An "almost all" result for the binary Goldbach problem I have a question. My professor in the lecture said that Vinogradov's method by applying the Hardy-Littlewood circle method (minor and major arc) for the ternary Goldbach problem can be used to prove an "almost all" result for the Binary Goldbach problem. More precisely
Defining
$$ r(n) = \sum_{p_1 + p_2 = n} (\log p_1) (\log p_2) $$
and denoting $$ E(n) = \# \{ n \leq N: 2n \text{ is not a sum of two primes} \} $$
then
By the following theorem
$$ \sum_{\substack{ n=1 \\ 2 \mid n}}^{N} (r(n) - n \sigma(n) )^2 \ll N^3 (\log N)^{-A} $$
where
$$ \sigma(n) = \prod_{p \mid n } \left( 1 + \frac{1}{p-1} \right) \prod_{p \nmid n} \left(1 - \frac{1}{(p-1)^2} \right) $$
it will follows that $ E(N) = O(N (\log N)^{-A} ) $
I have three question:

*

*Where I can find a reference of this proof?

*For the proof of Vinogradov's theorem we use $ \Lambda $ instead of $\log $ in the definition of $r(n)$, and he prove that for all but finitely many odd integers are sums of three primes by showing that the contribution of $r(n)$ made by primes power is $O( N^{3/2} \log^2 N )$. Here what is the error term that we need to showing a similar results for the binary Goldbach problem ?

*I suppose that the minor and major arc method is too weak to solve this problem, why can solve for all but finitely many integers the Ternary Goldbach problem but it can't solve a similar result for the Binary Goldbach problem?

 A: *

*A classic reference is Vaughan's book "The Hardy-Littlewood Method".


*The contribution to $r(n)$ from prime powers is trivially $O(n^{1/2}(\log n))^3)$, say, since there are only $\ll n^{1/2}+n^{1/3}+\cdots+n^{1/\log n}$ many prime powers, that can possibly appear in $p_1+p_2=n$, each of which is counted with weight $\leq (\log n)^2$. Since $\sigma(n)\gg 1$ for all even $n$, to show that almost all even integers are the sum of two primes it would suffice to prove $r(n)> \tfrac{1}{2}n\sigma(n)$ for almost all integers $n$, which the bound you state implies.


*As a very short answer, we cannot show the minor arcs in the binary problem are a small error term compared to the major arcs. The reason this is possible for the ternary problem is that the minor arcs involve the 3rd power of the exponential sum - to show the minor arcs are a small amount we can then use the fact that the total average of the square is small (by e.g. Parseval) and the pointwise bound for each exponential sum in the minor arcs is small. So basically it's because $3=2+1$, and we can say a lot about $L^2$ averages and a little about the pointwise bound. For the binary problem, we can't do this - either $2=2+0$, and there is no pointwise information to exploit, or $2=1+1$, and we have to look at the $L^1$ average of the exponential sum, which doesn't cancel as nicely as the $L^2$ average.
