Timeline for On the shortest open cubic equation
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30 at 14:54 | vote | accept | Bogdan Grechuk | ||
| Jan 27 at 10:03 | answer | added | Denis Shatrov | timeline score: 2 | |
| Dec 18, 2023 at 19:54 | comment | added | Wlod AA | @te4, "𝑥 is odd and therefore 𝑦 is even." -- but BOTH x and y must be odd! | |
| Dec 18, 2023 at 16:47 | history | edited | Bogdan Grechuk | CC BY-SA 4.0 |
deleted 35 characters in body
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| Dec 12, 2023 at 11:04 | comment | added | Bogdan Grechuk | te4: Note that $yt=-(x^4+1)<0$, hence $y$ and $t$ have opposite sign. If $y>0>t$ then $t$ being $3$ mod $4$ is not a contradiction. If $t>0>y$, then $|y|$ is $2$ modulo $16$, and $y$ is $14$ modulo $16$, in which case $x$ and $t$ are $1$ modulo $8$, again no contradiction. | |
| Dec 11, 2023 at 20:28 | comment | added | te4 | The solution of $2y^3+xy+x^4+1=0$. Note that $x$ is odd and therefore $y$ is even. Considering the equation modulo 4, one obtains $4\mid(xy+2)$, so $y\equiv_42$. It follows from $y\mid(x^4+1)$ that $y=2p_1\cdot\ldots\cdot p_k$, for some necessarily distinct primes$p_i\equiv_81$, and thus $y\equiv_{16}2$. Modulo 8, we get $8\mid(xy+2)$. So, $x\equiv_43$. But $t\mid(x^4+1)$, where $0\ne t=x+2y^2\equiv_43$, so contradiction. | |
| Dec 9, 2023 at 12:35 | comment | added | Satan's Minion | OK, that's a fair answer. I apologize for my grumpy question. I guess solving individual diophantine equations isn't much to my taste, but that's a personal judgement. :) | |
| Dec 9, 2023 at 11:56 | comment | added | Bogdan Grechuk | In fact, this project is a fascinating journey of continuous learning! I know some methods, use them for solving equations in order, and find the simplest-looking equations for which these methods do not work. I then post these equations, and, if they are solved, I usually learn new method, apply it to the next equations, and then the whole process is iterated. As a result, I now have a large collection of methods for solving equations, and, for each method, a set of simplest-looking equations for which this method works but easier ones not. And a set of simplest-looking open equations. | |
| Dec 8, 2023 at 20:42 | comment | added | Denis Shatrov | $x + Y^2 = (z^2 - 2x^2)(z^2 + 2x^2)$. -x is a quadratic residue modulo $z^2 - 2x^2$ and $z^2 + 2x^2$. There are four cases depending on whether $y^2 - x > 0$ and whether $z$ is odd or even. I’m not sure if it’s possible to get a contradiction in all cases, but some cases are excluded by calculating Jacobi symbols. | |
| Dec 8, 2023 at 13:35 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |