<s>[INST] Integer solutions to $x^3=y^3+2y+1$? 
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$

My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
 [/INST] We have 
$$(y^3+2y+1)-(y-1)^3=3y^2-y+2=\frac{(6y-1)^2+23}{12}>0$$
and
$$(y+2)^3-(y^3+2y+1)=6y^2+10y+7=\frac{(6y+5)^2+17}{6}>0\,.$$
That is,
$$(y-1)^3<y^3+2y+1<(y+2)^3$$
for all $y\in\mathbb{Z}$.  If $y^3+2y+1$ is a cube, then either $y^3+2y+1=y^3$ or $y^3+2y+1=(y+1)^3$, which gives $y=-\dfrac12$, $y=-\dfrac13$, or $y=0$.  That is, $(x,y)=(1,0)$ is the only integer solution to $x^3=y^3+2y+1$.
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