4・235 Let $a, b, c, d$ be positive integers, satisfying $a^{5}=b^{4}, c^{3}=d^{2}$, and $c-a=19$. Find $d-b$.


[Solution] In $a^{5}=b^{4}$, since 5 and 4 are coprime, there exists a positive integer $m$ such that $a=m^{4}, b=m^{5}$.

Similarly, there exists a positive integer $n$ such that
$$
c=n^{2}, d=n^{3}.
$$

Thus, from $c-a=19$ we get
$$
n^{2}-m^{4}=19,
$$

which can be rewritten as
$$
\left(n-m^{2}\right)\left(n+m^{2}\right)=19.
$$

Noting that 19 is a prime number, and $n-m^{2}<n+m^{2}$, it follows that
$$
\left\{\begin{array}{l}
n-m^{2}=1, \\
n+m^{2}=19 .
\end{array}\right.
$$

Solving these equations gives $n=10, m=3$. Therefore,
$$
d-b=n^{3}-m^{5}=1000-243=757.
$$