Prove that $s_{n+1}=s_n^2-2$ diverges. Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.
1.If $\lim s_n = a$ then $a^2 -2 = a$.
$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$
2. If $\lim s_n = a$ then a has to be either $−1$ or $2$.
From (1), we have
$a=a^2-2$
$a^2-a-2=0$
$a=−1, 2$
3. Show that $\lim s_n$ can neither be $−1$ nor $2$. (Hint: write down the first few terms of $(s_n)$ to get some insights.)
$s_n\leq s_{n+1}$ for all $n$
Proof, by induction:

*

*$P_1$ is true; at $n=1$, $5=s_1\leq s_2=5^2-2=23$


*Suppose that $s_n\leq s_{n+1}$ for some $n$


*By the assumption we have $s_{n+1}=s_n^2-2\leq s_{n+1}^2-2=s_{n+2}$
As a result, $s_n$ is increasing, so $s_n\geq 5$ for all $n$. Now since $s_n\geq 5$, this implies that $\lim s_n \geq 5$ which implies that it cannot be $-1$ or $2$.
4. Combine these steps to show that $\lim s_n$ does not exist.
Is there any further step to do here or just combine the 2 and 3?
Are my answers correct, please? Thank you in advance.

More convenient to begin with an $s_0.$
Suppose we find the real number $A>1$ for which
$$ s_0 = A + \frac{1}{A}  $$
From $s_{n+1} = s_n^2 - 2$  we find
$$ s_1 = A^2 + \frac{1}{A^2} \; , \; \;   $$
$$ s_2 = A^4 + \frac{1}{A^4} \; , \; \;   $$
$$ s_3 = A^8 + \frac{1}{A^8} \; , \; \;   $$
generally
$$ s_n = A^{2^n} + \frac{1}{A^{2^n}} \; . \; \;   $$
This gets bigger and bigger, because $A> 1.$
With $ s_0 = 5 = A + \frac{1}{A}  $ we find $A = \frac{5 + \sqrt{21}}{2} \approx 4.79.$
If we were to change the $s_0$ to something between $0$ and $2,$  most things would be the same but $A = \frac{s_0 + \sqrt{s_0^2 -4}}{2}$  is now a  complex number of norm exactly one. So $s_n,$ which remains real, oscillates but always has absolute value less that two. Indeed, we have $s_n = 2 \cos \left(2^n T \right)$  where $T = \arctan \frac{\sqrt{4-s_0^2 }}{s_0} $
