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Mar 20, 2015 at 11:26 history edited Wolfgang CC BY-SA 3.0
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Feb 19, 2012 at 14:51 vote accept Wolfgang
Feb 19, 2012 at 1:07 answer added Noam D. Elkies timeline score: 11
Feb 5, 2012 at 13:16 comment added Wolfgang Thank you very much. I have edited the lower limit, as it is exactly examples like yours I'm interested in. I suppose the curve is uniquely determined by $d$, up to translation? So again the question: What about $n=-7$?
Feb 5, 2012 at 13:08 history edited Wolfgang CC BY-SA 3.0
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Feb 5, 2012 at 7:12 comment added Noam D. Elkies spanferkel's demand that the integral be over $(0,\infty)$ means that the curve must also have a rational 2-torsion point, which cuts the usual list of 13 CM discriminants down to seven: $$-3, -4, -7, -8, -12, -16, -28.$$ For example, the disc.-11 curve yields the complete elliptic integral of $dt / \sqrt{4t^3-4t^2-28t+41}$ which is a rational multiple of $$\pi^{-2} \phantom.\Gamma(1/11)\phantom.\Gamma(3/11)\phantom.\Gamma(4/11) \Gamma(5/11)\phantom.\Gamma(9/11),$$ but the cubic is irreducible so the integral cannot be written as $\int_0^\infty dt/\sqrt P(t)$.
Feb 5, 2012 at 4:27 comment added Will Sawin If I understand what is going on correctly, any polynomial with the same $j$-invariant should do: en.wikipedia.org/wiki/J-invariant#Algebraic_definition
Feb 5, 2012 at 0:03 comment added Wolfgang I didn't necessarily think of that restriction. So if we restrict them to that, which polynomial(s) would that be for the $d=-7$ case? And how to find them? Unique?
Feb 4, 2012 at 22:42 comment added Will Sawin Is this really about elliptic curves with complex multiplication? If so, then the answer is going to be yes if and only if the class number of $\mathbb Q(\sqrt{d})=1$.
Feb 4, 2012 at 22:30 history edited Wolfgang CC BY-SA 3.0
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Feb 4, 2012 at 22:20 history asked Wolfgang CC BY-SA 3.0