AUT_ABooklet_2020

$$(x,y)\in \{(-2,4),(-2,6),(0,10),(4,-2),(4,12),(6,-2),(6,12),(10,0),(10,10),(12,4),(12,6)\}.$$

An equivalent form of the given equation is $$\begin{array}{rl}{x}^2+y^2& =10x+10y\\ ⟺x^2-10x+y^2-10y& =0\\ ⟺(x-5{)}^2+(y-5{)}^2& =50\end{array}$$ with $x+y\ne 0$. We therefore have to solve the equation $a^2+b^2=50$ for integers $a$ and $b$. As $\max (a^2,b^2)\le 50$ we get $\max (|a|,|b|)\le 7$. Furthermore we obtain $\min (a^2,b^2)\le 25$ which yields $\min (|a|,|b|)\le 5$. Analyzing the individual cases, we obtain $(a,b)\in \{(\pm 1,\pm 7),(\pm 7,\pm 1),(\pm 5,\pm 5)\}$ as the only possible solutions. If $(a,b)\in \{(\pm 1,\pm 7),(\pm 7,\pm 1)\}$, we get that $x-5=\pm 1$ and $y-5=\pm 7$ (or $x$ and $y$ swapped), which yields $x\in \{4,6\}$ and $y\in \{-2,12\}$ (or $y\in \{4,6\}$ and $x\in \{-2,12\}$). The case $(a,b)=(\pm 5,\pm 5)$ gives $x-5=\pm 5$ and $y-5=\pm 5$ which is equivalent to $x\in \{0,10\}$ and $y\in \{0,10\}$. The pair $(x,y)=(0,0)$ is the only one violating $x+y\ne 0$. Therefore, we get the (eleven) different pairs listed in the answer.