1. Solve the equation in integers

$$
y^{2}(y-x+2)-y(x+4)+5 x+7=0
$$
Answer: $(15 ; 2),(10 ;-3),(-2 ; 1),(-5 ;-2)$.

Solution: Express $x$ from this equation:

$$
x=\frac{y^{3}+2 y^{2}-4 y+7}{y^{2}+y-5}=y+1+\frac{12}{y^{2}+y-5}
$$

Therefore, the number $y^{2}+y-5$ is a divisor of the number 12 and at the same time $x, y \in \mathbb{Z}$.

For the number 12, the divisors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$, but the conditions that $x, y \in \mathbb{Z}$ will ultimately be satisfied by:

$$
\left[\begin{array}{c}
y^{2}+y-5=1 \\
y^{2}+y-5=-3
\end{array} \Rightarrow \left[\begin{array}{l}
y^{2}+y-6=0 \\
y^{2}+y-2=0
\end{array} \Rightarrow \left[\begin{array}{l}
y=2, y=-3 \\
y=1, y=-2
\end{array} \Rightarrow \left[\begin{array}{l}
x=15, x=10 \\
x=-2, x=-5
\end{array}\right.\right.\right.\right.
$$