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 $y>0$, then $y^3< y^3+2y+1< (y+1)^3$, 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.

Or you can write $x=y+z$ for some $z\in Z$ (that is $z=x-y$). So we get a quadratic equation on $y$: $$ 3y^2z+y(3z^2-2)+z^3-1=0$$ which has a discriminant a perfect square: $$d^2 = (3z^2-2)^2-12z(z^3-1)=-3z^4-12z^2+12z+4$$

So we have $$3z^4+12z^2\leq 12z+4$$ and this can not be true for a lot of integers $z$ (in fact only for $0$ and $1$)...

Solution

Assume that $$x=y+t,~~~t \in \mathbb{Z}.$$

Then $$(y+t)^3=y^3+2y+1,$$ namely $$3ty^2+(3t^2-2)y+t^3-1=0.\tag1$$

If $t=0$, then $y=\dfrac{1}{2}$, which is absurd. Therefore, $t \neq 0$, which implies that $(1)$ could be seen as a quadratic equation with respect to $y$.

Consider the discriminant for $(1)$. $$\Delta=(3t^2-2)^2-4\cdot 3t \cdot (t^3-1)=-3(t^2+2)^2+12t+16.$$

If $|t|\geq 2$, then $$\Delta=-3(t^2+2)^2+12t+16<-3(t+2)^2+12t+16=4-3t^2<0.$$ Thus, $(1)$ has no real root.

Therefore, the possible values of $t$ are $t=\pm 1.$ Now, we may verify that $t=1$ is the only one solution. Under this case, $$x=1,y=0.$$

There are already a lot of solutions but I want to point out that your factorization can be completed to a proof:

Assume that y is nonnegative. From $$x^3=y^3+2y+1$$ follows that

$$x^3>y^3$$

and therefore $$x>y.$$ This implies $$3y\le 3y^2\le 1 (y^2+y^2+y^2)\lt (x-y)(x^2 + xy + y^2)=2y+1$$ and further $$y\lt 1 .$$

For negative $y$ a similar argument holds.