20180228, 20:04  #23 
Aug 2006
2^{2}·3^{3}·5·11 Posts 
If I didn't make it clear previously, this is where your reasoning falls apart. It is true that, for any x, there are infinitely many pairs (n, n+2) where neither n nor n+2 are divisible by any prime up to x, but this does not prove the twin prime conjecture.

20180302, 12:50  #24 
Feb 2018
2^{2}·3^{2} Posts 

20180302, 12:52  #25 
Feb 2018
2^{2}·3^{2} Posts 
What you said means or implies what?

20180302, 13:19  #26  
Feb 2018
2^{2}×3^{2} Posts 
Quote:
There can be NO prime such that it will end the possibilty of twin primes. There can be NO highest, lowest prime factor of a pair, because the equation l wrote would not equal zero. But EVERY pair is either twin prime or has lowest prime factor of 7 or above, out of (12+/1)+n30) (18+/1)+n30) and (30+/1)+n30) Contains ALL twin primes apart from 3/5 and 5/7. (2/4, 32/34, 62/64 etc) (3/5, 33/35, 63/65 etc) (4/6, 34/36, 64/66 etc) up to (28/30, 58/60, 88/90) contain 3/5 5/7 and absolutely zero other twin primes. The problem here is this that people are holding faith in methodoligies that have failed up to now and probably will continue to fail. Maybe square routes CAN help to solve that which l HAVE solved, but there is no necessity to include them in any proof. As long as l show 'all possibilities' and show that within all possibilities there is a continuance, that is all that is necessary. Most importantly, l am not making the mistake you think l am. Any ending of all twin primality is done with primes as lowest prime factor. But that cannot happen to completion because the equation which is 'valid and correct' would produce a result of zero, and it can't. The conjecture is proved. 

20180302, 13:31  #27  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20180302, 15:05  #28  
Feb 2017
Nowhere
3^{2}×433 Posts 
Quote:
(*) My understanding is, pairs of twin primes (apart from (3, 5) and (5, 7)) can only be of one of the forms 30*n  1, 30*n + 1 for some positive integer n; 30*n + 11, 30*n + 13 for some nonnegative integer n; or 30*n + 17, 30*n + 19 for some nonnegative integer n. It is also true that none of these possibilities allows 2, 3, or 5 as factors. The problem with the rest of your argument is that you can continue this approach, successively excluding 7, 11, etc. As you do so, you will wind up with a rapidly increasing number of "twin prime possibilities" defined modulo 2*3*5*7, 2*3*5*7*11, etc. If your argument were correct, the list of possibilities would collapse. It doesn't. Last fiddled with by Dr Sardonicus on 20180302 at 15:12 Reason: Fixing minor misstatements 

20180302, 19:13  #29 
Feb 2018
44_{8} Posts 

20180302, 19:37  #30 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 

20180302, 19:55  #31 
Nov 2008
2×3^{3}×43 Posts 
Here's a proof that there are infinitely many primes of the form n^{2}1:
Suppose there are finitely many primes of the form n^{2}1. Then there is some prime p which "ends the possibilities" of primes of this form. But then take N = p#, the product of all the primes up to p. Now N^{2}1 is a possibility for a prime of the form n^{2}1 that is not excluded by primes up to p. This contradicts the statement that p ends all possibilities for primes of this form. So there are infinitely many primes of the form n^{2}1. Of course this can't be a valid proof, as the only prime of this form is 3. The mistake in it is exactly the same as the mistake in your proof. 
20180302, 20:03  #32  
Feb 2018
2^{2}·3^{2} Posts 
Quote:
Please don't bring in modulo anything. Everything is in base 10. All that is necessary for you or anyone to ruin my proof is to prove that on one of those 3 series, pairs are NOT either twin prime or have a lowest prime factor of 7 or above. Or, that it is NOT true that if twin primes end(they don't) then there is a prime number that must be the final nail in the coffin as far as subsequent possibilies are concerned. Then, 5/7 × 9/11 × 11/13.....(prime(n)minus2)/prime(n) would equal zero, because after this prime(n) that closed final possibility, you would only be able to use 7 to (n) to close further possibilities, but the equation CAN'T produce zero. Therefore, there can be NO prime that, acting as lowest prime factor of a twin prime that ends the possibility of twin primes. On number line 1, 31, 61, 91 etc, prime(n) COULD, but isn't, at ((7×11×13......×(n))×30)+1. If the prime had closed all following possibilities, the equation would have to equal zero. But the only primes to use are 7 up to (n) which can't leave zero subsequent spaces/twin primes. Any subsequent prime used as lowest prime factor ONLY fills a proportion of spaces. Using just 7 leaves 5/7th open. Using 7 and 11 leaves 5/7 × 9/11 which = 45/77th open etc. Interestingly, 45/77 provides an answer to, "What numbers COULD close the possibility of future twin primes after using 7 and 11 . Answer: fourty five 77s. 

20180302, 20:07  #33  
Feb 2018
2^{2}·3^{2} Posts 
Quote:


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