75th Romanian Mathematical Olympiad

Let $a+a^2+a^3=n$, with integer $n$. If $a$ is rational, then $a=\frac{p}{q}$, where $p,q$ are coprime integers. It follows from here that $\frac{p}{q}+\frac{p^2}{q^2}+\frac{p^3}{q^3}=n$, so $q^{2}p+q{p}^2+p^3=q^{3}n$, which implies $q\mid p^3$. This is possible only if $q=\pm 1$, which shows us that $a=\pm p$ is an integer number.

If $a^2$ is rational, then $a+a^3=n-a^2$, so $a^2+2a^4+a^6=n^2-2a^{2}n+a^4$, which leads to $a^2(1+2n)+a^4+a^6=n^2$. An argument analogous to the previous one, applied to the last relation, shows us that $a^2$ is an integer and therefore $a$ is an integer.

If $a^3=r$ is rational, then $a+a^2=n-a^3$ is also rational; we get from here that the number $a^2+2a^3+a^4$ is rational, so $a^2+ra$ is also a rational number. Hence $a^2+ra-(a^2+a)=(r-1)a$ is rational. If $r\ne 1$, this would imply that $a$ is rational, so, due to the first part, it will follow from here that $a$ is an integer number. If $r=1$, then $a^3=1$, which means $a=1\in \mathbb{Z}$.