Croatian Mathematical Society Competitions

Obviously, all real numbers satisfy the given condition.
Let us, from now on, assume that $a$ is not a real number.
If $z$ is a root of $P$, then $\bar{z}$ is a root of $P$ as well. Hence $\bar{a} = a^2$, $\bar{a} = a^3$ or $\bar{a} = a^4$.

In the first case, we get $|a| = |a|^2$, and then $|a| = 1$ (since $a \neq 0$). Hence $a^3 = a^2 \cdot a = \bar{a} \cdot a = |a|^2 = 1$, so both factors $(x - a)(x - a^2)$ and $x - a^3$ of $P$ have real coefficients. This implies that $a^4 = a^3 \cdot a = a$ is a real number, which is a contradiction, and there is no solution in this case.

In the second case, we similarly get $a^4 = 1$, hence $a = \pm i$ and both can be verified as solutions.

In the third case, we similarly get $a^5 = 1$, hence
$$a = \cos \frac{2k\pi}{5} + i \sin \frac{2k\pi}{5}, \quad k = 1, 2, 3, 4.$$
Since $\bar{a}^2 = a^5 \cdot \bar{a}^2 = a^3 \cdot a^2 \cdot \bar{a}^2 = a^3 \cdot |a|^4 = a^3$, both factors $(x - a)(x - a^4)$ and $(x - a^2)(x - a^3)$ of $P$ have real coefficients, so has $P$.

$$a \in \{i, -i\} \cup \left\{ \cos \frac{2k\pi}{5} + i \sin \frac{2k\pi}{5} \mid k = 1, 2, 3, 4 \right\},$$
along with all real numbers.