Solve the equation $x^{3}+y^{3}+1=x^{2} y^{2}$ over the set of integers.
I. solution. Suppose that the integers $x, y$ satisfy the equation. Any common divisor of the two numbers must divide the number $x^{2} y^{2}-x^{3}-y^{3}=1$, so $x$ and $y$ are coprime. Due to symmetry, it is sufficient to consider the case $x \geq y$. If $x0$.

Rearrange the equation and factor both sides:

$$
\begin{aligned}
x^{3}+1 & =x^{2} y^{2}-y^{3} \\
(x+1)\left(x^{2}-x+1\right) & =y^{2}\left(x^{2}-y\right)
\end{aligned}
$$

1. First, consider the case $y \geq 0$. Then $x-1 \geq y \geq 0$, since $x, y$ are coprime, if $x=y$ then $x=y=1$ would follow, which does not provide a solution. Therefore, $x^{2}-y \geq x^{2}-x+1>0$, so necessarily $y^{2} \leq x+1$. If $y^{2} \leq x$ were true, then

$$
y^{2}\left(x^{2}-y\right) \leq x^{3}-x+1$, or

$$
x^{2}-y>x^{2}-x+1>0
$$

Thus, from the factorization of the equation, $y^{2}0$, and equation (1) becomes:

$$
-x z+1=\left(x^{2}+z\right)\left(z^{2}-x\right)
$$

Now the left side is certainly non-positive, the first factor on the right side is positive, so the second factor is either zero or negative. In the latter case, the absolute value of the right side is greater than $x^{2}$, while the absolute value of the left side is smaller, which is a contradiction. The only remaining case is if $z^{2}=x$ and $x z=1$. This gives $z^{3}=1$, so we get the solution $y=-1$ and $x=1$.

3. Both variables cannot be negative, because then the left side of the original equation would be negative, while the right side would be positive.
4. The only remaining case is if one of the variables is zero, for example, $x=0$. Then the equation is $y^{3}+1=0$, so we get the solution $x=0$, $y=-1$ (and vice versa, $x=-1$ and $y=0$). Taking symmetry into account, all solutions are:

$$
(3 ; 2), \quad(2 ; 3), \quad(1 ;-1), \quad(-1 ; 1), \quad(0,-1), \quad(-1 ; 0)
$$