<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] Here's one way (not sure if it is the easiest, though).

Let us start with a lemma which you may easily prove by induction.
Lemma: For each natural number $k$, $2^{2k+1}+1$ and $2^{2k}-1$ are divisible by $3$.

Next, we have the following
Claim:
For $n=2k+1$, one has
$$2^n+5^n+56\equiv-2(2^{2k+1}+1)(2^{2k}-1)\mod 9.$$
In fact,
$$2^n+5^n+56\equiv 2^{2k+1}+(-2^2)^{2k+1}+2\equiv 2(2^{2k}-2\cdot 2^{4k}+1) \mod 9.$$
Observe that the RHS is equal to the RHS of the claim.

By the lemma, the RHS of the claim is divisible by $3\cdot 3=9$, hence the LHS as well.
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