(solved by Dmitry Ivanov). Find all pairs of integers ( $x, y$ ) such that:

$$
x^{3}=y^{3}+2 y^{2}+1 .
$$
. Given that $2 y^{2}+1>0$ and the strict monotonicity of the cube function, we deduce that the equation implies $x>y$. On the other hand, we can rewrite the equation as:

$$
x^{3}=(y+1)^{3}-y^{2}-3 y
$$

Thus, if $y^{2}+3 y>0$, we deduce by the same argument as before that $x>y$ given that $x$ and $y$ are both integers. This leads us to consider when the quantity $y^{2}+3 y=y(y+3)$ is negative or zero. A quick sign table shows that this is for $y$ between -3 and 0. Since $y$ is an integer, only the values $-3,-2,-1$ and 0 remain, which we test one by one by substituting them into the equation. We obtain the list of solutions, which consists of the pairs $(1,0),(1,-2)$ and $(-2,-3)$.