Timeline for The "stubborn" solutions to sums of three cubes
Current License: CC BY-SA 4.0
19 events
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| Jan 4, 2022 at 13:56 | comment | added | Will Sawin | @GHfromMO Thanks! (And sorry for doing incomplete edits to fix it in a hurry.) | |
| Jan 4, 2022 at 8:20 | comment | added | GH from MO | I added the condition that $c>0$ is an integer, while I removed the condition that $n>0$ is sufficiently large. The point is that the fifth and sixth displays now hold exactly (not just asymptotically), while this was not true under the old conditions. (The old conditions were also fine, but one would need to tell the story slightly differently to take care of the $o(n)$ errors.) | |
| Jan 4, 2022 at 8:16 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Jan 4, 2022 at 8:10 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Jan 4, 2022 at 1:24 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Jan 4, 2022 at 0:13 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Jan 3, 2022 at 22:55 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Feb 3, 2021 at 14:52 | comment | added | Rosie F | @WillSawin Thanks for catching my mistake re number of solutions modulo 7, and for the Heath-Brown reference. | |
| Feb 3, 2021 at 14:50 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Feb 3, 2021 at 14:49 | comment | added | Will Sawin | @RosieF This is explained, and the solutions are counted, in the last two pages of Heath-Brown's paper The Density of Zeros of Forms for which Weak Approximation Fails. The specific densities for small $k$ were probably calculated along with the recent computer search efforts so you could look there. The "fake stubbon" examples you give are congruent to $8, 7,$ and $1$ mod $7$ - probably the fact that they are not congruent to $\pm 3$ mod $9$ is the source of their fallure to be stubborn. | |
| Feb 3, 2021 at 14:46 | comment | added | Will Sawin | @RosieF In fact $n=2 \mod 7$ and $n=3 \mod 7$ have the same number of solutions - there are $27$ $(x,y,z)$ with $x^3=y^3=z^3=1\mod 7$, and also $27$ where two are $1$ and the other is $0$ - the fewer number of choices that cube to $0$ exactly balances the choice of which one cubes to $0$. The correct thing is to multiply the the number of solutions mod $p$ divided by $p^2$ over all primes $p\neq 3$ times the number of solutions mod $9$ divided by $3^4$. | |
| Feb 3, 2021 at 14:44 | comment | added | Will Sawin | @DavidESpeyer Yes, good point, I forgot the $(p-1)$ factor in the Gauss sum terms. | |
| Feb 3, 2021 at 14:42 | comment | added | David E Speyer | It seems to me that there is a typo in your $p^3+6p^2$ formula? Shouldn't the answer be $1 \bmod p-1$, since this is a homogenous equation? I think the right answer is $p^3+6p^2-6p$. | |
| Feb 3, 2021 at 14:31 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| Feb 3, 2021 at 7:49 | comment | added | Rosie F | Why $n$ not a cube modulo many such primes $p$? I can see that $n=3\mod 7$ entails $x^3=y^3=z^3=1\mod 7$, but more possibilities exist for $(x, y, z)$ if $n=2\mod 7$; and I don't see how $n$ not being a cube mod $p$ suggests $n$ is stubborn if $p>7$. How are the primes weighted? I tried weighting $p$ by $1/p$; using the 3014 primes $p\le59971$ with $p=1\mod 6$, those $n<999$ seeming stubborn include 17, 205 and 982. Elsenhans and Jahnel's list of solutions suggest that they aren't all that stubborn. Perhaps a different weighting is better? | |
| Feb 3, 2021 at 5:30 | comment | added | Lucia | Excellent! My one nitpick is that "pigeon" doesn't have a d in it. | |
| Feb 2, 2021 at 22:02 | vote | accept | Alexander Kalmynin | ||
| Feb 2, 2021 at 14:15 | comment | added | Will Sawin | One can improve on this, maybe to $x^{1/3} e^{ \sqrt{ \log \log x}}$, by counting only points in a region like $|x|,|y|,|z|<m$, $|x^3+y^3+z^3|<n$. | |
| Feb 2, 2021 at 4:29 | history | answered | Will Sawin | CC BY-SA 4.0 |