THE 68th 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 $$\sqrt{x+\sqrt{x}}=\sqrt{\frac{p^2}{q^2}\left(\frac{p^2}{q^2}+1\right)}=\sqrt{\frac{p^2}{q^2}\left(\frac{p^2+q^2}{q^2}\right)}=\frac{pr}{q^2}\in \mathbb{Q}.$$