1822. * Let $a, b, c$ and $d$ be positive integers, and $a^{5}=b^{4}, c^{3}=d^{2}, c-a=19$, find $d-b$
Since $(4,5)=1,(2,3)=1$. According to the unique factorization theorem of integers, there exist two positive integers $m, n$, such that $a=m^{4}, b=m^{5}, c=n^{2}, d=n^{3}$. Thus, $19=c-a=n^{2}-m^{4}=$ $\left(n-m^{2}\right)\left(n+m^{2}\right)$, since 19 is a prime number, so $n-m^{2}=1, n+m^{2}=19$, hence $m=3$, $n=10, d=1000, b=243$. Therefore, $d-b=757$.