We say that a number is rational if it can be written in the form $\frac{p}{q}$, with $p$ and $q$ being integers. If a real number is not rational, we say that it is irrational. For example, $\frac{1}{4}$ is a rational number and $\sqrt{3}$ is irrational. Find all pairs of rational numbers $(a, b)$ such that

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

Squaring, we obtain the equivalent equations:

$$
\begin{aligned}
a+b+2 \sqrt{a b} & =2+\sqrt{3} \\
2 \sqrt{a b} & =2-a-b+\sqrt{3}
\end{aligned}
$$

Since $a$ and $b$ are rational numbers, $r=2-a-b$ is also a rational number. Square the last equation:

$$
\begin{aligned}
4 a b & =r^{2}+3+2 r \sqrt{3} \\
4 a b-r^{2}-3 & =2 r \sqrt{3}
\end{aligned}
$$

If $r \neq 0$, we have

$$
\sqrt{3}=\frac{4 a b-r^{2}-3}{2 r}
$$

Since $r$ is a rational number, the number $\frac{4 a b-r^{2}-3}{2 r}$ is also rational. However, this leads to a contradiction, as $\sqrt{3}$ is an irrational number. Therefore, $r=0$, and from the last equations, we have

$$
\begin{aligned}
a+b & =2 \\
a b & =3 / 4
\end{aligned}
$$

This shows that $a$ and $b$ are roots of the equation $x^{2}-2 x+3 / 4=0$, that is, $(a, b)=(1 / 2,3 / 2)$ or $(a, b)=(3 / 2,1 / 2)$. These numbers satisfy the first equation and, consequently, are solutions to the equation of the problem.