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I want to see a simple proof of a theorem that is weaker than chen's theorem.

Thus let $m,n$ be positive integers. An m-almost prime is a squarefree integer that is the product of at most $m$ primes. ( I added the squarefree condition to the " definition " )

Prove that there are an infinite amount of numbers $n,n+2$ such that both $n$ and $n+2$ are m-almost primes.

The case $m=2$ is known as Chen's theorem , or almost since I added the squarefree condition.

the case $m=1$ is the unsolved prime twins conjecture.

Since the proof of Chen's theorem is too complicated for a beginner like me , It seems larger $m$ should be simpler.

So I want to see a simple proof for a large $m$ so that I can get a first step into " this kind of number theory ".

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I think this is still a difficult question even for large $m$ like a million. For any fixed $m$, the numbers that are $m$-almost primes are rare, in the sense that $$ \lim_{N\to\infty} \frac{\#\{n\le N \colon n \text{ is an $m$-almost prime} \}}N = 0. $$ Furthermore, they're all roughly the same amount of rare: the numerator of that fraction (for fixed $m$) is roughly $\frac1{(m-1)!} N(\log\log N)^{m-1}/\log N$ when $N$ is large. So changing $m$ doesn't affect that rate of growth much.

It is certainly true that proving your assertion for $m$ large is easier than Chen's theorem - but that's because Chen added in a lot of technical complication to get $m$ down to $2$, not because it's easy for any given $m$.

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  • $\begingroup$ I am a number theory noob, but I thought en.wikipedia.org/wiki/Chen's_theorem is Chen's theorem. i.e. a weaker version of Goldbach conjecture. care to explain how this is related to product of m primes? i am slightly curious $\endgroup$ – Lost1 Nov 1 '13 at 1:55
  • $\begingroup$ @Greg I am aware of those two formula's. They follow from - for instance - the simplex method. I understand large $m$ can still be difficult but I desire a way to start understanding such things. Although I respect and thank you for your answer , from my viewpoint it is just some extra usefull comment because I already knew what you said. Of course your answer is usefull to others , but at this moment not to me , it is of course hard to say how much I know from your viewpoint. The point is I appreciate your answer but I cannot get started to understand things better at this point. $\endgroup$ – mick Nov 1 '13 at 13:28
  • $\begingroup$ @Lost1 Yes well that's simply Chen's OTHER proof :) $\endgroup$ – mick Nov 1 '13 at 13:30
  • $\begingroup$ I guess my short answer is: I don't think there is a simple proof for large $m$. $\endgroup$ – Greg Martin Nov 2 '13 at 10:03

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