Integer solutions of equation $x^3+10x-1=y^3+6y$?

Question

I want to find all integer solution of the following equation in integers$$x^3+10x-1=y^3+6y$$ I tried to factor it somehow, but failed. Also $x$ and $y$ are have different parity. How can one solve this?

Answer 1

Comparing terms $6y$ and $10x$ tells us that $x$ can not be greater than $y$. We'll find out that, let $y-x=z$ and equation becomes

$$x^3+10x-1=(x+z)^3+6(x+z)$$ $$-3zx^2-(3z^2-4)x-(z^3+6z+1)=0 $$

Discriminant of the obtained quadratic in $y$ must be non-negative, that is

$$(3z^2-4)^2-4(3z)(z^3+6z+1)\ge0,$$which gives $z=0$ only. Thus $y=x$, which gives $4x=1$. No integer solutions!