Timeline for Can an infinite sequence of integers generate integer-area triangles?
Current License: CC BY-SA 2.5
9 events
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| Jan 22, 2010 at 2:54 | comment | added | Bjorn Poonen | The corrected curve has no $2$-adic points. In fact, even the genus 1 curve it maps to has no $2$-adic points. Proof: The quartic is $-1$ mod 4 whenever $x$ is in $\mathbb{Z}_2$. Because of the $x \mapsto 1/x$ symmetry, there aren't any points with $x \in \mathbb{Q}_2$ either. | |
| Jan 21, 2010 at 17:18 | comment | added | Kevin Buzzard | Update: David fixed the sign error. Thanks David. So now Bjorn's comments don't apply because unfortunately the genus 1 curve y^2=f(x) has (for me at least) no obvious rational points at all. There are standard ways of analysing this curve but we have to wait until someone applies them. First thing to do is perhaps to check for no p-adic points for primes p of bad reduction. | |
| Jan 21, 2010 at 17:05 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Jan 21, 2010 at 16:31 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Jan 21, 2010 at 15:47 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Jan 20, 2010 at 22:01 | comment | added | Kevin Buzzard | Hold your horses everyone. Hasn't David made a sign error? y^2=-RHS is the relevant equation, right? @David: I don't think this will pan out, even with the sign fix. However trying something with small period bigger than 1 is probably worth doing. | |
| Jan 20, 2010 at 2:21 | comment | added | Bjorn Poonen | There are no rational points other than $(0,\pm 1)$. Your curve has the form $y^2=f(x^2)$. So it maps to the genus 1 curve $y^2=f(x)$, whose smooth projective model has 4 obvious rational points (above $x=0$ and $x=\infty$). There are no others since Magma says that its Jacobian (which is the same curve) has rank 0 and torsion subgroup of size 4. | |
| Jan 20, 2010 at 1:23 | comment | added | Mariano Suárez-Álvarez | apart from (0,1), I guess | |
| Jan 20, 2010 at 0:59 | history | answered | David E Speyer | CC BY-SA 2.5 |