Diophantine equation $x^3+z x^2-z y^2=0$ I'm not familiar with diophantine equations. At most my approaches doesn't give results. I need to solve the following equation
$$
x^3+zx^2-zy^2=0
$$
where $x,y,z\in\mathbb{Z}$
 A: We have $x^2 (x+z) = z y^2.$ If $x=0$ things simplify, so let us assume that $x,y,z \neq 0$ until such time as we can work those in.
$$  x+z = z \; \frac{y^2}{x^2}.$$
Define a rational number $r = \pm y / x,$ so
$$  x + z = r^2 z.  $$ Then
$$  x = (r^2 -1)z,  $$ and
$$  y = \pm x = \pm r (r^2 -1)z.  $$
ORIGINAL: All rational solutions are given by a rational parameter $r$ and any value of $z,$ with
$$ x = (r^2 - 1) z, \; \; y = \pm r (r^2 -1) z = \pm r x.   $$
Then
$$ x + z = r^2 z,    $$
$$  x^2 = (r^2 -1)^2 z^2,  $$
$$ y^2 = r^2 (r^2 -1)^2 z^2,    $$
$$  x^2 (x + z) =  (r^2-1)^2 z^2 \cdot r^2 z = r^2  (r^2-1)^2 z^3,   $$
$$ y^2 z = r^2 (r^2 -1)^2 z^2 \cdot z =  r^2 (r^2 -1)^2 z^3.   $$
So
$$ 0 = x^2 (x + z) - z y^2 = x^3 + z x^2 - z y^2.   $$
Next, how do we get this to come out as integers, without searching for a common denominator? taking the parameter $r = p/q$ with $\gcd(p,q) = 1,$ we get some integer $s$ with
$$  x = (p^2 - q^2 )q s, \; y = \pm (p^2 - q^2 )p s, \; z = q^3 s.  $$
Which says that we can multiply any time we like by some $s,$ and for the moment we might as well take it to be 1, resulting in a primitive two-parameter solution
$$  x = (p^2 - q^2 )q , \; y = \pm (p^2 - q^2 )p , \; z = q^3.  $$
Looking again, we can accomplish $\pm y$ by simply negating $p,$ so we get a prettier answer with
$$  x = (p^2 - q^2 ) \;q , \; \; \; y =  (p^2 - q^2 ) \; p , \; \; \; z = q^3.  $$
This takes care of nonzero $z.$ If $z=0,$ then $x=0$ but $y$ can be anything. However, having erased the factor $s,$ we are looking at only primitive solutions, so that this would force $y = \pm 1$ if $y$ is nonzero. As this situation is covered by $p = -1,0,1, \; \; q = 0,$ I would say we are in good shape.
