Let $x$ and $y$ be integers with $x+y \neq 0$. Determine all pairs $(x, y)$ for which

$$
\frac{x^{2}+y^{2}}{x+y}=10
$$

holds.

(Walther Janous)

Answer.

$$
(x, y) \in\{(-2,4),(-2,6),(0,10),(4,-2),(4,12),(6,-2),(6,12),(10,0),(10,10),(12,4),(12,6)\}
$$
The equation in equivalent form is

$$
\begin{aligned}
x^{2}+y^{2} & =10 x+10 y \\
\Longleftrightarrow \quad x^{2}-10 x+y^{2}-10 y & =0 \\
\Longleftrightarrow \quad(x-5)^{2}+(y-5)^{2} & =50
\end{aligned}
$$

with $x+y \neq 0$, i.e., we need to investigate $a^{2}+b^{2}=50$ for integers $a:=x-5$ and $b:=y-5$. Since neither of the two numbers can be greater in absolute value than 7, one obtains by checking the few cases $(a, b) \in\{( \pm 1, \pm 7),( \pm 7, \pm 1),( \pm 5, \pm 5)\}$ as the only solutions. With $(a, b) \in\{( \pm 1, \pm 7),( \pm 7, \pm 1)\}$, one gets $x-5= \pm 1$ and $y-5= \pm 7$ (or $x$ and $y$ swapped), leading to $x \in\{4,6\}$ and $y \in\{-2,12\}$ (or $y \in\{4,6\}$ and $x \in\{-2,12\}$). The pair $(a, b)=(5,5)$ leads to $x-5= \pm 5$ and $y-5= \pm 5$ with $x \in\{0,10\}$ and $y \in\{0,10\}$. The pair $(x, y)=(0,0)$ is the only one that does not satisfy the condition $x+y \neq 0$. Thus, the (eleven) solution pairs mentioned in the answer are obtained.

(Walther Janous)