8. Find all sequences $\left\{a_{1}, a_{2}, \cdots\right\}$ that satisfy the following conditions:
$a_{1}=1$ and $\left|a_{n}-a_{m}\right| \leqslant \frac{2 m n}{m^{2}+n^{2}}$ (for all positive integers $m, n$).
8. $\left|a_{2}-a_{1}\right|=\left|a_{2}-a_{n}+a_{n}-a_{1}\right| \leqslant\left|a_{2}-a_{n}\right|+\left|a_{n}-a_{1}\right| \leqslant \frac{2 \cdot 2 n}{2^{2}+n^{2}}+\frac{1 \cdot 2 n}{1^{2}+n^{2}}<\frac{6 n}{n^{2}}=\frac{6}{n}$, by making $n$ sufficiently large, we can make $\frac{6}{n}$ arbitrarily small, therefore $\left|a_{2}-a_{1}\right|$ must be zero, hence $a_{2}=a_{1}=1$, for any $i, k$, the same discussion can be applied to $\left|a_{i}-a_{k}\right|$, thus $a_{1}=a_{2}=\cdots=1$.