3・13 Let $a, b, c, d$ be positive integers, satisfying $a^{5}=b^{4}, c^{3}=d^{2}$, and $c-a$ $=19$. Find $d-b$.
(3rd American Invitational Mathematics Examination, 1985)
【Solution】Given $a^{5}=b^{4}, c^{3}=d^{2}$.
Noting that $(5,4)=1,(3,2)=1$, we know that there exist two positive integers $m$ and $n$, such that
$$
\begin{aligned}
a & =m^{4}, b=m^{5}, \\
c=n^{2}, d & =n^{3} .
\end{aligned}
$$

Thus, $\quad 19=c-a=n^{2}-m^{4}=\left(n+m^{2}\right)\left(n-m^{2}\right)$.
Since 19 is a prime number, and $n-m^{2}<n+m^{2}$, we have
$$
\left\{\begin{array}{l}
n-m^{2}=1, \\
n+m^{2}=19 .
\end{array}\right.
$$

Solving these equations gives $n=10, m=3$.
Therefore,
$$
\begin{array}{c}
d=n^{3}=1000, \quad b=m^{5}=243 . \\
d-b=1000-243=757 .
\end{array}
$$