Romanian Mathematical Olympiad

a) Since $\sqrt{x + \sqrt{x}} \in \mathbb{Q}$, $x$ must be a square. Let $x = n^2$, with $n \in \mathbb{N}$. We obtain $n(n + 1) = y^2$, and since $n^2 \le n(n + 1) < (n + 1)^2$, we deduce that $n = 0$, hence $x = y = 0$.

b) There are infinitely many Pythagorean triples $(p, q, r)$ with $p^2 + q^2 = r^2$. Choosing $x = \frac{p^4}{q^4}$, we obtain
$$\sqrt{x + \sqrt{x}} = \sqrt{\frac{p^2}{q^2} \left( \frac{p^2}{q^2} + 1 \right)} = \sqrt{\frac{p^2}{q^2} \left( \frac{p^2 + q^2}{q^2} \right)} = \frac{pr}{q^2} \in \mathbb{Q}.$$