2.3. (GDR, 73). Solve the equation

$$
x(x+1)(x+7)(x+8)=y^{2}
$$

in integers.
2.3. Let the numbers $x, y \in \mathbf{Z}$ satisfy the equation, then

$$
y^{2}=(x(x+8))((x+1)(x+7))=\left(x^{2}+8 x\right)\left(x^{2}+8 x+7\right)=z^{2}+7 z
$$

where we denote $z=x^{2}+8 x$. If $z>9$, then

$$
(z+3)^{2}=z^{2}+6 z+9<z^{2}+7 z=y^{2}<z^{2}+8 z+16=(z+4)^{2}
$$

which means that the number $y^{2}$ is trapped between the squares of two consecutive natural numbers, which is impossible. Therefore, $x^{2}+8 x=z \leq 9$, from which $-9<x \leqslant 1$. By sequentially checking the values

$$
x=-9,-8, \ldots ., 1
$$

we find that the number $x(x+1)(x+7)(x+8)$ is a square of an integer only when $x$ is $-9,-8,-7,-4,-1,0$ and 1. Thus, we obtain all solutions to the original equation: $(-9 ; 12)$, $(-9 ;-12),(-8 ; 0) ;(-7 ; 0),(-4 ; 12),(-4 ;-12),(-1 ; 0),(0 ; 0)$, $(1 ; 12),(1 ;-12)$