The twin prime conjecture states basically that it is possible to find two primes $p$, $p+2$ at a distance $2$ that are as big as wanted (Wikipedia).

I am learning about the basic properties associated to the distance of every $n \in \Bbb N$ to the next closest non-adjacent coprime ("non-adjacent" means that the distance to the coprime is the minimum possible strictly greater than 1, that distance is exactly the smallest prime number that does not divide $n$, here is a question about it) and I wonder if the following expression could be a valid way of expressing the twin prime conjecture by using the distance to the next non-adjacent coprime as the base of the expression, as follows:


$\ T=\{(n,n+2)\ (even)\ /\ \exists\ p,p+2 \in \Bbb P\ /\ (n,p)=1 \land (n+2,p+2)=1 \land (p-n)\gt 1\ \land \not\exists k\ /\ k\lt p\ \land (n,k)=1 \land (k-n)\gt1\}$

So the twin prime conjecture would be true only if the cardinality $\mid T\mid = \infty$


$n=38, n+2=40$, the next non-adjacent coprime to $n$ is $p=41$, and to $n+2$ is $p+2=43$, so $(38,40) \in T$

$T$ would be the set of pairs of even numbers $(n,n+2)$ whose closest non-adjacent coprimes are the pair $(p,p+2)$ being $p$ and $p+2$ twin primes. So only if the cardinality of that set is infinite, the conjecture is true.

Would the above expression be a valid equivalent way of defining the twin prime conjecture? If not, is it possible to fix it?

Thank you!


yes, i think they are equivalent.

If there are infinitely many twin primes, then for each $p,p+2$ we can find $n,n+2$ (just take $n=p-3$)such that your condition is satisfied, so $|T| = \infty$.

Conversely, if $T$ is infinite, then for each $(n,n+2) \in T$ there is a pair of primes $p,p+2$ which is greater than $n$. So there are infinitely many of them.

i hope i did not misunderstood your question

  • $\begingroup$ thank you for the review! Yes you express the same as I wrote in the question. I wanted to write the definition of the conjecture in terms of the cardinality of a set. $\endgroup$
    – iadvd
    Jul 29 '15 at 7:40

This is a clarification of the question, too long for a comment. I intend to turn this into an answer once I determine what the question is.

I believe your question is whether C is equivalent to the existence of infinitely many twin primes, where C is the statement

There are infinitely many pairs $(n,p)$ such that:

  1. $p$ and $p+2$ are prime,
  2. $\gcd(2n,p)=1=\gcd(2n+2,p+2)$,
  3. $p>2n+1$, and
  4. There is no $2n+1<k<p$ with $\gcd(2n,k)=1$.

I wonder if you've left out the condition $\gcd(2n,k+2)=1$ intentionally from #4?

  • $\begingroup$ hi, thank you for your review! yes my expression is the same one you wrote. Sorry about your last question... would it be gcd(2n,k+2)=1 as you wrote or gcd(2n+2,k+2)=1? if it is gcd(2n+2,k+2)=1 then I left it out on purpose because it is implied in the definition of $2n$ and $2n+2$ and in the condition #4 applied to $2n$. $\endgroup$
    – iadvd
    Jun 19 '15 at 0:40
  • $\begingroup$ hi, did my comment above answered your questions? please let me know if I can do something else :) $\endgroup$
    – iadvd
    Jun 28 '15 at 13:43

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