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 $a\ne 0$). 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 $$a=\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi }{5},k=1,2,3,4.$$ 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.