<s>[INST] Integer solutions of equation $x^3+10x-1=y^3+6y$? 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?  
 [/INST] 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!
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