74th Romanian Mathematical Olympiad

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: $y\ge 2x^2⟺x-\frac{1}{8}\ge 2x^2⟺(4x-1{)}^2\le 0⟺x=\frac{1}{4}$. 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: $y\ge 2x^2⟺x-\frac{1}{4}\ge 2x^2⟺8x^2-4x+1\le 0⟺4x^2+(2x-1{)}^2\le 0$, impossible. If $x-y=\frac{3}{8}$, that is $y=x-\frac{3}{8}$, we get: $y\ge 2x^2⟺x-\frac{3}{8}\ge 2x^2⟺16x^2-8x+3\le 0⟺(4x-1{)}^2+2\le 0$, impossible. The solutions are the pairs $(\frac{1}{4},\frac{1}{8})$, $(\frac{1}{8},\frac{1}{4})$ and $(a,a)$, with $a\in [0,\frac{1}{2}]$.