8. The sequence $\left\{a_{n}\right\}$ with all terms being positive integers is defined as follows: $a_{0}=m, a_{n+1}=a_{n}^{5}+487(n \in \mathbf{N})$, then the value of $m$ that makes the sequence $\left\{a_{n}\right\}$ contain the most perfect squares is $\qquad$.
8. 9 Detailed Explanation: When $a_{n}$ is a perfect square, it must be that $a_{n} \in\{0,1\}(\bmod 4)$.

If $a_{k} \equiv 0(\bmod 4)$, then $a_{k+i} \equiv\left\{\begin{array}{l}3(\bmod 4), i \text { is odd } \\ 2(\bmod 4), i \text { is even }\end{array}\right.$ Thus, when $n>k$, $a_{n}$ is not a perfect square; if $a_{k} \equiv$ $1(\bmod 4)$, then $a_{k+1} \equiv 0(\bmod 4)$, hence, when $n>k+1$, $a_{n}$ is not a perfect square. Therefore, the sequence $\left\{a_{n}\right\}$ contains at most two perfect squares, and they are consecutive terms, let $a_{k}$ and $a_{k+1}$ both be perfect squares.

Let $a_{k}=s^{2}$ (where $s$ is odd), then $a_{k+1}=s^{10}+487=t^{2}(2 \mid t)$. Set $t=s^{5}+r$, then $t^{2}=\left(s^{5}+r\right)^{2}=s^{10}+2 s^{5} r+r^{2}$, thus we have $2 s^{5} r+r^{2}=487$ (1). If $s=1$, then $r(r+2)=487$, there is no positive integer solution for $r$; if $s=3$, then $r^{2}+486 r-487=0$, solving gives the positive integer $r=1$; if $s>3$, then (1) clearly has no positive integer solution. Hence $a_{k}=9$. Also, when $n \geqslant 1$, $a_{n}>487>9, \therefore \quad m=a_{0}=9$. At this time, $a_{1}=9^{5}+487=244^{2}$ is a perfect square. Therefore, the required positive integer $m=9$, at this time, the sequence contains the most perfect squares, a total of two.