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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?

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  • $\begingroup$ Most likely, there will be ways to write this same proof down differently and more concisely, as has already been done by Misha, but it would seem to me that there is no way to circumvent the general technique that you are using, that $\gcd(7, 12) = 1$. Once you found one of these proofs, all others will probably be trivial, note, that I cannot prove that this is so, there might be some complicated solution that does not use $\gcd(7, 12) = 1$ under the hood but that seems unlikely to me. $\endgroup$ – Martin van IJcken Nov 17 '20 at 15:17
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Hint: $\gcd(7, 12) = 1$

so done.

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

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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.

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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.

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

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