Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$ $\ge$1 Prove your assertion. Find the maximum natural number $m$ such that $n^3-n$ is divisible by $m$ $\forall$ $n$$\ge$1 Prove your assertion.
I guess $m$$=$1 because if $m$ divides $n^3-n$ $\Rightarrow$ $m$ divides $n(n-1)(n+1)$ then $m$ divides at least one of them, but $\forall$$n$ natural number, $n$,$n-1$,$n+1$ are coprime, so $m$$=$1
Can anybody help correct or improve the answer?

We have $n^3 -n = n\cdot\left(n^2 - 1\right) = n\cdot(n-1)\cdot(n+1)$. These are three consecutive numbers so at least one of them has to be divisible by $3$ and one of them has to be divisible by $2$. Hence, $m$ has to be at least $2\cdot 3 = 6$.
However, for each prime $p \in \mathbb P$ with $p > 3$, we can find an $n \in \mathbb N$, so that $n^3 - n = (n-1)n(n+1)$ is not divisible by $p$, such as $n = p + 2$, since then the expression becomes $(p+1)(p+2)(p+3)$ and none of these factors can be divisible by $p$. We also cannot have $n^3 - n$ always being divisible by bigger powers of $2$ or $3$, any $n = 6k$ is a counterexample of that.
Therefore, $m$ has to be equal to $6$.
