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Trivially we know $2^2 = 4$, $3^3 = 27$ are integral solutions to the equation $x^x = n$ where $n$ is a positive integer. Most other solutions are likely to be irrational (eg $x^x = 5$).

But does there exist any positive integer $n$ that can be expressed as $x^x$ where $x$ is a rational number that is not an integer? Is it possible to prove that there can be no rational solutions?

Proof by reference to rational root theorem:
Let $x = p/q$ (with p and q coprime). Raise both sides of the equation to the power of $q/p$.
Then x is a solution to the equation $x = n$ ^ $(q/p)$ But the rational root theorem states that the only solutions to this equation are integers or irrational. This implies x is both rational (by definition) and not rational (by the theorem). Hence there can be no rational solution (proof by contradiction).

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Let $x = \frac{a}{b}$ where $a,b$ are coprime. Then we have $(\frac{a}{b})^{\frac{a}{b}} = n$. By lifting both sides to the power of $b$ we have: $$\left(\frac{a}{b}\right)^a = n^b \\ \implies a^a = n^b \cdot b^a $$ Assume $b\neq 1$. Then there exists a prime $p$ such that $p$ divides $b$. Therefore $p^a$ divides $b^a$, thus $p$ divides $a^a$. But if $a$ is not divisible by $p$, then neither is $a^a$. Therefore $a$ is divisible by $p$. As both $a$ and $b$ are divisible by $p$, they cannot be coprime. Contradiction.

Therefore $b = 1$. Thus integral solutions are the only rational solutions to $x^x = n$ .

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  • $\begingroup$ This seems only to go part way to a complete proof. To complete the proof it is necessary (and sufficient) to prove that if a and b are coprime then b^a does not divide a^a for any values of a and b other than b=1. It looks obvious, but proof seems tricky. $\endgroup$ Commented Aug 18, 2021 at 11:17
  • $\begingroup$ I've added more deiail to the proof. Tough in general it can be solved by knowing that $gcd(x^a,y^a) = gcd(x,y)^a$ . $\endgroup$ Commented Aug 18, 2021 at 16:38
  • $\begingroup$ Thats great. Thanks for your help.. $\endgroup$ Commented Aug 18, 2021 at 18:09

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