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Prove that there exist infinitely many primes that do not belong to any pair of twin primes. The book gives a hint as $21k+5$ for $k = 1, 2, \dots$

I'm not sure how I would do this because it would seem that eventually this sequence would run into a twin prime. I would assume that if you just omitted the twin primes from this sequence it would still be infinite.

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    $\begingroup$ Consider $60n+35,60n+37,60n+39$. The right one and the left one are composite for every value of $n$, the middle one is prime for infinitely many values of $n$ (according to Dirichlet's theorem on arithmetic progressions). $\endgroup$ – barak manos Nov 23 '16 at 13:17
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The numbers $21k+3$ and $21k+7$ are divisible by 3 and by 7 respectively, so are not prime. It follows that none of the infinitely many primes that occur in the arithmetic progression of numbers of the form $21k+5$ can be members of a twin prime pair.

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Consider:

  • $60n+35$ is composite for every value of $n$, since it is divisible by $5$
  • $60n+37$ is prime for infinitely many values of $n$, since $\gcd(60,37)=1$*
  • $60n+39$ is composite for every value of $n$, since it is divisible by $3$

*This is according to Dirichlet's theorem on arithmetic progressions.

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Here's another argument.

Consider the sum $$\sum_\text{p prime}\frac1p$$

Euler showed that this sum diverges. For an elementary proof, see here.

Now consider the sum $$\sum_{p \in T}\frac1p$$ where $T$ is the set of primes that belong to some twin-prime pair. Brun showed that the above sum converges to some finite value. See here for some details. A reasonable derivation can be found here.

What this, in particular, implies is that there must be infinitely many terms in the first sum that do not make an appearance in the second sum. This concludes the proof.

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