$a \in \mathbb{R}$ and $a^7, a^{12} \in \mathbb{Q}$. Prove that $a \in \mathbb{Q}$.

Question

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?

Answer 1

Hint: $\gcd(7, 12) = 1$

so done.

Answer 2

Your solution is fine. You could do it in one line ... \begin{eqnarray*} a=\frac{(a^{12})^3}{(a^7)^5}. \end{eqnarray*}

Answer 3

Well, we could the same things more directly: write $a$ as $\frac{(a^{12})^3}{(a^7)^5} = \frac{a^{36}}{a^{35}}$, or as $\frac{(a^7)^7}{(a^{12})^4} = \frac{a^{49}}{a^{48}}$, and conclude that it's in $\mathbb Q$. Your work probably simplifies to one of these after some substitutions.

Answer 4

Here is general way. You need to solve this linear equation

$$\dfrac {a^{12x}}{a^{7y}}=a \Longrightarrow 12x-7y=1$$ where $x,y\in \mathbb {Z^+}$

Now, $\gcd (12,7) =1$ , then we always have a solution. Thus, it is not even necessary to solve this equation with the Euclidean algorithm.

Answer 5

Using the Bezout relation $1=12\times3-7\times5$, we have $(a^{12})^3/(a^7)^5=a\in\mathbb Q$.