Romanian Mathematical Olympiad - Final Round

a) Any pair $(a, \sqrt{3-a^2})$, with $a \in [0, 1]$, is a solution.

b) By squaring the equality $y \cdot \sqrt{3-x^2} = 3 - x \cdot \sqrt{3-y^2}$, we deduce that $(\sqrt{3-y^2} - x)^2 = 0$, therefore $x^2 + y^2 = 3$. (1)
Assume that there are numbers $x, y \in \mathbb{Q}_+ \cap [0, \sqrt{3}]$, for which (1) is true.
It is obvious that $x \ne 0$ and $y \ne 0$. Consider the positive integers $a, b, c, d$, such that $(a, b) = (c, d) = 1$, $x = \frac{a}{b}$ and $y = \frac{c}{d}$. From (1) we obtain $a^2d^2 + b^2c^2 = 3b^2d^2$. (2)
From $(c, d) = 1$ and $d^2 \mid b^2c^2$, we deduce $d^2 \mid b^2$. Similarly, from $(a, b) = 1$ and $b^2 \mid a^2d^2$, we obtain $b^2 \mid d^2$. Therefore, $b^2 = d^2$ and (2) leads to $a^2 + c^2 = 3b^2$.
If the integers $a$ and $c$ aren't multiples of $3$, then $a^2 + c^2 = M_3 + 2 \ne 3b^2$, false. Consequently $3 \mid a$ and $3 \mid c$, therefore $3b^2 = M_9$, thus $3 \mid b$ and $(a, b) \ge 3$, false. This contradicts our assumption and the conclusion follows.