TITLE: Irreducible polynomials of real algebraic numbers
QUESTION [6 upvotes]: Suppose $\alpha$ is a real algebraic number with the property that its irreducible polynomial over $\mathbb{Q}$ is not a binomial, i.e., it is not of the form $x^n-q$ for some $n\geq 1$ and $q\in\mathbb{Q}$.
True or false: $\alpha^k\not\in\mathbb{Q}$ for all $k\geq 1$.
Example: $\sqrt{2}+\sqrt{3}$ has irreducible polynomial $x^4-10x^2+1$, and no integer power of $\sqrt{2}+\sqrt{3}$ is rational. 
Non-example when "real" assumption is dropped: $1+i$ has irreducible polynomial $x^2-2x+2$, but $(1+i)^4=-4\in\mathbb{Q}$.

REPLY [3 votes]: Let $\alpha$ be a real algebraic number. So that there exists some $k$ such that $\alpha^k \in \mathbb Q$. Let $t$ be the smallest positive number such that $\alpha^t \in \mathbb Q$. We claim that
$$p(x)=x^t-\alpha^t$$
is irreducible over $\mathbb Q$. Suppose not, then we know that 
$$p(x)=\prod_{i=1}^t (x-\alpha \zeta^i)$$
where $\zeta$ is a $t$-root of unity. Now if $p(x)$ is not irreducible we have that for some $a_1,\dots,a_n$ that
$$\prod_{i=1}^n \alpha \zeta^{a_i}=\alpha^n \prod_{i=1}^n \zeta^{a_i} \in \mathbb Q.$$
Since $\alpha$ is a real algebraic number this can only happen if
$$\beta=\prod_{i=1}^n \zeta^{a_i}$$
is also real. Since $\beta$ is a $t$-root of unity, $\beta$ is real if and only if $\beta= \pm 1$. Thereby $\alpha^n \in \mathbb Q$, but $t$ was the smallest such that this happened so $p(x)$ is irreducible. This proves the contrapositive of your statement.