Timeline for Which quaternary quadratic form represents $n$ the greatest number of times?
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| Oct 23, 2023 at 1:40 | comment | added | GH from MO | @MathqA Let me answer your second question. The last term does not depend on $Q$, because $\det(Q)\geq 1$. That is, $r_Q(n) \ll_\epsilon n^{k / 2-3 / 2+\epsilon}+n^{(k-2) / 2+\epsilon} \ll n^{k / 2-1+\epsilon}$. | |
| Oct 23, 2023 at 1:36 | comment | added | GH from MO | @MathqA Let me answer your first question. From $Q(x_1,\dots,x_k)=n$ we infer that $\sum_{i\leq j\leq k}c_{ij}x_j\ll\sqrt{n/h_i}$ for all $i$. For $i=k$ this means that $x_k\ll\sqrt{n/h_k}$. As $h_k\gg(\det Q)^{1/k}$, we conclude that $x_k\ll\sqrt{n}\det(Q)^{-1/(2k)}$. For $i=k-1$ the bound above says that $x_{k-1}+c_{k-1,k}x_k\ll\sqrt{n/h_{k-1}}$. Here $h_{k-1}$ might be as small as $1$, hence we can only conclude that $x_{k-1}+c_{k-1,k}x_k\ll\sqrt{n}$. However, at this point we already know that $c_{k-1,k}x_k\ll \sqrt{n}$, hence $x_{k-1}\ll\sqrt{n}$ also follows. And so on. | |
| Oct 22, 2023 at 18:42 | comment | added | MathqA | When you say "From the above, we see immediately that $x_k\ll\sqrt{n}\det(Q)^{-1/(2k)}$ and then also that $x_{k-1}\ll\sqrt{n}$...", could you please explain why for all the other $x_j$ there is no a factor $\det(Q)^{-1/(2k)}$ ? I also have a question in your formula $r_Q(n) \ll_\epsilon n^{k / 2-3 / 2+\epsilon}+n^{(k-2) / 2+\epsilon} \operatorname{det}(Q)^{-1 /(2 k)+\epsilon} \ll n^{k / 2-1+\epsilon}$. How is it that the last term does not depend on $Q$? | |
| Nov 6, 2019 at 0:18 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 6, 2019 at 0:00 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Dec 12, 2016 at 12:26 | comment | added | Vladimir Dotsenko | Wow. This is really cool. | |
| Dec 8, 2016 at 21:50 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 8, 2016 at 14:36 | vote | accept | Jeremy Rouse | ||
| Dec 8, 2016 at 14:36 | comment | added | Jeremy Rouse | This is fabulous! | |
| Dec 8, 2016 at 13:44 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 8, 2016 at 13:14 | comment | added | GH from MO | @JeremyRouse: I revised my response substantially, the bound is now independent of $Q$. | |
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| Dec 7, 2016 at 22:48 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 7, 2016 at 22:40 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 7, 2016 at 22:33 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Dec 7, 2016 at 22:26 | history | answered | GH from MO | CC BY-SA 3.0 |