Finding the general formula of $a_{n+1}=\frac{a_n^2+4}{a_{n-1}}$ with $a_1=1$ and $a_2=5$ 
Find the general formula of $a_{n+1}=\dfrac{a_n^2+4}{a_{n-1}}$ with $a_1=1$, $a_2=5$.

I have tried to write the recursion as a product, make summations, tried to look at patterns but its value grows very fast: $1,5,29,169,985,5741$… So I ran out of ideas.

If we start from $a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)$ we will get:
$$a_{n+1}^2+(a_n^2+4) = a_n^2+(a_{n+1}^2+4)\\ \Leftrightarrow\\ a_{n+1}^2+a_{n-1}a_{n+1} = a_n^2+a_{n}a_{n+2}\\ \Leftrightarrow\\ \frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{n}+a_{n+2}}{a_{n+1}}$$
Therefore we have $\frac{a_{n-1}+a_{n+1}}{a_n} = \frac{a_{0}+a_{2}}{a_1} = 6$. After that, we can follow the solution of JV.Stalker:
$$a_{n+2}-6a_{n+1}+a_n = 0 \Rightarrow r^2-6r+1=0 \Rightarrow r_{1,2} = 3\pm2\sqrt{2}\Rightarrow a_n = c_1(3+2\sqrt{2})^n+c_2(3-2\sqrt{2})^n$$
We can define $a_0$ as $a_0a_2=a_1^2+4\Rightarrow a_0 = 1$. Therefore we have $c_1+c_2=1$ and $3(c_1+c_2)+2\sqrt{2}(c_1-c_2)=1$ and hence $c_1=\frac{2-\sqrt{2}}{4}$ and $c_2=\frac{2+\sqrt{2}}{4}$. If we plug this into the equation for $a_n$ we receive $$a_n = \frac{2-\sqrt{2}}{4}(3+2\sqrt{2})^n+\frac{2+\sqrt{2}}{4}(3-2\sqrt{2})^n\\ \Leftrightarrow\\ a_n = \frac{2-\sqrt{2}}{4}(\sqrt{2}+1)^{2n}+\frac{2+\sqrt{2}}{4}(\sqrt{2}-1)^{2n}\\ \Leftrightarrow\\ a_n = \frac{\sqrt{2}}{4}(\sqrt{2}+1)^{2n-1}+\frac{\sqrt{2}}{4}(\sqrt{2}-1)^{2n-1}$$
