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There are infinitely many primes $p$ of the form $6m - 1$. $p + 2$ is of the form $6m + 1$. Since there is a one-to-one correspondence between $p$ and $p + 2$ there are infinitely many $p + 2$ of the form $6m + 1$. There are infinitely many primes $q = p + 2$ of the form $6m + 1$. Thus, there are infinitely many corresponding pairs of primes $\{p, q\} = \{p, p + 2\}$. Thus, there are infinitely many twin primes.

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    $\begingroup$ Find the error in this. There are infinitely many odd numbers $n$ so that $4|n-1$. And there are infinitely many odd numbers $n$ so that $4|n+1$. So there are infinitely many pairs of $(n-1, n-2)$ where $4|n-1$ and $4|n+1$. Is there any difference in your argument and mine. $\endgroup$ – fleablood Sep 6 '17 at 7:03
  • $\begingroup$ "There are infinitely many primes p of the form 6m−1". Okay, we are going to put those in a set called $P$. "There are infinitely many primes q=p+2 of the form 6m+1" Okay we are going to put all those q's in a set called $Q$. and we are going to put all the p=q-2 in a set called $P'$ So if $p\in P\cap P'$ then (p, p+2) is a twin prime. So we know that $P$ is infinite. And we know $Q$ is infinite. ANd $P'$ is infinite. But $P \ne P'$ and we don't know $P'\cap P$ is infinite. $\endgroup$ – fleablood Sep 6 '17 at 7:33
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Two infinite subsets of $\Bbb N$ need not intersect.

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  • $\begingroup$ "But out of these infinitely many p+2, there are infinitely many primes." You have absolutely no reason to believe that at all. $\endgroup$ – fleablood Sep 6 '17 at 7:05
  • $\begingroup$ I understand and agree with what you are saying. But what I am thinking is you start off with a set of primes $p$ that are of the form $6m−1$. There are infinitely many of these primes. Then with each of the primes $p$, associate with it a number $p+2$. There is a 1-1 correspondence between the prime number $p$ and $p+2$. But out of these infinitely many $p+2$, there are infinitely many primes. But these infinitely many primes $p+2$ are already linked with (or associated with) a corresponding prime $p$. Hence, there are infinitely many correspodning primes {$p$, $p+2$}. $\endgroup$ – T.C. Sep 6 '17 at 7:07
  • $\begingroup$ Out of these infinitely many $p+2$, there are infinitely many primes since $p+2$ are numbers of the form $6m+1$. But it can be proven that there are infinitely many primes of the form $6m+1$. $\endgroup$ – T.C. Sep 6 '17 at 7:11
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    $\begingroup$ The p's so that p = 6m-1 and p is prime form one set of numbers. That set is infinite. The p's so that p+2 = 6m+1 are prime form a different set of numbers. That set is infinite. BUT THEY ARE DIFFERENT SETS!!!!!! What the have in common is that i) they are infinite and ii) they are each some but not all of the numbers of the form 6m-1. Other than that, we have know nothing about what they have in common. Certainly not that their intersection is infinite. $\endgroup$ – fleablood Sep 6 '17 at 7:40
  • $\begingroup$ @fleablood How are they defined? $\endgroup$ – user1329514 Sep 8 '17 at 5:03
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(For each prime $p$ of the form $6m-1$) there is only one choice for $q = p+2$, and it might not even be prime. It will satisfy $q = 6m+1$, though.

There are infinitely many primes $q$ of the form $q = 6m+1$, but there is no (known provable) reason to expect many of those $m$ to be the values for which $6m-1$ is prime.

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  • $\begingroup$ $p$ is prime. $p+2$ may not be prime. But there are infinitely many of the $p+2$ that are prime. Since there is 1-1 correspondence between $p$ and $p+2$ (meaning they have the same value of $m$), then we have infinitely many corresponding pairs of primes {$p$, $p+2$}. $\endgroup$ – T.C. Sep 6 '17 at 6:31
  • $\begingroup$ There are indeed infinitely many $p+2$ that are prime. But why should many of them be the ones where $p$ is prime? $\endgroup$ – Hurkyl Sep 6 '17 at 6:33
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Here is a similar scenario:

There are infinitely many numbers of the form $6m-1$ which are divisible by $5$. And there are also infinitely many numbers of the form $6m+1$ which are divisible by $5$. But there are not infinitely many pairs $(6m-1,6m+1)$ which are both divisible by $5$. (In fact there are no such pairs, because $6m-1$ and $6m+1$ differ by $2$, so they can't have a common factor of $5$. But that isn't as important right now.)

Here is another similar one:

There are infinitely many numbers of the form $6m-1$ which are divisible by $7$. And there are also infinitely many numbers of the form $6m+1$ which are divisible by $7$. But there are not infinitely many pairs $(6m-1,6m+1)$ which are both divisible by $7$.

In general, for any property $P$, even if it is true that there are infinitely many $6m-1$ with property $P$, and infinitely many $6m+1$ with property $P$, it doesn't necessarily follow that there are infinitely many pairs $(6m-1,6m+1)$ that both have property $P$.

There could be "bad luck", if the values of $6m+1$ that have property $P$ are "interleaved" with the values of $6m-1$ that have property $P$, instead of "lining up".

Nobody knows if the property $P$ of primeness exhibits "interleaving" or "lining up" behavior. The Twin Prime Conjecture is that it "lines up" for infinitely many pairs $(6m-1,6m+1)$. But for now that is still open.

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Let $P$ but the set of all primes. Let $S$ be the set of all numbers that are $2$ less than a prime.

There are an infinite numbers of $p+2$ where $p \in P$. And there are an infinite numbers of $p+2$ where $p \in S$. But the $p$s that are in $P$ may be completely different than the $p$s that are in $S$.

Yes for ever $p \in P\cap S$ we will have $p, p+2$ are a prime pair. But we have utterly no reason to believe $P\cap S$ is infinite.

Replace "prime" in your argument with "is a multiple of 4":

There are infinitely many $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ p of the form $\color{blue}5m−1$. p+2 is of the form $\color{blue}5m+1$. Since there is a 1-1 correspondence between p and p+2 there are infinitely many p+2 of the form $\color{blue}5m+1$. There are infinitely many $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ q=p+2 of the form $\color{blue}5m+1$. Thus, there are infinitely many corresponding pairs of $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ {p, q} = {p, p+2}. Thus, there are infinitely many twin $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$.

But we no darn well there are $0$, not infinite, number of multiples of 4$ pairs.

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Actually here is a paraphrase of you argument.

Let $J =\{6m - 1|m \in \mathbb N\}$, $K=\{6m+2|m\in \mathbb N\}$ and let $j:J\to K$ via $j(n) = n+2$ by a bijection.

Let $P_1 = \{p\in J|p \text{ is prime}\}$ and $Q_1 = j(P_1) = \{j(p)|p \in P_1\}$. $P_1$ is infinite and in 1-1 corespondence with $Q_1$.

Let $P_2 = \{p \in J|j(p) = p + 2 \text{ is prime}\}$ and $Q_2 = j(P_2) = \{q \in K| q \text{ is prime}\}$. $Q_2$ is infinite and in 1-1 corespondence with $P_2$.

Let $P_3=\{p \in P_1|j(p) + 2 \in Q_2\}$ and $Q_3=\{q=j(p)=p+2|q \in Q_2; p \in P_1\}= j(P_3)$. $P_3$ and $Q_3$ are in $1-1$ corespondence and for any $p \in P_3$ the pair $\{p, j(p)\}$ are a pair of twin pairs

Your argument is as follows: $P_3 \subset P_1$ which is infinte; $Q_3 \subset Q_2$ which is infinite. $P_3$ and $Q_3$ are in 1-1 corespondence. And they are both subsets of infinite sets. Hence they are infinite.

Which simply doesn't work. In calling everything $p$ and $p+2$ and noting different sets wer infinite and in 1-1 corespondence of different sets, you simple got confused and wrongly assumed the wrong set coresponded to the wrong sets.

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All you've done is assume that $p+2$ is prime. So if you have $p := 6k - 1$, your assumption fails for $k = 4$, although $6k - 1 = 23 \in \mathbb{P}$ you should immediately see that $6k + 1 = 25 \not \in \mathbb{P}$. Those are two different sets you have that are formed independently of each other.

I might refer you to these:

Prime Gaps in Residue Classes

...De Polignac Sequence

Asymptotic Expression...

so that you understand that we define these two sets similarly. They have the same asymptotic, but the underlying structures are completely different.

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