$n^2(n^2-1)(n^2-4)$ is always divisible by 360 $(n>2,n\in \mathbb{N})$ How does one prove that $n^2(n^2-1)(n^2-4)$ is always divisible by 360? $(n>2,n\in \mathbb{N})$
I explain my own way:
You can factorize it and get $n^2(n-1)(n+1)(n-2)(n+2)$.
Then change the condition $(n>2,n\in \mathbb{N})$ into $(n>0,n\in \mathbb{N})$ that is actually equal to $(n\in \mathbb{N})$.
Now the statement changes into :
$$n(n+1)(n+2)^2(n+3)(n+4)$$
Then I factorized 360 and got $3^2 \cdot 2^3 \cdot 5$.
I don't know how to prove the expression is equal to $3^2 \cdot 2^3 \cdot5$.
Who can help me solve it?
 A: $n(n+1)(n+2)$, and $(n+2)(n+3)(n+4)$ are $2$ products of $3$ consecutive natural numbers hence each divisible by $3! = 6$, thus the product divisible  by $6\cdot 6 = 36$, hence it is divisible by $9$, and it is divisible by $5! = 120$ since the product contains $5$ consecutive natural numbers $n(n+1)(n+2)(n+3)(n+4)$, thus it is divisible by $\text{lcm}(36,120) = 360$.
A: Now you can just argue about each prime factor.  If $n$ is odd, $n+1$ and $n+3$ are even and one of them is a multiple of $4$, giving three factors of $2$.  One of your factors must be a multiple of $5$ and two of them must be multiples of $3$ (you need to separate out the case of $n+2$ being the multiple, but there are two of them.  So the product is a multiple of $360$
A: $$n^2(n^2-1)(n^2-4)=(n-2)(n-1)n(n+1)(n+2)\cdot n \\
=(n-2)(n-1)n(n+1)(n+2)(n+3)-3(n-2)(n-1)n(n+1)(n+2)$$
And
$$(n-2)(n-1)n(n+1)(n+2)(n+3)=720 \cdot \binom{n+3}{6}$$
$$3(n-2)(n-1)n(n+1)(n+2)=3\cdot 120 \binom{n+2}{5}=360 \binom{n+2}{5}$$ 
A: Since $\{0,1,4\}$ are the possible squares $\pmod{5},\pmod{8},$ your number is clearly divisible by $5\cdot 8=40$. By considering all the possibilities for $n\pmod 3$, we also have $9\mid n^2(n^2-1)(n^2-4)$, so $360\mid n^2(n^2-1)(n^2-4)$ as wanted.
A: For variety, I  prove a more general result, which sheds light on the innate structure.
${\rm Note}\,\ \ f = (n^3\!-\!n)g,\,\ \ g\, =\, n(\color{#c00}{n^2\!-\!4})\, =\, \color{#0a0}{n^3\!-\!n\!-\!3n}\, =\, (\color{blue}{n^2\!+\!1 -5})n\ $ and apply
Lemma $\ $ If $\ f = (n^3\!-\!n)g,\,\ \color{#c00}{n^2\!-\!4i}\mid g,\,\ \color{#0a0}{n^3\!-\!n\! -\! 3j}\mid g,\,\ \color{blue}{n^2\!+\!1\!-\!5k}\mid g$ then $\ 360 \mid f$
Proof $\ $ mod $5\!:\ f = n(n^2\!-\!1)(\color{blue}{n^2\!+\!1\!-5k}) \equiv\, n^5\!-\!n\equiv 0$ by little Fermat. 
Also $\ f = (n^3\!-\!n)(\color{#0a0}{n^3\!-\!n - 3j})\,$ so $\ 3\mid n^3-n\,\Rightarrow\, 3^2\mid f,\ $ by little Fermat.
If $\,n\,$ is odd then, $ $ mod $\,8\!:\ n^2\equiv 1\,$ so $\,f = (n^2-1)ng\equiv 0,\, $
else $\ 2\mid n\ $ so $\ 4\mid \color{#c00}{n^2\!-\!4i}\mid g\ $ so $\ 8\mid ng\mid f.\,$ In every case $\,8\mid f.$  
Therefore $\ 5,9,8\mid f\,\Rightarrow  {\rm lcm}(5,9,8)\mid f,\,$ i.e. $\ 360\mid f\ \ $ QED
