Let x mavropnevma wrote: Prove there exist infinitely many pairs $(x,y)$ of integers $1<x<y$, such that $x^3+y \mid x+y^3$. (Dan Schwarz) Let $y = x^4-x^3-1$ for some $x\ge 2$. Then $x^3+y = x^4-1$, $x+y^3 = x^{12}-3x^{11}+3x^{10}-x^9-3x^8+6x^7-3x^6+3x^4-3x^3+x-1 = (x^8-3x^7+3x^6-x^5-2x^4+3x^3-x+1)(x^4-1)$.

We want to find a pair $(x,y)$ such that $x|y$.Let $y=xt,t\in\mathbb{N^*}$.Then we have $x^3+xt|x^3t^3+x$ i.e. $x^2+t|x^2t^3+1$.After playing around with the divisibility realtions,we obtain $x^2+t|x^8-1$.Therefore we can take $t=x^8-x^2-1>1$(if we take $x\ge 2$),so we can take $y=x(x^8-x^2-1)$.Now it's not hard to verify that the pair $(x,x(x^8-x^2-1))$ respect the given condition.