Given a series defined by recursion. Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$ I have this problem from an old exam that I can't solve.

Let $\{a_i\}_{i \geq 0}$ be the series define by recursion as:
$a_0 = 2$
$a_1 = 3$
$a_{n+1} = \frac{a_n^2 +5}{a_{n-1}}$ , $\forall n \in \Bbb N$
Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$

I've been trying with induction but I guess that there must be a direct, and more algebraic, way of proving it.
Any thoughts?? Thanks!

Define a series $a_{n+1} = \frac{a_n^2 +k}{a_{n-1}}$ with some $a_0, a_1$ for $n\ge2$. Assume there are $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$.
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n + T a_{n-1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n a_{n-1} + T a_{n-1}^2 = a_n^2+k \tag{1}
$$
now
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
a_{n+1}a_{n-1} = a_n^2+k \\
a_{n+1}a_{n-1} - k = a_n^2
\tag{2}
$$
combine the two:
$$
S a_n a_{n-1} + T(a_n a_{n-2} - k) = a_n^2+k \\
a_n(Sa_{n-1} + Ta_{n-2}) - Tk = a_n^2+k \\
a_n^2 - Tk = a_n^2+k \\
T = -1
$$
now find $S$:
$$
a_{2} = \frac{a_1^2+k}{a_{0}} \\
S a_1 - a_{0} = \frac{a_1^2+k}{a_{0}} \\
S = \frac{a_0^2+a_1^2+k}{a_{0}a_{1}} \\
$$
and in this case: $$S = \frac{2^2+3^2+5}{2\times3} = \frac{18}{6} = 3$$
So $S$ is an integer if it exists.
Prove by induction that $$S a_n - a_{n-1} = \frac{a_n^2+k}{a_{n-1}}$$ (note that the base case $n=2$ is covered by how we set $S$). Assume the equality is correct for all $n+1\ge k\ge3$.
$$
a_{n+2} = \frac{a_{n+1}^2+k}{a_{n}} \\
a_{n+2} = \frac{(S a_n-a_{n-1})^2+k}{a_{n}} = \frac{S^2 a_n^2-2S a_n a_{n-1}+a_{n-1}^2+k}{a_{n}} = S^2 a_n-2Sa_{n-1}+\frac{a_{n-1}^2+k}{a_{n}} \\
a_{n+2} = S(S a_n-a_{n-1})-Sa_{n-1}+\frac{a_{n-2}}{a_{n}}\frac{a_{n-1}^2+k}{a_{n-2}} = S a_{n+1} -Sa_n + \frac{a_{n-2}}{a_{n}} a_n \\
a_{n+2} = S a_{n+1} -Sa_n + a_{n-2} = S a_{n+1} -Sa_n + (S a_{n-1} - a_n) = S a_{n+1} - a_n
$$
As required.
