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Old 2016-04-10, 03:16   #1
PawnProver44
 
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Post 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?
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Old 2016-04-10, 11:26   #2
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Quote:
Originally Posted by PawnProver44 View Post
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.
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Old 2016-04-10, 13:41   #3
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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
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Old 2016-04-10, 13:44   #4
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Quote:
Originally Posted by PawnProver44 View Post
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.
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Old 2016-04-10, 18:45   #5
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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)
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Old 2016-04-10, 18:47   #6
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Quote:
Originally Posted by PawnProver44 View Post
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.
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Old 2016-04-10, 18:51   #7
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If the twin prime conjecture is true, then so are these.
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Old 2016-04-10, 18:55   #8
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If the twin prime conjecture is true, then so are these.
okay. So go ahead, prove the twin prime conjecture ...
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Old 2016-04-10, 18:58   #9
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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.
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Old 2016-04-10, 19:10   #10
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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.

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Old 2016-04-10, 19:32   #11
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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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