20180302, 20:09  #34 
Feb 2018
2^{2}×3^{2} Posts 

20180302, 20:19  #35 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20180302, 20:38  #36  
Nov 2008
2×3^{3}×43 Posts 
Quote:
I don't have any problems with the argument you used in your proof. The problem is that what it proves isn't the twin prime conjecture. You seem to have ignored my "proof" that there are infinitely many primes of the form n^{2}1. I posted it because I thought it would be useful for you to look at it, work out why it doesn't actually prove what it sets out to prove, and then look at your own work in the same light. 

20180302, 23:51  #37  
Aug 2006
2^{2}×3^{3}×5×11 Posts 
Quote:
Quote:


20180303, 04:26  #38 
Romulan Interpreter
Jun 2011
Thailand
8963_{10} Posts 
I think you all go too far with this guy, his real problem is that he has no idea why you used a sharp sign after the p, because this is not a music class, and he also has no idea why you used small n here and big N there (possibly a typo?). And you are trying to teach him math...

20180303, 12:52  #39 
Feb 2018
2^{2}·3^{2} Posts 

20180303, 13:48  #40  
Feb 2018
44_{8} Posts 
Quote:
Imagine, 2 3 2 5 2 7 2 3 2 11 2 13 2 3 2 17 2 19 2 3 2 23 2 5 2 3 2 29 2 31 2 3 2 5 2 37 2 3 2 41 2 43 2 5 2 47 2 7 2 3 2 53 2 5 2 3 2 59 2 61 2 3 These are the lowest prime factors of 2 to 63. If you take out using 2s, 3s and 5s we get: (7s 11s etc 'become' lowest prime factors) 7 11 13 7 17 19 7 11 23 7 29 31 11 7 37 41 7 43 11 47 7 53 7 59 61 That is one 7, 11, 13, 17 etc repeated. Given the chance to use another 7, 11, 13, 17 etc, the same ""individual spacing""as is on (29/31 59/61 89/91 119/121 etc) to close twin prime possibilities, NEVER will the result produce zero gaps, not to 63 or any number, because (correct use of equation) 5/7 × 9/11 × 11/13 × 15/17.......× (n2)/n CANNOT equal zero. If twin primes ended at some point, then after the last twin prime, the equation, which DOES calculate spaces left open would have to, using primes 7 up to (n), equal zero. It can't. Ever! The equation can use a prime with 15 gazillion digits. It can't equal zero. 

20180303, 14:03  #41 
Feb 2017
Nowhere
7·557 Posts 

20180303, 14:07  #42  
Nov 2008
2×3^{3}×43 Posts 
Quote:


20180303, 15:20  #43 
Feb 2017
Nowhere
7·557 Posts 
For each prime P, there is a finite set of arithmetic progressions with common difference P# (the product of the primes up to P) giving all positive integers p for which neither p nor p + 2 is divisible by any prime less than or equal to P. You have given this for P = 5; there are three such pairs of AP's with common difference 2*3*5.
p = 30*k + 11, p + 2 = 30*k + 13 p = 30*k + 17, p + 2 = 30*k + 19 p = 30*k + 29, p + 2 = 30*k + 31 Each successive prime P multiplies the number of such pairs by P  2. So appending the prime 7 multiplies the number of pairs by 5, giving a total of 15 such pairs of AP's with common difference 2*3*5*7, or 210. A pair of positive integers p, p + 2 are both indivisible by 2, 3, 5, or 7 if and only if it is of one of the following 15 forms: p = 210*k + 11, p + 2 = 210*k + 13 p = 210*k + 17, p + 2 = 210*k + 19 p = 210*k + 29, p + 2 = 210*k + 31 p = 210*k + 41, p + 2 = 210*k + 43 p = 210*k + 59, p + 2 = 210*k + 61 p = 210*k + 71, p + 2 = 210*k + 73 p = 210*k + 101, p + 2 = 210*k + 103 p = 210*k + 107, p + 2 = 210*k + 109 p = 210*k + 137, p + 2 = 210*k + 139 p = 210*k + 149, p + 2 = 210*k + 151 p = 210*k + 167, p + 2 = 210*k + 169 p = 210*k + 179, p + 2 = 210*k + 181 p = 210*k + 191, p + 2 = 210*k + 193 p = 210*k + 197, p + 2 = 210*k + 199 p = 210*k + 209, p + 2 = 210*k + 211 This process can be continued. It may thus be shown that, for any prime P, there are infinitely many pairs p, p + 2, neither of which is divisible by any prime less than or equal to P. Unfortunately, "not divisible by any prime less than or equal to P" doesn't mean "prime." The smallest composite number which is indivisible by any prime less than or equal to P, is the square of the least prime that is greater than P. Thus, e.g. the smallest composite number indivisible by 2, 3, or 5, is 49. One can say more. It is well known (first proved by Dirichlet) that any individual arithmetic progression a*k + b where a and b are positive integers with gcd(a, b) = 1 contains infinitely many primes. Thus, for example, there are infinitely many primes of the form 30*k + 11. Thus, one can say with certainly that, in any of the pairs p, p + 2 as described above, there are infinitely many values of k for which at least one of the numbers p and p + 2 is prime. Unfortunately (for the sake of proving the twin primes conjecture) there is no result known about when both p and p + 2 are simultaneously prime for the same value of k. I would venture a guess that there may be known results that there are infinitely p for which neither p nor p + 2 have more than some fixed number of prime factors, perhaps as few as two prime factors. 
20180303, 18:57  #44 
Aug 2006
1734_{16} Posts 
Right  Chen proved this just before (and published just after) the Cultural Revolution. There are various strengthenings that have been published since, for example HeathBrown & Li show that i.o. p+2 has at most two prime factors and p+6 has a bounded number of prime divisors.

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