20160410, 03:16  #1 
"NOT A TROLL"
Mar 2016
California
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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? 
20160410, 11:26  #2  
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Jul 2009
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20160410, 13:41  #3 
"NOT A TROLL"
Mar 2016
California
197 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 20160410 at 13:41 
20160410, 13:44  #4  
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20160410, 18:45  #5 
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Mar 2016
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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) 
20160410, 18:47  #6  
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20160410, 18:51  #7 
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Mar 2016
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If the twin prime conjecture is true, then so are these.

20160410, 18:55  #8  
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20160410, 18:58  #9 
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Mar 2016
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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.

20160410, 19:10  #10  
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there are infinitely many whole numbers c not expressible as 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*k1)*(6*j1) = 36*j*k6*k6*j+1 = 6*(6*j*kkj)+1 (6*k1)*(6*j+1) = 36*j*k+6*k6*j1 = 6*(6*j*k+kj) 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*c1 is factorable and therefore not prime. Last fiddled with by science_man_88 on 20160410 at 19:10 

20160410, 19:32  #11 
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Mar 2016
California
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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 6p1 are all primes. I think that was going on in this thread. Last fiddled with by PawnProver44 on 20160410 at 19:34 
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