20180225, 18:21  #1 
Feb 2018
24_{16} Posts 
Twin Prime Conjecture Proof
TWIN PRIME CONJECTURE PROOF:
1. Unless 'twin prime' every twin prime possibilty, (11/13 41/43 71/73 etc) (17/19 47/49 77/79 etc)and(29/31 59/61 89/91) has a lowest prime factor of 7 or above. Eg. 119/121 has factors of 7 11 and 17, therefore it's lowest prime factor is 7. Lowest prime factors are all that we use. 2. If twin primes end, then necessarilly, there exists a prime(n) that closes the last possibility of a pair being twin prime: By applying this prime in the series as 'lowest prime factor' twin primes end, take it out, they go on. This would necessarilly be so. 3. PRIME(1) = 7: PRIME(2) = 11 etc. (Prime(1)  2) / Prime(1)) × ((Prime(2)  2)/ Prime(2).......... ... × (Prime(n)  2) / Prime(n)...would have to equal zero if a greatest prime number left zero gaps 'after' first position placing as final lowest prime factor of a twin prime possibilty. 5/7 × 9/11 × 11/13 × 15/17.....× (n  2)/n) will never equal zero. This is the number of 'gaps' (twin primes) left on 29/31 59/61 89/91 etc for example. 4. The equation has 'primes up to..' multiplied together, dividing 'primes minus 2, up to..' multiplied together. This cannot equal zero/leave zero gaps. 
20180225, 21:03  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20180226, 04:30  #3 
Romulan Interpreter
Jun 2011
Thailand
8,963 Posts 
1. How about "9 and 11", "13 and 15", "25 and 27", etc...?
2. Oh, you don't like numbers ending in 5? How about "31 and 33", or "51 and 53", etc...? 3. When you say "lowest prime factors", keep in mind that this can be huuuuuuge... How about consecutive odd numbers ending in 1, 3, 7, 9, which are not divisible by 3, 5, 7, 11, 17, and not prime either? (such pair can be easily computed with modular calculus, even with a "lowest prime factor" of thousands of digits). That product you compute, is an infinite product, and it is always zero. As is the product for all primes p, of (p2)/p which is a number under unit. 
20180226, 05:00  #4  
Jun 2003
4789_{10} Posts 
Quote:
Whoa, there! Not so fast. Even if there are only finite number of twin primes, that doesn't mean that there are only a finite number of twin prime "possibilities". In fact, there is an infinite number of them in each series after the "last" twin prime. Therefore, there is no last prime(n) that "closes the last possibility". We have an infinite number of primes to choose from to close the infinite number of possibilities. 

20180226, 12:05  #5 
Feb 2018
2^{2}×3^{2} Posts 
2, 3 and 5 multiples
Multiples of 2, 3 and 5 are not included because they are instantly known as not possible. It serves no purpose to include them. I did not say anything was finite, only contained within the 3 paired series.
Last fiddled with by Steve One on 20180226 at 12:40 
20180226, 12:30  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20180226, 12:41  #7 
Feb 2018
24_{16} Posts 
Primes under 1 million
I don't actually know what you mean by that.

20180226, 15:26  #8 
Aug 2006
13464_{8} Posts 
You're going to have to make your proof more formal before you discover that your mistake is that you've proved
Theorem: For any x, there are infinitely many pairs (p, p+2) such that neither p nor p+2 has a prime divisor <= x.instead of Conjecture: There are infinitely many pairs (p, p+2) such that neither p nor p+2 has a prime divisor <= sqrt(p+2).This is the usual mistake to make when attempting a proof of the twin prime conjecture, so don't feel too bad. 
20180226, 16:46  #9 
Feb 2018
2^{2}·3^{2} Posts 
I don't see where l have even mentioned square roots. Nothing yet has been said to disprove what l wrote. Please show where specifically an error occurs.
Last fiddled with by Steve One on 20180226 at 16:51 Reason: Missed out some writing 
20180226, 16:56  #10 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 

20180226, 17:29  #11 
Aug 2006
2^{2}×3^{3}×5×11 Posts 

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