Peter wrote:

Show that there is a unique sequence of integers $\{a_{n}\}_{n \ge 1}$ with $a_{1}=1, \; a_{2}=2, \; a_{4}=12, \; a_{n+1}a_{n-1}=a_{n}^{2}\pm1 \;\; (n \ge 2).$

It is obvious that $a_{3}=5$ and that $a_{n}\geq 3$ for all $n\geq 3$ .

Now let us prove that there is at most one sequence that satisfys the conditions of the problem.

If there are two such a sequences, then for some $n\geq 4$ we should have two ways to define $a_{n+1}$ . That means that both $\frac{a_{n}^{2}+1}{a_{n-1}}$ and $\frac{a_{n}^{2}-1}{a_{n-1}}$ are integers, then their sum should be an integer too. But $\frac{2}{a_{n-1}}<1$ can not be an integer.

Now we should only find a sequence of integers that satisfys the condition of the problem. We define $b_{n}$ as $b_{1}=1,b_{2}=2$ and $b_{n+1}=2b_{n}+b_{n-1}$ .(Obviously this is a sequence of integers.)

Let us prove that $b_{2n+1}b_{2n-1}=b_{2n}^{2}+1$ and $b_{2n+2}b_{2n}=b_{2n+1}^{2}-1$ .(This means that $b_{n}b_{n-2}=b_{n}\pm 1$ )

We have that $b_{n+1}=2b_{n}+b_{n-1}=2b_{n}+2b_{n-2}+b_{n-3}$ then $b_{n+1}-b_{n-3}=2(b_{n}+b_{n-2})$ $\Rightarrow$ $b_{n-1}(b_{n+1}-b_{n-3})=2b_{n-1}(b_{n}+b_{n-2})=(b_{n}-b_{n-1})(b_{n}+b_{n-1})=b_{n}^{2}-b_{n-1}^{2}$ . So we got that $b_{n+1}b_{n-1}-b_{n}^{2}=b_{n-1}b_{n-3}-b_{n-1}^{2}$ , thus $b_{n+1}b_{n-1}-b_{n}^{2}$ is a constant for any $n$ .

Finaly we have that $b_{2n+1}b_{2n-1}-b_{2n}^{2}=b_{3}b_{1}-b_{2}^{2}=1$ and $b_{2n+2}b_{2n}-b_{2n+1}^{2}=b_{4}b_{2}-b_{3}^{2}=-1$ , as we clamed above.