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Given an integer $n \geq3$ we consider $S(n)$ to be the set of all primes less than $n$ that do not divide $n$.
The question is: Are there two distincts numbers $n$ and $m$ such that $S(n)=S(m)$?

Another similar question is: Given a finite set of primes $P = \{p_1, ... , p_k\}$, does there exist a number $n > \underset{1 \leq i \leq k}{\max}p_i$ such that $p_i \nmid n$ for all $p_i \in P$, and that there are no primes between $\underset{1 \leq i \leq k}{\max}p_i$ and $n$?

For example is there a number $n$ between $53$ and $59$ such that none of the primes in $\lbrace 3, 17, 53\rbrace$ divide $n$? In this case the answer is yes because $n=55$ works.

Ignacio Henríquez
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    One way to look for such $n,m$ might be to consider numbers like $210\cdot 224,210\cdot 225$ or other similar values where consecutive numbers are made up of small factors, and then multiply these by the associated primorial... – abiessu May 01 '22 at 19:06
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    No solutions found with $m$ or $n$ less than 85 million. – Carl Schildkraut May 02 '22 at 03:28
  • Please specify more precise what "number" means here. If we allow positive integers, then we have $S(1)=S(2)$ – Peter May 02 '22 at 07:55
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    There is no trivial answer, but this question was asked before, in a different form though. – rtybase May 02 '22 at 11:00
  • @Peter Thanks for pointing that out, I have edited the question to specify that we are allowing $n$ to be an integer bigger than or equal to 3. – Ignacio Henríquez May 03 '22 at 03:38
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    @rtybase Thanks!. – Ignacio Henríquez May 03 '22 at 03:42
  • To get a counterexample, we have to find distinct positive integers $m,n\ge 2$ with the properties : $(1)$ $m$ and $n$ have the same prime factors $(2)$ There is no prime $p$ with $m<p<n$. – Peter May 03 '22 at 08:25
  • Here is how I would look, if I were willing to put in the work: Check the 3-smooth numbers, i.e. numbers which have only $2$ and $3$ as prime factors. Look for ones that are 'close' together; there are many that differ by less than 2 percent. If there are no primes between them, they satisfy your conditions. I am not optimistic that this search will actually yield any examples, as refinements to Bertrand's postulate suggest that those intervals are likely to contain primes. – Keith Backman May 27 '22 at 16:59

1 Answers1

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Assume n < n. If there is a prime n <= p < m then by definition m is a multiple of p. If there is exactly one prime p then m >= 2p >= 2n, if there are two or more primes then m >= product of those primes. That should be enough to prove there cannot be such a prime between n and m which means n and m are rather close.

So the primes less than n and the primes less than m are the same, the ones not dividing n and m are the same, therefore the primes dividing n and m are the same. Let N be the product of the distinct prime factors, then n = kN, m = k’N, where k and k’ are different products formed from these primes. We have (m - n) = N (k’ - k) >= N.

As an example, with N = 6, k, k’ could be 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27 etc. Unfortunately there is always a prime in between, for example n = 86, m = 96, withd the prime p = 53 in between.

It is known that the gap between consecutive primes < 1.8 * 10^19 is less than 1,550. So wo only need the examine N <= 1550, we can exclude N primes, and for each N < 1550 which is the product of two or more distinct primes, we can quickly find all products k of those distinct primes. So we can verify this quickly for n, m < 1.8 * 10^19.

We could also show reasonably quickly that a counter example must be for example <= 10^50 with a prime gap >= 10,000.

gnasher729
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