[Edited following Gerhard's answer. I had forgotten to state a crucial assumption; my apologies for the confusion. I have reworded slightly to try to make things clearer.]

Let $p$ be the $k$-th odd prime. Let $n$ be an even integer. Let $p_{1}$, $\ldots$, $p_{k+1}$ be consecutive odd primes. Suppose that $p$ divides $n - p_{i}$ for all $1 \leq i \leq k+1$. For instance, take $p=3$, take $n=100$, take $p_{1}=31$, and take $p_{2}=37$. Is there then at least one $i$ such that $n - p_{i}$ has a prime factor strictly greater than $p$?

Anything anybody can say (proof or reference) is welcome.

It is easy to prove that this holds when $k=1$. Indeed, using the fact that $n$ is even, so that both $n - p_{1}$ and $n - p_{2}$ are odd, the claim is immediate unless both $n-p_{1}$ and $n-p_{2}$ are powers of 3. In that case, the difference between $p_{1}$ and $p_{2}$ is $2p_{1}$, and this is impossible by Bertrand's postulate.

There are also plenty of examples when $k=2$. The one with lowest $n$ is $n=1662$, and the primes $1627$, $1637$, $1657$. Experimental evidence suggests that the question can be answered in the affirmative in this case too.

I would guess that there are examples for all $k$, but that very large numbers will be required; I do not actually have an example for any $k > 2$. Anything that can be said about the existence of examples is welcome too.

  • $\begingroup$ Please proofread: some of your $p$s are missing subscripts, and your example doesn't satisfy the property described. $\endgroup$ – Greg Martin Jun 12 '18 at 23:12
  • $\begingroup$ I think you are still leaving stuff out. Your proof for k=1 does not hold, and while the specific examples I provide (now at this writing) do not hold either, the general idea behind my construction still holds. Gerhard "Will Wait For Some Stability" Paseman, 2018.06.13. $\endgroup$ – Gerhard Paseman Jun 13 '18 at 21:15
  • $\begingroup$ By the way, n=550 is part of a counterexample for k=1. Gerhard "Because Primes Can Be Large" Paseman, 2018.06.13. $\endgroup$ – Gerhard Paseman Jun 13 '18 at 23:10

Consider $k=1$ (so $p=3$), $n=32$, $p_1=23$ and $p_2=29$. This is a counterexample to your statement.


I believe your example to consider is still out of sync with your stated problem. If you changed it to $n+ p_1,n+p_{k+1}$, it would line up.

Your result is not true for $k=1$. As an example , consider 24 and 36. All that is needed is to have a prime gap of 12 for your statement to fail, and indeed 199 , 211 is one such example.

For more counterexamples, consider smooth numbers whose relative differences are divisible by the $k+1$st primorial, then go hunting for consecutive primes whose differences match that. Much like searching for primes in arithmetic progressions, you will (eventually) be rewarded (with short examples).

Gerhard "Keep Hopes And Patience High" Paseman, 2018.06.13.

  • $\begingroup$ For an example coming after a prime gap, consider 48 and 54. It may be possible to prove that this is the only such example, and that other examples will have the first n less than the first p. Gerhard "Smooth Numbers Spread Out Quickly" Paseman, 2018.06.13. $\endgroup$ – Gerhard Paseman Jun 13 '18 at 16:53
  • $\begingroup$ I don't have time now to check if 240,250,270 is a counterexample for k=2, but it would not surprise me if you could find three consecutive primes less than a billion which had the same difference pattern. Gerhard "Likely Smaller Examples Also Exist" Paseman, 2018.06.13. $\endgroup$ – Gerhard Paseman Jun 13 '18 at 17:00

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