A cubic diophantine Equation While reading Diophantine equations I came across the following equation $$x^3+cy^3-3yx=0$$
Is there any known method to solve this equation for any $c$?
We have $y\mid x^3$ and $x\mid cy^3$. If we assume $y=ax$ for some integer $a$ then
substituting this in $x^3+cy^3=3xy$ yields
$$
x=a(3-ac), \quad y=a^2(3-ac).
$$
Conversely this is a solution of $x^3+cy^3=3xy$ for all integers $a$. This is a first step to solve this Diophantine equation (which is not homogeneous, i.e., not given by a binary cubic).
first we have (0,0,0),(0,a,0),(0,0,c) are solutions ∀ y=a  ,a,c∈Z

now,

by multiplying by xy we have

$yx^4-3x^2 y^2+cxy^4=0$
let,
$x^2=s   ,y^2=r ,   y=a   ,cx=k$

then 
$as^2-3sr+kr^2=0$

the discriminant  ,     $d=9-4ak=9-4cxy$

as d should be greater than or equal to zero and for Diophantine equations d is a perfect square 

$d&gt;=0  , d=u^2$ 

$cxy&lt;=2$, if we considered that cxy>=0

implies  $cxy=2$ , or $cxy=0$  because 1 makes d non-square

so both x,y >=0  or x,y&lt;=0 together

take c=∓2 ,

$xy=∓1,  x^3∓2y^3∓3=0 ,(∓1,1) -is-the-only-solution $

take c=∓1 ,

$xy=2,  x^3∓y^3=∓6 , -has-no-solution $

take c=0 ,

$  x^3-3xy=0 , -has-infinite-solutions $

if cxy&lt;0 ,

$  x^3+cy^3-3xy=0 , -has-infinite-solutions $

hint: d=u^2 and d is odd then if cxy&lt;0    2-cxy=m(m-1), $m∈z^+$

(m-2)(m+1)=-cxy  ,∀ m>2 &amp; either one of c,x,y is negative or all