<s>[INST] Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be correct to use this and try solving through induction?
 [/INST] The product $p(p+1)$ of two consecutive numbers is divisible by $2$ 
because either $p$ or $p+1$ is even.
The product $p(p+1)(p+2)$ of three consecutive numbers is divisible by $3$
because either $p,\ p+1$ or $p+2$ is a multiple of $3$.

$f(n)=n(n^2-1)=(n-1)n(n+1)$ is then divisible by $3$.
Also for $n=2p+1$ odd then $(n-1)(n+1)=(2p)(2p+2)=4p(p+1)$ is divisible by $4\times 2$
So $f(n)$ is divisible by $8\times 3=24$.
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