, 2019

Solution:
Note that if $(x, y)$ is a solution, so is $(-x,-y)$. Thus, we consider $x \geq 0$.

When $x \equiv 0 \pmod{4}$, $y = 3x/4$ by inequality 2. Inequality 1 gives $|x^{2}/9| \leq 1$, so $x \leq 3$, so $x = 0$.

When $x \equiv 1 \pmod{4}$, $y = (3x+1)/4$ by inequality 2. Beyond $x = 1$, $2y^{2} - x^{2} > 1$, so there are no more solutions.

When $x \equiv 2 \pmod{4}$, there are no solutions for $y$.

When $x \equiv 3 \pmod{4}$, $y = (3x-1)/4$ by inequality 2. Beyond $x = 7$, $2y^{2} - x^{2} > 1$, so there are no more solutions.

Thus, the solutions are $(0,0), (1,1), (3,2), (7,5)$, and the negations of the latter three, giving 7 solutions.