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Reformulation of Goldbach's conjecture

Upon the suggestion of another stackexchange user this question has been reformulated to address the comments Useful reformulation of Goldbach's conjecture?

If $J(x)$ is the form of only $B e^{2 \lambda x}$ where B is a positive coeffient and satisfies:

$$ \sqrt{\frac{e^{6x}}{1-e^{2x}} + J(x)} - P(-1)\int_{-\infty}^{x} \sqrt{\frac{e^{6x'}}{1-e^{2x'}} + J(x')} dx' $$ $$+ (\frac{P(-1)^2 - P(-2)}{2})\int_{-\infty}^{x} (\int_{-\infty}^{x'} \sqrt{\frac{e^{6x''}}{1-e^{2x''}} + J(x'')} dx'') dx' + \dots = 0$$

Where $P(k)$ is the prime zeta function and the boundary conditions:

$$ \lim_{x \to - \infty} e^{-4x} J(x) = 0 $$

And the function and any of it's integrals $\sqrt{\frac{e^{6x}}{1-e^{2x}} + J(x)} $ must vanish at $- \infty$

Then, Goldbach's conjecture is true.

Reasoning

Let us consider the following infinite order integral equations where $p_i$ is the $i$'th prime:

$$ y(x,s) - (\sum_{i=1}^\infty p_i^s)\int_{-\infty}^{x} y(x,s) dx + (\sum_{i<j}(p_i p_j)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} y(x'',s) dx'') dx' $$ $$+ (\sum_{i<j<k}(p_i p_j p_k)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} (\int_{-\infty}^{x''} y(x''',s) dx''') dx'' )dx' + \dots = 0 $$

To solve the above equation. We use a Fourier inverse transform in the variable $x$:

$$ y(x,s)= \int_{- \infty}^\infty \hat{y}(\omega(s)) e^{2 \pi i x \omega(s)} d \omega $$

Substituting the above in the first equation:

$$ y(x,s) - (\sum_{i=1}^\infty p_i^s)\int_{-\infty}^{x} \int_{- \infty}^\infty \hat{y}(\omega(s)) e^{2 \pi i x \omega(s)} d \omega dx + (\sum_{i<j}(p_i p_j)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} \int_{- \infty}^\infty \hat{y}(\omega(s)) e^{2 \pi i x'' \omega(s)} d \omega dx'') dx' $$ $$+ (\sum_{i<j<k}(p_i p_j p_k)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} (\int_{-\infty}^{x''} \int_{- \infty}^\infty \hat{y}(\omega(s)) e^{2 \pi i x''' \omega(s)} d \omega dx''') dx'' )dx' + \dots = 0 $$

Changing the order of integration and integrating:

$$ \implies \int_{- \infty}^\infty ((1 - \frac{(\sum p_i^s)}{(2 \pi i \omega)} + \frac{\sum (p_i p_j)^s}{(2 \pi i \omega)^2} - \frac{\sum (p_i p_j p_k)^s}{(2 \pi i \omega)^3} + \dots) \hat{y}(\omega(s)))e^{2 \pi i x \omega(s)} d \omega = 0 $$

Factorizing the above equation we get:

$$ \implies \int_{- \infty}^\infty (1 - \frac{2^s}{(2 \pi i \omega)})(1 - \frac{3^s}{(2 \pi i \omega)})(1 - \frac{5^s}{(2 \pi i \omega)}) \dots \hat{y}(\omega(s))e^{2 \pi i x \omega(s)} d \omega = 0 $$

Note: one can try expanding the first few terms and see why the above factorization holds.

Writing as an inverse Fourier transform:

$$ \implies F^{-1} [(1 - \frac{2^s}{(2 \pi i \omega)})(1 - \frac{3^s}{(2 \pi i \omega)})(1 - \frac{5^s}{(2 \pi i \omega)}) \dots \hat{y}(\omega(s))] = 0 $$

Taking Fourier transform both sides:

$$ (1 - \frac{2^s}{(2 \pi i \omega)})(1 - \frac{3^s}{(2 \pi i \omega)})(1 - \frac{5^s}{(2 \pi i \omega)}) \dots \hat{y}(\omega(s)) = 0 $$

Hence,

$$ y= A_1 e^{2^s x} + A_2 e^{3^s x} + \dots $$

As, if $y_1$ and $y_2$ is a solution then so will $y_1 + y_2 $ be a solution.

Special Cases

Hence, we can conclude for:

$$ y(x,s) - (\sum_{i=1}^\infty p_i^s)\int_{-\infty}^{x} y(x,s) dx + (\sum_{i<j}(p_i p_j)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} y(x'',s) dx'') dx' $$ $$+ (\sum_{i<j<k}(p_i p_j p_k)^s)\int_{-\infty}^{x} (\int_{-\infty}^{x'} (\int_{-\infty}^{x''} y(x''',s) dx''') dx'' )dx' + \dots = 0 $$

The solution is given by:

$$ y= A_1 e^{2^s x} + A_2 e^{3^s x} + \dots $$

We "analytically continue" this integral equation for positive values of $s$ as well with the help of the prime zeta function $P(s')$.

$$ P(-s) = 2^s + 3^s + 5^s + \dots = \sum_{i=1}^\infty p_i^s $$

$$ \frac{P(-s)^2 - P(-2s)}{2}= \sum_{i<j}(p_i p_j)^s $$

Now, consider the below equation

$$ \sqrt{\frac{e^{6x}}{1-e^{2x}} + J(x)} - P(-1)\int_{-\infty}^{x} \sqrt{\frac{e^{6x'}}{1-e^{2x'}} + J(x')} dx' $$ $$+ (\frac{P(-1)^2 - P(-2)}{2})\int_{-\infty}^{x} (\int_{-\infty}^{x'} \sqrt{\frac{e^{6x''}}{1-e^{2x''}} + J(x'')} dx'') dx' + \dots = 0$$

We know for $s=1$:

$$ y= A_1 e^{2 x} + A_2 e^{3 x} + \dots $$

Hence, $J(x)$ must satisfy:

$$ J(x) = y^2 - \frac{e^{6x}}{1-e^{2x}}$$

Now, if we add the additional boundary condition:

$$ \lim_{x \to - \infty} e^{-4x} J(x) = 0 $$

Hence, this implies:

$$ y= A_2 e^{3 x} + A_3 e^{5 x} + \dots $$ $$ \implies y^2 = A_2^2 e^{6x} + 2A_3 A_2 e^{8x} + \dots $$

Hence, Goldbach's conjecture has been reduced to proving there exists a solution for $J(x)$ with only positive coefficients with in the basis $e^{\lambda x}$.

Questions

Can I swap the order of integration? Wikipedia says " Under suitable conditions, ... ( one may express $y$ as)":

$$ y(x,s)= \int_{- \infty}^\infty \hat{y}(\omega(s)) e^{2 \pi i x \omega(s)} d \omega $$

What are these suitable conditions?

Is there any general pattern for the coefficients of the integral $P(-1)$, $\frac{P(-1)^2 - P(-2)}{2})$, etc?

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