For integer $n\ge2$, Is there always a prime p such that $v_p(n^3-n)=1$?
For example,
$n=2$: $p=2$ $(6=2\times3)$
$n=3$: $p=3$ $(24=2^3\times3)$
$n=9$: $p=5$ $(720=2^4\times 3^2\times5)$
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Well, $\ n^3-n\ = (n-1)\cdot n\cdot(n+1).\ $ First of all, I'd think, one should know as much as possible about the products of two consecutive natural numbers -- is one of them divisible by prime $\ p\ $ but not by $\ p^2?\ $ Of course $\ 8\cdot 9\ $ case is special. $\endgroup$– Wlod AACommented Oct 28, 2022 at 7:22
-
$\begingroup$ Case of $\ n^2-1\ =(n-1)\cdot(n+1)\ $ is interesting too! $\endgroup$– Wlod AACommented Oct 28, 2022 at 7:27
-
5$\begingroup$ It's a famous conjecture that there do not exist three consecutive powerful numbers. $\endgroup$– mathworker21Commented Oct 28, 2022 at 7:41
-
$\begingroup$ If there is an $n$ such that $\nu_p(n^3-n) \ge 2$, for all prime dividing $n^3-n$, then $n$ is powerful and all odd prime $p$ dividing $n^2-1$ must be Wieferich prime to the base $n$, so are very rare. $\endgroup$– CHUAKSCommented Oct 30, 2022 at 6:57
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
If the answer is yes, then it will be difficult to prove. Indeed, in this case there is no $n\geq 2$ such that $n-1$, $n$, $n+1$ are simultaneously powerful, but this is only a conjecture of Erdős (1976) and Mollin-Walsh (1986).
answered Oct 28, 2022 at 7:45
-
-
1$\begingroup$ @mathworker21 No, this is an answer. Let us call the conjecture of Erdős (1976) Conjecture 1. One can similarly conjecture that the odd parts of $k$, $k+1$, $2k+1$ cannot be simultaneously powerful. Let us call this Conjecture 2. Then Conjectures 1-2 are equivalent to the statement that the answer to the OP's question is affirmative. $\endgroup$ Commented Oct 28, 2022 at 8:12
-
1