AUT_ABooklet_2023

Answer. There are three such pairs, $(x,y)=(2,2)$, $(x,y)=(2,3)$ and $(x,y)=(3,2)$.

For $x=1$, we get $d=1$ and the given equation becomes the contradiction $y=y+2$. This works analogously for $y=1$. Therefore, we can assume $x\ge 2$ and $y\ge 2$.

We start with the case $d=1$ which gives the equation $$xy=x+y+1⟺(x-1)(y-1)=2.$$ The possible factorizations $2=1\cdot 2$ and $2=2\cdot 1$ give the pairs $(x,y)=(2,3)$ and $(x,y)=(3,2)$, respectively, because $\gcd (x,y)=1$ is satisfied.

Now, we treat the case $d\ge 2$. The given equation is equivalent to $$\frac{1}{xd}+\frac{1}{yd}+\frac{d}{xy}=1.$$ Because of $xd\ge 4$ and $yd\ge 4$, we get $$1\le \frac{1}{4}+\frac{1}{4}+\frac{d}{xy}⟺xy\le 2d.$$ Together with $xy\ge d^2$, we obtain $d=2$, $x=y=2$ which gives indeed the third pair $(x,y)=(2,2)$ with $\gcd (2,2)=2$.