Here is a sketch of my approach for the first part (i.e. that all numbers can't be positive): Note that $x_{i+1}=\frac {x_i^2-a}{x_{i+2}}=\frac {x_{i-1}^2-a}{x_i}$, so $x_i^3-ax_i=x_{i-1}^2x_{i+2}-ax_{i+2}$ and summing cyclically, we obtain $\sum x_i^3 = \sum x_{i-1}^2x_{i+2}$ and using AM-GM, we have that $\frac {2x_{i-1}^3+x_{i+2}^3}{3} \geq x_{i-1}^2x_{i+2}$, and summing cyclically, we have that equality in the above inequality holds, so $x_{i-1}=x_{i+2}$ for all $i$. Now either the sequence is constant (if $n$ is not divisible by 3 - but constant sequence is obviously impossible), or there are exactly three values the sequence has, so now it is easy to finish considering three equations from the given.

First assume that $x_{i}>0\forall i$. Then notice that $x_{i}^2=a+x_{i+1}x_{i+2}>x_{i+1}x_{i+2}$ and if we multiply all such inequalities for $1\leq i\leq n$ we get that $\prod\limits_{i=1}^{n}x_{i}^2>\prod\limits_{i=1}^{n} x_{i}x_{i+1}=\prod\limits_{i=1}^{n}x_{i}^2$, contradiction. This means that there is at least one negative number among the $x_{i}$-s. WLOG $x_{1}<0$ as the equations are cyclic. Assume that $x_{i}\geq 0\forall i\neq 1$. As the equations are homogenous, we can assume that $x_{1}=-1$ (we divide every $x_{i}$ by $|x_{1}|$ and $a$ by $x_{1}^2\neq 0$). Now we have that $1=x_{1}^2=a+x_{2}x_{3}>a\Longrightarrow a\in(0,1)$ and $x_{2}^2=a+x_{3}x_{4}>a\Longrightarrow x_{2}\geq \sqrt{a}$. However, notice that now we have a contradiction because: $$0\leq x_{n}^2=a+x_{1}x_{2}=a-x_{2}<a-\sqrt{a}<0$$

Quickest approach perhaps? If all numbers are positive and without loss of generality $x_1$ is minimal, we get $x_1^2 = a + x_2x_3 \geq a + x_1^2 > x_1^2$, contradiction! Now suppose without loss of generality that $x_1 < 0$ and all others are non-negative. Then $x_1^2 = x_2x_3 + a \geq a$, so $x_1 \leq -\sqrt{a}$; also $x_2^2 = x_3x_4 + a \geq a$, i.e. $x_2 \geq \sqrt{a}$. This implies $x_n^2 = x_1x_2 + a \leq 0$ and hence $x_n = 0$, $x_1 = -\sqrt{a}$, $x_2 = \sqrt{a}$. But then $a = x_1^2 = x_2x_3 + a$, giving $x_3 = 0$ and $0 = x_3^2 = x_4x_5 + a \geq a$ when $n\geq 5$ (so that $x_5 \neq x_1 < 0$), and also for $n=4$ (from $x_n = 0$), contradiction! Remark: I invite the ones who are interested to look for significantly better lower bounds for the number of negative terms in the sequence.

https://dgrozev.wordpress.com/2022/04/23/bulgarian-national-math-olympiad-2022-part-3/

Here is my solution: https://calimath.org/pdf/BulgariaNMO2022-4.pdf And I uploaded the solution with motivation to: https://youtu.be/3f7e8sW52og