<s>[INST] Find the value of $b-d$ If it's given that ->
$a^{3} = b^{2}$ ;
$c^{5} = d^{4}$ ;
$a - c = 19$
Find $b-d$
PS- I have given 4 hours of my life to this question and even then I am NOT able to get even the SLIGHTEST hint . So , i request you all to explain the solution of this question . Thanks in Advance !!
 [/INST] Put $a=x^2, b=x^3, c=y^4, d=y^5$:
$$\begin{align}
a^3=b^2\;&\Rightarrow (x^2)^3=(x^3)^2\\
c^5=d^4\; &\Rightarrow (y^4)^5=(y^5)^4\\
\text{Given} \qquad\qquad a-c&=19\\
x^2-y^4&=19\\
(x-y^2)(x+y^2)&=19\end{align}
$$
As $19$ is prime and $y^2>0$, and also given that $x,y$ are integers, possible combinations are
$$
x-y^2=1; x+y^2=19 \;\Rightarrow x=10, y=\pm3\\
\text{or}\\
x-y^2=-19; x+y^2=-1\;\Rightarrow x=-10, y=\pm 3\\$$
Solutions are:
$$
(x,y)=(\pm 10, \pm 3) \\
\Rightarrow b-d=\pm 1000 \pm243\\=\pm757, \pm 1243\\
$$
__
$$\color{lightgrey}{\begin{align}(x,y)=(10,3)\;\Rightarrow b-d&=757\\
(x,y)=(10,-3)\;\Rightarrow b-d&=1243\\
(x,y)=(-10,3)\;\Rightarrow b-d&=-1243\\
(x,y)=(-10,-3)\;\Rightarrow b-d&=-757\end{align}}$$
</s>