<s>[INST] Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$. I found this problem in a book, I can't solve it unfortunately.
Prove that for all integer values $n$, $n^9 - 6n^7 + 9n^5 - 4n^3$ is divisible by $8640.$
So far I've noticed that $8460 = 6! \times 12$, also I've tried to simplify that expression and I've found that it's equal to this $n^3(n^3-3n-2)(n^3-3n+2)$, but I can't move on after that.
 [/INST] We can further note that $-1$ is a root of $n^3-3n-2$, and that $1$ is a root of $n^3-3n+2$.

You can find these roots by the Rational Root theorem:

Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.

Other techniques can also be found in this question and its answers.

Then we simplify it to \begin{align*} n^3(n+1)(n^2-n-2)(n-1)(n^2+n-2) &= n^3(n+1)(n-2)(n+1)(n-1)(n+2)(n-1)\\ &=n^3(n+1)^2(n-1)^2(n-2)(n+2) \end{align*}

There are 6 factors two in 8640. 


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*If $n$ is even, $n, n-2$ and $n+2$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2. 

*If $n$ is odd, $n-1$ and $n+1$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2. 



There are 3 factors three in 8640. 


*

*If $n \equiv 0 \mod 3$, then $n$ is divisible by 3, and then $27 \mid n^3$.

*If $n \equiv 1 \mod 3$, then $n-1$ and $n+2$ are divisible by 3. 

*If $n \equiv 2 \mod 3$, then $n+1$ and $n-2$ are divisible by 3. 



The factor 5 can be taken care of since the product is a multiple of five consecutive integers, namely a product of $(n-2)(n-1)n(n+1)(n+2)$.
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