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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 $(1+i)^4=-4\in\mathbb{Q}$.

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Maybe this works. Suppose $\alpha^k=a/b$ with $a,b$ integers, not both $d$th powers for some $d\gt1$ dividing $k$. Then $\alpha$ is a root of the polynomial $bx^k-a$. If you can prove this polynomial is irreducible, you're done. Now Eisenstein may not apply to this polynomial, since it could be that for every prime $p$ dividing $a$ you also have $p^2$ divides $a$. But it may be that if you try to mimic the steps in the proof of Eisenstein's Theorem you will find that you have enough extra information to make it work. –  Gerry Myerson Oct 25 '12 at 3:09
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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.

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