2. Find all pairs of integers $(x, y)$ that satisfy the equation $x^{2}-x y-6 y^{2}-11=0$. For each pair $(x, y)$ found, calculate the product $x y$. In the answer, write the sum of these products.
Solution. $x^{2}-x y-6 y^{2}-11=0,(x-3 y)(x+2 y)=11$. Since $x$ and $y$ are integers, we have four cases:
$\left\{\begin{array}{c}x-3 y=11 \\ x+2 y=1\end{array} \Leftrightarrow\left\{\begin{array}{c}y=-2 \\ x=5\end{array}\right.\right.$
2)
$\left\{\begin{array}{c}x-3 y=-11 \\ x+2 y=-1,\end{array} \Leftrightarrow\left\{\begin{array}{c}y=2 \\ x=-5\end{array}\right.\right.$
$\left\{\begin{array}{c}x-3 y=1, \\ x+2 y=11,\end{array} \Leftrightarrow\left\{\begin{array}{l}y=2 \\ x=7\end{array}\right.\right.$
$\left\{\begin{array}{c}x-3 y=-1 \\ x+2 y=-11\end{array} \Leftrightarrow\left\{\begin{array}{l}y=-2 \\ x=-7\end{array}\right.\right.$

Answer: 8.

$$
\begin{aligned}
& \text { 3. Let } \\
& g(x)=\frac{2}{x^{2}-8 x+17} . \text { Find all possible } \\
& a^{2}+6 a+\frac{727}{145} \leq g\left(g^{4}(x)\right) \leq 10 a^{2}+29 a+2
\end{aligned}
$$