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Why does each integer $x$ between two consecutive primes have at least one non-trivial divisor that unique on set of all integers between these two consequtive primes except $x$?

We call a divisor $d$ of a integer $x$ unique on a set of integers $Q$, if there is no number from $Q$ divisible by it.

Perhaps this is a question for math.stackexchange, but this fact (or, may be, false empirical observation?) seemed to me interesting.

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    $\begingroup$ This would follow if it were known that there is always a prime between $x$ and $x+\sqrt{x}$ (which I believe is conjectured to be true). Indeed, if $m$ had no unique divisors, then in particular its largest divisor $d$ would be nonunique. We have $d\ge \sqrt{m}$ so either the interval $m-\sqrt{m}$ to $m$ or $m$ to $m+\sqrt{m}$ would have no primes. $\endgroup$ – Sameer Kailasa Mar 6 at 18:25
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    $\begingroup$ @SameerKailasa Conjectured to be true for large enough $x$. There are no primes between $116$ and $116 + \sqrt{116}$: I think this is conjectured to be the largest counterexample. See OEIS sequence [A127441. $\endgroup$ – Robert Israel Mar 6 at 18:54
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If we consider divisors to mean prime divisor, then this holds for most short intervals (less than length 6) and fails for some long intervals. 120 and 125 and 126 are smooth numbers between the same two consecutive primes; 24 and 27 are two others. In general, if you have an interval of 2k+1 consecutive composite numbers which contains a number whose prime factors are all at most k, this will spoil your conjecture.

Although it is not guaranteed to find all such examples, algorithm S from other MathOverflow questions (243490,248042) can be modified to find some other examples (such as 30600,30625).

If we are talking arbitrary divisors, then every number n has a divisor of n. If we are talking arbitrary proper divisors, then it is likely to be true as most numbers n have a divisor larger than the observed prime gap (larger than (log n)^2) and most n have a divisor larger than the asymptotic upper bound for prime gaps (O(n^(21/40)). This would mean (as observed in a comment above) that a spoiler would have to be a number with two equal or nearly equal nontrivial prime divisors, placed in the middle of an unexpectedly large prime gap.

Gerhard "It's Also About Smooth Numbers" Paseman, 2019.03.06.

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  • $\begingroup$ Gerhard, thank you! But I do not undestand Why interval between primes 113 and 127 is counterexample? I have output "[x - number, u - min unique divisor] all divisors": [x:114 u:19]2,3,6,19,38,57| [x:115 u:23]5,23| [x:116 u:29]2,4,29,58| [x:117 u:13]3,9,13,39| [x:118 u:59]2,59| [x:119 u:17]7,17| [x:120 u:8]2,3,4,5,6,8,10,12,15,20,24,30,40,60| [x:121 u:11]11| [x:122 u:61]2,61| [x:123 u:41]3,41| [x:124 u:31]2,4,31,62| [x:125 u:25]5,25| [x:126 u:14]2,3,6,7,9,14,18,21,42,63| So, number 120 has unique divisor 8, 125 - 25 and 126 - 14. $\endgroup$ – Dmitry Pyatin Mar 6 at 20:40
  • $\begingroup$ I misread the problem, and added a clarification. Because of its tie to another conjecture, I prefer the prime divisor version of the problem. Your version is likely as hard as (and almost the same as) a conjecture like Legendre's or Opperman's. Gerhard "Maybe It Will Be Proved" Paseman, 2019.03.06. $\endgroup$ – Gerhard Paseman Mar 6 at 20:52
  • $\begingroup$ Also, the asymptotic result on prime gaps talks about n sufficiently large. It is remotely possible that there is a superlarge prime gap of size bigger than n^2/3, in which case spoilers might have three prime divisors. However, there would be only finitely many of those. Gerhard "Remember To Accept The Exceptions" Paseman, 2019.03.06. $\endgroup$ – Gerhard Paseman Mar 6 at 21:11

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