6. (5 points) Find all pairs of integers $(x, y)$ for which the following equality holds:

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

#
# Solution:

$$
\begin{array}{ll}
x(x+1)(x+7)(x+8)=y^{2} \\
\left(x^{2}+8 x\right)\left(x^{2}+8 x+7\right)=y^{2}
\end{array}
$$

Let $z=x^{2}+8 x+\frac{7}{2} \Rightarrow 2 z=2 x^{2}+16 x+7 \in Z$. From (2) we find

$$
\left(z-\frac{7}{2}\right)\left(z+\frac{7}{2}\right)=y^{2} \Leftrightarrow z^{2}-\frac{49}{4}=y^{2}
$$

which is equivalent to $(2 z-2 y)(2 z+2 y)=49$. Since $(2 z \pm 2 y) \in Z$, and $49=1 \cdot 49=7 \cdot 7=49 \cdot 1=-1 \cdot(-49)=(-7) \cdot(-7)=(-49) \cdot(-1)$, only the following six cases are possible:

1) $\left\{\begin{array}{c}2 z-2 y=1 \\ 2 z+2 y=49\end{array} \Rightarrow y=12,2 z=25 \Rightarrow\left[\begin{array}{c}x=1 \\ x=-9\end{array}\right.\right.$
2) $\left\{\begin{array}{l}2 z-2 y=7 \\ 2 z+2 y=7\end{array} \Rightarrow y=0,2 z=7 \Rightarrow\left[\begin{array}{c}x=0 \\ x=-8 .\end{array}\right.\right.$
3) $\left\{\begin{array}{c}2 z-2 y=49 \\ 2 z+2 y=1\end{array} \Rightarrow y=-12,2 z=25 \Rightarrow\left[\begin{array}{c}x=1 \\ x=-9 \text {. }\end{array}\right.\right.$
4) $\left\{\begin{array}{c}2 z-2 y=-1 \\ 2 z+2 y=-49\end{array} \Rightarrow y=-12,2 z=-25 \Rightarrow x=-4\right.$.
5) $\left\{\begin{array}{l}2 z-2 y=-7 \\ 2 z+2 y=-7\end{array} \Rightarrow y=0,2 z=-7 \Rightarrow\left[\begin{array}{l}x=-1 \\ x=-7\end{array}\right.\right.$
6) $\left\{\begin{array}{c}2 z-2 y=-49 \\ 2 z+2 y=-1\end{array} \Rightarrow y=12,2 z=-25 \Rightarrow x=-4\right.$.

Answer:

$(1 ; 12),(1 ;-12),(-9 ; 12),(-9 ;-12),(0 ; 0),(-8 ; 0),(-4 ;-12)$,

$(-4 ; 12),(-1 ; 0),(-7 ; 0)$

(10 pairs).