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Old 2018-02-28, 20:04   #23
CRGreathouse
 
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Originally Posted by Steve One View Post
If so, IF twin primes end, which they don't, one must reach starting with 7, then 11, then 13, then 17 etc a prime number that closes the last possiblity of a pair being twin prime.
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.
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Old 2018-03-02, 12:50   #24
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7 is not quaranteed to be lowest ...
Show a lower one than 7 and l will send you 100 pounds sterling.
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Old 2018-03-02, 12:52   #25
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What you said means or implies what?
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Old 2018-03-02, 13:19   #26
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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.
No! You've missed the point.
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.
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Old 2018-03-02, 13:31   #27
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No! You've missed the point.
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.
The word is roots. Maybe read up on the dunning kruger effect.
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Old 2018-03-02, 15:05   #28
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The numbers l wrote 'should' read (12+/-1)+n30), (18+/-1)+n30) and (30+/-1)+n30). Sorry!
Apology accepted. Restatement understandable(*), and applicable to all pairs of twin primes other than (3, 5) or (5, 7).

(*) 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 non-negative integer n; or

30*n + 17, 30*n + 19 for some non-negative 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 2018-03-02 at 15:12 Reason: Fixing minor misstatements
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Old 2018-03-02, 19:13   #29
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The word is roots. Maybe read up on the dunning kruger effect.
What(?) word is roots?
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Old 2018-03-02, 19:37   #30
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What(?) word is roots?
It's square roots not square routes.
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Old 2018-03-02, 19:55   #31
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Here's a proof that there are infinitely many primes of the form n2-1:

Suppose there are finitely many primes of the form n2-1.
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 N2-1 is a possibility for a prime of the form n2-1 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 n2-1.

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.
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Old 2018-03-02, 20:03   #32
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Apology accepted. Restatement understandable(*), and applicable to all pairs of twin primes other than (3, 5) or (5, 7).

(*) 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 non-negative integer n; or

30*n + 17, 30*n + 19 for some non-negative 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.
Yes you can continue getting rid of 7s then 11s then 13s, but why bother? We now have just 3 paired number sequences. The difference is that the other series were gotten rid of because EVERY member of 1 or both series combined as pairs is divisible by 2, 3 or 5. Whereas, in remaining 3 series only SOME are divisible by 7, then 11, then 13 etc. It would screw up a perfectly good starting point.
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.
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Old 2018-03-02, 20:07   #33
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Originally Posted by 10metreh View Post
Here's a proof that there are infinitely many primes of the form n2-1:

Suppose there are finitely many primes of the form n2-1.
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 N2-1 is a possibility for a prime of the form n2-1 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 n2-1.

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.
I keep having to repeat that the "if primes end, then....," is not something l believe. It is used to show that if it were true then the equation l wrote would have to equal zero. The equation CAN'T equal zero, therefore there can't be a prime that does this. How many times do l have to say this?
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