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?
 A: Induction on odd numbers is easier when expressed as $P(n_0)$ is true, $P(n) 
\implies P(n+2)$.
Let $f(n)=n(n^2-1)$. Then $f(1)=24$ and $f(n+2)-f(n)=6 (n + 1)^2$.
If $n$ is odd, then $n+1$ is even and so $(n+1)^2$ is a multiple of $4$. 
Therefore, $f(n+2)-f(n)$ is a multiple of $24$ and so is $f(n+2)$ by induction.
For fun, here is a different approach using finite differences.
Write $n=2t+1$. Then
$$
n(n^2-1) = 8 t^3 + 12 t^2 + 4 t
= 24 \binom{t}{1} + 72 \binom{t}{2} + 48 \binom{t}{3} 
$$
clearly a multiple of $24$.
A: 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 $8\times 3=24$.
A: Easier with congruences:


*

*$n^3-n\equiv 0\mod 3$ for all $n$ (that's Little Fermat's theorem), 

*If $n$ is odd, $n\equiv \pm 1,\pm3\mod 8$, so $\;n^2\equiv 1\mod8$,

*last, use the Chinese remainder theorem.

A: Let $n=2m+1$.
Among $2m(2m+1)(2m+2)$, one of the extreme factors is a multiple of $4$ and the other is even, and one of the three factors is a multiple of $3$.

Check with
$$2\cdot3\cdot4,\\4\cdot5\cdot6=4\cdot5\cdot2.3,\\6\cdot7\cdot8=2.3\cdot7\cdot2.4,\\8\cdot10\cdot12=2.4\cdot10\cdot2.2.3,\\\cdots$$
A: Because $N=n(n+1)(n-1)$ is a product of three consecutive numbers it is a multiple of both $2$ and $3$ so a multiple of $6$. Besides $N=2n+1$ gives
$$N=(2n+1)(2n+2)(2n=4(2n+1)(n+1)n$$ and $N$ is also a multiple of $4$ so $N=24t$ for some integer $t$.(Note that $(n+1)n$ is even).
