What I want to ask is about the structure of the Goldbach function that defined by $$ R(x)=\#\{ p \mid x-p \in \mathbb{P} , \ p\leq x/2\}$$ for $x\in 2\mathbb{N}$, where $\mathbb{P}$ is the set of prime numbers. As I understand, the methods which are developed to prove the Goldbach type problems have difficulties in the case of the binary Goldbach problem and it seems that there have to be more essential principle to prove the problem that $R(x)$ follows (though it is still unsolved that $R(x)>0$) $$ c\frac{x}{\log^2 x}\prod_{p|x, \ \ p>2} \frac{p-1}{p-2} $$ with a positive constant $c$.

If I make an explanation for my question, we would consider a correspondence between the unary problems and binary problems (or a natural extension of the unary problem under given specific condition), for example, the prime counting function $\pi(x)$ can be expressed as $$ -\sum_{p\in \mathbb{P}}\mu(p), $$ and $R(x)$ can be expressed as $$ \sum_{p,q\in \mathbb{P} \atop p+q=x}\mu(p)\mu(q).$$ Except the difference of signs in the point of view that the importance is to estimate these moduluses, we would say that there is a correspondence under the condition $a+b=x$ which seems as one of the most simple conditions we can add. And then a question rises, whilst there is a logical equivalence between the estimates of $\pi(x)$ and $\sum_{n\leq x}\mu(n)$, are there any known relation between $R(x)$ and the sum $$\sum_{ns+mt=x \text{ for some } s,t\in\mathbb{2N}-1}\mu(n)\mu(m)? $$ (where $n,m\in 2\mathbb{N}-1$ such that $(x,n)=1=(x,m)$) On the function above, the odd numbers $s,t$ be put as a condition via some deductions and because it seems more meaningful than $a+b=x$ at least as I thought.

  • $\begingroup$ the link between $Mertens(x) = \sum_{n \le x} \mu(n)$ and $\pi(x) = \sum_{p \le x} 1$ is through $\psi(x) = \sum_{p^k \le x} \ln p = \sum_{n \le x} Mertens(x/n) \ln n$. can you do the same with your double sums ? $\endgroup$ – reuns Mar 20 '16 at 21:26
  • $\begingroup$ I would appreciate if you see my reply on the answer below. $\endgroup$ – B . O Mar 22 '16 at 14:57

Many analytic number theory questions about the primes are really questions about estimating sums involving the von Mangoldt function $\Lambda$. For instance, the Goldbach conjecture (when viewed from the perspective of analytic number theory) is basically asking for a good estimate on the quantity $$ \sum_{a+b=x} \Lambda(a) \Lambda(b) \quad (1).$$ Now this quantity is similar, but not quite the same as, the quantity $$ \sum_{a+b=x} \mu(a) \mu(b) \quad (2)$$ mentioned in your post. There are a number of useful combinatorial identities relating $\Lambda$ to $\mu$, e.g. $$ \Lambda(n) = \sum_{n=dm} \mu(d) \log m.$$ If one inserts this identity directly into (1), one gets the expression $$ \sum_{m_1, m_2} \log m_1 \log m_2 \sum_{m_1 a + m_2 b = x} \mu(a) \mu(b).$$ Roughly speaking, this calculation (as well as similar calculations using more sophisticated combinatorial identities which are truncated in various analytically convenient ways) suggests that in order to estimate (1) well, it would suffice to estimate sums of the form $$ \sum_{m_1 a + m_2 b = x} \mu(a) \mu(b) \quad (3) $$ to reasonably good accuracy (in particular, one would probably want the error terms to be smaller than the main term by a couple factors of $\log x$, in order to absorb various losses coming from the other logarithmic terms appearing in the above expansion). The sum (2) is one such member of this family of sums (3), but it is not the only one. Just knowing a bound on (2) alone is unlikely to be directly useful for controlling (1), but it is likely to be indirectly useful by suggesting a way to control the related sums (3), which can then be used to control (1).

This can be compared with the case of the prime number theorem, which is basically about controlling the sum $$ \sum_{n \leq x} \Lambda(n).$$ Using the same sort of analysis as above, we come to the conclusion that to control the above sum, it would be desirable to control the sums $$ \sum_{mn \leq x} \mu(n) $$ for various $m$. But this sort of sum can be rewritten as $\sum_{n \leq x/m} \mu(n)$, and so there is a fairly direct link between the Möbius summatory function and the von Mangoldt summatory function, which makes the equivalence of the prime number theorem with the bound $\sum_{n \leq x} \mu(n) = o(x)$. There is still the issue of not losing logarithmic factors when going back and forth; there are some clever ad hoc tricks in the case of the prime number theorem, exploiting Möbius inversion, that end up not losing these log factors, but they unfortunately do not seem to generalise to other settings. For instance, I do not know how to derive Goldbach-type results (or twin prime type results) from Chowla type conjectures unless the error term is a couple log-factors better than the main term in the latter (see e.g. Section 3 of this blog post of mine for some discussion along these lines).

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    $\begingroup$ And thank you for your kind answer! $\endgroup$ – B . O Mar 23 '16 at 2:32

To find an answer to your question I suggest to take moment of your time to read this link: https://vixra.org/abs/2007.0090
Here you find a deep connection between the Goldbach conjecture and the binary version of the Mobius sum, using linear Diophantine equations and sieve methods.

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  • $\begingroup$ Only you can make such a short edit but you might want to paste this umlaut o in to Möbius: ö $\endgroup$ – samerivertwice Aug 1 at 10:36
  • $\begingroup$ this should be deleted as it is a link to a fake Goldbach conjecture proof on vixra and as such against MO rules $\endgroup$ – Conrad Aug 1 at 14:31
  • $\begingroup$ why is a fake proof , you should read it and after you can comment ,don't judge anything from its apparence. $\endgroup$ – Mohamed Zouhal Aug 1 at 16:07

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