Timeline for Are there consecutive integers of the form $a^2b^3$ where $a$, $b$ > 1?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2010 at 20:47 | comment | added | Ian Agol | ... where $x+\sqrt{8}y=(3+\sqrt{8})^{9+18k}$. | |
| Jul 23, 2010 at 13:54 | vote | accept | Ken Fan | ||
| Jul 23, 2010 at 13:47 | comment | added | shreevatsa | @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): $$\begin{align*} 5425069447 &= 7^3 \cdot 41^2 \cdot 97^2\\ 5425069448 &= 2^3 \cdot 26041^2\end{align*}$$ | |
| Jul 23, 2010 at 5:56 | comment | added | Ken Fan | 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? | |
| Jul 23, 2010 at 5:50 | comment | added | Mohammad Alaggan | That is 15061377048201 and 15061377048200. Nice job. | |
| Jul 23, 2010 at 5:44 | comment | added | shreevatsa | Nice. (3880899, 1372105) is such a solution, so $143737^29^3 − 1372105^22^3 = 1$. | |
| Jul 23, 2010 at 5:09 | history | answered | user631 | CC BY-SA 2.5 |