8. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for any $x, y \in \mathbf{R}$,
$$
f\left(x^{2}\right)+f\left(y^{2}\right)=f^{2}(x+y)-2 x y .
$$

Let $S=\sum_{n=-2020}^{2020} f(n)$. Then $S$ has $\qquad$ possible values (answer with a number).
8. 2041211.

By setting $y=-x$ in the conditional equation, we know that for any real number $x$, we have
$$
2 f\left(x^{2}\right)=f^{2}(0)+2 x^{2} \text {. }
$$

By setting $x=0$ in equation (1), we get $f(0)=0$ or 2.
If $f(0)=2$, then equation (1) indicates that for all non-negative real numbers $x$, we have
$$
f(x)=x+2 .
$$

Thus, the conditional equation becomes for all real numbers $x, y$ satisfying $x+y \geqslant 0$,
$$
x^{2}+y^{2}+4=x^{2}+y^{2}+4 x+4 y+4 \text {. }
$$

But when $x+y>0$, this cannot hold.
If $f(0)=0$, then equation (1) indicates that for all non-negative real numbers $x$, we have
$$
f(x)=x \text {. }
$$

By setting $y=0$ in the conditional equation, we get
$$
\begin{array}{l}
f\left(x^{2}\right)=f^{2}(x) \\
\Rightarrow f(x)= \pm x(x \in \mathbf{R}) .
\end{array}
$$

It is easy to verify that for any non-negative real number $x$, in the conditional equation, $f(x)=x$ and $f(x)=-x$ are both allowed, and the function values for different $x$ are independent of each other.
$$
\begin{array}{l}
\text { Hence } S=f(0)+\sum_{i=1}^{2020}(f(i)+f(-i)) \\
=2 \sum_{i=1}^{2020} i \delta_{i},
\end{array}
$$

where, $\delta_{1}, \delta_{2}, \cdots, \delta_{2020} \in\{0,1\}$.
Thus, the range of $\frac{S}{2}$ is from 0 to $1+2+\cdots+2020=\frac{2021 \times 2020}{2}=2041210$ of all integers.
Therefore, $S$ has a total of 2041211 possible values.