# Rational Solutions to $x^x = n$ where $n$ is a positive integer

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).

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 $$x^x = n$$ .
• I've added more deiail to the proof. Tough in general it can be solved by knowing that $gcd(x^a,y^a) = gcd(x,y)^a$ . Commented Aug 18, 2021 at 16:38