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.
Hint: if $y&gt;0$, then $y^3&lt; y^3+2y+1&lt; (y+1)^3$, so the RHS expression cannot be a perfect cube. A similar idea works if $y$ is a small enough negative number, but some negative numbers close to $0$ (or indeed $0$ itself) can provide a solution. 

Try to find a lower bound, and then check the remaining possible values.
$$x^3=y^3+2y+1\tag{1}$$
We are after integer solutions of $(1)$. On setting $y=x-a$, where $a\in\mathbb{Z}$, we have from $(1)$
$$x^3=(x-a)^3+2(x-a)+1=x^3-3a x^2+(3a^2+2)x-a^3-2a+1\tag{2}$$
which rearranges to
$$3a x^2-(3a^2+2)x+a^3+2a-1=0\tag{3}$$
Solve for $x$
\begin{align*}
x&amp;=\frac{(3a^2+2)\pm\sqrt{(3a^2+2)^2-12a(a^3+2a-1)}}{6a}\\
&amp;=\frac{(3a^2+2)\pm\sqrt{-3a^4-12a^2+12a+4}}{6a}\tag{4}
\end{align*}
The discriminant is
\begin{align*}
-3a^4-12a^2+12a+4&amp;=-3a^4-4(3a^2-3a-1)\\
&amp;=-3a^4-4[3\left((a-\tfrac12)^2-\tfrac7{12}\right)]\\
&amp;=-3a^4-12(a-\tfrac12)^2+7\\
&amp;=7-(3a^4+3(2a-1)^2)
\end{align*}
Now check various integral $a$ for solutions: $a=0$ gives division by zero in $(4)$; $a=1$ gives $x_+=1$ or $x_-=\frac23$; $a=-1$ gives  discriminant of $-23$; and for $|a|\ge2$ we get the discriminant $7-(3a^4+3(2a-1)^2)&lt;0$ (since $3a^4+3(2a-1)^2$ is nonnegative).

Hence $x=1$, $y=0$ is the only solution.