Find all pairs $(a, b)$ of positive rational numbers such that

$$
\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}} .
$$

Answer: $(a, b)=\left(\frac{1}{2}, \frac{3}{2}\right)$ or $(a, b)=\left(\frac{3}{2}, \frac{1}{2}\right)$.
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 $(a, b)=\left(\frac{3}{2}, \frac{1}{2}\right)$.