Integer pairs satisfying $(y-a)(y-b) = x^3$ Given the equation in title where $a$ and $b$ are known constants. In how many ways we can find a pair of integers satisfy that equation? For instance we can let $(y-a)^2 = (y-b)$ then if we can find some $y$ satisfies the prior equation we get $(y-a)^3 = x^3$. What else one can do?
 A: Given $a$ and $b$, let $c = a-b$ and consider the equation:
$$(y-c)y = x^3\tag{1}$$
If we can find a solution $(x',y')$ to (1), then we can find a solution of the original equation by putting $y=y'+b$ since:
$$(y-a)(y-b) = ((y'+b)-(c+b))((y'+b)-b) = (y'-c)y' = (x')^3\tag{2}$$
Clearly the converse also holds.  So the problem reduces to finding values of $c$ for which (1) has solutions.
A solution of (1) exists where $c=m^2-m$ for some integer $m$, since we can put $y=m^2$ so that:
$$(y-c)y=(m^2-(m^2-m))m^2= m.m^2 = m^3\tag{3}$$
This can be generalised as follows. Suppose $c=g-h$, and $p_1,...,p_n$ are all the prime factors of either $g$ or $h$, and their respective non-negative integer exponents are $s_i$ and $t_i (i=1,...,n)$ so that:
$$g = p_1^{s_1}p_2^{s_2}....p_n^{s_n}\tag{4}$$
$$h = p_1^{t_1}p_2^{t_2}....p_n^{t_n}\tag{5}$$
and suppose that, for each of $i=1,...,n$, $s_i+t_i$ is divisible by 3.  Then $y=g$ provides a solution of (1) since:
$$(y-c)y=(g-c)g=gh=p_1^{s_1+t_1}p_2^{s_2+t_2}....p_n^{s_n+t_n}\tag{6}$$
which is a cube because of the divisibility condition on the exponents.
