Suppose there exists $i \in \{1, 2, \dots, n\}$ for which $a_i = 0$. Clearly $a_{i+1} = 0$, and consequently $a_1 = a_2 = \dots = a_n = 0$.

Suppose now that $a_i \ne 0$ for all $i \in \{1, 2, \dots, n\}$. Choose $p, q \in \{1, 2, \dots, n\}$ such that $a_q \le a_i \le a_p$ for all $i \in \{1, 2, \dots, n\}$.

Now $a_p^3 = a_{p+1}^2 + a_{p+2}^2 + a_{p+3}^2 \le 3a_p^2$ and $a_q^3 = a_{q+1}^2 + a_{q+2}^2 + a_{q+3}^2 \ge 3a_q^2$ yield $a_p \le 3 \le a_q$, therefore $a_1 = a_2 = \dots = a_n = 3$.