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I know the classic direct n!+k style proof. I'm looking for an alternate proof that illustrates exactly why we run into trouble if we assume that there are only finitely many numbers appearing in the prime gap sequence. As a simple example, if you assume that the sequence becomes 4,2,4,2,4,.... (it is impossible for either 2 or 4 to appear twice in a row, so this is the only possibility) then we are secretly adding 6 repeatedly, and some prime number will be divisible by 7. I think a version of this issue happens regardless of the finite set you choose. I'm hoping to generalize this argument without using analysis.

  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 21 '23 at 01:58
  • It is well known that there arbitrarily large prime gaps... – paw88789 May 21 '23 at 02:01
  • @paw88789 Please see edit. –  May 21 '23 at 02:04
  • Maybe you seek a proof something like this? If the only gaps are $,\color{#c00}2,\color{#0a0}4,$ then adding a gap to any solution of $x\equiv \color{#c00}{-2}\equiv 1\pmod{!3},\ x\equiv \color{#0a0}{-4}\equiv 1\pmod{5},$ yields a multiple of $,3,$ or $,5,$ so is composite (when $> 5.),$ But by Dirichlet the CRT solution progression $,x \equiv 1\pmod{!15}$ contains a (large enough) primes $,p,,$ so $,p+2,p+4,$ are composite, so the gap after $,p,$ is not in ${2,4},,$ so those are not all the prime gaps. The argument clearly generalizes any finite set of gaps. – Bill Dubuque May 21 '23 at 03:23
  • @BillDubuque Yes, this is much closer to the spirit of the proof I'm looking for. I'm hoping to avoid analysis, however (which, if I recall, is what the proof of Dirichlet requires). The other question I asked that you commented on would help toward such a proof, but the one other comment on the question refers me to Green-Tao, which seems to suggest that asking for such a proof is maybe ambitious. –  May 21 '23 at 03:29
  • @BillDubuque Also, in the case of 2,4, I can make exactly the argument I'm hoping to generalize: 2 cannot appear twice in a row, because otherwise one prime would be divisible by 3. The same reasoning yields that 4 cannot appear twice in a row. Therefore, if the only gaps are 2,4, then the sequence becomes 2,4,2,4,... . Then we are secretly adding 6 repeatedly, and some prime number will be divisible by 7. –  May 21 '23 at 03:38

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