Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1? - MathOverflow

Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?

2010-07-23T03:51:30Z

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 http://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 = $61^25^3$.)

Answer by Lavender Honey for Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?

2010-07-23T05:09:27Z

As once remarked by Mahler, $x^2 - 8 y^2 = 1$ has infinitely many solutions with $27 | x$.

Answer by Max Alekseyev for Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?

2010-07-23T17:20:50Z

See also http://oeis.org/classic/A076445 and this thread on the search for consecutive odd powerful numbers: http://www.mersenneforum.org/showthread.php?t=3474 Similar technique can be used for search for just consecutive powerful numbers (i.e., without the oddness restriction).

P.S. And of course, http://oeis.org/classic/A060355 is relevant.