When is this sequence of positive integers a square? 
I have two sequences below, and I would like to know for which $n$ the number $k_n$ is a square.
  $$
\begin{align}
k_1 &= 9\\
t_1 &= 1\\
k_{n+1} &= 9k_n + 80t_n\\
t_{n+1} &= k_n + 9t_n
\end{align}
$$

I'm conjecturing that there are no $n>1$ such that $k_n$ is square based on computational evidence, however I'm having a hard time proving it. My main idea that I've been trying to use is induction, and saying that if $k_n$ is not a square, then $k_{n+1}$ is not a square. I've tried using modular arithemtic on the sequence with modulars such as $4$, $5$, $8$, $9$, $16$, $80$ and other obvious choices, however none of them avoided the quadratic residues in the cycle started from $k_2=161$.
If you can find an explicit formula for $k_n$, please do give it.
 A: We have:
$$\left(\begin{array}{c}k_{n+1}\\t_{n+1}\end{array}\right)=\left(\begin{array}{cc}9&80\\1&9\end{array}\right)\left(\begin{array}{c}k_{n}\\t_{n}\end{array}\right)$$
and since the eigenvalues of the characteristic matrix are $9\pm 4\sqrt{5}$, we have:

$$ k_n = \frac{1}{2}\left((2+\sqrt{5})^{2n} + (2-\sqrt{5})^{2n}\right),\tag{1} $$

with $k_{n+2}= 18 k_{n+1}-k_n$. It is interesting to notice that, by $(1)$,
$$ k_n = \frac{1}{2}\left((2+\sqrt{5})^n + (2-\sqrt{5})^n\right)^2-(-1)^n=\frac{1}{2}K_n^2-(-1)^n\tag{2}$$
so we can detect which elements of $\{k_n\}_{n\in\mathbb{N}}$ are squares by intersecting  the sequence $\{K_n\}_{n\in\mathbb{N}}$ with the two sequences giving the solutions of the Pell equations:
$$2A_n^2 \pm 1 = \square. $$
It is worth to try to adjust Cohn's proof of the fact that the only squares in the Fibonacci sequence are $0,1,144$. It can be found here.
A: Automorph of $$ k^2 - 80 t^2. $$ By Cayley-Hamilton, both
$$  k_{n+2} = 18 k_{n+1} - k_n   $$ and
$$  t_{n+2} = 18 t_{n+1} - t_n   $$
So this is for that quartic you made up with the multiples of 10. Anyway, characteristic polynoial is $\lambda^2 - 18 \lambda + 1$ and the roots are irrational. That is the only thing available that gives a "formula" for $k_n.$ 
Let's see, $$ \lambda = \frac{18 \pm \sqrt {320}}{2} = \frac{18 \pm 8 \sqrt {5}}{2} = 9 \pm 4 \sqrt 5 = 9 \pm  \sqrt {80}.    $$ Both are positive, so
there are real constants $A,B$ such that
$$  k_n = A \left(9 +  \sqrt {80} \right)^n + B \left(9 -  \sqrt {80} \right)^n. $$ 
