From (i) and (ii) follows $x \ge 0$ and $y \ge 0$. Moreover, $x = 0$ if and only if $y = 0$. This points to the solution $(0, 0)$ and the other solutions $(x, y)$ have $x > 0$ and $y > 0$.
Let $(x, y)$ be a solution with $x > 0$ and $y > 0$.
From (i) and (ii) follows $x \ge 2y^2 \ge 8x^4$, therefore $x(8x^3 - 1) \le 0$ and, since $x > 0$, this leads to $8x^3 \le 1$, hence $0 < x \le \frac{1}{2}$. In the same way, $0 < y \le \frac{1}{2}$.
It is enough to look at the case $x \ge y$. We have $0 \le 8(x-y) < 8x \le 4$ and, from (iii), follows $8(x-y) \in \{0, 1, 2, 3\}$, therefore $x-y \in \{0, \frac{1}{8}, \frac{1}{4}, \frac{3}{8}\}$ and the following situations may rise.
* If $x - y = 0$, that is $x = y$, all the conditions from the statement are fulfilled. So, the pairs $(a, a)$, with $a \in (0, \frac{1}{2}]$, are solutions.
* If $x - y = \frac{1}{8}$, that is $y = x - \frac{1}{8}$, we get: $y \ge 2x^2 \iff x - \frac{1}{8} \ge 2x^2 \iff (4x - 1)^2 \le 0 \iff x = \frac{1}{4}$. This yields $y = \frac{1}{8}$, and the pair $(\frac{1}{4}, \frac{1}{8})$ also fulfills (i), therefore it is a solution.
* If $x - y = \frac{1}{4}$, that is $y = x - \frac{1}{4}$, we get: $y \ge 2x^2 \iff x - \frac{1}{4} \ge 2x^2 \iff 8x^2 - 4x + 1 \le 0 \iff 4x^2 + (2x - 1)^2 \le 0$, impossible.
* If $x - y = \frac{3}{8}$, that is $y = x - \frac{3}{8}$, we get: $y \ge 2x^2 \iff x - \frac{3}{8} \ge 2x^2 \iff 16x^2 - 8x + 3 \le 0 \iff (4x - 1)^2 + 2 \le 0$, impossible.
The solutions are the pairs $(\frac{1}{4}, \frac{1}{8})$, $(\frac{1}{8}, \frac{1}{4})$ and $(a, a)$, with $a \in [0, \frac{1}{2}]$.