Rational Solutions to a^a = b^b

Here's a problem that I encountered on this site:

Find all pairs of rational numbers $(a,b)$ such that $a^a=b^b$ for $0 < a < b$.

One solution that was proposed runs as follows:

Write $b=ax$ for some $x>1$. Then $x=b/a$ so $x$ is rational.

Then taking logs, we get $$\begin{align} a\log a&=b\log b\\&=ax\log(ax) \\ \log a&= x \log(ax) \\&=x \log(x)+x \log(a)      \\&=\frac{x\log x}{1-x} \\  a&=x^{\frac{x}{1-x}}\\b&=x^{\frac{1}{1-x}} \end{align} \\ \text{since } x\neq1 \text{ and } a\neq0$$ Since $x$ is rational, it suffices to find the values of $x>1$ for which $x^{\frac{1}{1-x}}$ is rational.

We claim that the only such values of $x$ are $x=\dfrac{n+1}{n}$ where $n$ is an integer.

We may write $x=1+\dfrac1n$ where $n\in\mathbb{Q}$, $n>0$.

It is easy to see if $n$ is a positive integer that $x^{\frac{1}{1-x}}$ is rational.

If $n$ is not an integer, write $n=p/q$ where $p,q\in\mathbb{Z}^+$ are coprime with $q>1$.

We have $x^{\frac{1}{1-x}}=\left(\dfrac{n}{n+1}\right)^n=\left(\dfrac{p}{p+q}\right)^{p/q}$.

Hence $p$ and $p+q$ must be $q \,$th powers.

No two $q \,$th powers can differ by $q$ since for positive integers $u,v,q$, we have by the Binomial Theorem, $$(u+v)^q-u^q=qu^{q-1}v+\ldots+v^q>q\,.$$ Therefore, if $n$ is not an integer, $x^{\frac{1}{1-x}}$ is not rational, so the only rational solutions are given by $$a=\left(\dfrac{n}{n+1}\right)^{n+1} \text{ and } \, b=\left(\dfrac{n}{n+1}\right)^n \text{ with } n\in\mathbb{Z}^+$$ The initial values of $a$ and $b$ are shown in Figure 1.

Figure 1: permalink