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

Question

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.

Answer 1

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.