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