# Tag Info

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The real period can be computed very efficiently using the AGM. See my paper http://dx.doi.org/10.1016/j.jnt.2013.02.002

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HINT#1: If $$(x^2+xy+y^2)(x^2-xy+y^2)=z^2,\quad\gcd((x^2+xy+y^2,x^2-xy+y^2)=1,$$ then $$\begin{cases} x^2+xy+y^2=u^2\\ x^2-xy+y^2=v^2\\ z=uv. \end{cases}$$ $$\begin{cases} (u+x+y)(u-x-y)=xy,\\ (x-y+v)(x-y-v)=xy,\\ z=uv. \end{cases}$$ Then use the fact that equation $$mn=pq$$ has solutions $$m=ac,\quad n=bd,\quad p=ad,\quad q=bc.$$ HINT#2: If ...

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The following is an idea based on Cassel's Lectures on Elliptic Curves (see chapter 20, exercise 3). Consider the real third-root of $a$, say $a^{1/3}$ (the other roots are $\rho{a^{1/3}}$ and $\rho^{2}{a^{1/3}}$ where $\rho$ is a primitive third-root of unity). Define $\phi:C\rightarrow{E}$ by sending $(x,y,z)$ to ...

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[Edited to add the citations of Euler and others from Dickson] Each of $E_\pm: x^4 \pm x^2 y^2 + y^4 = z^2$ is an elliptic curve. It turns out that in each case a Fermat-style "descent" suffices to find all solutions; in practice these days such questions are solved by reducing the curve to standard form and then either using software such as J.Cremona's ...

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COMMENT.- I feel but I have no proof that there are only the trivial solutions $(x,y,z)= (t,0,t^2),(0,t,t^2)$. I give here an outline of what could perhaps lead to a proof. $$x^4-x^2y^2+y^4=z^2\iff (x^2-y^2)^2+x^2y^2=z^2\qquad (1)$$ Hence, as it is well known, $$\begin{cases}x^2-y^2=t^2-s^2\\xy=2ts\\z=t^2+s^2\end{cases}\qquad (2)$$ It follows in particular ...

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Just so this has an answer, let me expand upon what I said in the comments slightly. So, let us begin with the following trivial observation: Observation: If $f:E\to E'$ is an isogeny of elliptic curves over $k$, with $\mathrm{char}(k)=p>0$, then $f$ and $\widehat{f}$ are separable if and only if $p\nmid\deg(f)$. Indeed,this follows immediately ...

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If I am reading your question correctly, then you are asking how to come up with the formula for the line $\bar{C}$. Well, one down-to-earth method to gain some intuition would be to look at explicit instances of curves $C$ together with three colinear points $P_1, P_2, P_3$ and then apply $\phi$ to the points. Then one can calculate $\bar{C}$ explicitly, ...

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Your curve $y^2 + 7xy = x^3 + 16x$ is isomorphic to $y^2 = x^3 - 44091x + 3304854$. Maybe you can find the transformation between these two models yourself?

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You can use that there is a complex number $\omega \in \mathbb C$ s.t. $|\omega| = \sqrt{q}$ and $$\#E(\mathbb F_{q^n}) = q^n +1 - (\omega^n+\overline\omega^n)$$ where $\overline\omega$ is the complex conjugate of $\omega$. (Weil conjecture applied to elliptic curves.) Applying this formula makes it essentially trivial. $a_n = q^n +1 - \#E(\mathbb ... 3 Let$K=\Bbb{F}_4=\{0,1,\alpha,\alpha+1\}$with$\alpha^2=\alpha+1$. For all$x\in K*$we have$x^3=1$. Therefore the cubic$x^3+\alpha$only takes values in$K\setminus\Bbb{F}_2$. On the other hand,$y^2+y\in\Bbb{F}_2$for all$y\in K$. Therefore the equation $$y^2+y=x^3+\alpha$$ has no solutions$(x,y)\in K^2$. The point at infinity is thus the only ... 1 Because$E(\Bbb{F}_{p^n})$is a subgroup of$E(\Bbb{F}_{p^m})$whenever$n\mid m$we have "necessary" conditions:$r$has to be a prime, and$|E(\Bbb{F}_p)|=1$. A rare exception might be when$E(\Bbb{F}_p)=E(\Bbb{F}_{p^r})$, but this is difficult to arrange (may be impossible - I haven't tried). Let's try$p=2$. If we have$|E(\Bbb{F}_p)|=1$, then the zeros ... 4 By Hasse's bound we know that$1\le |E(\mathbb{F}_3)|\le 7$; and indeed there is an elliptic curve with$E(\mathbb{F}_{3})=\{\mathcal{O}\}$, given by $$y^2=x^3-x-1.$$ Actually, since we know that all such curves are given by the long Weierstrass equation$y^2=x^3+ax^2+bx+c$with nonzero discriminant, we can just try all possibilities for$a,b,c$. There are ... 1 Let$f(x,y)=y^2-x^3-x^2$. Make$f(x,y)=0$,$\dfrac{df}{dx}=0$and$\dfrac{df}{dy}=0$. The singularities are the common points of the equations. 3 What you do is plug the parameterization into the equation. You want to find that if you plug$t^2-1$in for$x$you get the given expression for$y. Unfortunately, it fails. So \begin {align} x^3-x^2 &=(t^2-1)^3-(t^2-1)^2\\ &=(t^2-1)^2(t^2-1-1)\\ &=(t^2-1)^2(t^2-2)\\&=y^2-2x^2 \end {align} Could it be your equation is supposed to ... 0 For both curves, the2$-torsion points are the points$(x_0,0)$such that$x_0^3+ax_0+b=0$(together with the point at infinity of course). 2 First, I would suggest determining$End(E_1)$. To do this, you might note that if$\alpha\in\mathbb{C}$and$\alpha(\mathbb{Z}+i\mathbb{Z})\subset\mathbb{Z}+i\mathbb{Z}$if and only if$\alpha\in\mathbb{Z}+i\mathbb{Z}$and hence$End(E_1) = \mathbb{Z}+i\mathbb{Z}$. Then, the automorphisms will be the endomorphisms of norm 1, that is the units of ... 2 Since you know that${\rm End}(\Bbb C/\Lambda)=\{\lambda\in\Bbb C\mid\lambda\Lambda\subseteq\Lambda\}$it should be clear that $${\rm Aut}(\Bbb C/\Lambda)=\{\lambda\in\Bbb C\mid\lambda\Lambda=\Lambda\}.$$ When$\Lambda=\Bbb Z\otimes\Bbb Zi=\Bbb Z[i]$the latter set is identified with the set of invertible elements in the ring$\Bbb Z[i]$i.e.$$... 1 By definition, an endomorphism of a complex elliptic curve$E$is a holomorphic map$E\rightarrow E$, which fixes the origin. It turns out that this condition is enough to force it to be a homomorphism of abelian groups in the usual sense, i.e., it is a$\mathbb{Z}$-module homomorphism. This definition makes sense for every algebraically closed field$k, ... 2 Per your request, I am turning my comment into an answer: Consider the change of variables \begin{align*} x' & = x\\ y' &= \sqrt{c}y \end{align*} Then in the new variables we haveE_c: (y')^2 = (x')^3 + ax' + b$. The map$y\mapsto y'$is multiplication by a non-zero element of$\overline{k}$which gives a$\overline{k}$isomorphism between ... 2 Multiply by$y(x^2+1)$so your map is$((x^4+3y)y, (x+1)(x^2+1), y(x^2+1))$. Now further you want$X,Y,Z$to be homogeneous so set$x=\frac{X}{Z}$and$y=\frac{Y}{Z}$and now multiply by the lowest power of$Z$to clear the denominators. 3 The ring of endomorphisms of$E$is the ring of complex numbers$\alpha$such that$\alpha\Lambda\subseteq \Lambda$. Thus, to prove that$\beta$defines an endomorphism you just need to show that$\beta\Lambda\subseteq \Lambda$. The kernel of$\beta$is the set of all$P\in E(\mathbb C)$such that$\beta(P)=O$. Choosing a fundamental parallelogram for ... 2 If you look at a neighbourhood of a point$(t,0)$, and you project it on the$x$-axis, you see that the projection is$2$-to-$1$. This means that$(x-t)$is not a uniformizer for the local ring at that point. The projection map$x : E \to \Bbb P^1(k)$locally looks just like the map$z \mapsto z^2$around$0$(if you work over$\Bbb C\$ there are actual ...

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