For $n,m \geq 3$, define $ P_n = \{ p : p$ is a prime such that $ p\leq n$ and $ p \nmid n \}$ .
For example : $P_3= \{ 2 \}$ $P_4= \{ 3 \}$ $P_5= \{ 2, 3 \}$, $P_6= \{ 5 \}$ and so on.
Claim: $P_n \neq P_m$ for $m\neq n$.
While working on prime numbers I formulated this problem and it has eluded me for a while so I decided to post it here. I am not sure if this is an open problem or solved one. I couldn't find anything that looks like it. My attempts haven't come to fruition though I have been trying to prove it for a while. If $m$ and $n$ are different primes then it's clear. If $m \geq 2n$, I think we can find a prime in between so that case is also taken care of. My opinion is that it eventually boils down to proving this statement for integers that share the same prime factors. My coding is kind of rusty so would appreciate anybody checking if there is a counterexample to this claim. Any ideas if this might be true or false? Thanks.
Update: Lucia has provided a conditional proof of the claim for large integers on mathoverflow. Please see
https://mathoverflow.net/questions/287011/a-conjecture-regarding-prime-numbers/287039#287039
Let $m,n\ge 3$ be given and suppose that $m\neq n$. Without loss of generality, we can say that $n>m$.
Assume there are primes between $m$ and $n$. Let $p$ be the largest prime number with $n>p\ge m$. It is clear that $p\not\in P_m$. If $p\not\in P_n$, we must have $p\mid n$, so $n\ge 2p$. However, Betrand's postulate tells us there is at least one prime inbetween $p$ and $2p$, contradicting the definition of $p$. We conclude that $p\in P_n$ and $P_n\neq P_m$.
Now, say there are no primes between $n$ and $m$. Now, we have $P_n=P_m$ if and only if $rad(n)=rad(m)$. Say $rad(n)=r$. So now we have $k,l\in\mathbb{N}$ with $n=kr$ and $m=lr$. This is the hard part. (In fact it's very hard, as shown in an answer to the repost on Mathoverflow)