Baltic Way mathematical team contest, 2002

Solution:
Squaring both sides of the equation gives
$$a + b + 2 \sqrt{a b} = 2 + \sqrt{3}$$
so $2 \sqrt{a b} = r + \sqrt{3}$ for some rational number $r$. Squaring both sides of this gives $4 a b = r^2 + 3 + 2 r \sqrt{3}$, so $2 r \sqrt{3}$ is rational, which implies $r = 0$. Hence $a b = 3 / 4$ and substituting this into (8) gives $a + b = 2$. Solving for $a$ and $b$ gives $(a, b) = \left(\frac{1}{2}, \frac{3}{2}\right)$ or $(a, b) = \left(\frac{3}{2}, \frac{1}{2}\right)$.