Let $S$ = { $a^2b^3$ : $a, b \in \mathbb{Z}\_{>1}$ }.


Does there exist $n$ such that $n$, $n+1 \in S$?


Motivation: I was thinking about [Question on consecutive integers with similar prime factorizations](https://mathoverflow.net/questions/32412/question-on-consecutive-integers-with-similar-prime-factorizations), wondering whether any such pair had to have prime signatures with at least one 1. This would follow if the answer to the above question is negative. (This would also follow from weaker versions of the above question too, such as taking out perfect $n$th powers from $S$.)


Please note that $a$ and $b$ in the set definition are not allowed to be equal to 1. Otherwise, there'd be solutions like 8, 9 or 465124, 465125. (465124 = $(2\cdot 11 \cdot 31)^2$ and 465125 = $61^25^3$.)



Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?
As once remarked by Mahler, $x^2 - 8 y^2 = 1$ has infinitely many solutions with $27 | x$.


