When is this sequence of positive integers a square?

Question

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.

Answer 1

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 $$ k_n = A \left(9 + \sqrt {80} \right)^n + B \left(9 - \sqrt {80} \right)^n. $$

Answer 2

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.