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I'm interested in number theory, and everyone seems to be saying that "It's all about the Riemann hypothesis (RH)". I started to agree with this, but my question is:

  • Why then doesn't RH imply the (asymptotic) Goldbach conjecture?

By "asymptotic" here I mean that any $n\in\mathbb N$ big enough can be written as $p+q$, with $p,q$ primes. I already asked some experts, and they told me that "RH is rather about the distribution of primes".

But look at this table,

(number of ways to write an even number n as the sum of two primes, 4 ≤ n ≤ 1,000,000) isn't that saying that the asymptotic Goldbach conjecture is also about the distribution of primes? I don't understand.

Any help would be very welcome.

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The Riemann hypothesis is an asymptotic statement. In particular, it can't rule out something silly like all primes being congruent to $1 \bmod 3$ (which would prevent any number divisible by $3$ from being a sum of two primes). – Qiaochu Yuan Dec 29 '12 at 6:56
I suspect RH might give you an asymptotic formula for the number of ways to write the numbers $1$ through $n$ as sums of $2$ primes, and so imply that generally there are lots of ways to write $n$ as a sum of two primes. But it wouldn't rule out there being an infinite (but not very dense) set of $n$ that could not be written in this way. – Alex Becker Dec 29 '12 at 7:00
Thank you very much both - my question is stupid indeed. – Brian Dec 29 '12 at 9:36
@Alex, Qiaochu: There has been major work towards the Goldbach conjecture which was conditional on the Riemann hypothesis. Assuming RH, Kaniecki proved that every odd integer is the sum of at most $5$ prime numbers. (this is now known unconditionally) I am not sure I understand what you mean by RH only working in an asymptotic sense. The error term comes with effective constants, and this is part of the reason why we can prove the Ternary Goldbach conjecture under GRH, but only for sufficiently large $N$ unconditionally. – Eric Naslund Dec 30 '12 at 1:14
You might be interested in my blog where I give both heuristics and rather rigorous arguments to establish the asymptotic Goldbach's conjecture unconditionnaly. Please don't hesitate to leave a comment there. – Sylvain Julien Dec 10 '14 at 15:32

1 Answer 1

Currently, it is has not been proven that RH implies the Goldbach conjecture, but there are partial results in this direction.

Here is a paper which outlines why GRH implies the ternary Golbach conjecture, that is the statement that every odd integer greater than five is the sum of three primes. This theorem has now been proven unconditionally by Helfgott.

Additionally, here is an answer I posted to this question which I have added here:

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.

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It might be worth updating this to include more recent results on the weak Goldbach conjecture (even though they are unconditional on RH). – Donkey_2009 Dec 20 '14 at 12:17

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