$ (3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n] $ Proof that for all $n\in \mathbb{N}$ : 
$$
(3+\sqrt{5})^n+(3-\sqrt{5})^n\equiv\; 0 \; [2^n]
$$

Another view on this:
$\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$
Is the trace of the $n$-th power of
$\left(\begin{array}{cc} 2 & 1 \\ 1 & 1\end{array}\right)$
And another:
$\dfrac{3+\sqrt 5}{2}$ and $\dfrac{3-\sqrt 5}{2}$ are algebraic integers (they're roots of $x^2 - 3x + 1$). Then since $\left(\dfrac{3+\sqrt 5}{2}\right)^n +\left(\dfrac{3-\sqrt 5}{2} \right)^n$ is rational, it's an integer.
All of these methods are closely related.
