then only primes larger than 30 would "perpetually hop over" all candidate pairs. But we don't have to stop there! Now write down the number line with all factors of 2,3,5, and 7 removed. By similar arguments only primes greater than 210 would "perpetually keep hopping over uneliminated pairs". Removing 2,3,5,7 and 11 and the number becomes 2310. And so on. Surely you can see that the magnitude of the number required to "perpetually keep hopping over uneliminated pairs" grows much faster than the magnitude of the factors you removed to create that number line. Therefore it will always be possible to write down a number line where any number you pick will be too small to always skip over possible twin prime pairs. I.e., your argument doesn't work and is not a proof. 
20190208, 18:47  #32 
If I May
"Chris Halsall"
Sep 2002
Barbados
2·4,657 Posts 
Thank you for this. Sincerely. Awojobi is probably just a troll (possibly an AI being "trained" with the help of a human).
But having such serious diligent analysis is likely helpful for those people who have legitimate questions about the maths. 
20190208, 19:10  #33 
Aug 2006
3×5^{2}×79 Posts 

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