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).
Best Answer
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 $x^x = n$ .