MHB Solve for $d-b$: $a^5=b^4,\,c^3=d^2,\,c-a=19$

[anemone]

Assume that $a,\,b,\,c$ and $d$ are positive integers such that $a^5=b^4,\,c^3=d^2$ and $c-a=19$.

Determine $d-b$.

 

[kaliprasad]

Assume that $a,\,b,\,c$ and $d$ are positive integers such that $a^5=b^4,\,c^3=d^2$ and $c-a=19$.

Determine $d-b$.

Spoiler: my solution

As $c^3=d^2$ So there exists x such that $c=x^2$ and $d = x^3$
Further as $a^5=b^4$ so there exists y such that $a=y^4$ and $b=y^5$
Now
c-a=19
$=> x^2-y^4=19$
$=> (x-y^2)(x+y^2)=19$
As 19 is prime and $x+y^2 > x- y^2$ we have
$x-y^2=1$ and $x+y^2=19$
Solving these we get $x=10$ and $y = 3$
So $d-b = 10^3 - 3^5 = 1000 - 243= 757$