<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] It still involves cases, but here's a somewhat slicker proof.
By Euler's theorem, $a^{\varphi(n)}\equiv 1$ mod $n$, where $\varphi$ is Euler's totient function, which count the number of positive integers less than $n$ coprime to $n$.
As $\varphi(9)=6$, this implies that the residue of
$$2^n+5^n+56$$
mod $9$ depends only on $n$ mod $6$. If we assume $n$ is odd, then the only possible residues are $1,3,5$. As such it is sufficient to verify the claim for $n=1,3,5$.
We can further reduce the number of calculations we have to do by noting that $2$ and $5$ are multiplicative inverses mod $9$. Then, as $2^5\equiv 5$, we must have $5^5\equiv 2$, so the $n=5$ case follows from the $n=1$ case. Similarly, as $2^3\equiv -1$, we have $5^3\equiv -1$, letting us easily verify the $n=3$ case.
Though it is arguably simpler to just directly verify these three cases
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