3. In the set of integers, solve the equation:

$$
y^{4}-x(x+1)(x+2)(x+3)=1
$$
Solution. The given equation is successively equivalent to the equations:

$$
\begin{aligned}
& \left(x^{2}+3 x+1\right)^{2}-y^{4}=0 \\
& 16\left(x^{2}+3 x+1-y^{2}\right)\left(x^{2}+3 x+1+y^{2}\right)=0 \\
& \left((2 x+3)^{2}-(2 y)^{2}-5\right)\left((2 x+3)+(2 y)^{2}-5\right)=0 \\
& (2 x+3-2 y)(2 x+3+2 y)-5)\left((2 x+3)+(2 y)^{2}-5\right)=0
\end{aligned}
$$

The ordered pair $(x, y)$ of integers is a solution to the last equation if and only if one of the following eight cases is satisfied:

1) $2 x+2 y+3=1,2 x-2 y+3=5$,
2) $2 x+2 y+3=5,2 x-2 y+3=1$,
3) $2 x+2 y+3=-1,2 x-2 y+3=-5$,
4) $2 x+2 y+3=-5,2 x-2 y+3=-1$,
5) $2 x+3=1,2 y=2$,
6) $2 x+3=1,2 y=-2$,
7) $2 x+3=-1,2 y=2$,
8) $2 x+3=-1,2 y=-2$.

Solving these eight systems of equations, we obtain the following solutions to the given equation:

$$
(0,-1),(0,1),(-3,1),(-3,-1),(-1,1),(-1,-1),(-2,1),(-2,-1)
$$