Task B-1.6.

Determine all pairs of integers $x$ and $y$ such that $x^{2}+6 x y+8 y^{2}+3 x+6 y=2$.
## Solution.

The expression $x^{2}+6 x y+8 y^{2}+3 x+6 y$ can be written in the form $x^{2}+6 x y+8 y^{2}+3 x+6 y=$ $(x+3 y)^{2}-y^{2}+3 x+6 y$ from which, by factoring, we obtain the following sequence of equalities:

$$
(x+3 y)^{2}-y^{2}+3 x+6 y=(x+2 y)(x+4 y)+3(x+2 y)=(x+2 y)(x+4 y+3)
$$

Thus, we have $(x+2 y)(x+4 y+3)=2$.

Since $x$ and $y$ are integers, we conclude that one of the factors of the expression $(x+2 y)(x+4 y+3)$ is equal to 1, and the other is 2, or one of the factors is equal to -1, and the other is -2.

By solving the system $\left\{\begin{array}{l}x+2 y=2 \\ x+4 y+3=1\end{array} \quad\right.$ we get $x=6$ and $y=-2$.

From $\left\{\begin{array}{l}x+2 y=1 \\ x+4 y+3=2\end{array} \quad\right.$ we get $x=3$ and $y=-1$.

Similarly, by solving the system $\left\{\begin{array}{l}x+2 y=-2 \\ x+4 y+3=-1\end{array}\right.$ we find that $x=0, y=-1$,
and from the system $\left\{\begin{array}{l}x+2 y=-1 \\ x+4 y+3=-2\end{array} \quad\right.$ we get $x=3$ and $y=-2$.

Thus, we have four solutions: $(6,-2),(3,-1),(0,-1),(3,-2)$

Note: If a student guesses the solutions, they get one point for each correct solution.