Skip to main content
7 events
when toggle format what by license comment
Sep 16, 2016 at 0:24 vote accept Andrew Poelstra
Sep 15, 2016 at 23:08 comment added Felipe Voloch @JeremyRouse I was just thinking along the same lines and I think that does it.
Sep 15, 2016 at 23:07 comment added Jeremy Rouse Here's an idea. The trace of Frobenius on a $j=0$ elliptic curve (for $p \equiv 1 \pmod{3}$) is given by the representation of $p$ by the form $x^{2} + xy + y^{2}$, which is related to the representation of $4p$ in the form $x^{2} + 3y^{2}$. Since $(p+1-q)^2 + 3u^{2} = 4p$, then $(q+1-p)^{2} + 3u^{2} = 4q$.
Sep 15, 2016 at 23:07 comment added Felipe Voloch @QiaochuYuan In the prime field case every value in the Hasse interval corresponds to a curve, by a result of Deuring.
Sep 15, 2016 at 23:03 comment added Qiaochu Yuan Surely being in the Hasse interval is necessary but not sufficient, right?
Sep 15, 2016 at 22:47 history edited Felipe Voloch CC BY-SA 3.0
added 125 characters in body
Sep 15, 2016 at 22:30 history answered Felipe Voloch CC BY-SA 3.0