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 $$\begin{gathered} \begin{array}{rl}a\log a& =b\log b\\ & =ax\log (ax)\\ \log a& =x\log (ax)\\ & =x\log (x)+x\log (a)\\ & =\frac{x\log x}{1-x}\\ a& =x^{\frac{x}{1-x}}\\ b& =x^{\frac{1}{1-x}}\end{array} \\ \text{since }x\ne 1\text{ and }a\ne 0 \end{gathered}$$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 $x={\displaystyle \frac{n+1}{n}}$ where $n$ is an integer.

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

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, $$(u+v{)}^q-u^q=q{u}^{q-1}v+\dots +v^q>q.$$Therefore, if $n$ is not an integer, $x^{\frac{1}{1-x}}$ is not rational, so the only rational solutions are given by $$a={\left({\displaystyle \frac{n}{n+1}}\right)}^{n+1}\text{ and }b={\left({\displaystyle \frac{n}{n+1}}\right)}^n\text{ with }n\in {\mathbb{Z}}^{+}$$The initial values of $a$ and $b$ are shown in Figure 1.

Figure 1: permalink