Timeline for On the equation $9x^3+y^3=z^2+3$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22 at 21:27 | comment | added | Denis Shatrov | A typo in second sentence: If $p \mid y$ then $p \mid x$. | |
| Feb 22 at 21:15 | comment | added | Denis Shatrov | @TimothyChow If $y$ is not divisible by $p$, then $3 \equiv (\frac{-3x}{y})^3 \pmod{p}$ which is impossible. If $y \mid p$, then $x \mid p$. $9(x/p)^3 + (y/p)^3 = (z^2 + 3)/p^3$. We can continue this process until we get $9(x/p^k)^3 + (y/p^k)^3 = (z^2 + 3)/(p^{3k}) $ where $(z^2 + 3)/(p^{3k})$ is not divisible by $p$. Hence, $\nu_p(z^2 + 3) = 3k$. | |
| Feb 22 at 19:31 | comment | added | Timothy Chow | Can you say more about why $3$ being a cubic nonresidue mod $p$ implies that $3\mid \nu_p(z^2+3)$? I don't follow that step. | |
| Feb 20 at 10:40 | history | edited | Denis Shatrov | CC BY-SA 4.0 |
fixed typo
|
| Feb 19 at 12:43 | vote | accept | Bogdan Grechuk | ||
| Feb 19 at 10:30 | history | answered | Denis Shatrov | CC BY-SA 4.0 |