Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e.,
Positive integers $(x,y)$ such that
$x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?
Finding infinite number of pairs is no problem, as in:
$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$ 
But how would you determine the exhaustive list?
 A: $ x^2 - y^2 = a^3 $ is fairly trivial, because clearly for every integer solution there are integers $ p, q, r, s $ with $p$ and $q$ squarefree and coprime such that $ x + y = p q^2 r^3 $ and $ x - y = p^2 q s^3 $, so that:
$ 2 x = p q (q r^3 + p s^3) $
$ 2 y = p q (q r^3 - p s^3) $
$ a = p q r s $
and these satisfy the equation identically, with $x, y$ integers iff either $p q$ is even or $p, q, r, s$ all odd.
(or $r, s$ both even, although for "reduced" solutions with no integer $t > 1$ such that t^3 divides $x, y$ and t^2 divides $a$ we can assume that $gcd(r, s) = 1$, which we do hereafter ..)
Plugging these relations into $(2 x)^3 - (2 y)^3 = 8 b^2 $ gives $ p^4 q^3 (3 q^2 r^6 s^3 + p^2 s^9) = 4 b^2 $.
Noting that $p$ and $q$ are coprime and squarefree we then conclude $2 b = p^2 q^2 s B$ for some integer $B$, so that $ s (3 q^2 r^6 + p^2 s^6) = q B^2 $.
From this we see that $q$ divides $p^2 s^7$ and hence, since $gcd(p, q) = 1$ and $q$ is squarefree, that $q$ divides $s$.
Thus with $s = q S$ and $B = q C$ we obtain $ S (3 r^6 + p^2 q^4 S^6) = C^2 $.
Then, taking $S = u v^2$ with $u$ squarefree, we conclude that $C = u v D$ for some integer $D$ so that finally $3 r^6 + (p q^2 u^3 v^6)^2 = u D^2 $
In this $u$ divides $3 r^6$; but since $gcd(r, s) = 1$ and $u$ divides $s$ we conclude that $u$ divides 3.
Thus, since $u$ > 0, we must have $u = 1$ or $u = 3$, which reduces the problem to one of the following respectively:
$3 r^6 + M^2 = N^2$   ($u = 1$)
$ r^6 + 3 M^2 = N^2$  ($u = 3$)
Now $X^2 + 3 Y^2 = Z^2$ has general integer solution $X, Y, Z = k(m^2 - 3 n^2), 2 k m n, k(m^2 + 3 n^2)$ with $gcd(m, n) = 1$ and $m + n$ odd.
So the cases require either $2 k m n = r^3$ or $k(m^2 - 3 n^2) = r^3$, each of which is trivial by suitable choice of $k$ (although arguably if you want explicit values of m, n the second is not so trivial).
A: Not an answer, just thinking out loud about the question of whether we can ever take $x,y$ relatively prime. 
We solve $x^2-y^2=a^3$ by writing $a^3=a_1a_2$ with $a_1,a_2$ of the same parity, then $x-y=a_1$, $x+y=a_2$, $2x=a_2+a_1$, $2y=a_2-a_1$, and we see any common divisor of $a_1,a_2$ other than perhaps $2$ is a common divisor of $x,y$. So we have $a_1=c^3$, $a_2=d^3$ with $c,d$ relatively prime, or else $a_1=2c^3$, $a_2=4d^3$, or else $a_1=4c^3$, $a_2=2d^3$. I'll just look at the first case. 
We have $2x=d^3+c^3$, $2y=d^3-c^3$, $c,d$ relatively prime. Then $$8b^2=(2x)^3-(2y)^3=(d^3+c^3)^3-(d^3-c^3)^3=6d^6c^3+2c^9$$ so $(2b)^2=c^3(3d^6+c^6)$. Since $c,d$ are coprime, the only possibilities for $\gcd(c^3,3d^6+c^6)$ are $1$ and $3$. I'll just look at the first case. 
Now $c^3$ and $3d^6+c^6$ must be squares, so $c=e^2$ and $3d^6+e^{12}=f^2$. So $$(f-e^6)(f+e^6)=3d^6$$ If $f-e^6,f+e^6$ are coprime (and they must be pretty nearly coprime, if you trace back through to the coprimality of $c,d$), then $f-e^6=g^6$, $f+e^6=3h^6$, or else $f-e^3=3g^6$, $f+e^6=h^6$. Taking, as usual, just the first case, we get $$g^6+2e^6=3h^6$$ 
That's as far as I go. There is the trivial solution, $e=g=h=1$. That rules out using congruences to show there aren't any solutions, but I suspect there aren't any other coprime solutions. Maybe someone can actually prove this, and clean up the cases I haven't discussed. 
