If $a \in \mathbb{C}$ and $\exists n \in \mathbb{N}$ s.t. $\{ a^n, a^{n+1} \} \in \mathbb{N}$, prove $a \in \mathbb{N}$ [closed]

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – darij grinberg, John Omielan, blub, Shailesh, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.

• What have you tried? Is there some way to make $a$ out of $a^n$ and $a^{n+1}$? Does it help? – darij grinberg Apr 21 at 6:11
• 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? – Lexi S. Apr 21 at 6:18

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.
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.