Timeline for $x^2+7y^2=2^n$ and sums of four squares
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 16, 2020 at 0:43 | comment | added | Jack L. | Okay, missed that. Thanks | |
| Oct 16, 2020 at 0:33 | comment | added | Zhi-Wei Sun | Note that I define $\mathbb N$ in the posting as the set of nonnegative integers. | |
| Oct 15, 2020 at 23:22 | comment | added | Jack L. | @Zhi-Wei Sun: Why do you claim that $4\times 3+1 =2^0(2\times 6+1)$ —that is, $a=0$ and $b=6$—when your conjecture explicitly requires that $a,b\in\mathbb{N}$? (Or did you rather mean $\mathbb{Z}$—or $\mathbb{N}_0$—instead of $\mathbb{N}$? | |
| Oct 15, 2020 at 18:18 | comment | added | Zhi-Wei Sun | In my 2017 JNT paper I proved that any positive integer can be written as $x^2+(2xy)^2 +(xz)^2+w^2$ with $y,z,w\in\mathbb N$ and $x$ a power of two. | |
| Oct 15, 2020 at 16:10 | comment | added | Lucia | This is a typical conjecture of OP's: one that has an obvious probabilistic justification, but with little hope of saying a second sentence beyond that. The values allowed for $x^2+y^2$ form a set with about $(\log n)^2$ elements (note that $x$ and $y$ need not be coprime), and one wants $n-x^2 -y^2$ to be a sum of two squares, which happens with probability about $1/\sqrt{\log n}$. So one fails with probability $(1-c/\sqrt{\log n})^{d (\log n)^2} = \exp(-C (\log n)^{\frac 32})$. Since this is very small, "Borel--Cantelli" would indicate that there are at most finitely many exceptions. | |
| Oct 15, 2020 at 9:34 | comment | added | Zhi-Wei Sun | See also oeis.org/A338162 for related data. | |
| Oct 15, 2020 at 9:31 | comment | added | Zhi-Wei Sun | @joro I mean that even if the solution structure for $x^2+7y^2=2^n$ is as clear as $x+24y=z^2$, the related conjecture on sums of four squares might still be quite challenging. | |
| Oct 15, 2020 at 9:27 | comment | added | Zhi-Wei Sun | @Toni $4\times3+1=2^0(2b+1)$ with $b=6$ even. | |
| Oct 15, 2020 at 7:55 | comment | added | joro | I don't understand the relation with sum of four squares, and your comments in the answer are confusing. If you can deterministically express $n$ as sum of 4 squares, this will be new result (probabilistic solution is possible). | |
| Oct 15, 2020 at 5:48 | comment | added | Toni Mhax | $4\times 3+1=2^a(2b+1)?$ | |
| Oct 15, 2020 at 4:06 | answer | added | KConrad | timeline score: 16 | |
| Oct 15, 2020 at 0:17 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |