Determine d-b if $a^5=b^4,c^3=d^2,c-a=19$ if $a,b,c,d$ are integers. My Attempt:
Let $a^5=b^4=x^{20}$ and let $c^3=d^2=y^6$. Therefore, $a=x^4$ and $c=y^2$. $ \implies 19 = c-a=y^2-x^4=(y-x^2)(y+x^2)$. Since $a,\ b,\ c,\ d$ are integers, $x,\ y$ are also integers $\implies y-x^2$ and $y+x^2$ are also integers. And since $-x^2<x^2$ for all integers $x$, $y-x^2<y+x^2$. 
Therefore there are only two cases:
1) $\ y-x^2=1,\ y+x^2=19$. In this case, x=3 and y=10. 
2) $\ y-x^2=-19,\ y+x^2=-1$. This case is not possible since $y+x^2$ is always positive because $y,\ x$ are both positive. 
So $b=x^5=3^5=243$ and $d=y^3=1000$. Therefore, $d-b=1000-243=757$.
Is my attempt to the problem correct?

The comments basically sum it up, the answer is correct.  And yes, it is the only solution.  The numbers $a$ and $c$ are squares differing by an odd prime, so must be consecutive squares, and $a$ must further be a fourth power.  This forces $a=81=9^2=3^4, c=100=10^2$, which gives the quoted solution only.
