8. The number of non-negative integer solutions $\left(x_{1}, x_{2}, \cdots, x_{25}\right)$ that satisfy the equation
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{25}^{2} \\
=2+x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{24} x_{25}
\end{array}
$$

is $\qquad$ groups.
8. 29900 .

Let $x_{0}=0$. Then
$$
\begin{array}{l}
x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+\cdots+x_{25}^{2} \\
=2+x_{0} x_{1}+x_{1} x_{2}+\cdots+x_{24} x_{25}+x_{25} x_{0} \\
\Rightarrow\left(x_{0}-x_{1}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\cdots+ \\
\quad\left(x_{24}-x_{25}\right)^{2}+\left(x_{25}-x_{0}\right)^{2}=4 .
\end{array}
$$

Since the integers $x_{0}-x_{1}, x_{1}-x_{2}, \cdots, x_{24}-x_{25}$, $x_{25}-x_{0}$ sum to 0 and their squares sum to 4, these 26 numbers can only be $\pm 1, \pm 2, 0$.

If one of these 26 integers is $\pm 2$, then the rest must all be 0, which contradicts their sum being 0. Therefore, these 26 numbers can only be 0 and $\pm 1$. Since their sum is 0, they must consist of 2 ones, 2 negative ones, and 22 zeros.

Noting that $x_{0}=0, x_{1}, x_{2}, \cdots, x_{25}$ are all non-negative integers, the order in which 1 and -1 appear can only be $-1,-1,1,1$ or $-1,1,-1,1$.
Thus, the number of non-negative integer solutions is
$$
2 \mathrm{C}_{26}^{4}=29900 \text {. }
$$