Timeline for Analysis of a quadratic diophantine equation
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2010 at 22:56 | comment | added | apples | Thanks, this has helped provide insight into a problem I'd otherwise be entirely lost with. I'll look into del Pezzo surfaces a bit further (hopefully being able to utilize David's link that he posted in his answer), and see what I can come up with. | |
| Oct 25, 2010 at 22:49 | vote | accept | apples | ||
| Oct 25, 2010 at 15:17 | history | edited | Robin Chapman | CC BY-SA 2.5 |
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| Oct 25, 2010 at 9:56 | comment | added | Robin Chapman | Ouch - then it's the intersection of two quadrics in $P^4$. I'm not sure what that is: maybe a del Pezzo surface? | |
| Oct 25, 2010 at 9:00 | comment | added | Fedor Petrov | It looks like you lost some variable after homogenizing. It should be $X^2+Y^2=Z^2+T^2$, $X^2-Y^2=W^2-T^2$. | |
| Oct 25, 2010 at 8:28 | comment | added | Robin Chapman | Dylan, by Siegel's theorem there are only finitely many integer points on an elliptic curve with Weierstrass model over $\mathbb{Z}$. I admit that I haven't worked this problem through to the extent that I am certain it reduces to a problem of this nature, but I suspect it does. | |
| Oct 25, 2010 at 8:14 | comment | added | Dylan Thurston | @Gerry Myerson: What? The points on an elliptic curve form an abelian group (once you fix a basepoint), which is often infinite. | |
| Oct 25, 2010 at 7:50 | comment | added | Gerry Myerson | And if it does come down to finding integer points on an elliptic curve, then there won't be any "properties of $a$ and $b$ that lead to solutions," because there will only be finitely many solutions. | |
| Oct 25, 2010 at 6:59 | history | answered | Robin Chapman | CC BY-SA 2.5 |