If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question

Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that  $a_{n+1}=Sa_n+Ta_{n-1}$.

Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?

$7=3S+2T\\
18=7S+3T$
is a linear system of two equations with two unknowns.
It has a unique solution :
by multiplying the first line by $3$ and the second by $2$, then substracting, you get $21-36 = 9S-14S$, so $-15 = -5S$, then $S=3$, and then $T=-1$.
Let $b_n$ be the sequence defined by $b_0=a_0, b_1=a_1, b_{n+2}=Sb_{n+1}+Tb_n$.
You want to show that the recurrence relation defining $a_n$ is also valid for $b_n$, i.e. $b_nb_{n+2} - b_{n+1}^2 - 5 = 0$
Since $(b_n)$ satisfies a linear recurrent equation of order $2$, $(b_n^2)$ and $(b_nb_{n+2})$ satisfy a recurrent linear equation of order $3$ (if I'm not mistaken, $c_{n+3} = 8c_{n+2}-8c_{n+1}+c_n$)
The constant sequence $5$ also satisfies this recurrence relation, so the whole $(b_nb_{n+2} - b_{n+1}^2 - 5)$ is a linear recurrent sequence of order $3$ (it would be of order $4$ if it wasn't the case). 
To show that it is the $0$ constant you only need to show that its first three terms are $0$. Since $a_0=b_0$ and $a_1=b_1$ by definition and $S$ and $T$ were chosen so that $a_2=b_2$ and $a_3=b_3$, then you only need to check that $a_4=b_4$.
since $a_4 = 47 = 3*18-7 = b_4$, this is true, and so $a_n = b_n$ forall $n$.
