TITLE: The Circle Method and the binary Goldbach Problem
QUESTION [10 upvotes]: I've been learning about the Circle Method (at the level of the book 
"An Invitation to Modern Number Theory," by Miller and Takloo-Bighash). The arguments in the book show how the 
Circle method can be applied successfully to the ternary Goldbach problem, but fails for the binary problem (since the minor arc 
contribution can't be bounded sufficiently well). Is this the final word on the Circle method's application to this problem? Is there any hope 
that more sophisticated arguments/estimates might make it work? 
I have seen some numerical experiments that seem to indicate that the minor arc contribution is small - for example, see the undergraduate project at 
http://web.williams.edu/go/math/sjmiller/public_html/mathlab/public_html/jr02spring/goldbach/goldbach5.pdf .
Also, what book and/or set of notes would be good for me to continue studying the Circle Method? I have looked at 
Vaughn's book, it is too dense for me, without enough motivating material. Is the book by Nathanson ("Additive Number Theory: The Classical Bases") a 
good place to continue? 
Thanks,
Tom

REPLY [4 votes]: 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.