Romanian Mathematical Olympiad - District Round

If $c = 0$, then $b^2 + d^2 \le 2 \cdot 7^2$, which means that $\overline{ab}$ must be a prime number such that $1924 \le \overline{ab}^2 < 2022$, false. Therefore, $c \ne 0$. As $\overline{ab}$ and $\overline{cb}$ are primes, from the given equality we deduce that $d = 2$. We obtain $\overline{ab}^2 + \overline{cb}^2 = 2018 < 47^2$, thus $\overline{ab}$ and $\overline{cb}$ are at most 43. Since $b$ is an odd digit, and the last digit of $\overline{ab}^2 + \overline{cb}^2$ is 8, we deduce that $b \in \{3, 7\}$.

If $b = 7$, then $\overline{ab}, \overline{cb} \in \{17, 37\}$, therefore $\overline{ab}^2, \overline{cb}^2 \in \{289, 1369\}$ and $\overline{ab}^2 + \overline{cb}^2 \ne 2018$.

If $b = 3$, then $\overline{ab}, \overline{cb} \in \{13, 23, 43\}$, and we verify that only for $(\overline{ab}, \overline{cb}) \in \{(13, 43), (43, 13)\}$ the given equality holds, therefore the solutions are $\overline{abcd} = 1342$ and $\overline{abcd} = 4312$.