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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.

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  • $\begingroup$ Do you have any reference to a "nonelementary" proof? $\endgroup$
    – Wojowu
    Commented Oct 31, 2015 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
    Commented Oct 31, 2015 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
    Commented Oct 31, 2015 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
    Commented Nov 2, 2015 at 18:18
  • $\begingroup$ @Gerhard That is very interesting, thanks. Probably above my current level though. $\endgroup$
    – René Gy
    Commented Nov 3, 2015 at 17:47

1 Answer 1

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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

In

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".

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  • $\begingroup$ If the numbers are composite, can we give a constant lower bound on $n+k-1$, as long as $n>2k$? $\endgroup$
    – Daniel W.
    Commented Jun 6, 2020 at 11:45

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