Recurrence relation $a_{n+1}=a_n^2-2$ Sequence $a_n$ is defined $a_{n+1}=a_n^2-2, a_0=\alpha$. 
I know that a closed form for $a_n$ is $a_n=\beta^{2^n}+\frac{1}{\beta^{2^n}}$, where $\beta$ satisfies $\beta+\frac{1}{\beta}=\alpha$ and I can prove it by induction. 
But this way, we have to know answer. Is there any other solution that leads us to this form?

So far as I know, there is no general method for solving non-linear difference equations, even quadratic ones like this.  The equation may suggest the clever transformation $$
\begin{align}
a_n&=b_n+\frac1{b_n}\\
a_n^2&=b_n^2+\frac1{b_n^2}+2\\
a_{n+1}&=b_n^2+\frac1{b_n^2}\\
b_{n+1}&=b_n^2
\end{align}$$
but I don't know that I'd have come up with that if I hadn't seen the answer.
