67th Romanian Mathematical Olympiad

Denote $m=\sqrt{n+3}+\sqrt{\sqrt{n+3}}$. Then $n+\sqrt{n+3}=(m-\sqrt{n+3}{)}^2$, whence $(2m+1)\sqrt{n+3}=m^2+3$. Then there exists $p\in \mathbb{N}$ so that $n+3=p^2$ and, since $p+\sqrt{n+p}\in \mathbb{N}$, there exists $q\in \mathbb{N}$ so that $n+p=q^2$. Eliminating $n$ yields $p^2-3=q^2-p$, that is $4p^2+4p-12=4q^2$, or $(2p+1{)}^2-(2q{)}^2=13$, which is equivalent to $(2p+1-2q)(2p+1+2q)=13$. This gives $p=3$, so $n=6$.