Winter Mathematical Competition

Since $x{y}^2+2y$ divides $$(2x+y)(x{y}^2+2y)-y(2x^{2}y+x{y}^2+8x)=2y^2-4xy,$$ we conclude that $xy+2$ divides $2y-4x$. We consider two cases.

Case 1. Let $2y-4x\ge 0$. Then we have two possibilities:

1.1) If $x\ge 2$ then $xy+2>2y-4x$. Hence $2y-4x=0$, i.e. $x=a$ and $y=2a$, where $a\ge 2$ is an integer. Since $x{y}^2+2y=4a(a^2+1)$ divides $2x^{2}y+x{y}^2+8x=8a(a^2+1)$, this gives a solution.

1.2) If $x=1$ then $y^2+2y$ divides $8$, i.e. $y=2$.

Hence in this case the solutions are $(a,2a)$, where $a$ is a positive integer.

Case 2. Let $2y-4x<0$, i.e. $4x-2y>0$. If $y\ge 4$, then $xy+2>4x-2$. Therefore $y=1,2$ or $3$.

2.1) If $y=1$ then $\frac{2x^2+9x}{x+2}=2x+5-\frac{10}{x+2}$ is an integer. Hence $x+2$ divides $10$ and this gives the solutions $x=3,y=1$ and $x=8,y=1$.

2.2) If $y=2$ then $\frac{x^2+3x}{x+1}=x+2-\frac{2}{x+1}$ is an integer which gives $x=1$.

2.3) If $y=3$ then $\frac{6x^2+17x}{9x+6}$ is an integer. This implies that $3|x$, i.e. $x=3k$ for some positive integer $k$. After simplifications we obtain that $\frac{18k^2+17k}{9k+2}=(2k+1)+\frac{4k-2}{9k+2}$ is an integer, which is impossible for $k\ge 1$.

Finally, the solutions are $x=a,y=2a$ for all positive integers $a$ and $x=3,y=1,x=8,y=1$.