Romanian Mathematical Olympiad

Suppose $x$ is an integer with the given properties. Then $x$ is a multiple of $y = a \cdot \overline{bc} \cdot \overline{cb}$. Since $y$ has eight divisors, it follows that $x = y$.

The numbers $\overline{bc}$ and $\overline{cb}$ are distinct primes, so $\overline{bc} \in \{13, 17, 37, 79\}$. The condition that $a + \overline{bc} + \overline{cb}$ is a square and $a$ is a prime forces $a = 5$, $\overline{bc} = 13$ and $\overline{cb} = 31$, implying $x = 2015$.