Suppose that $K$ is an extension field of $\mathbb{Q}$ and that $\alpha \in K$ is a non-zero algebraic integer (i.e. it is integral over $\mathbb{Z}$). I am trying to show that there are only finitely many natural numbers $n$ such that $\alpha/n$ is integral over $\mathbb{Z}$. Here is my attempt so far:
Since $\alpha$ is integral over $\mathbb{Z}$, it follows that $\alpha$ is a complex number, so let $\alpha = a + bi$ for some $a, b \in \mathbb{R}$. Let $|\alpha| = \sqrt{a^2 + b^2}$ be the modulus of $\alpha$. I have the idea of showing that if $N$ is any natural number such that $N > |\alpha|$, then $\alpha/N$ is not integral over $\mathbb{Z}$, which would prove the desired claim. So suppose for reductio that $\alpha/N$ were integral over $\mathbb{Z}$, being a root of the monic polynomial $f \in \mathbb{Z}[x]$ given by $$ f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_{1}x + a_0. $$ Then we have that $f(\alpha/N) = 0$, from which it follows by some short calculations that
$$ \alpha^n + Na_{n-1}\alpha^{n-1} + N^{2}a_{n-2}\alpha^{n-2} + \ldots + N^{n+1}a_{1}\alpha + N^{n}a_0 = 0. $$ But from here I am not sure how to proceed (assuming that this idea will even work). Any suggestions would be appreciated!
Take the norm of all the algebraic integers in the ring. There is a minimal non-zero norm and the norms of the sequence you describe is monotonic decreasing and converging to zero. Thus all but finitely many of them would have norm less than the minimal norm and so can't be integral.