<s>[INST] $n^2(n^2-1)(n^2-4)$ 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?
 [/INST] $$n^2(n^2-1)(n^2-4)=(n-2)(n-1)n(n+1)(n+2)\cdot n \\
=(n-2)(n-1)n(n+1)(n+2)(n+3)-3(n-2)(n-1)n(n+1)(n+2)$$
And
$$(n-2)(n-1)n(n+1)(n+2)(n+3)=720 \cdot \binom{n+3}{6}$$
$$3(n-2)(n-1)n(n+1)(n+2)=3\cdot 120 \binom{n+2}{5}=360 \binom{n+2}{5}$$ 
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