75th Romanian Mathematical Olympiad

Since $\sqrt{\overline{ab}-4}+\sqrt{\overline{cd}}<10+10=20$, we obtain $p=2$ or $p=3$. If $p=2$, then $\overline{cd}=\overline{ab}+4$ and $\sqrt{\overline{ab}-4}+\sqrt{\overline{ab}+4}=4$, therefore $\overline{ab}=5$, false. If $p=3$, then $\overline{cd}=\overline{ab}+5$ and $\sqrt{\overline{ab}-4}+\sqrt{\overline{ab}+5}=9$. We obtain the unique solution $\overline{ab}=20$, thus $\overline{abcd}=2025$.

Alternative solution: Since $\overline{ab}-4$, $\overline{cd}$ and $\sqrt{\overline{ab}-4}+\sqrt{\overline{cd}}$ are integers, the natural numbers $k,n$ exist, such that $\overline{ab}-4=k^2$ and $\overline{cd}=n^2$, hence $k+n=p^2$. (1) We obtain $p+6=\overline{cd}-\overline{ab}+4=n^2-k^2=(n-k)(n+k)$. Using (1), we deduce that $p+6=(n-k)\cdot p^2$, therefore $p\mid (p+6)$, so $p\in \{2,3\}$. If $p=2$, we obtain $k=1$ and $n=3$, therefore $\overline{ab}=5$, which is false. If $p=3$, we obtain $k=4$ and $n=5$, so $\overline{ab}=20$ and $\overline{cd}=25$, which satisfies the hypothesis. Therefore, the solution is 2025.