Doubling Down on a Sequence

A sequence is defined by $a_1=3$ and $a_{n+1}=a_n^2-2$ for $n\ge 1$. If the $n$-th term can be explicitly expressed as:$$a_n = \alpha^{2^{n-1}} + \frac{1}{\alpha^{2^{n-1}}}$$for a fixed number $\alpha\in\mathbb{R}^+$, find the value of $\alpha$. The answer is in the form $\frac{a+\sqrt{b}}{c}$ for where $a$, $b$, $c$ are positive integers and $\gcd(a,c) = 1$. Input the answer as the sum $a+b+c$.