Timeline for find the minimum difference between the factors of a number
Current License: CC BY-SA 3.0
2 events
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| Jun 2, 2012 at 15:51 | comment | added | Noam D. Elkies | And conversely, given a hard instance $S \subset {\bf R}^{> 0}$ of the partition problem, one can find $C>0$ and primes close enough to $\exp C\cdot S$ that their product is a hard instance of the nearest-factor problem at hand. | |
| Jun 2, 2012 at 15:10 | history | answered | Zander | CC BY-SA 3.0 |