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This is a major weakening of many prime sum / difference existence questions.

Let $a \in \Bbb{Z}$ and $(a)$ the ideal generated by $a$. Then do there exist two primes $p, q$ such that $p - q \in (a)$ at least?



marked as duplicate by Bill Dubuque abstract-algebra Nov 2 '18 at 19:27

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    $\begingroup$ Sure. Pick a prime $q$ not dividing $a$. Then Dirichlet tells us there are infinitely many primes $p\equiv q \pmod a$. $\endgroup$ – lulu Nov 1 '18 at 15:31
  • $\begingroup$ @lulu It seems you forgot the simple answers the last time around (in the dupe), so here is a comment to remind you. $\endgroup$ – Bill Dubuque Nov 2 '18 at 19:28
  • $\begingroup$ @BillDubuque Thanks! Definitely missed it. $\endgroup$ – lulu Nov 2 '18 at 19:30

This can be done in an elementary way; Dirichlet is extreme overkill.

To see this, note that there are infinitely many primes, but only finitely many remainders on division by $a$. By the pigeonhole principle, there are two primes $p$ and $q$ with the same remainder, and we are done.

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    $\begingroup$ But it's still Dirichlet! (box principle). $\endgroup$ – Bill Dubuque Nov 2 '18 at 19:17

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