# 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 $x,y,z\in\mathbb{Z}$

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One important fact from this equation that we need to notice is that it is homogeneous : all the monomials have same degree. Therefore should we find that a solution must satisfy $x \equiv y \equiv z \equiv 0 \mod n$ for some integer $n$, we could divide through and show that $(x/n, y/n, z/n)$ also is a solution and use an infinite descent argument. I'm trying to find such an $n$ at the moment. –  Patrick Da Silva Feb 5 '12 at 3:22
I give up on this one for tonight after noticing several things : $$x^3 + zx^2 - zy^2 = x^3 + z(x+y)(x-y)$$ and trying many things mod $2$, mod $5$... mod $3$ is not very successful. Went nowhere up to now. Sorry. –  Patrick Da Silva Feb 5 '12 at 3:37
Thanks for your attempt –  Norbert Feb 5 '12 at 13:14

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.

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Can you prove the first statement (that all rational solutions are in that form)? –  sdcvvc Feb 5 '12 at 5:00
@sdcvvc, yes. Beginning typing. Why not. –  Will Jagy Feb 5 '12 at 5:04
It is a rare case for me when a diophantine equation of order greater then 2 have infinitely many solutions. Thanks! –  Norbert Feb 5 '12 at 14:21
Excellent. I did not expect solutions ; I guess there is "more than 1". –  Patrick Da Silva Feb 5 '12 at 17:07