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9

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 ...


7

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 ...


5

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 $\deg(g)\neq1$ ...


4

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 ...


4

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 $\equiv 3 \pmod 8$. ...


3

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 $4$, ...


2

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


2

[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 $\phi$. ...


2

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 ...


1

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 ...


1

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. $G\oplus G = G \times G$


1

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 $x$.


1

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.


1

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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