8. Find $f: \mathbf{N}^{\top} \rightarrow \mathbf{N}^{\top}, f(m+n) f(m-n)=f\left(m^{2}\right)$ for all $m, n \in \mathbf{N}^{\top}$.
8. $f(n)=1 . n \in \mathbf{N}^{*}$. Solution: Let $m=n^{\prime}, n=n^{\prime}-1$, substitute into the condition to get: $f(1) f\left(2 n^{\prime}-1\right)=$ $f\left(n^{\prime 2}\right)$, similarly, $f(3) f\left(2 n^{\prime}-1\right)=f\left(\left(n^{\prime}+1\right)^{2}\right)$. Here $n^{\prime} \geqslant 3$. Thus $\frac{f(3)}{f(1)}=\frac{f\left((n+1)^{2}\right)}{f\left(n^{2}\right)}$. Let $\frac{f(3)}{f(1)}=$ $k$, then $f\left(n^{2}\right)=k^{n-3} f(9)$, here $n \geqslant 3$. Similarly, for $h \geqslant 1$, $\frac{f(h+2)}{f(h)}=\frac{f\left((m+1)^{2}\right)}{f\left(m^{2}\right)}$. Therefore, $f(2 h)=$ $k^{h-1} f(2), f(2 h+1)=k^{h} f(1)$.
$$
\text { But } \frac{f(25)}{f(9)}=\frac{f(25)}{f(23)} \cdots \cdots \frac{f(11)}{f(9)}=k^{8}, \frac{f(25)}{f(9)}=\frac{f(25)}{f(16)} \cdot \frac{f(16)}{f(9)}=k^{2} \text {. Thus } k=1, f(16)=f(9) \text {. }
$$

This implies that for all $j, h$ we have $f(2 h+1)=f(1)=f(2)=f(2 j)$. Therefore, $f$ is a constant. It is easy to see that $f(n)=1, n \in \mathbf{N}^{*}$.