<s>[INST] Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer 
Prove that $9 \mid2^n + 5^n + 56$ where n is odd

I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
 [/INST] Hint: $ $ let $\ \begin{align}n &= 2k\!+\!1\\ a&=(-2)^k\end{align}. $ $\bmod 9\,$ it's $\:\!f(a)\equiv\, 2a^2\!+\!5a\!+\!2\,\equiv 2\,(a\!-\!1)^2\equiv 0\,$ by $\, 3\mid a\!-\!1\!\!$
Or, conceptually $\,a\equiv 1\,$ is double root of $\,f(a)\,$ by  $\,f(1)\equiv 0\equiv f'(1)\,$ by $\,f'(a) \equiv 4a\!+\!5$
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