Bulgaria competition problems, 2022

Notice that $x_i^2 x_{i+1} = a x_{i+1} + x_{i+1}^2 x_{i+2}$ holds for all $i = 1, 2, \dots, n$. After summing these equations we get $\sum_{i=1}^n x_i^2 x_{i+1} = a \sum_{i=1}^n x_i + \sum_{i=1}^n x_i^2 x_{i+1}$ and therefore $\sum_{i=1}^n x_i = 0$. Since $a > 0$ at least one of the numbers $x_i$ is not equal to $0$ and therefore at least one of them is negative. Assume that there is at least one negative and let it be $x_1$ without loss of generality. So $x_2^2 = a + x_3 x_4 \ge a$ and $|x_1| = x_2 + \dots + x_n$. Hence we get that $(x_2 + \dots + x_n)^2 = x_1^2 = a + x_2 x_3$ and therefore $x_2^2 - a + \sum_{i=3}^n x_i^2 + x_2 x_3 \le 0$. So $x_2 = \sqrt{a}$ and $x_3 = \dots = x_n = 0$. Now, after substituting in the equality of the condition for $i = 3$, we get $0 = x_3^2 = a + x_4 x_5 = a$, which is contradiction. Therefore, at least two of the numbers are negative.