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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.

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  • $\begingroup$ 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. $\endgroup$ – lulu Jul 12 '16 at 12:15
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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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  • $\begingroup$ Oh, thanks (+1). I was greatly over-thinking the point. $\endgroup$ – lulu Jul 12 '16 at 13:36
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By the Lemma below, $\,x\,$ is not rational if $\,x^x$ is an integer not of the form $\,n^n.\,$ Therefore, in particular, $\,x^x - 2\,$ has no rational roots, since $\,2\neq n^n\,$ for some integer $\,n.$

Lemma $ $ If $\:r = a/b\:$ is rational then $\,r^r\,$ an integer $\,\Rightarrow\, r\,$ an integer.

Proof $\ $ If $\ r^r = n\,$ is an integer then $\:r\:$ is a root of $\,x^x = x^{a/b} = n,\,$ so $\:r\:$ is a root $\,x^a = n^b.\,$ Thus $\,r\,$ is an integer by the Rational Root Test.

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