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Let $\alpha $ be a real number such that for all prime numbers $p$, the number $p^ \alpha$ is an integer. Is it necessarily true that $\alpha$ is an integer? If $\alpha$ is a rational, say $m/n$, one gets that $p^m$ equals an integer $k$ raised to the power of $n$. It is infered immediately that $k$ is itself a power of $p$, leading to the conclusion that $m/n$ should be an integer. Also, the hypothesis of the problem guarantees that every integer $i$ raised to the power of $\alpha$ is an integer.....

karparvar
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    https://math.stackexchange.com/a/573337/78967 – mathlove Nov 24 '19 at 12:06
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    By the Gelfond-Schneider theorem, if $p\in\mathbb{P}\subset\mathbb{A}$, $p\neq 0,1$ and $\alpha\in\mathbb{A}/\mathbb{Q}$ then $p^\alpha$ is transcendental so $\alpha$ can't be algebraic irrational. This still leaves the case where $\alpha$ is transcendental though. – Jam Nov 24 '19 at 12:30
  • If $p$ is prime, and $\alpha=m/n$ is in lowest terms then because $p$ is prime it's $n$th root is not an integer. This means that (per Wikipedia), it's irrational, if $m$ and $n$ are coprime raising to the $m$th power does not bring the expression back into the rationals, therefore only if $n=1$ is the expression a rational and therefore can be a real – user1543042 Nov 24 '19 at 13:40
  • @user1543042 This is effectively a rephrasing of Karparvar's proof. The case for ruling out numbers such as $2^{\ln3/\ln2}=3$ is surprisingly tricky and is covered here. – Jam Nov 24 '19 at 13:45

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