Example 27 (20th Nordic Mathematical Contest) Given a sequence of positive integers $\left\{a_{n}\right\}$ satisfying $a_{0}=m, a_{n+1}=$ $a_{n}^{5}+487(n \geqslant 0)$. Find the value of $m$ such that the number of perfect squares in $\left\{a_{n}\right\}$ is maximized.

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Note that, if $a_{n}$ is a perfect square, then
$\qquad$ $a_{n} \equiv 0$ or $1(\bmod 4)$.
Ou
Lin
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Mathematics
Problem
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 $a_{k+1} \equiv 0(\bmod 4)$. Thus, when $n>k+1$, $a_{n}$ is not a perfect square.
Therefore, the sequence $\left\{a_{n}\right\}$ contains at most two perfect squares, say $a_{k}$ and $a_{k+1}$.
Let $a_{k}=s^{2}$ (where $s$ is odd), then $a_{k+1}=s^{10}+487=t^{2}$.
Let $t=s^{5}+r$, then $t^{2}=\left(s^{5}+r\right)^{2}=s^{10}+2 s^{5} r+r^{2}$.
Thus, $2 s^{5} r+r^{2}=487$.
If $s=1$, then $r(2+r)=487$, this equation has no integer solutions.
If $s=3$, then $486 r+r^{2}=487$, solving gives $r=1, r=-487$ (discard).
If $s>3$, equation (1) clearly has no positive integer solutions.
Thus, $a_{k}=9$.
And when $n>0$, $a_{n}>487$, so $m=a_{0}=9$.
On the other hand, when $a_{0}=9$, $a_{1}=9^{5}+487=244^{2}$ is a perfect square.
In summary, the required $m=9$.