## Task 13/68

All integer solutions of the equation $\sqrt{y^{2}-80}=2(x+2)$ are sought.
Since the right term of the equation should be an integer, the left term must also be an integer. Therefore, it holds that $\sqrt{y^{2}-80}=z$ with integer $z$ or

$$
y^{2}-80=z^{2}, y^{2}-z^{2}=(y-z)(y+z)=80
$$

Obviously, $y$ and $z$ are either both even or both odd, since otherwise the factors $y-z$ and $y+z$ would both be odd, and thus their product would be odd. Therefore, one decomposes the number 80 into its even factors:

$$
\begin{array}{r}
80=|40| \cdot|2|=|21+19| \cdot \mid 21-19] \\
80=|20| \cdot|4|=|12+8| \cdot \mid 12-8] \\
80=|10| \cdot|8|=|9+1| \cdot \mid 9-1]
\end{array}
$$

Thus, the only possible values for $y$ are $\pm 9 ; \pm 12 ; \pm 21$. For $\pm 12$, an integer value for $x$ is obtained: $x=2$. The integer solutions are therefore $x_{1}=2 ; y_{1}=12$ and $x_{2}=2 ; y_{2}=-12$.