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Inspired by this question on the "Sum of a odd prime and a odd semiprime as sum of two odd primes?", I wonder, if it is possible to show, that every even number $2n\ge 12$, can be written as a sum of a prime and semiprime $$p_1+p_2p_3=2n?$$

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I don't know if it is proven for all even numbers $\geq 12$, but for sufficiently large numbers it is true. See Chen's theorem. –  Michalis Mar 22 '12 at 22:29
    
@Michalis Can you give a sketch of the proof? The link doesn't. –  draks ... Mar 22 '12 at 22:33
    
Unfortunately I don't know anything about the proof... –  Michalis Mar 22 '12 at 22:39
    
Why is Chen's Theorem restricted to "$p_1+p_2p_3$, where $p_1,p_2,p_3$ are all distinct primes"? –  draks ... Mar 22 '12 at 22:49
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Short summary of the proof: reduction to sieving, estimation of sieving functions, search for upper bounds using the Jurkat-Richert theorem, using a bilinear form inequality, and joining together of all these results to create a function that counts the number of representations of a given number as either $p_1+p_2$ or $p_1+p_2p_3$, and showing that that function always returns a positive integer when the given number is sufficiently large. (from here) –  draks ... Mar 22 '12 at 23:08

1 Answer 1

up vote 9 down vote accepted

Take a look at section 1, "Combinatorial Sieving", of this blog post of Terry Tao. He builds you up to the following weakening of the twin prime conjecture:

There are infinitely many $n$ such that both $n$ and $n+2$ have the property that their every prime divisor is greater than $\sqrt[20]{n}$.

One can rewrite this into an even weaker statement:

There are infinitely $n$ such that both $n$ and $n+2$ are products of no more than $19$ primes

Roughly, the proof is as follows: Let $z = \sqrt[20]{N}$. Heuristically, for $n$ odd, the probability that neither $n$ nor $n+2$ is divisible by a prime less than $z$ is $\prod_{3 \leq p < z} (1-2/p) \approx c/(\log z)^2$ where the value of the constant $c$ is unimportant. So one would expect that there are about $c N/(\log z)^2$ integers less than $N$ meeting the given conditions. If you try to make this argument rigorous directly, you get huge error terms, but if you are very careful you can make it work. See Terry's post for the careful details.

I don't have a big picture understanding of why you couldn't use the same argument to actually prove the twin prime theorem, taking the product $\prod_{3 \leq p < \sqrt{N}} (1-2/p)$ instead of $\prod_{3 \leq p < z} (1-2/p)$. It's just that when you get into the nitty gritty of the proof, $\sqrt[20]{N}$ is small enough and $\sqrt{N}$ isn't.

I have never looked at the proof of Chen's theorem, but sieving methods refer to arguments like this one, so you might want to read this example of a sieving proof (already quite challenging, in my opinion), before taking on what Eisenstein describes as a "highly technical application of sieving methods".

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