Question 7. Let $a, b, c, d$ be positive integers, satisfying $a^{8}=b^{4}, c^{3}=d^{2}$, and $c-a=19$. Find $d-b$.

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Since the prime factorization of a positive integer is unique, and because 4 and 5 are coprime, 2 and 3 are coprime, it follows that there exist two positive integers \( m \) and \( n \) such that
$$
a=m^{4}, b=m^{6}, c=n^{2}, d=n^{3}.
$$

Thus, \( 19=c-a=n^{2}-m^{4} \):
$$
=\left(n-m^{2}\right)\left(n+m^{2}\right) \text{. }
$$

Since 19 is a prime number, and \( n-m^{2} < n+m^{2} \), it must be that
$$
n-m^{2}=1, n+m^{2}=19 \text{. }
$$

Therefore, \( m=3, n=10, d=1000, b=243 \).
$$
d-b=1000-243=757
$$