Timeline for On the equation $9x^3+y^3=z^2+3$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 20 at 12:21 | comment | added | YCor | (In my previous comment I'm referring to rational solutions of the OP's equation $9x^3+y^3=z^2+3$, not the follow-up equation proposed in the comments for whatever reason. Or equivalently integral solutions to the homogeneized equation $9x^3+y^3=z^2w+3w^3$, with $w\neq 0$. With denominator $9$ it thus refers to integral solutions of $9x^3+y^3=9z^2+27$.) | |
| Feb 19 at 13:03 | comment | added | YCor | Note that there are rational solutions, e.g. $(4,15,14)/9$, $(7,0,10)/9$, $(10,-9,26)/9$, $(17,-9,38)/18$, $(19,-15,79)/9$, $(-15,48,28)/27$, $(-13,39,35)/18$, $(-3,39,1)/27$, $(-2,27,44)/9$, $(-1,27,11)/18$, $(-53,129,299)/9$, etc. (Just with denominator $9$ one might wonder whether there are infinitely many solutions.) | |
| Feb 19 at 12:43 | vote | accept | Bogdan Grechuk | ||
| Feb 19 at 10:30 | answer | added | Denis Shatrov | timeline score: 17 | |
| Feb 18 at 20:51 | comment | added | Will Jagy | ran a little program, lots of those... | |
| Feb 18 at 20:45 | comment | added | Bogdan Grechuk | The simplest solution to $9x^3+y^3=z^2+3w^2$ with last 2 variables coprime is (3,1,1,9). Or, if you like all variables pairwise coprime, (3,1,13,5). | |
| Feb 18 at 20:36 | comment | added | Will Jagy | Good. Next restriction, with $\gcd(z,w) = 1.$ If this is difficult or impossible it may explain some things | |
| Feb 18 at 20:34 | comment | added | Bogdan Grechuk | Of course $9x^3+y^3=z^2+3w^2$ is solvable. The simplest solution is (0,0,0,0). Another example is (1,3,3,3). | |
| Feb 18 at 20:29 | comment | added | Will Jagy | well, what about $9 x^3 + y^3 = z^2 + 3 w^2 $ | |
| S Feb 18 at 20:26 | history | suggested | mathworker21 | CC BY-SA 4.0 |
fixed typo
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| Feb 18 at 20:23 | review | Suggested edits | |||
| S Feb 18 at 20:26 | |||||
| Feb 18 at 20:07 | comment | added | JoshuaZ | Possibly useful note for extending the search range: Mod 9 shows that $z \not \equiv 0\pmod{3}$, and one can use that to then show that $y \equiv 1 \pmod{3}$. | |
| Feb 18 at 18:28 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |