Divisible by?

$(\sqrt{3})^{4n}+(\sqrt{2})^{4n}=9^n+4^n$

$\textbf{Claim:}$

$a+b|a^n+b^n$ when $n$ is odd

$\textbf{Proof:}$

Since $n$ is odd, write $n=2k+1\quad k\in\mathbb{N}$

$\begin{aligned} a^n+b^n &=a^{2k+1}+b^{2k+1}\\ &=(a+b)(a^{2k}-a^{2k-1}b+a^{2k-2}b^2-\ldots -ab^{2k-1}+b^{2k}) \end{aligned}$

$\phantom{cccccccccccccccccccccccccccccccccccccccccccccc}\square$

In this case $a=9$ and $b=4$

$9+4|9^n+4^n$

$\Longrightarrow \boxed{13}|9^n+4^n$

$\begin{aligned} \left(\sqrt 3\right)^{4n} + \left(\sqrt 2\right)^{4n} & = 9^{\color{#3D99F6}n} + 4^{\color{#3D99F6}n} & \small \color{#3D99F6} \text{Let }n = 2m+1 \\ & = (9+4)\left(9^{2m} - 9^{2m-1}\cdot 4 + 9^{2m-2}\cdot 4^2 - \cdots + 4^{2m} \right) \\ & = {\color{#3D99F6}13}\left(9^{2m} - 9^{2m-1}\cdot 4 + 9^{2m-2}\cdot 4^2 - \cdots + 4^{2m} \right) \end{aligned}$

Therefore, it is always divisible by $\boxed{13}$ .