Slovenija 2008

If $(x,y)$ satisfies the equation then so does $(-x,y)$. We can thus limit ourselves to only non-negative values of $x$.

Multiply the equation by $2y$ to get $x^{2}y+10=14y$. This implies $y=\frac{10}{14-x^2}$.

We see that $14-x^2$ has to be a divisor of $10$, so $-10\le 14-x^2\le 10$, $-24\le -x^2\le -4$ and $24\ge x^2\ge 4$. Since we have assumed that $x\ge 0$, we get $5>x\ge 2$.

When $x=2$ we have $y=1$, when $x=3$ we have $y=2$ and when $x=4$ we have $y=-5$.

Hence, the only integral solutions of the equation are $(2,1)$, $(3,2)$, $(4,-5)$, $(-2,1)$, $(-3,2)$ and $(-4,-5)$.