TITLE: How are the integral parts of $(9 + 4\sqrt{5})^n$ and $(9 − 4\sqrt{5})^n$ related to the parity of $n$?
QUESTION [11 upvotes]: I am stuck on this question,

The integral parts of $(9 + 4\sqrt{5})^n$ and $(9 − 4\sqrt{5})^n$ are:  

even and zero if $n$ is even;
odd and zero if $n$ is even;
even and one if $n$ is even;
odd and one if $n$ is even.


I think either the problem or the options are wrong. To me it seems that answer should be odd irrespective of $n$. Consider the following:
$$
\begin{align*}
(9 \pm 4 \sqrt{5})^4 &= 51841 \pm 23184\sqrt{5} \\
(9 \pm 4 \sqrt{5})^5 &= 930249 \pm 416020\sqrt{5}
\end{align*}$$
Am I missing something?

REPLY [6 votes]: To prove that 
$$
a_n:=\left(9+4\sqrt5\right)^n+\left(9-4\sqrt5\right)^n
$$ 
is an even integer, it suffices to observe that $a_0=2,a_1=18$ and 
$$
a_{n+2}=18\ a_{n+1}-a_n
$$
for $n\ge0$.
Variation. 
Note that $a_n$ is an integer. Indeed $a_n$ is of the form $b_n+c_n\sqrt5$ with $b_n,c_n$ integers, and $a_n$ doesn't depend on the choice of the square root of $5$, so $c_n=0$.
Once we know that $a_n$ is an integer, we can compute it in the field $\mathbb F_2$ with two elements (in which $\sqrt5$ exists).
EDIT. The above arguments show this: If $a,b,d,n$ are integers, and if $n\ge0$, then 
$$
\left(a+b\sqrt d\right)^n+\left(a-b\sqrt d\right)^n
$$
is an even integer.