Prove that $n^3(n^2-1)$ is divisible by 24 for all n I thought of trying induction but got stuck 
$P(n)=24 \mid n^3(n^2-1)$ for $n_0 =1$. 
For the base case if $n=1$ then $24\mid 1(1-1)$ or $24 \mid 0$, which is true. 
Now, assume $P(n)$ is true I want to show it holds for $n+1$.
$24 \mid (n+1)^3((n+1)^2-1)$
$24 \mid n^5+5n^4+9n^3+7n^2+2n$
Now I can't see any way to use my assumption with factoring. I thought of trying a system of linear congruence's to use the Chinese remainder theorem, but it didn't go anywhere. Any hints of tips would be appreciated. 
 A: This is a proof that uses elementary number theory:
Since $n^3(n-1)(n+1)$ contains a product of three consecutive integers it is divisible by $3!=6.$
If $n=2k$ is even $n^3=8k^3$  is divisible by $8.$
If $n=2k+1$ is odd $n^2-1=4k(k+1)$ is divisible by $8.$
Therefore in either case $n^3(n^2-1)$ is divisible by $\text{lcm} (6,8)=24.$

Moreover there are many single line proves for this fact. One is: $$n^3(n^2-1)=24(n-2)\dbinom{n+2}{4}+24\dbinom{n+1}{3}.$$

A: $2\mid n \,\Rightarrow\, 8\mid \color{#c00}{n^3},\,$ else $\,(n,2)=1 \,\Rightarrow\, 8\mid \color{#0a0}{n^2-1}\ $ by $\,{\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1 \pmod{\!8}$
$3\mid n \,\Rightarrow\, 3\mid\color{#c00}{ n^3},\,$ else $\,(n,3)=1 \,\Rightarrow\, 3\mid \color{#0a0}{n^2-1}\ $ by $\,n\not\equiv0\,\Rightarrow\, n\equiv\pm1\,\Rightarrow\,n^2\equiv 1\pmod{\!3}$
So in all cases $\,\color{#c00}{n^3}(\color{#0a0}{n^2-1})\,$ is divisible by $8$ and $3$ so it is divisible by their lcm $= 24$.

Remark $ $ The same idea works for any primes if we use Euler vs. brute-force case analysis.
Theorem $\ $ For primes $\rm\:p \ne q\:,\:$ naturals $\rm\:e,\:$ and $\rm\ j,\ k \:\le\: d\ $
$$\rm\quad\quad\ \phi(p^j),\ \phi(q^k)\ |\ e\ \ \Rightarrow\ \ p^j\ q^k\ |\ n^d(n^e - 1)\ \ \ \forall\ n\in \mathbb N $$
Proof $\ $ If $\rm\ p\ |\ n\ $ then $\rm\ p^j\ |\ n^d\ $ by $\rm\ j\le d\:.\:$ Else $\rm\:n\:$ is coprime to $\rm\: p\:,\:$ so by Euler's little theorem we have $\rm\bmod p^j\!:\ n^{\phi(p^j)}\equiv 1\ \Rightarrow\ n^e\equiv 1\ $ by $\rm\ \phi(p^j)\ |\ e\:.\ $  Thus $\rm\  n^d\ (n^e - 1)\ $ is divisible by $\rm\ p^j\ $ and, similarly it is divisible by $\rm\ q^k\:,\ $ hence it is also divisible by their lcm = product. $\quad$ QED
In fact for $\rm\ p = 2,\ j > 2\ $ we can use $\rm\ \phi(2^j)/2\ $ vs. $\rm\ \phi(2^j)\ $ because $\rm\ \mathbb Z/2^j\ $ has multiplicative group $\rm\ C(2)\times C(2^{j-2})\ $ for $\rm\ j> 2\:$. For more see a post on the Fermat-Euler-Carmichael theorem.
Generalization: $\ 24\,$ is the largest $m$ such that $\,m\mid n^2-1\,$ for all $\,n\,$ coprime to $\,m.\,$ This theorem is at the heart of many deductions in (elementary)  number theory (and also in some more advanced exotic contexts, e.g. see  Moonshine beyond the Monstor).
A: It is easier just to use the factorization you are given.  $n^2-1=(n+1)(n-1)$ so you have a factor $3$.  The if $n$ is even $n^3$ is divisible by $8$ and if $n$ is odd $n^2-1$ is divisible by $8$ because one of the factors is divisible by $4$.
A: Use congruences and Fermat's small theorem: it suffices to prove it is divisible by $3$ and by $8$.
Now,  $n^3\equiv n\mod 3$, so $n^3(n^2-1)\equiv n(n^2-1)=n^3-n\equiv 0\mod 3$.
Modulo $8$:


*

*either $n$ is even, so $n^3\equiv 0\pmod 8$,

*or $n$ is odd: $n=2m-1$ for some $m$, so 
$\;\dfrac{n^2-1}4=m^2-m\equiv 0\pmod 2.$

A: Because
$$n^3(n^2-1)=n^5-5n^3+4n+4n^3-4n=$$
$$=(n-2)(n-1)n(n+1)(n+2)+4(n-1)n(n+1).$$
A: You may use Congruences & the difinition of $lcm(a,b)$ to prove that $24\mid n^{3}(n^{2}-1)$
1) $ n^3 (n^2 -1)$=$ n^2[(n-1)n(n+1)]$
$[(n-1)n(n+1)]\equiv 0[mod 3!]$
$[(n-1)n(n+1)]\equiv 0[mod 6]$
2) Now, what about $n^3$? it may be even or odd?
$n$ Even, than $n=2m$, so $n^3 =8m^3$ and $8m^3 \equiv 0[mod8]$
$n$ Odd, than $n=2m+1$, so $n^3 =4m(m+1)$ and $4m(m+1) \equiv 0[mod8]$
3) $lcm(6,8)|n^3 (n^2 -1)$ so $24\mid n^{3}(n^{2}-1)$
