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 January 1st, 2017, 07:15 AM #1 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 Twin prime conjecture proof TWIN PRIME CONJECTURE PROOF Primes > 3 and their odd multiples that are not multiples of 3 are of the form 6n Â± 1, where n is a positive integer. Consider a sieving process operating on numbers of the form 6n Â± 1 that are arranged in ascending order i.e. 5,7,11,13,17,19,23,25,29,31,35,37, â€¦ In this sieving process, a number will be eliminated only once. So 5 eliminates its multiples â‰  5 in ascending order i.e. 5x5, 5x7, 5x11, 5x13, 5x17, 5x19, 5x23, 5x5x5, 5x29, 5x31, 5x5x7, 5x37, 5x41, 5x43, 5x47, 5x7x7, 5x53, 5x5x11, â€¦ and then 7 eliminates its multiples â‰  7 that are not multiples of 5 i.e. 7x7, 7x11, 7x13, ... and then 11 eliminates its multiples â‰  11 that are not multiples of 5 and 7 starting from 11x11 and so on. So the numbers that will not be eliminated in this process are the prime numbers > 3. Consider a finite part of the infinite sequence described above. Let the last number in this sequence be N, where N is some multiple of 5 of the form 6n Â± 1. In the sieve described here, it should be clear that the number of multiples of 5 eliminated in the number sequence > the number of non-multiples of 5 eliminated for any N. Also the minimum difference between any 2 multiples of 5 is 10 and there will be a pair of numbers, 6n-1 and 6n+1, in between these 2 multiples of 5. For instance, 5x23 =115 and 5x5x5 =125. 119 and 121 are of the form 6n-1 and 6n+1 respectively and in between 115 and 125. Larger differences mean more than 1 pair of 6n Â± 1 e.g. consecutive multiples 5x7=35 and 5x11=55 in the sequence above have a difference of 20 between them and will have 41, 43 (of the form 6n-1, 6n+1 respectively) and 47, 49 (of the form 6n-1, 6n+1 respectively) in between them. This means that, for any N > 5x13=65 (see sequence above), the number of multiples of 5 of the form 6n Â± 1 < the number of pairs of 6n Â± 1. This means that the non-multiples of 5 of the form 6n Â± 1 eliminated < the number of pairs of 6n Â± 1. This therefore means that there will always be pairs of 6n Â± 1, the twin primes, that will not be eliminated. The twin prime conjecture is true i.e. there are infinitely many twin primes.
 January 1st, 2017, 07:25 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. Paragraphs! Slabs of text are hard to read.
 January 1st, 2017, 10:27 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,038 Thanks: 423 I doubt merely paragraphing this particular text would provide any help. There is no concise overview of the proof. Indeed, nothing is concise. For example it takes over five lines to say: Consider the sequence of all primes greater than 3 taken in ascending order, all of which can be expressed in the form $6n \pm 1.$
 January 1st, 2017, 11:13 AM #4 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 So why do you have a problem with the fact that I have made my proof accessible to college kids by stating 5 lines that you have compressed into 1 line. At least you didn't have a problem with the mathematics. Make comments about the mathematics instead of criticizing the presentation.
 January 1st, 2017, 11:31 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,045 Thanks: 1618 MrAwojobi considers divisors such as 3 and 5, but higher divisors can turn up. For example, the pair (2117, 2119), where 2117 = 29*73 and 2119 = 13*163.
 January 1st, 2017, 11:37 AM #6 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 3 and its multiples never featured in the proof. What is the problem with the higher divisors you talk about? I don't really get your point.
 January 1st, 2017, 12:04 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,045 Thanks: 1618 You implicitly omitted multiples of 3, because the numbers 6n Â± 1 cannot have 3 as a divisor. The pair (2117, 2119) isn't eliminated by the process you describe, but isn't a prime pair (neither is prime in that example). You haven't shown that it isn't the case that every sufficiently large pair turns out not to be a prime pair. You're effectively saying "it should be clear" that isn't the case, but it isn't clear.
 January 1st, 2017, 12:10 PM #8 Senior Member   Joined: Aug 2010 Posts: 158 Thanks: 4 They are eliminated by the process I describe. You have answered this query in your previous post already i.e. 13 will eliminate 2119 and 29 will eliminate 2117.
 January 1st, 2017, 03:40 PM #9 Global Moderator   Joined: Dec 2006 Posts: 19,045 Thanks: 1618 Eliminating multiples of 5 leaves some pairs. Likewise, eliminating multiples of 7 leaves some pairs. It's not clear, though, that this can be continued indefinitely, as you haven't proved that some pairs always remain.
January 1st, 2017, 03:50 PM   #10
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Math Focus: Yet to find out.
Quote:
 Originally Posted by JeffM1 I doubt merely paragraphing this particular text would provide any help.
Just a pet peeve of mine, regardless of the contents validity. In this case it would aid in the readers ability to easily correct any mistakes made by OP. Unless that was intentional.

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