2105/111) Given 4 positive integers a,b,c and d such that $a^5=b^4, c^3=d^2$ and $c−a=19$ what is $d−b$

as $c^3 = d^2$ so c will be a square let $c = x^2$

as $a^5 = b^4$ so $a = y^4$

now
$c-a = 19$
=> $x^2 - y^4 = 19$
=> $(x-y^2)(x+y^2) = 19$
hence $x - y^2 =1$ and $x+y^2 = 19$ as 19 is a prime
so $x = 10$ and $y = 3$

so $a = y^4$ or $a^5 = y^20 = b^ 4$ or $b= y^5 = 243$

$c= x^2 => c= 100$ and hence $d^2 = 10^6$ and so $d = 1000$
$d-b = 1000 - 243 = 757$