Question

Let's solve the Goldbach Conjecture (Part 28, probably)

Answer 1

I call the 'Goldbach partition function' \(G(n)\) to be the number of ways you can represent an even number as the sum of two primes:

\[G(8) = 1\] since 3+5=8 is the only way to represent it as the sum of primes.
\[G(10)=2\] since it can be written in two ways, 3+7=5+5=10.

If \[G(2n) >1\] for all n, then that's the same thing as the Goldbach conjecture being true.

This sounds cute, but so what? Well what I found was the generating function for G(n):

\[\sum_{n=3}^\infty G(2n)x^{2n} = \left( \sum_{n=2}^\infty x^{p_n} \right)^2\]

Answer 2

The most general way I could have written the generating function is this way:
\[\sum_{n=2}^\infty G(n)x^{n} = \left( \sum_{n=1}^\infty x^{p_n} \right)^2\]

That tells you G(n) for all numbers how many ways you can uniquely write it as a sum of two primes, we could evaluate G(5)=1 with that if we were interested in odd numbers, since 2+3=5. However since Odd+Odd=Even and Even+Even=Even and I only want to look at even numbers, and since there's only one way to make Even+Even=Even with primes, I decided to just completely remove \(p_1=2\), and subsequently G(4) along with all the G(2n+1) terms that are unnecessary to the Goldbach Conjecture.

Ok, I realize I never actually calculated anything, so I'll calculate some of it and discover a slight problem (I'm just double counting, which really doesn't affect the GC if that's what we want to prove with this anyways, but it can be fixed pretty easily):

\[\sum_{n=2}^\infty x^{p_n} = x^3+x^5+x^7+x^{11}+ \cdots\]

\[\left( \sum_{n=2}^\infty x^{p_n}\right) = ( x^3+x^5+x^7+ \cdots ) ( x^3+x^5+x^7+ \cdots )\]
(Technically I wrote the sum wrong so I should write \(x^{3+5}+x^{5+3}=2x^8\) and I am double counting so I will officially fix it below, for now I'll just omit the double terms)
\[= x^{3+3}+x^{3+5}+x^{3+7}+x^{5+5}+x^{5+7}+x^{7+7}+ \cdots \]
\[=x^6+x^8+2x^{10}+x^{12}+ \cdots\]
\[=G(6)x^6+G(8)x^8+G(10)x^{10}+G(12)x^{12}+ \cdots\]

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Alright here's the revised CORRECT version without double counting: (Although the other one is sufficient for trying to prove the GC since it gives \(G(n) >1\) anyways)

\[\sum_{n=3}^\infty G(2n)x^{2n} = \frac{1}{2} \left( \sum_{n=2}^\infty x^{p_n} \right)^2 + \frac{1}{2}\sum_{n=2}^\infty x^{2p_n} \]

Answer 3

Also the 2n is just a matter of convenience since all the \[G(2n+1)x^{2n+1}=0\], I could just as well write this:
\[\sum_{n=3}^\infty G(2n)x^{2n} = \frac{1}{2} \left( \sum_{n=2}^\infty x^{p_n} \right)^2 + \frac{1}{2}\sum_{n=2}^\infty x^{2p_n} \]
as this:

\[\sum_{n=3}^\infty G( n)x^{n} = \frac{1}{2} \left( \sum_{n=2}^\infty x^{p_n} \right)^2 + \frac{1}{2}\sum_{n=2}^\infty x^{2p_n} \]

To make it into a simpler looking generating function... Anyways, not sure if this helps us in actually solving it, but now that primes are in exponents I was thinking maybe Fermat's little theorem or some substitution for x would let us do something nice, or maybe there is some kind of closed form we can find (highly doubtful lol).

Answer 4

wow i need time to read this :O X_X