mersenneforum.org Goldbach's Conjecture
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 2007-03-22, 10:59 #1 Patrick123     Jan 2006 JHB, South Africa 157 Posts Goldbach's Conjecture Hi all, At the risk of getting the , would this constitute as a proof of the Goldbach conjecture?? Code: It can be shown that all primes greater than 3 have the form 6n+1 or 6n-1 where n is any positive integer > 1. It can be shown that any even integer of 10 and greater has one of three forms, namely 6n-2, 6n and 6n+2 The sum of two primes > 3 will have one of 3 forms. (6a-1) + (6b-1) = 6(a+b) -2 (6a-1) + (6b+1) = 6(a+b) or (6a+1) + (6b-1) = 6(a+b) (6a+1) + (6b+1) = 6(a+b) +2 Letting a and b be any number > 0, it can be shown that all even numbers > 10 are the sum of two primes. It can be shown manually for the even numbers 4,6 and 8 as the sum of their primes include the primes 2 and 3. Regards Patrick
 2007-03-22, 11:40 #2 Wacky     Jun 2003 The Texas Hill Country 32×112 Posts Patrick, Although your "proof" deserves the "crank" label, I will give you the benefit of doubt and simply point out that you have made a typical error in logic. By your first axiom, IF p is prime, then p is of the form …. You then claim IF n is of the form …, then n is prime. http://en.wikipedia.org/wiki/Affirming_the_consequent Last fiddled with by Wacky on 2007-03-22 at 11:42
 2007-03-22, 11:59 #3 Patrick123     Jan 2006 JHB, South Africa 157 Posts Thanks Wacky, I had a block but just could not see where, at this stage all I've shown is that all even numbers can be split in the form (6a +or- 1) + (6b +or- 1). I've not shown that they are prime. Self-inflicted Regards Patrick
 2007-04-13, 05:35 #4 David John Hill Jr     Jun 2003 Pa.,U.S.A. 22·72 Posts A little more on Goldbach Greetings to Johannesburg. I come from a family (one branch)of american coal miners,and happen to have attended nursery school in saxonwald, before retreating to east london as my father was in the nylon business: I am unfortunately a heavy coffee drinker, but have managed to keep the yellowing hands down to a minimum . I want to point out something I am convinced is a must for a proof of Goldbach, that is not generally accepted: It must be proven by parts, a) for those evens divisible by 2 once b) for those evens divisible by 4 or more This divides all evens into exactly two seperate types, which behave totally differently, so as to try for a proof without seperation, is totally ludicrous in my opinion. I can go into more detail if you wish. Its been a long time since I witnessed the mine dances, by the way.
2007-04-13, 17:01   #5

"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 Posts

Quote:
 Originally Posted by Patrick123 Code: It can be shown that all primes greater than 3 have the form 6n+1 or 6n-1 where n is any positive integer > 1.
So the primes greater than 3 are congruent to either +1 or -1 mod 6. Okay.

Quote:
 Code: It can be shown that any even integer of 10 and greater has one of three forms, namely 6n-2, 6n and 6n+2
So the even integers, 10 or greater, are congruent to either -2, 0, or +2 mod 6. Okay.

Quote:
 Code: The sum of two primes > 3 will have one of 3 forms. (6a-1) + (6b-1) = 6(a+b) -2 (6a-1) + (6b+1) = 6(a+b) or (6a+1) + (6b-1) = 6(a+b) (6a+1) + (6b+1) = 6(a+b) +2
So the sum of two primes, each greater than 3, is congruent to either -2, 0, or +2 mod 6. Okay.

Quote:
 Code: Letting a and b be any number > 0, it can be shown that all even numbers > 10 are the sum of two primes.
Not okay.

Just because two sets of numbers (the set of even integers each 10 or greater, and the set of sums of two primes each greater than 3, in this case) map to the same set of equivalence classes doesn't mean that the sets of numbers are equal ... or that they have a single member in common!

Now, by examining specific cases, you can show that the sets share at least one member (e.g., 10 is in both sets). But showing that the sets are equal (i.e., that the sets share all of their members) is an (*ahem*) infinitely greater task.

Last fiddled with by cheesehead on 2007-04-13 at 17:04

 2010-07-16, 15:19 #6 science_man_88     "Forget I exist" Jul 2009 Dartmouth NS 2·3·23·61 Posts I'm going to add my 2 cents (and yes I know it won't necessarily be counted in the dollars given). number freak talks of something similar and then says "it remains unsolved as of this writing" (about Goldbach's conjecture) could we use a similar method to the one on pg. 124 (with one major alteration) to help Goldbach's conjecture ?
 2010-07-16, 17:20 #7 science_man_88     "Forget I exist" Jul 2009 Dartmouth NS 2·3·23·61 Posts .5x±n= prime for a certain n and a even x is basically what Goldbach's conjecture says if i understand it right.
 2010-07-16, 17:29 #8 3.14159     May 2010 Prime hunting commission. 24×3×5×7 Posts Goldbach's conjecture states that any even number that is greater than 2 can be decomposed into the sum of two prime numbers. Ex: 100 = 53 + 47 In most cases, there is more than 1 decomposition into prime numbers. Last fiddled with by 3.14159 on 2010-07-16 at 17:33
2010-07-16, 18:24   #9
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by 3.14159 Goldbach's conjecture states that any even number that is greater than 2 can be decomposed into the sum of two prime numbers. Ex: 100 = 53 + 47 In most cases, there is more than 1 decomposition into prime numbers.
I realize that but .5x±n= prime just means that they have to be the same distance away from the mean of all the 2 number sums to a number which is obvious.

2010-07-16, 23:11   #10
3.14159

May 2010
Prime hunting commission.

24×3×5×7 Posts

0.5x ± n = p? If you're referring to the Goldbach conjecture, it would be: p1 + p2 = 2n.

If I didn't understand your statement, apologies.

Quote:
 .5x±n= prime for a certain n and a even x is basically what Goldbach's conjecture says if i understand it right.
How did you deduce that?

Last fiddled with by 3.14159 on 2010-07-16 at 23:13

2010-07-17, 00:37   #11
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by 3.14159 0.5x ± n = p? If you're referring to the Goldbach conjecture, it would be: p1 + p2 = 2n. If I didn't understand your statement, apologies.
don't worry i'm used to nobody knowing what i say lol. even my family doesn't understand me some days.

Quote:
 Originally Posted by 3.14159 How did you deduce that?
my equation is logical as any 2 numbers that add to a third must have and average of half that third number. this in turn means to average correctly to that .5x that they are equally displaced (hence the +/-n) . so in my mind my equation is what would have to be solved for all even x and a given n (which seems to be 0 or an odd number)

Last fiddled with by science_man_88 on 2010-07-17 at 00:38

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