Does $x^{-x}$ have rational solutions when $x$ is a non-integer rational? I want to know, given a rational number $x=\frac ab$ where $a$ and $b$ are distinct positive integers with $b\ne1$ and $\gcd(a,b)=1$, whether $x^{-x}$ can be rational as well.
What I achieved so far is as follows.
$$
\begin{align*}
x^{-x} &= \bigg(\frac{a}{b}\bigg)^{\big(-\frac{a}{b}\big)} \\
       &= \bigg(\frac{b}{a}\bigg)^{\big(\frac{a}{b}\big)} \\
       &= \frac{\sqrt[b]{b^a}}{\sqrt[b]{a^a}} \\
\end{align*}
$$
For this to be rational, either (i) both the numerator and the denominator must be rationals or (ii) they must share an irrational factor $r$ which will be canceled out by the division.
I can see that the first condition is unsatisfiable: As $a$ and $b$ have no common prime factors, both $\sqrt[b]{a}$ and $\sqrt[b]{b}$ must be rationals but $\sqrt[b]{b}$ cannot be a rational.
I cannot, however, prove or disprove whether the second condition is also unsatisfiable. If you can provide me a simple proof or point me to a complicated one, I would be grateful.
 A: Assuming the contrary, let $x^{-x}$ be rational. Hence, it can be expressed as $\dfrac m n$, with $m$ being an integer and $n$ being a positive integer. Moreover, we can assume that $m$ and $n$ are co-prime.
$$\begin{array}{rcl}
x^{-x} &=& \dfrac m n \\
\left(\dfrac b a\right)^{a/b} &=& \dfrac m n \\
\dfrac{b^a}{a^a} &=& \dfrac{m^b}{n^b} \\
b^a n^b &=& a^a m^b \\
\end{array}$$
Let $p^k$ be a prime power factor of $b$ with maximum $k$. Then, $p^{ka}$ is a prime power factor of $b^a$ with maximum power. Since $b$ and $a$ are co-prime, the factorization of $a^a$ cannot contain $p$ at any power. Therefore, $p^{ka}$ must be a prime power factor of $m^b$.
Let $p^h$ be a prime power factor of $m$ with maximum $h$. Then, $p^{hb}$ is a prime power factor of $m^b$ with maximum power. Since $m$ and $n$ are co-prime, by the same argument, $p^{hb}$ must be a prime power factor of $b^a$.
Hence, $ka \le hb$ and $hb \le ka$, so $ka=hb$. Since $a$ and $b$ are co-prime, $k$ is divisible by $b$.
Since $p$ is arbitrary, this holds for any prime factor of $b$. In other words, if for any prime $p$, $p^k$ is a maximum prime power factor of $b$, then $k$ is divisible by $b$. This means that $b$ can be expressed as $c^b$ where $c$ is a positive integer.
We know that $c$ cannot be $1$, or else $b$ would be $1$. If $c$ is any larger, then we must have $c^b > b$. Therefore, it is impossible.
A: Suppose $b>1$ and that $x^{-x}=\frac{\sqrt[b]{b^a}}{\sqrt[b]{a^a}}=\frac pq$ for some $p,q\in\Bbb Z$. Cross-multiplying and raising to the power of $b$ we get
$$a^ap^b=q^bb^a$$
Now consider factors of $a$ and $b$ in $q^b$ and $p^b$:


*

*Because $\gcd(a,b)=1$, $a$ and $b$ occur at least $a$ times in $q^b$ and $p^b$ respectively.

*To keep $p^b$ and $q^b$ perfect powers we need at least $-a\bmod b$ more factors of $a$ or $b$ on both sides.

*To preserve the equality we need $-a\bmod b$ more factors each of $b$ in $q^b$ and $a$ in $p^b$ respectively.

*To keep $p^b$ and $q^b$ perfect powers we need at least $+a\bmod b$ more factors of $a$ or $b$ on both sides.

*To preserve the equality we need $+a\bmod b$ more factors each of $b$ in $p^b$ and $a$ in $q^b$ respectively.


Because $\pm a\not\equiv0\bmod b$, steps 2-5 are repeated infinitely, so $a^ap^b=q^bb^a$ is never satisfied if $b>1$. Since $b=1$ trivially yields a rational number, $x^{-x}$ is rational iff $x$ is an integer.
A: Hint: if $\,x\!=\!\frac{a}b,\ (a,b)\!=\!1\,$ & $\,x^{x}\! = \frac{m}n,\ (m,n)\!=\!1\,$
then $\,\large \frac{a^a}{b^a} \!=\! \frac{m^b}{n^b}\,$ so $\,\color{#0a0}b^a \!= n^{\color{#c00}b}$ (or $\,m^\color{#c00}{b}\,$ if $\,a<0),\,$ thus the coprime expts $\,(a,\color{#c00}b)\!=\!1\Rightarrow \color{#0a0}b\,$ too is a $\,\color{#c00}b$'th power, i.e. $\exists\:\! n\!\in\!\Bbb Z\!:\ b = n^b\, [\:\!\ge 2^b > b\,\Rightarrow\!\Leftarrow]$
