Timeline for A heuristic for the density of solutions to Diophantine equations
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 13, 2013 at 2:23 | history | edited | George Lowther | CC BY-SA 3.0 |
fix latex
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| Jun 21, 2011 at 22:28 | answer | added | George Lowther | timeline score: 4 | |
| Jun 21, 2011 at 20:52 | comment | added | George Lowther | @Dan: Yes, I was careful to keep track of the various factors of 2, and the answer agrees with the formula given by Lehmer in 1900 for the number of triples with hypotenuse less than N (mathworld.wolfram.com/PythagoreanTriple.html). Also, after a bit of a search using the keywords mentioned by JSE, I now see that it is correct that I was out by a factor of 2. | |
| Jun 21, 2011 at 20:48 | history | edited | George Lowther | CC BY-SA 3.0 |
typo
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| Jun 21, 2011 at 20:18 | vote | accept | George Lowther | ||
| Jun 21, 2011 at 20:17 | comment | added | George Lowther | @Qiaochu: It does seem that it is more likely to work when the number of variables is large relative to the degree. I wasn't sure how large but, for quadratic forms, it seems that you need at least 5 variables. I'm still not sure why it fails for fewer variables. Maybe because, as you suggest, the number of integer points in $\{x\colon\vert f(x)\vert < 2a\}$ differs from the estimate used in the heuristic, whether it is because they are not evenly distributed mod N, or whether conditional on $\vert f(x)\vert < 2a$ and $f(x)=0$ (mod N), the probability that $f(x)=0$ is different from $2a/N$. | |
| Jun 21, 2011 at 10:53 | comment | added | Dan Petersen | Sorry for this very naive comment, but did you take into account that Euclid's parametrization only gives you half of all primitive Pythagorean triples (those with $x$ odd and $y$ even)? | |
| Jun 21, 2011 at 10:32 | answer | added | Daniel Loughran | timeline score: 12 | |
| Jun 21, 2011 at 2:39 | comment | added | Qiaochu Yuan | Here's a possible failure point of the heuristic: if the sets $\{ x : |f(x)| < 2a \}$ are sufficiently irregular there's no reason to expect their volume to be a good estimate for the number of integer points they contain. As JSE says it seems like this might be less of a problem if the number of variables is large relative to the degree. | |
| Jun 21, 2011 at 2:09 | answer | added | JSE | timeline score: 15 | |
| Jun 20, 2011 at 22:13 | history | edited | George Lowther | CC BY-SA 3.0 |
added 5 characters in body
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| Jun 20, 2011 at 22:07 | history | asked | George Lowther | CC BY-SA 3.0 |