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Martin Brandenburg
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Douglas Zare
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$Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$

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Denis Serre
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AssumeObserve that we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$.

More generally, assume that $K$ is a finite extension of Q. Is there any $\alpha \in K$ such that $K=Q(\alpha^n)$ for every $n \in N$. For example we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$.?

Assume that $K$ is a finite extension of Q. Is there any $\alpha \in K$ such that $K=Q(\alpha^n)$ for every $n \in N$. For example we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$.

Observe that we have $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$.

More generally, assume that $K$ is a finite extension of Q. Is there any $\alpha \in K$ such that $K=Q(\alpha^n)$ for every $n \in N$?

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Sina
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