Timeline for Is the following consequence of the Lang conjecture known?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2015 at 18:09 | vote | accept | R.P. | ||
| Mar 13, 2015 at 22:41 | |||||
| Jan 27, 2014 at 19:02 | history | edited | ACL | CC BY-SA 3.0 |
added 1042 characters in body - corrections explaining why this is a non-answer, adding a second reference
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| Jan 27, 2014 at 19:01 | comment | added | GH from MO | According to Hardy-Wright: An introduction to the theory of numbers, Chapter XXI, we should have $f(n)\leq 16$, but this is not known. | |
| Jan 27, 2014 at 18:58 | comment | added | GH from MO | An easy upper bound is $c_\epsilon n^\epsilon$ for any $\epsilon>0$, where $c_\epsilon$ is a suitable constant. In fact even the number of solutions of $x^2+y^2=n$ satisfies this bound. | |
| Jan 27, 2014 at 18:54 | comment | added | R.P. | @ACL: no problem. I missed my own point as well when I initially accepted your answer (having failed to notice the $n$ at the base of the tree of iterated exponents). :) | |
| Jan 27, 2014 at 18:52 | comment | added | ACL | @René and Joël: Sorry, you're right. I missed your point. I give some hints in the comments to your question. | |
| Jan 27, 2014 at 18:51 | comment | added | Joël | ACL, I am not following you. When you say "the bound given by pgadey is enough", enough for what ? The question asked for a bound independent of $n$. | |
| Jan 27, 2014 at 18:48 | comment | added | ACL | @MattF.: Oops, of course this much smaller bound given by pgadey is enough. The point of using Rémond's theorem is that it does not rely on the specific form of the equation, only on its degree and on the size of its coefficients. | |
| Jan 27, 2014 at 18:46 | comment | added | ACL | @pgadey: rather $2(2\sqrt[4]n+1)$ ? | |
| Jan 27, 2014 at 17:51 | comment | added | pgadey | Sorry -- I dropped some constants, we get that $f(n) \leq 4\sqrt{n}$. | |
| Jan 27, 2014 at 17:41 | comment | added | pgadey | Can't we do much better than this? If we're only looking at integral solution, then certainly $$ |\{ (x,y) \in \mathbb{Z}^2 : x^4 + y^4 = n \}| \leq |\{ (x,y) \in \mathbb{Z}^2 : |x|, |y| \leq \sqrt[4]{n}\}| \leq n^{2/4} $$ | |
| Jan 27, 2014 at 17:30 | comment | added | user44143 | @ACL I see what you mean -- but in that case does Remond's theorem help at all here? There are obviously at most $n^{1/4}$ positive solutions. | |
| Jan 27, 2014 at 17:16 | comment | added | ACL | @MattF. No, his $M$ has to be an upper bound for the coefficients. By the way, in a previous work, he had obtained the bound $\exp(5^{4^4}(\log (n))(\log\log(n)))$ which is better. (The first bound I copied is better in $n$, but worse in the degree.) | |
| Jan 27, 2014 at 16:55 | comment | added | user44143 | Looking at the abstract, it seems that $3^{2^{3^{16}}}$ will do. | |
| Jan 27, 2014 at 16:13 | comment | added | R.P. | Wait, this bound depends on $n$, right? I am asking for a bound that is independent of $n$. | |
| Jan 27, 2014 at 15:50 | vote | accept | R.P. | ||
| Jan 27, 2014 at 16:12 | |||||
| Jan 27, 2014 at 15:46 | history | answered | ACL | CC BY-SA 3.0 |