## Task $9 / 84$

Determine all pairs $(x ; y)$ of integers that satisfy the equation:

$$
x^{2} y+x y^{2}=2 x y+x+y+1
$$
It is

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

If we set $x y=a, x+y=b$, the equation takes the form

$$
a b=2 a+2 b+1 \quad \text { or } \quad b=\frac{2 a+1}{a-1}=2+\frac{3}{a-1}
$$

(since obviously $a=1$ does not provide a solution). For $b$ to be an integer, only the values $a_{1}=-2, a_{2}=0, a_{3}=2$ and $a_{4}=4$ are possible. From these, the values for $b$ are $b_{1}=1, b_{2}=$ $-1, b_{3}=5$ and $b_{4}=3$. Now, the four systems of equations

$$
x y=a_{i} \quad ; \quad x+y=b_{i}
$$

with $i=1 ; 2 ; 3 ; 4$ need to be solved for integers $x ; y$. Only the first two have integer solutions $(x ; y)$, and they are

$$
\begin{aligned}
& \text { 1.) } x_{11}=y_{12}=-1 \quad ; \quad x_{12}=y_{11}=2 \\
& \text { 2.) } \quad x_{21}=y_{22}=0 \quad ; \quad x_{22}=y_{21}=-1
\end{aligned}
$$

Thus, the four pairs $(-1 ; 0),(0 ;-1),(2 ;-1),(-1 ; 2)$ are solutions to the given equation. No further solutions can exist based on the solution method.