THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) We have $m=p^2+2p\sqrt{n}+n$, hence $\sqrt{n}$ is a rational number, and therefore, $n$ is a square. Similarly $m$ is a square, as well.

b) We deduce from a) that $\overline{abcd}$ and $\overline{acd}$ are both squares. Since $11|\overline{bb}$, it follows that $11|\overline{abcd-acd}$, hence $11|100(\overline{ab}-\overline{a})$, and, finally, $11|9a+b$. This yields $(a;b)\in \{(1;2),(2;4),(3;6),(4;8),(6;1),(7;3),(8;5),(9,7)\}$.

By inspection, we find that $\overline{abcd}=1296$.