# The Goldbach Conjecture and Hardy-Littlewood Asymptotic

A source I am reading refers to the Goldbach conjecture (that every even number is the sum of two primes), and then immediately follows with the "Hardy-Littlewood conjecture" that

$\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n) = 2 \prod \limits_{p \geq 3} \left(1-\frac{1}{(p-1)^2}\right) \left( \prod \limits_{p | N} \frac{p-1}{p-2}\right)N^{1+o(1)}$

which is termed "Goldbach for almost all even numbers", and apparently this also

$\Longleftrightarrow \prod \limits_{p | N} (\frac{p-1}{p-2}) N^{1+o(1)} = O(\log N)^c$. Here, $\Lambda(n)$ denotes the Von-Mangoldt function.

It then states the following theorem:

Let $A \in \mathbb{R}$. Then for all but $\frac{x}{\log^A x}$ even numbers $N \leq x$, we have $\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n) = 2 \prod \limits_{p \geq 3} \left(1-\frac{1}{(p-1)^2}\right) \left( \prod \limits_{p | N} \frac{p-1}{p-2}\right)N (1+o_A(1))$, where $o_A$ denotes the fact that the constant in the limiting behaviour may depend on $A$.

Now I can't see how this is the same as Goldbach at all really. I can see that if $N= p+q$ is the sum of 2 primes then the corresponding term on the LHS will be nonzero, but the Von-Mangoldt function is also nonzero for powers of primes, so it might be nonzero for some $N= P^i + Q^j$. I am beginning to think the RHS may be some sort of probabilistic slant on the conjecture, but can't quite see it.

It may be the case that the Von-Mangoldt function is just negligible on all prime powers except the primes themselves (this is certainly often the case), but at the least it seems like this should have been stated somewhere since it certainly doesn't seem like a trivial deduction to me. Or is this simply an "approximation to Goldbach"; namely a relationship which seems to imply that Goldbach might well be true, but as I have said doesn't remove the problem of the prime powers? (Obviously I am aware that Goldbach is unproved, but the text doesn't clarify whether this is a proved statement weaker than Goldbach, or an unproved statement equivalent to Goldbach: I suspect the latter.)

I also can't see how the "$\Longleftrightarrow$" follows from the first conjecture; I'd be very grateful if someone could help me understand what's going on here, at least heuristically if not formally.

Next, the text goes on to "prove" Vinogradov from the latter theorem which has been stated (I say "prove" because I can't see how the proof works). It says:

Corollary (Vinogradov) Every sufficiently large odd number is the sum of 3 primes. Proof: Let N be odd. Then taking $A=2$ in the theorem, there is some prime $p\leq N/2$ for which the Hardy-Littlewood asymptotic holds. In particular, $N-p$ is the sum of 2 primes.

Now this time I really can't see what's happening: the sum over the Von-Mangoldt function on the LHS will surely just go to zero almost every time $N-n$ is a sum of 2 primes (unless this sum of 2 primes is a prime power of course), and I don't see how the RHS tells us nothing about the LHS except that it is not "extremely small". Could anyone explain what's going on here to me?

It is possible I transcribed some of this material wrong, though I do not see where I might have made an error which could have caused all my confusions simultaneously. Again, any insight you could provide would be desperately appreciated; many thanks in advance.

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There are a few irregularities in your opening statements that seem to be typos. The statement of the Hardy-Littlewood conjecture should be

$\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n) = 2 \prod \limits_{p\ge3} \left(1-\frac{1}{(p-1)^2}\right) \left( \prod \limits_{p \mid N, p\ge 3} \frac{p-1}{p-2}\right)N(1+o(1)),$

rather than ending with $N^{1+o(1)}$, as originally stated. The previous version, while not any less true, is no more precise than just saying $\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n) = N^{1+o(1)}$. In particular, I claim that

$2 \prod \limits_{p\ge3} \left(1-\frac{1}{(p-1)^2}\right) \left( \prod \limits_{p \mid N, p\ge 3} \frac{p-1}{p-2}\right) = N^{o(1)}.$

To see this, note that the infinite product ($p\ge 3$) converges to some positive constant, so the factor $2 \prod \limits_{p \geq 3} \left(1-\frac{1}{(p-1)^2}\right)$ is $N^{c/(\log N)} = N^{o(1)}.$ The remaining factor is at least 1, and smaller than $\prod\limits_{p \mid N, p\ge 3} \left(1 + \frac{2}{p-1}\right) \le \prod\limits_{p\mid N} \left(1 + \frac{1}{p-1}\right)^2 = \left(\frac{N}{\phi(N)}\right)^2,$ which is well-known to be $N^{O(1/\log\log N)} = N^{o(1)}.$

The “equivalent” formulation in the second equation is a bit baffling: it is very much false as written (the left-hand side is certainly larger than $\sqrt{N}$ and therefore much larger than $O(\log N)^c$), but I don't see an obvious correction. I also cannot understand why you write “the sum over the Von-Mangoldt function on the LHS will surely just go to zero” — can you explain? I think I can address your remaining questions.

• The relation between $\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n)$ and Goldbach problem: you are correct in that the summation also includes solutions of the form $p^r + q^s = N$. However, the number of solutions containing a prime power is very small, certainly less than $O(\sqrt{N})$ (to see this, note that there are fewer than $\log N/\log 2$ choices for $r > 1$, and only $\pi(\sqrt{N})$ choices for $p$.

Therefore even if we subtract all prime power solutions from $\sum \limits_{n \leq N} \Lambda(n) \Lambda(N-n)$, we lose at most $O(\sqrt{N} \log^2 N) = N^{1/2 + o(1)}$, which is negligible compared to $N^{1+o(1)}$. The prime powers, as usual, have no effect here (though they do play a role in prime number races).

• Every large odd number is the sum of 3 primes. The Theorem shows that there are very few exceptions to Goldbach's conjecture. In particular, for sufficiently large $N$ there are at most $N/(\log N)^2$ even numbers between $N/2$ and $N$ which are not the sum of two primes. But there are about $N/(2 \log N)$ odd primes $p$ between $1$ and $N/2$, and for each one, $N-p$ is a distinct even number between $N/2$ and $N$.

If you can find just one prime $p$ such that $N-p$ is the sum of two primes $q+r$, you can write $N = p+q+r$. By comparing $N/(2 \log N)$ to $N/(\log N)^2$, we can see that there are more primes $p$ than there are ineligible values of $N-p$, so this is always possible once $N$ is large enough.

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Erick - many thanks for responding. I should say, I was ill when these notes were taken in a lecture course so they were from a friend: it appears he may have transcribed the statement incorrectly from the board when the material was being lectured, particularly as the "theorem" stated immediately after is almost in the form you provide. When I wrote about the sum over the Von Mangoldt function, I was merely referring to the fact that it's defined as zero on everything but prime powers, so I couldn't see what summing over $\Lambda(n)\Lambda(N-n)$ would do except for give us a very small sum. –  Spyam Jun 6 '12 at 11:09
However, I should add that the first statement is "Littlewood conjectured that..." so it may simply be that Littlewood's conjecture was slightly weaker than what we are able to prove. Having spoken to the friend, he says he meant to write $\prod_{p | N} \frac{p-1}{p-2} = O(\log N)^c$ for the "equivalent" formulation, though it turns out he isn't sure $\Leftrightarrow$ is meant to mean equivalent, possibly just some related statement (he didn't pay much attention unfortunately, sorry!) - would this make any more sense? I will digest the rest of your answer now, thanks again. –  Spyam Jun 6 '12 at 11:14
Erick - 2 last things. Firstly, do you think you could explain your comment "note that there are fewer than $\log N / \log 2$ choices for $r>1$, and only $\pi(\sqrt(N))$ choices for $p$"? I can't quite see why that's the case. Secondly, is there any heuristic reason for what the 2 products in the formulation of the theorem mean? They seem a little arbitrary to me, is there an intuition behind them? –  Spyam Jun 6 '12 at 11:34
If $p^r$ is a prime power (but not a prime) less than $N$, then we must have $p,r \ge 2$. Therefore, $p^2 \le p^r \le N$ and $2^r \le p^r \le N$. The first inequality gives $p \le \sqrt{N}$, and the second gives $r \le \log N/\log 2$. –  Erick Wong Jun 6 '12 at 16:36
The products taken over primes, taken together, form what are called a “singular series”. It is slightly more intuitive to write the constant as $2 \prod \limits_{p\nmid N, p\ge 3} \left(\frac{p(p-2)}{(p-1)^2}\right) \left( \prod \limits_{p \mid N, p\ge 3} \frac{p}{p-1}\right)$. Perhaps someone can recommend a good reference on singular series? –  Erick Wong Jun 6 '12 at 16:45