# Why does each integer between two consecutive primes have at least one “unique” non-trivial divisor?

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.

• 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. – Sameer Kailasa Mar 6 at 18:25
• @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. – Robert Israel Mar 6 at 18:54

• 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. – Dmitry Pyatin Mar 6 at 20:40