Every term in a recurrence is an integer

Question

Let $a_1:=1$, $a_2:=10$. For $n > 2$, define: $$a_n:=\frac{a_{n-1}^2-1}{a_{n-2}}$$ Show that every $a_n \in \mathbb Z$.

I computed a couple terms and indeed they all seem to be integers, but I'm at a complete impasse. I am told that this somehow has something to do with determinants.

Answer 1

Maybe this will help: $$a_n=\frac{a_{n-1}^2-1}{a_{n-2}}\iff a_n\cdot a_{n-2}=a_{n-1}^2-1\iff a_{n-1}\cdot a_{n-1}-a_n\cdot a_{n-2}=1\iff \begin{vmatrix} a_{n-1} & a_{n} \\ a_{n-2}&a_{n-1} \end{vmatrix}=1$$ In other words, you could say that for any $n>1$: $$\begin{vmatrix} a_{n} & a_{n+1} \\ a_{n-1} & a_{n} \end{vmatrix}=1$$

Answer 2

Hint

Since, computing the first terms, you observed that $a_n$ is an integer, you could also observe that $$a_n=10\,a_{n-1}-a_{n-2} \qquad \text{with}\qquad a_0=0\quad \text{and}\quad a_1=1$$ Now the problem is more than simple.