Does Riemann Hypothesis imply strong Goldbach Conjecture?

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  • Why the Riemann hypothesis doesn't imply Goldbach?

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Let $$G(2N)=\sum_{p+q=2N}\log p\log q$$ be the weighted count of the number of representations of $2n$ as a sum of two primes, and let $$J(2N)=2NC_2 \prod_{p|N,\ p>2}\left(\frac{p-1}{p-2}\right)$$ where $C_2$ is the twin prime constant. In his paper, refinements of Goldbach’s conjecture and the generalized Riemann hypothesis, Granville proves that:

Theorem: The Riemann hypothesis is equivalent to the statement that $$\sum_{2N\leq x} (G(2N)-J(2N))\ll x^{3/2-o(1)}.$$

Note that this is not equivalent to the Goldbach conjecture as one of these terms could be of size $N$. Here is a proof of this theorem:

Proof: First, we have that $$\sum_{2N\leq x} J(2N)=\frac{x^2}{2}+O(x\log x).$$ Next, since $$\sum_{n\leq x} G(2N) =\sum_{p+q\leq x}\log p\log q = \sum_{p\leq x}\theta(x-p)$$ where $\theta(x)=\sum_{p\leq x}\log p$, and since the Riemann hypothesis is equivalent to the statement that $\theta(x)=x+O(x^{1/2+o(1)})$ we see that $$\sum_{p+q\leq x}\log p\log q=\frac{x^2}{2}+O\left(x^{3/2-o(1)}\right)$$ if and only if the Riemann hypothesis holds. Combining these two facts proves the theorem.

Lastly, it is important to note that there are no absolute value bars in the statement of Granville’s theorem. This means that even if the Riemann hypothesis is true, this theorem does not imply that $$G(2N)=J(2N)+O(N^{1/2+o(1)})$$ for any $N$ – it could be that the error term is always of size $N$ and there is magical cancellation in the above sum.