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

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, 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 = $61^25^3$.)

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Have you tried a computer search? – Qiaochu Yuan Jul 23 '10 at 3:57
A computer search up to 100 for a and b shows a minimum difference of 4 amongst elements. – Steve Huntsman Jul 23 '10 at 4:04
The equality $a^2b^3-c^2d^3=1$ implies that the quadratic irrationality $(d/b)^{3/2}$ is "too well approximated" by the rational $a/c$. More precisely, $|(d/b)^{3/2}-a/c|<\operatorname{const}/c^4$. – Wadim Zudilin Jul 23 '10 at 4:37
I am too optimistic, the magnitude is of size $\operatorname{const}/c^2$ (and a similar one for approximating $(d/b)^{1/2}$ by $ab/cd$) but at least it's clear that the examples could only come from continued fractions of quadratic irrationalities. – Wadim Zudilin Jul 23 '10 at 4:52
Numbers of the form $a^2b^3$ (without the restriction $a>1$, $b>1$) are known as powerful numbers, or squarefull numbers. Knowing this vocabulary may help you find something in the literature. It's known that every integer can be expressed as a difference of powerful numbers in infinitely many ways. It's unknown whether there are three consecutive powerful numbers. (The two preceding sentences are not directly related to your question, I just thought I'd throw them in.) – Gerry Myerson Jul 23 '10 at 6:37

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

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Nice. (3880899, 1372105) is such a solution, so $143737^29^3 − 1372105^22^3 = 1$. – shreevatsa Jul 23 '10 at 5:44
That is 15061377048201 and 15061377048200. Nice job. – M. Alaggan Jul 23 '10 at 5:50
Wow...neat...thanks! I'm still interested in the same question with perfect $n$th powers removed though because that still would imply that consecutive numbers with the same prime signature have to have a 1 in the signature. Should the question about prime signatures be asked separately? – Ken Fan Jul 23 '10 at 5:56
@Ken: Following up a comment in the Wikipedia article, here's a solution by Jaroslaw Wroblewski from primepuzzles.net/problems/prob_053.htm (without perfect powers, but not with the same prime signature though): \begin{align*} 5425069447 &= 7^3 \cdot 41^2 \cdot 97^2\\ 5425069448 &= 2^3 \cdot 26041^2\end{align*} – shreevatsa Jul 23 '10 at 13:47
... where $x+\sqrt{8}y=(3+\sqrt{8})^{9+18k}$. – Ian Agol Jul 31 '10 at 20:47

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.

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