3. Solve the equation in integers:

$$
x^{2}-x y-2 y^{2}=7
$$
Answer: $\{(3 ;-2),(5 ; 2),(-3 ; 2),(-5 ;-2)\}$

## Solution:

Let's factor the left side of the equation, for example, by grouping:

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

From this, we get the following form of the original equation:

$$
(x+y)(x-2 y)=7
$$

Considering that $x$ and $y$ are integers, and the number 7 is a prime number, solving the equation reduces to solving four systems:

$$
\left\{\begin{array}{l}
x+y=1 \\
x-2 y=7
\end{array},\left\{\begin{array}{c}
x+y=7 \\
x-2 y=1
\end{array},\left\{\begin{array}{l}
x+y=-1 \\
x-2 y=-7
\end{array},\left\{\begin{array}{l}
x+y=-7 \\
x-2 y=-1
\end{array}\right.\right.\right.\right.
$$

Solving these systems of equations, we get four pairs of solutions:

$$
(3 ;-2),(5 ; 2),(-3 ; 2),(-5 ;-2)
$$

Note: 1 point for each correct solution found by trial and error.