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The Goldbach Conjecture: Every even number greater than 2 is the sum of 2 primes

Consider P(n) = (n is congruent to 3mod4)

Then g(n) ={0 if n = 1 or 2n is a sum of 2 primes

1 if 2n is not a sum of 2 primes and P(n) is true

-1 if 2n is not a sum of 2 primes and P(n) is false}

N

Next we define G{N}= ∑((g(k))/(2^k))

k=1

And now we can define the following family of rational intervals:

G={[G{N}-(1/(2^{N-1})),G{N}+(1/(2^{N-1}))], N=1,2,...}

Prove that the Goldbach Conjecture is true if and only if G = 0

2. Relevant equations

3. The attempt at a solution

I'm okay with the first half of the biconditional. If the Golbach Conjecture is true, g(n) will of course always be 0.

The other half is really giving me some trouble - If G = 0, then the Goldbach Conjecture is true. I figure it will suffice to show that if there is a J so that g(n) = 0 for ll n < J and g(J) != 0, then G{N} must b bounded away from 0 for all n>J. But I can't get there. Anybody have some suggestions?

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# Homework Help: Goldbach Number

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