Timeline for The product of n radii in an ellipse
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2010 at 13:27 | comment | added | Kevin Buzzard | So you're doing the $n=8$ case by writing the relevant curve as a degree two cover of the curve Fermat used to prove no non-triv solutions to $x^4+y^4=z^4$. So there is a chance that the problem for even $n$ is equivalent to FLT after some cunning re-arrangement. | |
| Jun 26, 2010 at 11:56 | comment | added | Sergei Ivanov | $2a^3b^3=a^2+b^2$ does not have nontrivial solutions either. Indeed, the equation implies that $(a^2+b^2)/2ab$ is a square of a rational, hence $(x^2+1)/2x=y^2$ for $x=a/b$ and $y$ rational. The discriminant of this as of a quadratic equation in $x$ is $4(y^4-1)$. It is never a square of a rational unless equals zero (this quickly reduces to the equation $x^4-y^4=z^2$ over integers). | |
| Jun 26, 2010 at 9:17 | comment | added | Kevin Buzzard | Because of the coincidence above, there's a non-zero chance that the question for general even $n$ is equivalent to FLT for $n/2$. This is just a stab in the dark though. | |
| Jun 26, 2010 at 9:15 | comment | added | Kevin Buzzard | I didn't check Will's claim that the $n=6$ problem is equivalent to the cubic he gives above, but I can solve $4a^3b^2=3a^2+b^2$. Clumping the b^2 terms together one gets $(4a^3-1)b^2=3a^2$ so one is seeking rationals $a$ with $3a^2(4a^3-1)$ a square. Now $a^2$ is a square so we just want $3(4a^3-1)$ a square; this is an elliptic curve, and multiplying up by 144 and setting $x=12a$ we get $y^2=x^3-432$, which is $X_0(27)$ and IIRC birational to the Fermat cubic, which has rank 0 and torsion leading to $a=1$. In particular the question for $n=6$ is basically equivalent to FLT for $n=3$. | |
| Jun 26, 2010 at 3:17 | history | edited | Will Jagy | CC BY-SA 2.5 |
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| Jun 26, 2010 at 3:11 | history | answered | Will Jagy | CC BY-SA 2.5 |