
#19
Nov1711, 01:20 PM

P: 3

I can prove the goldbach conjecture with a simple method
but first I prove square of any prime 1 can be divided by 12 primexprime 1 = 12k k is an integer . 



#20
Nov1711, 09:49 PM

PF Gold
P: 161

Let p be a prime number. p[itex]^{2}[/itex]1 = (p+1)(p1) Since p is prime, it is not divisible by 2 (unless p is 2 but 2[itex]^{2}[/itex]1=3 is not 12k). So p = 2n+1 for some positive integer n. Any integer number can be written in the form 3k, 3k+1 or 3k+2. Try n = 3k. Then p = 6k+1 Try n = 3k+1. Then p = 6k+3 = 3(2k+1) = 3m is impossible since p is prime. (unless k = 0 then p = 3 is prime, but 3[itex]^{2}[/itex]1 = 8 is not 12k) Try n = 3k+2. Then p = 6k+5. We see that p must be of the form 6k+1 or 6k+5. Then (p+1)(p1) = (6k+2)(6k) = 12(3k+1)k = 12m or (p+1)(p1) = (6k+6)(6k+4) = 12(k+1)(3k+2) = 12m Therefore, for any prime number p greater than 3, p[itex]^{2}[/itex]1 is divisible by 12. 



#21
Nov2011, 10:52 AM

P: 94

The complication of all of this: is proving there is no space between primes through infinity: such that for the binary part, the last prime before the spacing is such that it added to itself ( or times 2): is greater than or equal to the spacing been it and the next prime.
Of course this comes down to prime numbers distribution, and the maximum spacing between primes. 



#22
Nov2011, 06:31 PM

P: 69

FYI, someone submitted a proof of Goldbach Conjecture and Riemann's Hypothesis in Arxiv. It's under General Mathematics section.




#23
Nov2011, 06:45 PM

P: 94

ode_to_joy
got a link? . . 



#24
Nov2111, 04:54 AM

P: 124

http://arxiv.org/ftp/math/papers/0005/0005185.pdf
Is this the paper you were talking about ode_to_joy? 



#25
Nov2111, 08:05 PM

P: 69

http://arxiv.org/abs/1110.3465
goldbach conjecture http://arxiv.org/abs/1110.2952 riemann hypothesis 



#26
Nov2111, 09:36 PM

Mentor
P: 16,690





#27
Nov2211, 12:43 PM

Sci Advisor
PF Gold
P: 4,862





#28
Nov2211, 10:45 PM

Sci Advisor
PF Gold
P: 4,862

According to:
http://mathworld.wolfram.com/GoldbachConjecture.html what Vinogradov proved for sufficiently large odd n was the ternary form of odd numbers. Restricted to odd numbers only, the ternary does not imply the binary. However, it is correct, as I argued above, that the ternary for all numbers implies the binary; and the binary trivially implies the ternary. 



#29
Nov2311, 05:27 AM

P: 48

the proof of the riemann hypothesis is pretty interesting, although at first glance I'm not sure about one of his derivatives, and there seems to be ALOT of assumptions...




#30
Nov2311, 06:10 AM

Mentor
P: 16,690

From the first glance, the proof of the Riemann hypothesis suffers from the same problem. He checks it for the first 23 trillion values and he proves it for sufficiently large numbers. That is not enough...




#31
Dec2611, 02:30 PM

P: 15

I have a question concerning Goldbach's Conjecture. First, it is easy to demonstrate that the sum of any two odd integers will always be even.
For example, let an odd integer q=2k+1 and an odd integer p=2m+1 It then follows that q+p= 2(k+m+1) = 2n which is even. Now, it is true that any prime number >2 is odd. So, couldn't you simply use this fact to prove a substantial amount of Goldbach's conjecture? I don't understand people who sit down and add two arbitrary primes >2 to find an even number as the sum thinking that they may find an exception to the conjecture, but they won't because of what I just proved. I know that prime numbers don't have a neat little general form like 2k+1 or 2n, but they are a subgroup of the odd numbers. All prime numbers >2 are odd but not all odd numbers are prime. Primes >2 are odd simply by definition of a prime number. Then after you prove that the only time you will get an even number as a sum by using the prime number 2 is when you add 2 to itself. I'm probably missing some subtle logical step, and if so, please enlighten me. In conclusion, wouldn't it be simpler to focus on parity instead of the primality of numbers to prove the conjecture? 



#32
Dec2611, 06:22 PM

P: 688

Hi, fibonacci235,
there are many ways of adding two primes and obtaining an even number. The point of the conjecture is, can *all* even numbers (> 2) be written as the sum of two primes? In other words, given *any* arbitrary even number greater than 2, say, 123456, why is it mandatory that there exists a prime "p" such that 123456p is also prime? That's the problem. 



#33
Dec2711, 08:11 PM

P: 15





#34
May2412, 03:04 AM

P: 1

Goldbach Conjecture is 2n = Prime (a+n)+ Prime (an), 1 is here prime
2 = (0+1)+(10) 4 = (1+2)+(21) 6 = (2+3)+(32) 8 = (3+4)+(43) 10= (2+5)+(52) 12= (1+6)+(61) 14 =(4+7)+(74) 16 =(3+8)+(83) 18 =(4+9)+(94) 20=(3+10)+(103) 22=(6+11)+(116) . . . 2n=(a+n)+(na) Proof: (a+n) = 2n+(an)=2n(na) q.e.d. 



#35
May2412, 06:40 AM

Sci Advisor
PF Gold
P: 4,862

Forget proof, you don't even posit any argument at all that an 'a' with right properties exists. Listing that it does for N cases is irrelevant, unless you list all infinity cases. Certainly, there is no induction in your (total non) argument.




#36
May2412, 08:09 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,895

Proof: (a+n) = 2n+(an)=2n(na) q.e.d.[/QUOTE] So, essentially, you are telling us that you do not know what a "proof" is. 


Register to reply 
Related Discussions  
Goldbach’s Conjecture and the 2Way Sieve  General Math  6  
euler product and Goldbach conjecture..  Linear & Abstract Algebra  1  
Are Riemann hypothesis and Goldbach conjecture related?...  Linear & Abstract Algebra  2  
a partial solution to the Goldbach Conjecture  Linear & Abstract Algebra  7  
Goldbach conjecture  Linear & Abstract Algebra  20 