$a_{n+1}=a_{n}^{2}-2 $

Question

I know $$ a_{n+1}=a_{n}^{2}-2 $$ has a closed form.

If a_1 is given in complex number and solve this recurrence relation . How solve this without using induction.

Answer 1

Put $a_n=a^{2^n}+a^{-2^n}$. Then \begin{align} a_n^2-2 &=(a^{2^n}+a^{-2^n})^2-2\\ &=a^{2^{n+1}}+a^{-2^{n+1}}\\ &=a_{n+1} \end{align} Given $a_1=a^2+a^{-2}$ we get for $|a_1|\geq 2$ $$a=\pm\sqrt{\frac{a_1\pm\sqrt{a_1^2-4}}{2}}$$