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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.2012-10-25

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