2
$\begingroup$

I was thinking about the usage of Dirichlet's theorem in proving some facts about the Goldbach's conjecture.
I will start with an example. Using Dirichlet's theorem, we know that there are infinitely many primes in the form of, let's say, $6k+1$. This means there are infinitely many primes in the form of $6k+4-3$. Let's say $p=6k+4-3$ and so $p+3=6k+4$. Now tell me if my reasoning is not true. From this, can we say there are infinitely many numbers in the form of $6k+4$ which the Goldbach's conjecture is true about them? I may assume this is true for the next things im gonna say.
Now, we are going to prove something more general, by using one special case of the Dirichlet's theorem: $p=2ak+(b-q)$ where $gcd(2a,b-q)=1$ and so $2ak+b=p+q$. If we find a number $a$ and a prime number $q$ so that $gcd(2a,2-q)=1$, $gcd(2a,4-q)=1$, ... and $gcd(2a,2a-q)=1$ ,we may say there are infinitely many numbers in the form of $2ak+2$ which the Goldbach's conjecture is true about them, same for the forms of $2ak+4$, $2ak+6$, ... and $2ak+2a$. From this we can say Goldbach's conjecture is true for infinitely many numbers in any even form.
Can you help me find such a numbers (that $q$ and $a$ i mean)? What can it lead to?

$\endgroup$
  • $\begingroup$ Unfortunately $6k+4$ is always divisible by $2$ $\endgroup$ – Andrea Mori Jun 18 '13 at 17:21
  • 1
    $\begingroup$ @AndreaMori And Goldbach's conjecture is talking about even numbers. doesn't it? $\endgroup$ – CODE Jun 18 '13 at 17:24
  • $\begingroup$ infinitely many does not mean all numbers - I cannot see what you want to reach $\endgroup$ – mau Jun 18 '13 at 17:26
  • $\begingroup$ Goldbach's conjecture is already known to be true for the vast majority of even numbers (the set of exceptions has density $0$). So not only is it true for infinitely many numbers of a given (linear) form, it is true for the vast majority of those. The hard part is ruling out the few exceptions. $\endgroup$ – Erick Wong Jun 18 '13 at 17:27
  • 2
    $\begingroup$ @CODE : Sorry, I think I misread your question. Yes, Dirichlet implies that the Goldbach property holds for infinitely numbers in any arithmetic sequence of the form $ak+(b+p)$ where $gcd(a,b)=1$ and $p$ is an odd prime. But I don't see how this would help in tackling the general conjecture (even not considering ErickWong's comment!) $\endgroup$ – Andrea Mori Jun 18 '13 at 17:33
2
$\begingroup$

You are correct that this approach will show that there are an infinite number of even numbers in each of these series that can be expressed as the sum of two primes. $a=2,q=3$ seems to work for what you want.

$\endgroup$
  • $\begingroup$ q must be a prime number. $\endgroup$ – CODE Jun 18 '13 at 17:46
  • $\begingroup$ I think with $a=2$, any odd prime will work. $\endgroup$ – Ross Millikan Jun 18 '13 at 17:53
  • $\begingroup$ Oh, thanks. Now this means there are infinitely many numbers in the forms $4k+4$ and $4k+2$ which the conjecture is true about them. Am i right? $\endgroup$ – CODE Jun 18 '13 at 17:57
  • 2
    $\begingroup$ The Dirichlet theorem says already that there are an infinite number of primes of any form you want. If you apply it to $4k+1$ and add $3$, you can conclude that an infinite number of numbers of the form $4k+4$ can be expressed as the sum of two primes. This will be true for any form you want. Applying it to $124k+1$ tells you there are infinitely many numbers of the forms $124k+4, 124k+6, 124k+8,$ etc that can be expressed as the sum of two primes. I'm not sure how useful this is. $\endgroup$ – Ross Millikan Jun 18 '13 at 18:08
  • $\begingroup$ Thanks. that was enough information i think. $\endgroup$ – CODE Jun 18 '13 at 18:20
1
$\begingroup$

Vinogradov showed that if $A(x)$= # of $n\leq x|n\in2N$ and $n$ can not be written as sum of 2 odd primes, then $\lim_{x\to\infty}\frac{A(x)}{x}=0$ in other words $A(x)=o(x)$. That tells us that Goldbach conjecture is true for 'almost all' even numbers, like your result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.