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Let $a$ be a complex number. If it exists a natural number $n$ (different of $0$), such that $a^n$ and $a^{n+1}$ are integers, prove that $a$ is an integer.

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closed as off-topic by darij grinberg, John Omielan, blub, Shailesh, Cesareo Apr 21 at 8:27

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    $\begingroup$ What have you tried? Is there some way to make $a$ out of $a^n$ and $a^{n+1}$? Does it help? $\endgroup$ – darij grinberg Apr 21 at 6:11
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    $\begingroup$ Thank you, so $a= \frac{a^{n+1}}{a^{n}}$, which is rational, because $a^{n+1}$ and $a^n$ are integers. Furthermore, in order for $a$ to be not only rational, but integer even, $a^{n}$ must divide $a^{n+1}$, which is by all means true. Is this correct? $\endgroup$ – Lexi S. Apr 21 at 6:18
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We have that $a^{n+1}$ and $a^n$ are integers. Suppose $a^n = 0$. In this case, since $a^{n} = 0 \implies a = 0$, hence $a$ is an integer.

Suppose $a^n \neq 0$: hence, $a = \frac{a^{n+1}}{a^n}$ is a rational number (because it is of the form $\frac{p}{q}$ where $p,q \in \mathbb{Z}, q \neq 0$).

Therefore, let $a = \frac{p}{q}$ where $p,q$ are co-prime. Since $a^{n} \in \mathbb{Z}$, therefore $\frac{p^n}{q^n} \in \mathbb{Z}$. Since $p,q$ are co-prime, hence $p^n$ and $q^n$ are also co-prime. Since $\frac{p^n}{q^n} \in \mathbb{Z} \implies q = 1$. Hence, $a = p/1 = p \in \mathbb{Z}$. Therefore, $a$ is an integer, proved.

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  • $\begingroup$ Why do we need to consider that? We can get the desired result without any such considerations. $\endgroup$ – Ekanshdeep Gupta Apr 21 at 7:32
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If both $a^n$ and $a^{n+1}$ are integer, then $\frac{a^{n+1}}{a^n}=a$ is rational. Also because it's power ($a^n$) is an integer, then $a$ itself is an integer.

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