Return to Revisions

There is an infinite number of rational solutions $$a=\left(\frac{n+1}{n}\right)^n,\;\;b=\left(\frac{n+1}{n}\right)^{n+1},\;\;n\in\mathbb{Z},\;\;0\neq n\neq -1.$$ For a proof that these are all the rational solutions of $a^b=b^a$ with $a\neq b$, see this posting.