factor into $n^2(n-1)(n+1)(n^2+n+1)(n^2-n-1)$ this is always dividing by 9, you can try it yourself with 0, 1, 2 mod 3 now we consider mod 8 first we see $n$ has to be even since if it is odd we have so many even terms with + and - 1 then we see $n$ cannot be 0 mod 4 since then $n^2$ would be divisible by 8 so we get $n$ is 2 mod 4 that is all Peace Br Morgan