2. Find all solutions to the equation

$$
x^{4}=y^{2}+2 y+2
$$

where $x, y$ are integers.
Solution. Let's factor out the square of the sum on the right side, move the square to the left, and factor the left side.

$$
\begin{gathered}
x^{4}=y^{2}+2 y+1+1 \quad x^{4}=(y+1)^{2}+1 \\
x^{4}-(y+1)^{2}=1 \\
\left(x^{2}-y-1\right)\left(x^{2}+y+1\right)=1
\end{gathered}
$$

The product of two integer factors equals 1 only when both factors are simultaneously 1 or -1. Let's solve two systems of equations.

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

Add the two equations.

$$
2 x^{2}=2
$$

$$
\left\{\begin{array}{l}
x^{2}=1 \\
y=-1
\end{array}\right.
$$

We obtained two pairs of solutions $(-1,-1)$ and $(1,-1)$. Now, let's solve the second system.

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

This system has no solutions, as it leads to the equation $2 x^{2}=-2$.

Answer. $(-1,-1)$ and $(1,-1)$.