Find all negative integer solutions of x and y such that $$ x^3 + 3y^2 = xy^2 + 27 $$
To begin, I isolated $x$ and $y$ to see if there is anything I could do and got $$ y = \pm \sqrt{x^2+3x+9} \\ x = \dfrac{1}{2} \left(-3 \pm \sqrt{y^2-27}\right) $$
Since $x$ and $y$ are negative here I thought I could remove the positive branch and got $$ y = - \sqrt{x^2+3x+9} \\ x = - \dfrac{1}{2} \left(3 + \sqrt{y^2-27}\right) $$
And I really don't know what to do at this point. I don't have much knowledge on finding integer solutions, any help is appreciated.
The equation $x^3 + 3y^2 = xy^2 + 27$ can be written as $$x^3-27+3y^2-xy^2=0$$ $$(x-3)(x^2+3x+9)+y^2(3-x)=0$$ $$(x-3)(x^2+3x+9-y^2)=0$$
Now, $x^2+3x+9-y^2=0$ can be written as $$\bigg(x+\frac 32\bigg)^2+\frac{27}{4}-y^2=0$$ Multiplying the both sides by $4$ gives $$(2x+3)^2+27-(2y)^2=0$$ and so $$(2y-2x-3)(2y+2x+3)=27$$ I think that you can continue from here.