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Assuming $A$ and $B$ are the semi-axes AND the point with coordinates $(x,y) = (C, E)$ lies on the ellipse, which is given by $$ \left( \frac{x}{A} \right)^2 + \left( \frac{y}{B} \right)^2 = 1 $$ so $$ \left( \frac{C}{A} \right)^2 + \left( \frac{E}{B} \right)^2 = 1 $$ holds, it is simply $$ F = A - C = A - A \sqrt{1 - (E/B)^2} \\ D = B - E = B - B \sqrt{1 - ...


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When mathematical equations/ models distinctly trifurcate on basis of discriminant /characteristic sign of its principal arguments (< 0, 0, > 0) in a group ( i.e., when there is a transformation possible between positive and negative sign types ) we have three classes/types (elliptic, parabolic, hyperbolic ) which need not generally be associated with an ...


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Your question covers a vast amount of material and without specifying further it is hard to say exactly what you're looking for. Also, you did not state the level of material you were looking for. Almost any text on Algebraic Geometry (or more specifically Arithmetic Algebraic Geometry), Elliptic Curves, or Number Theory related to these areas would contain ...


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A partial answer, going from "easy" to "hard": "Elliptic hyperboloid" and "Elliptic paraboloid" these are of course quadric surfaces with the cross sections implied by their respective names; one could rightly ask why they weren't "hyperbolic ellipsoids" and "parabolic ellipsoids", tho. "Elliptic integral" The "second kind" studied by Legendre is ...


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For the sake of any future readers, I thought I would extend coffeemath’s excellent answer a little to give a self-contained proof, since it’s quite easy. Primitive Pythagorean triples We briefly recall the classification of primitive Pythagorean triples. Let $x^2+y^2=z^2$ with $x$, $y$ coprime natural numbers. An odd square is congruent to $1$ modulo $4$, ...


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If the equation $y^2=x^3-x$ has rational soloutions in which $x,y$ are each nonzero, we may put $y=tx$ and after division by $x$ we have the quadratic equation for $x$ $$x^2-t^2x-1=0 \tag{1}$$ whose discriminant $t^4+4$ must be square. Put $t=m/n$ with $\gcd(m,n)=1.$ Here $m,n$ are nonzero and $m \neq n$ since $m=n$ means $t=1$ in which case (1) has no ...


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This paper talks a lot about Mordell's equations. I haven't read through all of it, but on Page 2, Theorem 2.3, there is a proof of how $y^2=x^3-6$ has no integral solutions. Here's an outline of the proof: Prove $y$ is odd and $x \equiv 7 \pmod 8$. Get $y^2-2=x^3-8=(x-2)(x^2+2x+4)$. Note that the latter factor is always positive and is $\equiv 3 \pmod 8$. ...


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HINT.-Put the equation of the cubic as $$f(x)=y^2-(x-\alpha)(x-\beta)(x-\gamma)=0$$ where $\alpha,\beta,\gamma$ are supposed reals in the affin plan. We must have $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$. Since $\frac{\partial f}{\partial y}=2y=2\sqrt{(x-\alpha)(x-\beta)(x-\gamma)}$ then $$(x-\alpha)(x-\beta)(x-\gamma)=0$$ On the ...


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Let $f$ be the polynomial $x^3+ax^2+bx+c$ in $\mathbf Q[x]$ and suppose that it has a double root. Then the gcd $g$ of $f$ and its derivative $f'$ is a nonconstant polynomial in $\mathbf Q[x]$. If $\deg(g)=1$ then $g$ has a root in $\mathbf Q$. This root is a root of $f$ and $f'$ in $\mathbf Q$, i.e., $f$ has a double root in $\mathbf Q$. If $\deg(g)\neq1$ ...


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First, the form $y^2=x^3+ax^2+bx$ is close enough to the classical form to do all the necessary computations in, as long as the characteristic is not $2$. Second, when an elliptic curve is in this form, I hope it’s obvious that the points of order two are exactly those on the $x$-axis. Third, on this particular curve, the origin is thus a point of order ...


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It's a direct sum, i.e. the elements in the products with finitely many non-zero coefficients. For finite sums this requirement is trivially satisfied, i.e. $G\oplus G = G \times G$


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You have done an error in your discriminant which should begin by a "4": $$\Delta=4(B-\epsilon ^2L)^2-4(\epsilon ^2-1)(\epsilon ^2L^2-B^2)$$ Expanding it, and then factoring it, you will obtain the following result: $$\Delta = 4 \epsilon^2 (B-L)^2$$ which is a perfect square ; then you will find the (simple) desired formula for $x$.


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If you use for $ax^2+bx+c=0$ $$\Delta=b^2-4ac$$ expand and simplify; you should get $$\Delta=4 \epsilon ^2 (B-L)^2$$ I am sure that you can take it from here.


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[Corrected (mistake in characteristic 2: multiplication by $-1$ always fixes $\omega_E$)] The converse is true, except in characteristic 2 and for some supersingular $E/k$ in characteristic 3. Composing $\phi$ with translation by $-\phi(0)$ yields an automorphism $\psi: E \to E$ sending $0$ to $0$ and inducing the same action on $\omega_E$ as the $\phi$. ...


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The fibre product $\operatorname{Spec} R[[x]] \times_R \operatorname{Spec} R[[y]]$ is not isomorphic to $\operatorname{Spec} R[[x,y]]$. Indeed, the map \begin{align*} \phi \colon R[[x]] \otimes_R R[[y]] &\to R[[x,y]]\\ f \otimes g &\mapsto fg \end{align*} is injective (I think), but not surjective, even when $R = k$ is a field (and probably one can ...


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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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Updated We can try to find solutions ourself. Equation 1. $\quad x^4+x^2y^2+y^4 = z^2.$ Let $$\gcd(x,y) = m,\quad x=mX,\quad y=mY,\quad z=m^2Z,$$ then $$X^4+X^2Y^2+Y^2=Z^2,\quad\gcd(X,Y)=1.$$ $$(X^2+XY+Y^2)(X^2-XY+Y^2)=Z^2,\quad\gcd(X,Y)=1.$$ Multipliers in the left part are the same parity. If they are even, then both $X$ and $Y$ are even and ...


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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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