Take $k$ consecutive composite integers from a prime gap. What is known about the largest prime divisor of their product?

It seems to me that except for the triplet $(8,9,10)$ and the pair $(8,9)$ , this largest prime divisor is always larger than $2k$, but I could not find an elementary (my level) proof.

As $k$ grows it seems that the sharpness of $2k$ as lower bound is lost, therefore I expect that there should exist a more suitable lower bound, from which an "easy" argument for the minoration by $2k$ would follow.

The question is essentially the same as this one from MSE but it did not get any answer. I could only find messy $k$-specific arguments for $k$ up to $4$ which I hope are correct.

  • $\begingroup$ Do you have any reference to a "nonelementary" proof? $\endgroup$ – Wojowu Oct 31 '15 at 13:09
  • $\begingroup$ @Wojowu. No, but I am not professional mathematician, so my knowledge of (and access to) math literature is very limited. If you know of relevant papers, I 'd be interested. $\endgroup$ – René Gy Oct 31 '15 at 13:21
  • $\begingroup$ I was asking because when you said "...I could not find an elementary (my level) proof.", I thought you might have found a hard proof and you were just looking for a simpler one. $\endgroup$ – Wojowu Oct 31 '15 at 13:22
  • $\begingroup$ You might be interested in a paper of Filip Najman at arxiv.org/abs/1108.3710 . The largest prime divisor of the product n+1 to n +f(k) is looked at and shown to be larger than k when n is larger than k. f(k) is pretty small and conjectured to be O(log(k)^2). So you don't need as many as k consecutive integers. Gerhard "Sees This As Smooth Intervals" Paseman, 2015.11.02 $\endgroup$ – Gerhard Paseman Nov 2 '15 at 18:18
  • $\begingroup$ @Gerhard That is very interesting, thanks. Probably above my current level though. $\endgroup$ – René Gy Nov 3 '15 at 17:47

There have been several investigations into the largest prime factor of a product of consecutive integers; the Sylvester--Schur theorem is an early example. Here is a survey by Shorey and Tijdeman: https://www.math.leidenuniv.nl/~tijdeman/shoretij.pdf


Laishram, Shanta(6-TIFR-SM); Shorey, T. N.(6-TIFR-SM) The greatest prime divisor of a product of consecutive integers. Acta Arith. 120 (2005), no. 3, 299–306

it is shown that $$ P(n(n+1) \cdots (n+k-1)) > 2k $$ as long as $n > \max\{k+13, \frac{279}{262}k\}$.

Note that for large $k$, the lower bound condition on $n$ is implied by your hypothesis that all of $n, n+1, \dots, n+k-1$ are composite; so this answers your question for large $k$. (And for bounded $k$, it reduces checking your conjecture to a finite computation.)

There is more good news: the MathSciNet review records that "the proof is elementary".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.