Despite the rather recent progress in prime number theory (see the proof of the ternary Goldbach conjecture by H.A. Helfgott, and the striking result of Yitang Zhang improved by Tao, Maynard and others), to my knowledge, we're still unable to prove rigorously that there exists $N>0$ such that for all integer $n>N$, $2n$ is the sum of two primes. What are the main reasons for that? Lack of knowledge about exponential sums? Interesting heuristics that no one is able to turn into a rigorous proof? If so, what are the obstructions to getting a rigorous version of such arguments?
Thanks in advance for any insight.

5$\begingroup$ I think one of the main obstructions is the "parity problem" in sieve theory, which all of the results above depend on to some extent. This is summarized by Terry Tao as follows: "If $A$ is a set whose elements are all products of an odd number of primes (or are all products of an even number of primes), then (without injecting additional ingredients), sieve theory is unable to provide nontrivial lower bounds on the size of $A$. Also, any upper bounds must be off from the truth by a factor of 2 or more." $\endgroup$ – Stanley Yao Xiao Mar 31 '14 at 11:53

1$\begingroup$ To state the obvious, one possible reason is that it's false... $\endgroup$ – Tom Leinster Apr 1 '14 at 20:11
As far as I know there are two approaches to Goldbach type problems, the circle method and sieve methods. In the sequel I will restrict myself to the circle method, hoping that someone else writes something about sieves.
Define the exponential sum $S(\alpha)=\sum_{n\leq N}\Lambda(n) e(n\alpha)$, where $e(x)=e^{2\pi i x}$. By orthogonality of $e(x)$ we have that the (weighted) number of representations of $N$ as the sum of 3 primes is $$ \int_0^1 S(\alpha)^3e(N\alpha)\;d\alpha. $$ If $\alpha=\frac{p}{q}$ is rational, then $$ S(\alpha)=\sum_{a=0} ^{q1} e \left( \frac{ap}{q} \right) \underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n), $$ which can be approximated using the prime number theorem for arithmetic progressions. If $\alpha$ is very close to a rational number, evaluation can still be done by partial summation. Hence define the major arcs $\mathfrak{M}\subseteq[0,1]$ as the set of integers very close to rational numbers with small denominator. Then $$ \int_\mathfrak{M}S(\alpha)^3e(N\alpha)\;d\alpha $$ can approximately be evaluated, it turns out that this integral essentially equals the expected main term. Let $\mathfrak{m}=[0,1]\setminus\mathfrak{M}$ be the so called minor arcs. To prove that every sufficiently large odd integer is the sum of three primes, Vinogradoff showed that for suitably defined arcs the integral over the major arcs is $\sim\mathfrak{S}(N)N^2$, and that for $\alpha\in\mathfrak{m}$ we have $S(\alpha)\ll\frac{N}{\log^2 N}$. Hence \begin{eqnarray*} \int_\mathfrak{m}S(\alpha)^3e(N\alpha)\;d\alpha & \leq &\int_\mathfrak{m}S(\alpha)^2\;d\alpha\max_{\alpha\in\mathfrak{m}}S(\alpha)\\ & \leq & \int_0^1S(\alpha)^2\;d\alpha\max_{\alpha\in\mathfrak{m}}S(\alpha)\\ & = & \sum_{n\leq N}\Lambda(n)^2\max_{\alpha\in\mathfrak{m}}S(\alpha)\\ & \ll & \frac{N^2}{\log N} \end{eqnarray*} So for large $N$ the error coming form $\mathfrak{m}$ is of smaller order then the main term, and we obtain an asymptotic for the number of representations. Since proper powers are pretty rare, passing from $\Lambda$ to primes is no problem.
All this was known and used by Hardy and Littlewood some 20 years before Vinogradov, however, they could not give an unconditional bound for $S(\alpha)$, and had to assume the generalized Riemann hypothesis. Among the things Helfgott did, was a numerical bound which is nontrivial for rather small values of $N$, which in analytic number theory is usually pretty difficult. However, no matter what progress may come, the argument used for odd $N$ can never prove binary Goldbach. The crucial point in the argument above was that the integral over the major arcs is asymptotically equal to the conjectural main term, and the integral over the minor arcs is of smaller order. The integral over the major arcs can still be evaluated, and is $\mathfrak{S}(N)N$, as it should, however, $$ \int_0^1 S(\alpha)^2\;d\alpha = \sum_{n\leq N}\Lambda(n)^2\sim N\log N $$ is bigger then the main term. To prove Goldbach for even integers we have to show that $\left\int_\mathfrak{m} S(\alpha)^2e(N\alpha)\;d\alpha\right$ is considerably smaller than $\int_\mathfrak{m} S(\alpha^2)\;d\alpha$, that is, that there is some cancelation within the integral.
Still, one can obtain interesting results in this way. For example, Montgomery and Vaughan (The exceptional set in Goldbach's problem, Acta Arith. 27) have shown that all even integers up to $x$ with at most $x^{1c}$ exceptions can be written as the sum of two primes. They used the fact that $\int_\mathfrak{m} S(\alpha)^2e(N\alpha)\;d\alpha$ cannot simultaneously be large for many different $N$.
Somewhat surprisingly Brüdern, Granville, Perelli, Vaughan, and Wooley, (On the exponential sum over kfree numbers, R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356) showed that there are interesting binary additive problems which can be solved by the circle method, e.g. problems involving squarefree integers. However, as they explicitly mention, their method fails for sets of density 0, the reason being the same as above.

$\begingroup$ Ok, thank you for your answer. I know I still think like a physicist, but to me these approaches look rather technical, involving a quite heavy machinery for the results one can get through it. I wish someone brilliant and both intuitive and rigorous like Terry Tao could turn my heuristic arguments developped in threads like "About Goldbach's conjecture" into a proper proof. Referring to this last question, a rigorous proof that the quantity denoted by $\alpha_{n}$ is an $o(n)$ would already be interesting. I have no idea of how to achieve this though. $\endgroup$ – Sylvain JULIEN Apr 1 '14 at 18:16

3$\begingroup$ @SylvainJULIEN: Green and Tao have a major theorem devoted to proving essentially best possible results about solving linear equations in primes (such as Goldbach, which is the equation $p_1 + p_2 = N$) which don't (explicitly or implicitly) include single equations in two variables (such as Goldbach or twin primes) inside them (so their results, in the end, don't address Goldbach or twin primes). So I think it's safe to assume that Terry has devoted a lot of effort to understanding the difficulties in Goldbach and twin primes ... . $\endgroup$ – tracing Oct 29 '14 at 16:36

$\begingroup$ I don't doubt it. I just meant to point out the huge technical aspect of mathematicians' ideas, while the physicists' ones, though more simple, lack rigor. $\endgroup$ – Sylvain JULIEN Oct 29 '14 at 16:46

$\begingroup$ Sorry for digging out this old question, but I would have something to ask that may serve as an insight to the "(expected) cancellation inside the integral". Isn't the isobarycenter of the values $e(N\alpha)$ close to 0 ? I guess the set of those values has a non trivial symmetry group, and that at least a pairwise cancellation of the considered exponential can be reached for most values of $N$. Does such a rough idea have any interest? $\endgroup$ – Sylvain JULIEN Oct 31 '19 at 14:37

$\begingroup$ A curios question about the sum $S(\alpha)=\sum_{a=0} ^{q1} e \left( \frac{ap}{q} \right) \underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n)$ Before Alex M. edited Jan's answer the sum was given by $\sum_{(a,q)=1} e(\frac{ap}{q})\underset{n\equiv a\pmod{q}}{\sum_{n\leq N}} \Lambda(n)$ $\endgroup$ – Isak the XI Dec 23 '20 at 0:24