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Find an irrational $n$ such that $n^n$ is a rational number.

I have some tries to find this... I have tried so much numbers but no success. How can I find them.

Rodrigo de Azevedo
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Taha Akbari
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  • Pick a desired rational, say $2$. We can then solve $2^{1/x}=x$ to get $x= \frac {\ln(2)}{W(\ln 2)}$. Where $W$ denotes the Lambert function. – lulu Jul 12 '16 at 12:15

2 Answers2

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An implicit solution:

Let $n$ be such that $$n^n=2,$$ and let $n$ be the irreducible fraction $\dfrac rs$.

Then

$$\left(\frac rs\right)^{r/s}=2$$

or

$$r^r=2^ss^r.$$

So $r$ is even and $s$ odd, and calling $\rho$ the multiplicity of the factor $2$ in $r$,

$$r\rho=s$$ and $r=1$, a contradiction.

As a corollary, $\dfrac{\ln 2}{W(\ln 2)}$ is indeed irrational.

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By below $\,x^x = 2\,$ has no rational root, by $\, n^n\neq 2\,$ for $\,n\in\Bbb Z$.

Lemma $ $ If $\,0 <r = a/b\in\Bbb Q\,$ then $\,r^r = n\in\Bbb Z\,\Rightarrow\, r\in\Bbb Z$

Proof $\ r\:$ is a root $\,x^a = n^b\,$ so $\,r\in\Bbb Z\,$ by the Rational Root Test.

Bill Dubuque
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