My Math Forum New Goldbach Conjecture foundation

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 November 6th, 2015, 01:34 PM #1 Newbie   Joined: Apr 2015 From: Mountain View Posts: 8 Thanks: 0 New Goldbach Conjecture foundation (Cut & Paste to Notepad to get columns to line up) Goldbach Conjecture Proof (almost) 1. For all primes > 5 there exist positive 0< m<=k integers such that p1 n=3(m+k) => r=(3(k-m)-1) b. 6(m)+1 + 6(k)+1 = 6(m+k)+2 = 2(3m+3k+1) => n=3(m+k)+1 => r=3(k-m) c. 6(m)-1 + 6(k)-1 = 6(m+k)-2 = 2(3m+3k-1) => n=3(m+k)-1 => r=3(k-m) d. 6(m)-1 + 6(k)+1 = 6(m+k) = 2(3m+3k) => n=3(m+k) => r=(3(k-m)-1) e. 6(k)+1 - 2(r) = p1 (Can be used where p1=3) f. 6(k)-1 - 2(r) = p1 (Can be used where p1=3) 3. Except for 2, any two primes add to an even integer. Note: p1+p2 add to an even integer in all 4 equations. 4. 2(n) can express any even integer >= 10 and "n" can express any integer >= 5 looking at all 4 equations 5. From Bertrand's Postulate we know at least one p2 exists in n < p2 < 2n (I suspect many for large n) 6. From the Prime Number Theorem we know at least n/log(n) p1 primes exist in 0 < p1 < n. 7. Chose any even number >= 10. Chose m and k such that a.,b.,c. or d. is satisfied above. (a. and d. may both work) 8. Now choose the prime numbers that match the row(s). Goldbach is satisfied. 9. To disprove this, simply find "n" such that a.,b.,c. or d. cannot produce prime numbers, i.e. p1+p2 != 2(n) Note: Integers 4, 6, 8 can be proved by hand. 10. If this does not lead to a proof, it may at least lay the groundwork for a proof since it deals with all the elements of Goldbach including multiple partitions. For Example: 2(n) = 22, so n=11 (prime) Already solved, but let's find more. a./d. 6(m+k) = 2(3m+3k) = 2(n) => n=11 => 11=3m+3k => 11/3 = m+k (Impossible - Integers can't add to fraction ) b. 6(m+k)+2 = 2(3m+3k+1) = 2(n) => n=11 => 11=3m+3k+1 => 10/3 = m+k (Impossible) c. 6(m+k)-2 = 2(3m+3k-1) = 2(n) => n=11 => 11=3m+3k-1 => 12 = 3(m+k) => 4=m+k => m=1,k=3;m=2,k=2 22 22 p1=5, p2=17, r=6 p1=11, p2=11, r=0 p1=3, p2=19, r=8 works for p2 of b.& c. but otherwise fails. But 19 - 2(r) = 3 2(n) = 18, so n=9 (odd but not prime) a./d. 6(m+k) = 2(3m+3k) = 2(n) => n=9 => 9=3m+3k => 3=m+k => m=1,k=2 b. 6(m+k)+2 = 2(3m+3k+1) = 2(n) => n=9 => 9=3m+3k+1 => 8/3 = m+k (Impossible) c. 6(m+k)-2 = 2(3m+3k-1) = 2(n) => n=9 => 9=3m+3k-1 => 10/3 = m+k (Impossible) 18 18 p1=7,p2=11,r=2 p1=5,p2=13,r=4
 November 12th, 2015, 07:11 AM #2 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 "5. From Bertrand's Postulate we know at least one p2 exists in n < p2 < 2n (I suspect many for large n)" I don't think (n, 2n) suffices. Seems you need a smaller interval. {I worked that out many, many years ago but I could have been wrong then - which of course means I could be wrong now.} _________ "I know just enough math to be dangerous." (Anne N. Onymus)
November 12th, 2015, 07:34 AM   #3
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Quote:
 Originally Posted by rlrandallx 7. Chose any even number >= 10. Chose m and k such that a.,b.,c. or d. is satisfied above. (a. and d. may both work)
OK... but a, b, c, and d are all true for all choices of m and k.

Quote:
 Originally Posted by rlrandallx 8. Now choose the prime numbers that match the row(s). Goldbach is satisfied.
What makes you think that you can do this? It seems like you tried to prove the Goldbach conjecture by assuming the Goldbach conjecture.

 November 13th, 2015, 08:50 AM #4 Newbie   Joined: Apr 2015 From: Mountain View Posts: 8 Thanks: 0 Bertrand's postulate, also called the Bertrand-Chebyshev theorem or Chebyshev's theorem, states that if n>3, there is always at least one prime p between n and 2n-2. Equivalently, if n>1, then there is always at least one prime p such that n
 November 13th, 2015, 09:04 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Yes, I think we're all familiar with that -- what is your point?
 November 13th, 2015, 09:11 AM #6 Newbie   Joined: Apr 2015 From: Mountain View Posts: 8 Thanks: 0 For a. n=3(m+k) => 3,6,9,12,15, ... set A b. n=3(m+k)+1 => 4,7,10,13,16 ... set B c. n=3(m+k) -1 => 5,8,11,14,17, ... set C d. n=3(m+k) => same as a. n can only be in set A or D, B, or C so m and k are considered at most twice. If 2n=32 and n=16, p2 = 25 if m=1 and k=4 but then p2 !=prime so you must select another m,k pair so that n=16, e.g. m=2,k=3. Thus p1=13, p2=19. GC satisfied.
 November 13th, 2015, 09:21 AM #7 Senior Member   Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 Yes, I've confused myself with this problem many times and finally concluded that if 'at least one prime in (n, 2n)' sufficed then the the Conjecture would have been long-ago proved.
 November 13th, 2015, 10:00 AM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms There seem to be communication difficulties. Would you walk me through an example? Let's say I chose 2n = 3325581707333960528.
 November 14th, 2015, 04:09 AM #9 Member   Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 There is a big fundamental error in this proof. Yes, you can write an even number in one or more of those forms. Yes, all primes > 5 can be written as 6k - 1 or 6k + 1. You have to remember 6k - 1 and 6k + 1 aren't always prime. Just because there exists a prime between n and 2n, all that means is you can find one prime of the form 6k - 1 or 6k + 1. It doesn't mean that the corresponding number (or second term in the equation) will be prime. You basically assumed what you were trying to prove. I'm not sure how clear that sounded, but I tried my best.

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