## Task A-3.2.

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

$$
x^{2}(y-1)+y^{2}(x-1)=1
$$
## Solution.

Given the equation is equivalent to $s$

$$
\begin{gathered}
x^{2} y - x^{2} + x y^{2} - y^{2} = 1 \\
x y(x + y) - (x^{2} + y^{2}) = 1 \\
x y(x + y) - ((x + y)^{2} - 2 x y) = 1
\end{gathered}
$$

If we introduce new variables $u = x + y, v = x y$, we get the equation:

$$
u v - (u^{2} - 2 v) = 1
$$

from which it follows that $u v + 2 v = u^{2} + 1$, or

$$
v = \frac{u^{2} + 1}{u + 2} = \frac{u^{2} - 4 + 5}{u + 2} = u - 2 + \frac{5}{u + 2}
$$

Now we see that $u + 2$ must be a divisor of the number 5, so we have four cases:

$$
\begin{array}{llll}
u + 2 = 5 & u + 2 = 1 & u + 2 = -1 & u + 2 = -5 \\
\frac{5}{u + 2} = 1 & \frac{5}{u + 2} = 5 & \frac{5}{u + 2} = -5 & \frac{5}{u + 2} = -1 \\
u = 3 & u = -1 & u = -3 & u = -7 \\
v = 2 & v = 2 & v = -10 & v = -10
\end{array}
$$

Now, due to Viète's formulas, we know that $x, y$ are the solutions of the quadratic equations:

$$
\begin{gathered}
z^{2} - 3 z + 2 = 0 \\
z^{2} + z + 2 = 0 \\
z^{2} + 3 z - 10 = 0 \\
z^{2} + 7 z - 10 = 0
\end{gathered}
$$

The solutions of the first equation are 1 and 2, and the solutions of the third equation are -5 and 2.

The solutions of the second equation are not real because its discriminant is negative, while the solutions of the fourth equation are irrational because its discriminant 89 is not a perfect square.

Finally, the solutions we are looking for are $(x, y) \in \{(1,2), (2,1), (2,-5), (-5,2)\}$.