Let $a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.
my solution:
$a^7 \in \mathbb{Q} \Rightarrow a^{14} \in \mathbb{Q}$
$a^{14}=a^{12}\cdot a^{2}\in \mathbb{Q}$, but $a^{12} \in \mathbb{Q} \Rightarrow$ $a^2 \in \mathbb{Q}$.
Using the same rule: $a^{21} \in \mathbb{Q}$ and $a^{24}\in \mathbb{Q} \Rightarrow$ $a^{24}=a^{2} \cdot a \cdot a^{21} \Rightarrow a \in \mathbb{Q} $.
Is there any other solution?