\section*{Problem 2 - 121242}

All pairs \((x, y)\) of integers are to be specified for which the equation is satisfied:

\[
x(x+1)(x+7)(x+8)=y^{2}
\]
}

With \(x=z-4\), the equation becomes more manageable:

\[
\begin{array}{r}
(z-4)(z-3)(z+3)(z+4)=y^{2} \\
\left(z^{2}-16\right)\left(z^{2}-9\right)=z^{4}-25 z^{2}+144=\left(z^{2}-12\right)^{2}-z^{2}=y^{2}
\end{array}
\]

From the last equation \(\left(z^{2}-12\right)^{2}-z^{2}=y^{2}\), it becomes clear that for a pair \((z, y)\) that satisfies the equation, on the one hand,

\[
\left(z^{2}-12\right)^{2}>y^{2}
\]

holds (for \(z \neq 0\)), but also

\[
\left(z^{2}-13\right)^{2}=z^{4}-26 z^{2}+16925\). This means that \(\left(z^{2}-12\right)^{2}-z^{2}\) lies strictly between two consecutive squares and therefore cannot be equal to a square \(y^{2}\). Consequently, one only needs to check the cases for \(z^{2} \leq 25\).

One quickly sees that exactly \(z \in\{-5,-4,-3,0,3,4,5\}\) yields a square number. The solution to the problem is thus \((x, y) \in\{(-9, \pm 12),(-8,0),(-7,0),(-4, \pm 12),(-1,0)(0,0),(1, \pm 12)\}\).