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I was working through some basic number theory questions , when I came across :

Show that there are infinitely many primes that are not one of the primes in a pair of twin primes

How can I go about solving it ? I have absolutely no idea ...

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    $\begingroup$ Similar question with an answer to your question in the comments (it is equivalent to Micah's posted question there and you can see the comment by Ross Millikan, which answers the question. There are infinitely many primes of the form $p=15n+7$ by Dirichlet's theorem on arithmetic progressions and $5\mid p-2, 3\mid p+2$, which is similar to John's answer). $\endgroup$ – user26486 Feb 19 '15 at 5:56
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All primes $p$ and $q$ of the form $p = 23+30n$, $q = 37 + 30n$ where $n$ is natural, are unpaired.

To proof this, think of all of the factors of 30 and why 25 and 35 will not be prime. Know also, all prime pairs greater than 3, like 5,7 use the forms 6n-1, 6n+1 to make a pair.

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    $\begingroup$ $$\begin{cases}5\mid p+2=25+30n\\ 3\mid p-2=21+30n\\ 3\mid q+2=39+30n\\ 5\mid q-2=35+30n\end{cases}$$ You didn't mention that there are infinitely many of those primes by Dirichlet's theorem on arithmetic progressions. $\endgroup$ – user26486 Feb 19 '15 at 5:53
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    $\begingroup$ Yes, I did not mention that. But, only one form which fails make a pair for all n is all that is needed here. $\endgroup$ – John Nicholson Feb 19 '15 at 6:12
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With the exception of 7, numbers ≡ {7, 23} mod 30, which sequence as 7 {+16+14} {repeat ... ∞} cannot possibly be twin primes given their closest possible proximity to other prime numbers is +4 and -4, respectively. A modulo 30 factorization wheel, which is populated by n ≡ {1, 7, 11, 13, 17, 19, 23, 29} (mod 30) (aka, natural numbers not divisible by 2, 3, or 5), parses all twin prime candidates > (5,7) into 3 distribution channels, and makes this clear:

  1. n ≡ {11,13} mod 30 (at 132° & 156°)
  2. n ≡ {17,19} mod 30 (at 204° & 228°)
  3. n ≡ {29, 1} mod 30 (at 348° & 12°).

Ref: primesdemystified.com/twinprimes

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    $\begingroup$ Essentially, what the accepted answer already says - but in a more confusing way. No need to link to you own site, really. $\endgroup$ – Alex M. May 15 '18 at 21:02

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