8. The number of non-negative integer solutions satisfying $x_{1}{ }^{2}+x_{2}{ }^{2}+\cdots+x_{2023}{ }^{2}=2+x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2022} x_{2023}$ is $\qquad$ groups
(You can use binomial coefficients to represent the result)
Answer: $2 C_{2024}^{4}$
Supplement the definition $x_{0}=0$ so the condition transforms into $\left(x_{0}-x_{1}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\cdots+\left(x_{2023}-x_{0}\right)^{2}=4$

Since the solutions are non-negative integers, and $x_{0}=0, x_{i}=\left(x_{1}-x_{0}\right)+\left(x_{2}-x_{1}\right)+\cdots+\left(x_{i}-x_{i-1}\right)$
Therefore, $\left(x_{0}-x_{1}, x_{1}-x_{2}, \cdots, x_{2023}-x_{0}\right)$ can only be $-1,-1,1,1$ or $-1,1,-1,1$
Thus, there are $2 C_{2024}^{4}$