Let n be a natural number, show that at least one of $(n^5)-1$, $(n^5)$ and $(n^5)+1$ is divisible by $11$.
I started by considering two cases: $n$ is even. $n=2k$, so the three numbers become $32(k^5)-1$, $32k^5$, and $32(k^5)+1$ In mod $11$, it is equal to $-(k^5)-1$, $-k^5$, and $-(k^5)+1$
If $n$ is odd, $n^5=32k^5+80k^4+80k^3+40k^2+10k+1$ in mod $11$ it is $-k^5+3k^4+3k^3+7k^2-k+1$
And I want show that one of them eventually become $0$ in mod $11$, but it doesn't seem to help...
If $11\mid n$, then $11\mid n^5$.
Otherwise, $11\mid n^{10}-1$ by Fermat's little theorem. And since $n^{10}-1=(n^5-1)(n^5+1)$, it follows from this that $11$ divides one of the numbers $n^5-1$ or $n^5+1$.