, 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 $x^2y + 10 = 14y$. This implies $y = \frac{10}{14 - x^2}$.

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

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 $(-4, -5)$.