<s>[INST] Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$ 
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$

My attempt
I can factor the polynomial 
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so: 
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)

Note: I correct my "solution" after a correction... I made a mistake :\
 [/INST] We have $3$ consecutive numbers so one of these will give a factor of $3$.
If $n$ is even then $n$ will give a factor of $16$. If $n$ is odd then $(n-1)$ and $(n+1)$ will both be even and if $(n+1)^2$ only gives $2$ factors of $2$ then $(n-1)$ will be divisible by $4$, so $(n-1)(n+1)^2$ will be divisible by $16$.
So $\color{red}{48}$ will always divide $n^7+n^6-n^5-n^4$.
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