What is the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ is an integer? 
Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit. 

I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any pattern. I coudn't find any pattern but I eventually solved the problem to be $n=6$, but is there any easier more practical approach to this problem?

$$
\begin{align}
(-\sqrt{2} + i \sqrt{6})^n
& = \left(\sqrt{2} \left(-1 + i \sqrt{3} \right)\right)^n
& = 2^{n/2} \left(-1 + i \sqrt{3} \right)^n\\
& = 2^{n/2} \left( 2 \left( - \frac12 + i \frac{\sqrt{3}}{2} \right) \right)^n
& = 2^{n/2} 2^n \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n \\
& = 2^{3n/2} \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n
& = 2^{3n/2} \left(\cos \left(\frac{2 \pi}{3} \right) + i \sin \left(\frac{2 \pi}{3} \right)\right)^n
\end{align}
$$
$$(-\sqrt{2} + i \sqrt{6})^n = 2^{3n/2} \left(\cos \left(\frac{2 n\pi}{3} \right) + i \sin \left(\frac{2 n\pi}{3} \right)\right)$$
For $2^{3n/2}$ to be an integer, we need $\frac{3n}{2} \in \mathbb{Z}$. We also need $\sin \left(\frac{2 n\pi}{3} \right) = 0$. This gives us $ \frac{2n}{3} \in \mathbb{Z}$.
Hence, $n = \frac{3k_1}{2} = \frac{2k_2}{3}$ where $k_1,k_2 \in \mathbb{Z}$. Hence, $n = 6k$ where $k \in \mathbb{Z}$.
