abstract algebra - How are the integral parts of $(9 + 4sqrt{5})^n$ and $(9 − 4sqrt{5})^n$ related to the parity of $n$?

I am stuck on this question,

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

1. even and zero if $n$ is even;


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


3. even and one if $n$ is even;


4. 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?

Answer

The idea is to see that $(9+4\sqrt{5})^n+(9-4\sqrt{5})^n=2d_n$ is an even number for every $n$. This can be seen using the binomial expansion formula, and seeing that odd terms appear once with $+$ once with $-$, and even terms are always with $+$ and integers.

Moreover, $0<(9-4\sqrt{5})=9-\sqrt{80}=\frac{1}{\sqrt{81}+\sqrt{80}}<1$. This means that the integer part of the second term is zero. And for the other one, think as this

$$2d_n-1<(9+4\sqrt{5})^n<2d_n$$ so the integer part of the first term is always odd. The correct answer would be the second one (although this happens for every $n$).

[edit] As Arturo Magidin wrote in his comment, the integer part of a real number
$x$, often denoted $\lfloor x \rfloor$ is the unique integer $\lfloor x \rfloor=k$ such that $k \leq x < k+1$, and it does not equal $a$ from the expansion $(9\pm 4\sqrt{5})^n=a\pm b\sqrt{5},\ a,b \in \Bbb{Z}$.