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Are there an infinite number of primes which are any multiple of $n$ apart? That is take $n\in \mathbb{N}$, then is there an infinite number of primes which are separated by $\textbf{any}$ of the numbers in the below set $$ \{n,2n,3n,4n,\ldots \}. $$

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    $\begingroup$ I‘m not entirely sure if this is what you‘re asking but Dirichlet‘s Theorem states that there are infinitely many primes in any arithmetic progression $a+bn$ for $a, b$ coprime. $\endgroup$
    – bsbb4
    Aug 12, 2019 at 22:35
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    $\begingroup$ Let $n$ be any odd number. There aren't infinitely many primes separated by $3n$. Or by $5n$. Or by $mn$ with $m$ odd. $\endgroup$ Aug 12, 2019 at 22:45
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    $\begingroup$ Can you clarify your quantifiers? Are you asking whether for each $n$, there exist infinitely many pairs primes whose difference is a multiple of $n$? Or are you asking whether for each $n$, there exists $k$ such that there exist infinitely many pairs of primes whose difference is $kn$? Or something else? $\endgroup$ Aug 12, 2019 at 22:56
  • $\begingroup$ for each n, there exists an infinite number of prime gaps of lengths in the set : $n,2n,3n,4n,...$. So your first case @EricWofsey. i.e for each n, there exist infinitely many pairs primes whose difference is a multiple of n $\endgroup$
    – Monty
    Aug 12, 2019 at 23:15

2 Answers 2

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Prime numbers are infinite, so, by the pigeonhole principle $\forall n \exists k $ s.t. there is an infinite number of primes $p\equiv k \pmod n$. Any couple among these primes has a difference which is a multiple of $n $.

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    $\begingroup$ Nice. Very clever that you did that with just the pigeonhole principle. $\endgroup$
    – subrosar
    Aug 12, 2019 at 22:57
  • $\begingroup$ Primes are infinite but it does not directly, without using other results, imply that there are infinitely many primes in every modulo class. E.g. what if all the primes after a certain stage are all of the form $100n+3$. So we need to show that this is not the case. Infinitude of primes only implies that there is at least one modulo class having infinite number of primes. However this one class alone is sufficient to prove our required result. $\endgroup$ Aug 13, 2019 at 18:24
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Yes. Dirichlet's Theorem provides the proof. Let $n$ be any positive integer. Let $a$ be some number which is relatively prime to $n$. Then by Dirichlet's Theorem there are infinitely many primes in the arithmetic progression $a+kn$, $k\in \mathbb{N}.$ Any two of these primes are separated by a multiple of $n$, therefore there are infinitely many values of $k\in \mathbb{N}$ for which there is a prime gap of size $kn$.

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    $\begingroup$ -1 It does not follow that there is a $k\in\Bbb{N}\backslash\{0\}$ such that there are infinitely many prime gaps of size $kn$. $\endgroup$
    – Servaes
    Aug 13, 2019 at 8:37
  • $\begingroup$ I guess that wasn't clear. I didn't mean that for some $k$ there are infinitely many gaps of size $kn$. I meant that there are infinitely many values of $k$ for which there is a gap of size $kn$. I'll try to clarify that. $\endgroup$
    – subrosar
    Aug 13, 2019 at 17:11
  • $\begingroup$ +1 It is clear now. $\endgroup$
    – Servaes
    Aug 13, 2019 at 23:44

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