$a^3=b^2$
$c^3=d^2$
$c-a=64$
I tried this way :
$c-a=64$
$\Rightarrow \frac{d^2}{c^2}-\frac{b^2}{a^2}=(\pm 8)^2$
$\Rightarrow (\frac{d}{c})^2-(\frac{b}{a})^2=(\pm 8)^2$
$\Rightarrow (\frac{d}{c}-\frac{b}{a}) (\frac{d}{c}+\frac{b}{a})=(\pm 8)^2 $
$\Rightarrow\frac{d}{c}-\frac{b}{a}=\pm 8 $ & $\frac{d}{c}+\frac{b}{a}=\pm 8$
So, now I can get $4$ equations each having $4$ variables but how can I know the number of solutions without actually solving those?Is there any way or I have to solve!!
Big Hint: For integers,$a,b$, having $a^3=b^2$ means that $a$ is a perfect square and $b$ is a perfect cube. Likewise for $c$ and $d.$ Then if $a=a_0^2,$ you get $b=a_0^3,$ and $c=c_0^2,d=c_0^3.$ Now you need:
$$c-a=(c_0-a_0)(c_0+a_0)=64.$$
There are only so many ways to factor $64,$ and for each you can solve for $c_0$ and $a_0$ and get a solution $(a,b,c,d)=(a_0^2,a_0^3,c_0^2,c_0^3)$ when $a_0,c_0$ are both positive integers.
Note that my $c_0=\frac{d}{c}$ and $a_0=\frac{b}{a}$ from your beginnings.
You are assuming the factorization of $64$ is $(\pm 8)(\pm 8).$ But there are other factorizations of $64,$ and just because $UV=8\cdot 8$ doesn't mean that $U=\pm 8,V=\pm 8.$ Indeed, given that $U=c_0-a_0\neq c_0+a_0=V,$ you probably want other factorizations. When $U=V$ here, you get $a_0=0$ and hence $a=0,$ which violates the condition that $a$ is positive.
