Question Regarding Goldbach's Conjecture

Edwin

1*1
1*3 2*2 3*1
1*5 2*4 3*3 4*2 5*1
1*7 2*6 3*5 4*4 5*3 6*2 7*1
*
*
*
etc.

If we go on constructing the pattern of numbers above, will each row contain atleast 1 product of two odd prime numbers? If the answer to this can be proven to be yes, then would this prove also Goldbach’s conjecture considering the fact that the sum of any numbers in a row adds up to twice the square root of the perfect square of that row, and that all even numbers greater then 4 are accounted for in the matrix?

For example, in the row that contains the perfect square 2*2, we have 3+1 = 4, with 3 and 1 odd prime numbers. In the row containing the perfect square 3*3 we have 5+1 = 6, with 5 and 1 odd prime numbers…etc.

Might this be related to Goldbach’s conjecture?

Inquisitively,

Edwin G. Schasteen

shmoe

That's the same as goldbach, rather trivially. Goldbach just asks if for any integer n if one of the pairs (2,2n-2), (3,2n-3), ... (n,n) is made up of two primes (yours is actually just slightly weaker, and considers (1,2n-1) as well, which won't help goldbach). It's no easier (actually probably harder) to determine if (k)*(2n-k) is a product of two primes than it is to determine if k and 2n-k are both prime on their own (this is assuming you don't know the factors of (k)*(2n-k) on hand, if you do it's equivalent).

Office_Shredder

Edwin, 1 isn't a prime number

Edwin

Oh yea, one isn't prime, I overlooked that.

That makes sense. So then to solve Goldbach's conjecture might be a step towards solving the conjecture above, not the other way around. I take it that one would probably have to prove Goldbach's conjecture to prove the concept above. Does this seem correct?

Inquisitively,

Edwin

matt grime

No, they are equivalent. Solve one and you solve the other. The thing is that as stated your version of the problem is going to be harder to prove.

lokofer

In fact..didn't Vinogradov gave the "exact" formula explaining the way in which an Odd number could be represented as the sum of 3 primes:

[tex] I(N)= \int_{0}^{1}dx( \sum_{p<N}e^{2i p \pi x})^{3}e^{-2i \pi Nx } [/tex] of course we could use some "lower bound" for the prime number counting function to calculate the sum over primes inside the integral.

shmoe

Why the quotes around "exact"? Hardy and Littlewood used the circle method to prove 3 primes given the generalized riemann hypothesis, so no Vinogradov didn't come up with this expression. He was the first to get 3 primes unconditionally though.

If you had a paper for every "of course" you've written, you would fill volumes. I'd love to see how you hope to do this calculation given a lower bound for pi(x).

CRGreathouse

I'd like to see that too. In the interests thereof, I'll provide the sharpest lower bound I know for pi(x), due to Pierre Dusart:

[tex]\pi(x)>\frac{x}{\ln x}\left(1+\frac{1}{\ln x}+\frac{1.8}{\ln^2x}\right)[/tex]

for x >= 32299.

lokofer

Sorry..i didn't know if was vinogradov or other who gave that formula for I(N).. the formul is exact but the problem is that you need to evaluate a sum over primes of the exponential i think that :

[tex] \sum_{p<N} e^{2i p \pi x} \sim \int_{2}^{N} dt \frac{e^{2i \pi t}{log(t)} [/tex]

for big N.. in that case you could prove that any N N-->oo can be expressed as the sum of 3 primes.

By the way..could someone indicate me where could i find "circle method" approach to the Goldbach conjecture?..thanks.

matt grime

Jose, could you please write one post where you don't say 'we could' and then write something you totally fail to justify as a valid method of attack on a problem? (Not sure whatt the latex code is supposed to be).

shmoe

The left side depends on x, the right doesn't, did you want another "x" in the exponential on the right? This is going to be a terrible approximation in any case, trying to convert a sum like this into an integral doesn't really work when the summand is oscillatory, it's derivative won't be small.

hmm... http://www.physicsforums.com/showthread.php?t=127122

CRGreathouse

Actually, I think the left hand side is [tex]\pi(\lceil N-1\rceil)[/tex] by Euler's formula.

shmoe

There's an "x" in the exponent. See the I(N) integral he gave earlier. When you cube this sum, the coefficient of the term with 2*Pi*i*n*x in the exponent will give the number of ways to express n as a sum of 3 primes, the integral over x picks off the desired coefficient by orthogonality. The goal is to then show for large enough N this integral is always positive (for the ternary goldbach problem). The difficulty (one of them at least) is dealing with this sum over primes.

CRGreathouse

I know there's an x in the exponent.

It appears that the left side is the sum of one raised to the power of x * p.

shmoe

No, it's e to the power 2*Pi*i*x*p. This isn't the same thing as e^(2*Pi*i) raised to the power x*p. Branches of logarithms and all that jazz.

lokofer

Then we could try with the "best" known possible lower boun for the prime number counting function:

[tex] \pi(x) >Ag(x) [/tex] g(x) is a known function and A a real constant (of course i don't know much about the upper and lower bounds for the prime number counting function) in that case we have that:

[tex] \sum_{p<N}e^{2i \pi px} > \sum_{k=2}^{N}[g(k)-g(k-1)]e^{2i \pi kx} [/tex]

then insert the sum on the left and get I(N) of course if we had that the integral:

[tex] \int_{0}^{1}dx(\sum_{k=2}^{N}[g(k)-g(k-1)]e^{2i \pi kx})^{3} e^{-2i \pi N}=\int_{0}^{1}dx e^{-2 \pi N x}\int_{2}^{N}e^{2i \pi xt}g'(t)dt >0 [/tex]

then Goldbach conjecture would be true....if this isn't possible then we should look another way to prove it.

CRGreathouse

I rather suspect that eljose was not considering branches of logs.