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 2016-04-10, 03:16 #1 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts Gaps of Primes? It may be proven soon that for every even integer n, there exist infinitely many primes p such that n+p is prime, as well as there are infinitely many primes gaps the length of n. I've been looking for a page to show that if for 2 even numbers n, and m: n and m not congruent to both 1 (mod 3) n and m not congruent to both 2 (mod 3) There are infinitely many primes p such that p+n is prime and (p+n)+m is prime. (And the gaps between three successive primes are n, and m). A quicker generalization should be that for any two even numbers, n and m, there are infinitely many primes p such that: n+p and (n+p)+m are prime or n+p and ((n+p)+m)/3 is prime. (The same rule for gaps between (three) (at least 2) consecutive primes. Any ideas on how this would be?
2016-04-10, 11:26   #2
science_man_88

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Quote:
 Originally Posted by PawnProver44 It may be proven soon that for every even integer n, there exist infinitely many primes p such that n+p is prime, as well as there are infinitely many primes gaps the length of n. I've been looking for a page to show that if for 2 even numbers n, and m: n and m not congruent to both 1 (mod 3) n and m not congruent to both 2 (mod 3) There are infinitely many primes p such that p+n is prime and (p+n)+m is prime. (And the gaps between three successive primes are n, and m). A quicker generalization should be that for any two even numbers, n and m, there are infinitely many primes p such that: n+p and (n+p)+m are prime or n+p and ((n+p)+m)/3 is prime. (The same rule for gaps between (three) (at least 2) consecutive primes. Any ideas on how this would be?
your congruences don't cover all even numbers.

 2016-04-10, 13:41 #3 PawnProver44     "NOT A TROLL" Mar 2016 California 110001012 Posts They do because given two even numbers both congruent to 1 or 2 (mod 3): n, m, are both congruent to 1 (mod 3) assuming p > 3 if n, m = 1 (mod 3): (p+n) is 2 (mod 3); (p+n)+m is 0 (mod 3) if n, m = 2 (mod 3): (p+n) is 1 (mod 3); (p+n)+m is 0 (mod 3) Any other cases other than the ones I listed is true for all even numbers. Last fiddled with by PawnProver44 on 2016-04-10 at 13:41
2016-04-10, 13:44   #4
science_man_88

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Quote:
 Originally Posted by PawnProver44 They do because given two even numbers both congruent to 1 or 2 (mod 3): n, m, are both congruent to 1 (mod 3) assuming p > 3 if n, m = 1 (mod 3): (p+n) is 2 (mod 3); (p+n)+m is 0 (mod 3) if n, m = 2 (mod 3): (p+n) is 1 (mod 3); (p+n)+m is 0 (mod 3) Any other cases other than the ones I listed is true for all even numbers.
well you said even numbers can be 1 mod 3 or 2 mod 3 they can also be 0 mod 3. so the categories n and m can be in have not been covered.

 2016-04-10, 18:45 #5 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts The case n = 2, m = 4 would work since there is a prime p such that p+2 is prime and (p+2)+4 is prime. On the other hand n = 2, m = 8 would not work since there is not a prime p (greater that 3) such that p+2 is prime and (or) (p+2)+4 is prime. (One condition may hold, but not both) To make the second condition true, one example could be that n = 2, m = 8, and there is a prime p such that: p+2 is prime and (or) ((p+2)+4)/3 is prime. (Both of these hold infinitely many times)
2016-04-10, 18:47   #6
science_man_88

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Quote:
 Originally Posted by PawnProver44 The case n = 2, m = 4 would work since there is a prime p such that p+2 is prime and (p+2)+4 is prime. On the other hand n = 2, m = 8 would not work since there is not a prime p (greater that 3) such that p+2 is prime and (or) (p+2)+4 is prime. (One condition may hold, but not both) To make the second condition true, one example could be that n = 2, m = 8, and there is a prime p such that: p+2 is prime and (or) ((p+2)+4)/3 is prime. (Both of these hold infinitely many times)
LIke Batalov told me in the other thread that can't prove infinitely often it happens.

 2016-04-10, 18:51 #7 PawnProver44     "NOT A TROLL" Mar 2016 California 3058 Posts If the twin prime conjecture is true, then so are these.
2016-04-10, 18:55   #8
science_man_88

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Quote:
 Originally Posted by PawnProver44 If the twin prime conjecture is true, then so are these.
okay. So go ahead, prove the twin prime conjecture ...

 2016-04-10, 18:58 #9 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts I don't know exactly how to prove the twin prime conjecture, but the proof will most likely be related to the proof of infinitely many primes.
2016-04-10, 19:10   #10
science_man_88

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Quote:
 Originally Posted by PawnProver44 I don't know exactly how to prove the twin prime conjecture, but the proof will most likely be related to the proof of infinitely many primes.
maybe, maybe not, there will typically be more than one way to state a thing like the twin prime conjecture we know that primes greater than 3 will be 1 or 5 (aka -1) mod 6 ( aka 6*k+1 or 6*k-1) you can restate the statement about there being infinitely many twin prime pairs in another way related to the c values as:

there are infinitely many whole numbers c not expressible as $6*a*b\pm{a}\pm{b}$ and this comes from the facts that:
(6*k+1)*(6*j+1) = 36*j*k+6*j+6*k+1 = 6*(6*k*j+k+j)+1
(6*k-1)*(6*j-1) = 36*j*k-6*k-6*j+1 = 6*(6*j*k-k-j)+1
(6*k-1)*(6*j+1) = 36*j*k+6*k-6*j-1 = 6*(6*j*k+k-j) -1

and for that last one reverse the signs so if a whole number c is of these forms either 6*c+1 or 6*c-1 is factorable and therefore not prime.

Last fiddled with by science_man_88 on 2016-04-10 at 19:10

 2016-04-10, 19:32 #11 PawnProver44     "NOT A TROLL" Mar 2016 California 110001012 Posts It is an easy proof to show that there are finitely many prime triplets (p, p+2, p+4) since 1 of 3 numbers with consecutive differences of j (j = 1 or 2 (mod 3)) are divisible by 3. It is also reasonable to show that there many be infinitely many primes p such that (not the first case however)*. p/3, p+2, p+4 are primes.* p, (p+2)/3, p+4 are primes. p, p+2, (p+4)/3 are primes. And again, this is still unproven, but conjectured. Meanwhile it is to show that if p and p+2 (twin primes) are prime, then p = 5 (mod 6), while p+2 = 1 (mod 6), so p+1 is at least divisible by 6, so to minimize this even further, there is a case of infinitely many primes p such that 6p+1, p, and 6p-1 are all primes. I think that was going on in this thread. Last fiddled with by PawnProver44 on 2016-04-10 at 19:34

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