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I am a little nervous writing this, because it is far from my field. But it was my understanding that there is a much more basic problem in proving the binary Goldbach conjecture by analytic methods.

Let $P$ be any set of positive integers and let $f_P(x) = \sum_{p \in P} x^p$. Then the behavior of $f_P$ near $1$ depends on the rate of growth of $\pi_P(n) := \# \{ k \in P: \ k \leq n \}$. Similarly, the behavior of $f_P$ near $e^{2 \pi i \ell/m}$ depends on the rates of growth of the functions $\pi_P(n: r,m) := \# \{ k \in P : \ k\leq n,\ k \equiv r \mod m \}$. The cricle method is all about making this dependence precise, and about approximating $f_P(e^{i \theta})$ by $f_P(e^{2 \pi i \ell/m})$ for some rational approximation $\ell/m$ to $\theta/(2 \pi)$.

However, there is a set of integers $P$ for which all the $\pi_{P}(n: r,m)$ have the same growth rates as for the primes, yet binary Goldbach is false for $P$. So, rather than improving the details of the circle method, one is left with a fundamentally new problem: Thinking of new properties of the primes to make use of.

So, what is $P$? I'll first do it with $\pi_P(n)$ having the correct growth, then I'll add in the more refined $\pi_P(n, r,m)$'s.

Divide $\mathbb{Z}_{\geq 0}$ up into intervals $[a_0, a_1) \sqcup [a_1, a_2) \sqcup \cdots$ where $a_0=1$ and $a_{k+1} = a_k + \sqrt{a_k} + O(1)$ where the $O(1)$ is some absolute cnstant bound for how much error we will permit (I think $2$ would work). Choose a subsequence $b_j$ of the $a_i$ such that $b_1=a_1$ and $b_{i+1} \geq 2 b_i$. We will make sure that none of the $b_i$ are in $P+P$.

Don't put any elements of $[a_0, a_1)$ into $P$. That ensures that $b_1=a_1$ is not in $P+P$. We now inductively construct $[a_k, a_{k+1}) \cap P$. Suppose that we have already constructed the part of $P$ in $\bigcup_{i < k} [a_i, a_{i+1})$, and that $P+P$ does not contain any $b_j$. Let $b_j$ be the unique $b$ such that $b_{j-1} \leq a_k < a_{k+1} \leq b_j$. When adding elements of $[a_k, a_{k+1})$ to $P$, we can't make a sum of the form $b_{j'}$ with $j' < j$, because $b_{j'}< a_k$. And we can't make a sum $b_{j''}$ with $j' > j$, because $b_{j''} > 2 b_j \geq a_{k+1} + a_{k+1}$. So our only concern is that we might create the sum $b_j$.

Choose $\sqrt{a_k}/\log a_k+O(1)$ elements of $[a_k, a_{k+1})$ to put into $P$, subject to the sole condition that we don't make $b_j$ land in $P+P$. Again, the $O(1)$ is a once and for all global choice. In order to avoid creating this sum, we must avoid putting in $b_j - p$, where $p$ is an element of $P$ lying in an interval of length $[\sqrt{a_k}]$. The size of $P \cap [x, x+L]$ is $\approx L/\log(x+L) \leq L/\log L$. So we have to avoid $\sqrt{a_k}/\log \sqrt{a_k} = 2 \sqrt{a_k}/\log a_k$ things, while choosing $\sqrt{a_k}/\log a_k$ things, all from an interval of length $\sqrt{a_k}$. Since $\log a_k$ is eventually bigger than $3$, this is not a problem.

Now, let's do this while, at the same time, making sure that the $\pi_P(n, r,m)$ grow correctly. This time, make sure to take the $b_j$ to be even, since you won't be impressed if they are odd. The $O(1)$ in the defining recursion for the $a$'s gives me enough room to make sure lots of $a$'s are even, so that's fine. This time, we are going to require that, in addition, for every $m$, we have $(\sqrt{a_k}/\log a_k)/\phi(m!)+O(1)$ elements of $P \cap [a_k, a_{k+1})$ in each invertible residue class modulo $m!$. Note that, for $m$ large, the $O(1)$ swamps the first term, so this condition is trivial. So for each $k$, this is in fact finitely many conditions to obey.

And, since $\log a_k$ does eventually get larger than $3 \phi(m!)$, this should be doable.

There is a lot of unchecked material here; I welcome corrections and citations.