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I was wondering if there exists a pair of natural numbers $n, m \in \mathbb{N}$, such that $e^n = m$? Of course excluding the simple solution (0,1).

Intuitively, I would say no. However, I am far from finding an argument, nor knowing which theory even treats these types of questions.

Cheers

NicFit_88
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1 Answers1

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There does not. Remember $e$ is a transcendental number which means it is not the root of any polynomial. In particular if $e^n=m$ was an integer then $e$ would be a root of the polynomial $x^n-m=0$.

andrew
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