# Why the Riemann hypothesis doesn't imply Goldbach?

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, http://en.wikipedia.org/wiki/File:Goldbach-1000000.png (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

There has also 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. Earlier this year, Tao proved this unconditionally.