Question
I think I managed to reformulate a stronger version of Goldbach's conjecture as an optimization problem:
$$ \frac{\partial F_n}{\partial a_n} = \frac{\partial F_n}{\partial \overline a_n} = \frac{\partial F_n }{\partial \lambda_n}=0$$
Where:
$$ F_n = |a_n|^2 - \lambda_n \frac{\sin(\pi |b_n|^2)}{|b_n|^2} $$
$$ (\sum_{r=1}^\infty |a_r|^2 x^{2r+1})^2 = \sum_{r=1}^\infty |b_r|^2 x^{2r+4}$$
Is this formulation correct? Is this stronger version convincing? Does it have any interesting implications?
Background
We are all familiar with Goldbach's Conjecture any even number greater then $2$ is expressible as the sum of primes. Expressing it combinatorically (without the case of $p_1$):
$$ (\sum_{r=2}^\infty x^{p_r})^2 = \sum_{r=1}^\infty b_r x^{2r+4} $$
where, $p_r$ is the r'th prime and $ b_r \neq 0 $.
I recently thought that a stronger version as a limit of $n \to \infty$:
"The minimum number of elements in a set required to construct the even numbers $<n$ as a sum of it's elements is the primes numbers"
Expressing the above as equations:
$$ \min(|a_r|^2) = \begin{cases} 1 \hfill & 2r+1=\text{prime number} \\ 0 \hfill & \text{ otherwise} \\ \end{cases} $$
Under the constraints:
$$ (\sum_{r=1}^\infty |a_r|^2 x^{2r+1})^2 = \sum_{r=1}^\infty |b_r|^2 x^{2r+4}$$
and, $b_r$ must be an a natural number
$$ \frac{\sin(\pi |b_n|^2)}{|b_n|^2} = 0 $$
For example
By comparing coefficients of $x$:
$|a_1|^4=|b_1|^2$
Using Lagrange multipliers:
$$ F_1= |a_1|^2 - \lambda_1 \frac{\sin(\pi |a_1|^4)}{|a_1|^4} = a_1 \overline a_1 - \lambda_1 \frac{\sin(\pi (a_1 \overline a_1)^2)}{(a_1 \overline a_1)^2} $$
Where $ \overline a_n $ is the complex conjugate of $a_n$.
Now, finding the minimum
$$ \frac{\partial F_1}{\partial a_1} = \frac{\partial F_1}{\partial \overline a_1} = \frac{\partial F_1 }{\partial \lambda_1}=0$$
This gives us the following equations: $$ \overline a_1 - \lambda_1 ( 2 \pi \frac{\cos(\pi( a_1 \overline a_1)^2)}{ a_1} - 2 \frac{\sin(\pi (a_1 \overline a_1)^2)}{(a_1^2) (\overline a_1)^3} ) = 0 $$
$$ a_1 - \lambda_1 ( 2 \pi \frac{\cos(\pi( a_1 \overline a_1)^2)}{ \overline a_1} - 2 \frac{\sin(\pi (a_1 \overline a_1)^2)}{(\overline a_1^2) ( a_1)^3} ) = 0 $$ $$ \frac{\sin(\pi (a_1 \overline a_1)^2)}{(a_1 \overline a_1)^2} = 0 $$
Solving the above:
$ |a_1|^2=1$ and $ \lambda_1 = -\frac{1}{2 \pi} $