20190201, 15:01  #1 
Feb 2019
7×13 Posts 
Proof of the twin prime conjecture (NOT!)
PROOF OF THE TWIN PRIME CONJECTURE
The twin prime conjecture states that there are an infinite number of twin primes. Twin primes are a pair of primes that differ by 2 e.g. 11 and 13, 17 and 19, 29 and 31. Consider the sieve of Eratosthenes acting on the infinite number line. 2 makes the first move and eliminates all its infinite number of multiples, including itself, (i.e. the even numbers) from the infinite number line. 3 then makes the next move and eliminates all its infinite number of multiples (including itself) that have not already been eliminated by 2, from the infinite number line. At this point of the sieving process, it is not difficult to see that all the infinite number of integers left on the infinite number line (after 1 is removed also) are in pairs of the form 6n1 and 6n+1, where n is some positive integer. Therefore odd primes are of the form 6n1 or 6n+1. So, for particular values of n, twin primes, 6n1 and 6n+1 will exist amongst the infinite integer pairs of the form 6n1 and 6n+1 remaining on the infinite number line. So 5 makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2 and 3 i.e. (in ascending order) 5x5, 5x7, 5x11, 5x13, 5x17, 5x19, 5x23, 5x5x5, 5x29, 5x31, 5x5x7, 5x37, 5x41, 5x43, 5x47, 5x7x7, 5x53, 5x5x11, … 7 then makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2, 3 and 5 i.e. 7x7, 7x11, 7x13, … 11 then makes the next move and 13 the next after 11 and the process carries on ad infinitum. So, following this process, the twin primes are simply those pairs of primes, 6n1 and 6n+1 that escape the elimination process. A little thought will show that for the twin prime conjecture to be false, it takes one and only one special prime to eliminate (at some point in its elimination process) the remaining infinite number of 6n1 and 6n+1 pairs of integers ahead of it on the infinite number line (i.e. those that have not already been eliminated by previous primes). Eliminating a pair of 6n1 and 6n+1 integers simply requires that one of them is eliminated i.e. one of them is a multiple of this special prime. It will now be shown that it is impossible for any special prime to achieve this i.e. the twin prime conjecture is true. In this part of the discussion, consider the infinite number line after 2, 3 and their infinite number of multiples have been eliminated (1 is removed also) as mentioned previously. Imagine a prime number hopping (infinitely) like a frog along this infinite number line consisting of integers of the form 6n1 or 6n+1 during its turn in the sieving process, as it eliminates its multiples as described above e.g. 17 takes its first hop and eliminates 17 x 17 = 289 as it lands on it. 17 then takes its second hop and then eliminates 17 x 19 = 323 as it lands on it etc. A little thought will show that each and every one of the infinite number of multiples of 17 with prime factors of 5, 7, 11, 13 and 17 only will have a pair of un  eliminated 6n1 and 6n+1 integers just before it (if the multiple of 17 is of the form 6n1) or after it (if the multiple of 17 is of the form 6n+1). So as 17 is hopping infinitely along the infinite number line described, it will hop over these infinite number of 6n1 and 6n+1 uneliminated pairs. So this will be true for all primes greater than 3. So 37 will hop over un eliminated pairs before or after the infinite number of multiples of 37 with prime factors of 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37 only. So no such special prime exists that can eliminate all the remaining uneliminated 6n1 and 6n+1 pairs ahead of it at some point in its infinite hopping. So the twin prime conjecture is true i.e. there are an infinite number of twin primes. 
20190201, 16:16  #2  
Feb 2017
Nowhere
F3B_{16} Posts 
Quote:


20190201, 16:35  #3  
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
Quote:


20190202, 11:20  #4 
Sep 2002
Database er0rr
5×701 Posts 
The word is spreading.
Last fiddled with by paulunderwood on 20190202 at 11:23 
20190202, 12:09  #5 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
4786_{10} Posts 
Aleluia

20190202, 12:14  #6 
Jun 2003
2×5×479 Posts 
The duplicate threads are nuked.

20190202, 14:06  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}×487 Posts 
There are infinite number of primes. Checked.
After sieving out multiples of 2 and 3 from the set of positive integers greater than 1, there are infinite pairs of twin (differencing by 2) integers left. Checked. After sieving out multiples of any finite set of primes greater than 3, there are infinite pairs of twin integers left, since any m.n# +/ 1 will be indivisible by any positive integer less than or equal to n and greater than 1. Checked. If there is no one "special prime" that can sieve out all the remaining infinite set of twin integers left, then there are infinitely many twin primes. False. There are a countable infinite number of primes left that will sieve the remaining countable infinite number of twin integers left. Both remaining twin integers and primes that multiples of which will sieve them out are countably infinite and have a one to one relationship. You can not apply a logic that would work for finite sets on infinite sets here. Last fiddled with by a1call on 20190202 at 15:00 
20190202, 16:37  #8 
Feb 2017
Nowhere
7473_{8} Posts 
I am taking to opportunity to show something with regard to "twin pairs" of residues, which may not be immediately obvious.
If you compute the number of "twin pairs" modulo 2*3, 2*3*5, 2*3*5*7, you might think that this number is significantly less than the number of primes up to the product. For example, there are 10 primes less than 30, but only 3 "twin pairs" modulo 30, namely +/1, 12+/1, and 18+/1. There are 46 primes up to 2*3*5*7 = 210, but only 15 "twin pairs" mod 210. But, if I've done my sums right, this is misleading. If p is prime, we let p# be, as usual, the "primorial" product of all the primes q less than or equal to p. We then have (Euler phifunction) phi(p#)/p# is given by This is of order 1/log(p). The number f(p) of residue classes r (mod p#) such that Mod(r1,p#) and Mod(r+1,p#) are both prime to p# (both multiplicatively invertible) is easily determined using the Chinese Remainder Theorem (CRT). For q = 2, there is 1 such residue (mod 2); the condition is satisfied when r is even. For q > 2, there are q  2 such residues (mod q); any residue other than 1 or q1 (mod q) satisfies the condition. If p > 2, we have by CRT that f(p)/p# is which is of order 1/log^{2}(p). By way of comparison, by PNT we have for the primecounting function, which (again by PNT) is of order 1/p for x = p#. Thus, if p is at all large, the number of residue classes Mod(r, p#) for which Mod(r1,p#) and Mod(r+1,p#) are both prime to p#, will completely swamp the number of primes up to p#, so the vast majority of the "twin pairs" up to p# will consist of two composite numbers. Whether any of the pairs consists of two primes is completely lost in the noise. A quick check using Li(x) as an approximation for pi(x) indicates that the number of "twin pair" residues will exceed the number of primes less than p# by p = 47. Unfortunately, by this time p# is way, way beyond the range of exact computations of the primecounting function. Last fiddled with by Dr Sardonicus on 20190202 at 16:41 Reason: gixnif spyto 
20190203, 16:55  #9 
Feb 2017
Nowhere
7×557 Posts 
It appears I may have been a bit pessimistic about existing tables of pi(x). The primorial 47# is less than 10^18.
Thomas R. Nicely's home page gives links to tables of pi(x) up to 10^{12}, and in intervals of 10^{12} to 10^{16}, then to 2*10^{16} pi(x) to 1e16 and pi(x) from 1e16 to 2e16 The 10^{16} table says, "The counts in this table were obtained by the author, employing a direct and explicit generation and enumeration of the primes." Isolated values of pi(10^n) are known for n up to 27. So, although pi(47#) may not be in any existing table, computation of pi(p#) is certainly feasible for p = 47 and probably a bit beyond. We have 10^{27} > 71# > 10^{26}, and 73# > 10^{28}. Last fiddled with by Dr Sardonicus on 20190203 at 16:56 
20190203, 19:55  #10 
Feb 2019
7·13 Posts 
Your replies have not explained clearly and concisely why if the twin prime conjecture is false, it will take one, and only one prime to achieve this. Just as any prime will eliminate all its infinite number of multiples, then why is it difficult to see that only one prime is needed if the twin prime conjecture is false.

20190203, 22:18  #11 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}×487 Posts 
Because there will be infinite pair of twin integers left unsieved after you sieve out any finite number of primes and all their multiples. After that sieving any additional finite number of primes (say 1 more) will still leave an infinite number of twin integers left. No single magicprime can exist that would be the factor of one halfpair of the infinite twin integers left. It would take an infinite number of primes and their multiples to reduce the infinite number of twin integers to a finite value, not a singleone.
Last fiddled with by a1call on 20190203 at 22:27 
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