Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in\mathbb{Z}$. Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in \mathbb{Z}$.
I tried to use induction here, but it seems like a dead end. Maybe I should somehow use binomial theorem?
I need a hint

For the divisibility, I am going to prove that  $a^{2 n+1}+(a-1)^{n+2}\equiv 0 \quad \pmod{a^2-a+1}$
\begin{aligned}
& a^{2 n+1}+(a-1)^{n+2} \\
\equiv & a^{2 n+1}+\left(a^{2}\right)^{n+2} \quad\left(\bmod a^{2}-a+1\right) \\
\equiv & a^{2 n}\left(a+a^{4}\right) \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & a^{2 n}\left[a+(1-a)^{2}\right] \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & a^{2 n}\left(a^{2}-a+1\right) \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & 0 \quad\left(\bmod a^{2}-a+1\right)\\
\end{aligned}
$\therefore a^{2 n+1}+(a-1)^{n+2} \text { is divisible by } a^{2}-a+1.$
