Timeline for Lower bound for the number of representations of integers as sum of squares
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| May 30, 2016 at 15:18 | comment | added | GH from MO | @Stabilo: For the Hardy-Littlewood circle method, the standard reference is Vaughan's book. You can find the asymptotic formula for $k\geq 5$ there (and also for the sum of $m$-th powers when $k\geq 2^m+1$). | |
| May 30, 2016 at 9:07 | vote | accept | Stabilo | ||
| May 30, 2016 at 9:06 | comment | added | Stabilo | Thanks a lot for your comments, I will check the Hardy-Littlewood circle method. @GregMartin, do you have some references for me? | |
| May 30, 2016 at 7:55 | comment | added | Greg Martin | @joro: for $k=4$ and $n$ an odd prime, the formula you linked to gives $r_k(n) = 8(n+1) \gg n^{1-\varepsilon}$. | |
| May 30, 2016 at 7:54 | comment | added | Greg Martin | For $k\ge5$, the circle method actually gives an asymptotic formula for $r_k(n)$ which is even stronger than the desired lower bound. | |
| May 29, 2016 at 19:58 | comment | added | George Shakan | Jacobis identity (and modular forms and dimension counting) is not the only way to approach this problem. See the Hardy Littlewood circle method, which can be much more flexible in certain situations. | |
| May 29, 2016 at 19:33 | history | edited | GH from MO |
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| May 29, 2016 at 19:32 | comment | added | GH from MO | @Lucia: Of course for $k=4$ there is a $2$-adic obstruction. The lower bound is fine if $n$ is not divisible by a fixed power of $2$. | |
| May 29, 2016 at 19:29 | answer | added | GH from MO | timeline score: 2 | |
| May 29, 2016 at 17:22 | answer | added | Douglas Zare | timeline score: 3 | |
| May 29, 2016 at 15:52 | comment | added | Lucia | For $k=4$ what you want follows from Jacobi; and then for all $k\ge 5$ follows easily by induction on $k$ (for example). For $k=3$ the number of representations of $n$ as a sum of three squares is a class number; and the inequality holds by Siegel's theorem. | |
| May 29, 2016 at 12:40 | comment | added | joro | This appears false for $k=4$ for primes. Check en.wikipedia.org/wiki/Jacobi%27s_four-square_theorem | |
| May 29, 2016 at 12:38 | comment | added | joro | For $k=2$, some positive integers are not the sum of two squares and some are in very few ways. | |
| May 29, 2016 at 11:55 | history | asked | Stabilo | CC BY-SA 3.0 |