Problem E : 17 052

             In GMB 11/2024 page 650,  author Mihaela BERINDEANU, Bucharest, proposed for 8th grade.

In translation:

"Solve in $\mathbb{Z}$ the equation $x^2+2y^2-4=2xy-2|x-y|.$"

ANSWER CiP

$$(x,y) \in \{(-2,-2),\;(-2,-1),\;(0,\pm 1),\;(2,1),\;(2,2)\}$$

                    Solution CiP

               Let us denote $z=x-y$. We have 

$$x=y+z \tag{1}$$

and the equation is written equivalently

 $$(y+z)^2+2y^2-4=2(y+z)y-2|z|\;\Leftrightarrow$$

$$\Leftrightarrow\;\;y^2+2yz+z^2+2y^2-4=2y^2+2yz-2|z|\;\Leftrightarrow\;y^2+z^2+2|z|-4=0\;\Leftrightarrow$$

$$\Leftrightarrow\;\;y^2+(|z|+1)^2=5. \tag{2}$$

In $\mathbb{Z}$ equation (2) is satisfied in the cases

$\begin{cases}y^2=1\\|z|+1=2; \end{cases} \tag{3}$

$\begin{cases}y^2=4\\|z|+1=1. \end{cases} \tag{4}$

Case (3) gives us \begin{cases}y=\pm1\\x-y=\pm1 \end{cases}

and we find the solutions $(2,1),\;(0,1),\;(0,-1),\;(-2,-1)$.

Case (4) give us \begin{cases}y=\pm 2\\x-y=0 \end{cases}

and we find the solutions $(2,2),\;(-2,-2)$. We got the answer.

$\blacksquare$