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This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Francesco Polizzi below. EDIT We show that the answer to the OP's question is yes. Thanks to Will Sawin for his comments. I use the notation of [Hartshorne, Algebraic Geometry, Chapter V Section 2]. Since $S$ is a ruled surface, there exists a ...


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This answer is for the concrete values of $p,a,b$ given in the question; it relies on $p\equiv 7\pmod{9}$, $a=0$, $\left(\frac{b}{p}\right)=-1$. The key property is that $p-1$ is divisible by $3$, but not by $3^2$. Instead $p+2$ is divisible by $3^2$. This allows us to follow an approach for cuberoots analogous to the one used for finding squareroots when ...


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A less direct explanation than that of Kevin Dong: your curve is elliptic, and any point of inflection may be taken to be the identity $\Bbb O$, after which choice you can use chord-and-tangent to describe the addition. Then, any point of inflection is a $3$-torsion point, and the set of all these is a group.


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Hint. Let $p, q \in C$ be two points of inflection, and let $L$ be the line between them. Let $L \cdot C = p + q + r$. Let $M$ the tangent line to $C$ at $r$, and $M \cdot C = 2r + s$. Show that $r = s$ by showing that $s$ lies on the line through $p$ and $q$. In the interests of completeness, we provide a complete solution. Let $M_1$ be the line tangent ...


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It is definitely not an elliptic curve: an equation $y^2=f(x)$ with $f$ having no repeated factors is "hyper-elliptic", giving an elliptic curve exactly when the degree of $f$ is $3$ or $4$. It is a curve of genus $g$ for $f$ of degree $2g+1$ and $2g+2$, generally, as one can see from the Riemann-Hurwitz formula, taking into account the points lying over ...


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I think I found the answer by looking at the errata here: http://www.math.brown.edu/~jhs/AEC/AECErrata.pdf Check out the modified page 93. This is how I interpret it: We know $\det(P^\sigma,Q^\sigma)$ means you take the determinant (using a certain basis) of the pair of points $P^\sigma$ and $Q^\sigma$, but $(P,Q)^\sigma$ apparently means you let $\sigma$ ...


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Let $x=A/p$ and $y=C/q$ in lowest terms. Cross-multiplying then gives us $$ p^3C^2 = q^2(A^3-2p^3)$$ Since $A/p$ is in lowest terms, $A$ and $p$ are coprime, and thus so are $A^3$ and $p^3$, and thus so are $p^3$ and $A^3-2p^3$. Therefore all of the prime factors in $p^3$ on the left must come from $q^2$ on the right. Similarly, $C^2$ and $q^2$ are coprime, ...


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This is part (a) Exercise 5.10 in Chapter V of Silverman's "The Arithmetic of Elliptic Curves". The proof of part (a), however, is implicit in the proof of Theorem 4.1 of the same chapter.


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As Álvaro Lozano-Robledo already showed, to prove that there is no $p$-torsion it is enough to find a prime $q \not\mid 6D$ such that $q \equiv 2 \bmod 3$ and $q \not\equiv -1 \bmod p$. The punchline he left unstated is that the existence of such $q$ is guaranteed by Dirichlet's theorem on primes in arithmetic progressions. Choose $m_0 \not\equiv 0, -1 ...


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After reading the following chapters of the book, here are some explanations that can help clarify why one can always transform those two Fermat equations to elliptic curves. Indeed, one can always transform a cubic equation in $P_K^2$ (2 dimensional projective space over $K$, $\operatorname{char}K \neq 2,3$) or a quartic equation of the form $v^2 = ...


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The following image shows half a torus mapping to a sphere via a scaled Weierstrass $\wp$ function. It was created in Mathematica. The four branch points lie on the two boundary circles of the half-torus, and map to the endpoints of the two "seams" on the resulting sphere. The other half of the torus maps to the sphere in the same way, with the deck ...


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I'm not quite sure what you mean. Given two curves $f(X,Y) = 0$ and $g(X,Y) = 0$ where $f$ and $g$ are polynomials, let $r_X(Y)$ and $r_Y(X)$ be the resultants of $f(X,Y)$ and $g(X,Y)$ with respect to $X$ and $Y$ respectively. A point $(x,y)$ common to both curves must satisfy $r_X(y) = r_Y(x) = 0$. In particular, if you want a common integer solution, a ...


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The key result: If $p \not \equiv 1 \mod{3}$ then every element of $\mathbb{F}_p$ has exactly one cube root. Prove this, then the problem should be easy.


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@Zilin J.:I have a method that sends all binomial addition $a+b=c$ (so, for example, $a^4+b^4=c^4$) to an elliptical curve, noted $V_A$, of the shape $$X^3+Y^3=AZ^3$$, where $A$ is a cube-free number, and this curve is birational equivalent to a Wierstrass one, the curve $$y^2z=x^3-432A^2z^2$$ (one can obviously work with affine curve making $z=1$), via ...


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Following the suggestion by @Lucian, we use the Euler's infinite product representation of the sine function and write $$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag 1$$ Next, taking the logarithmic derivative of $(1)$ yields $$\cot z=\frac1z-2z\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2-z^2} \tag 2$$ And then, taking the derivative of ...


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Another way (using properties of polygamma): $$A\equiv\sum_{m=0}^\infty\frac{1}{(z/\pi+m)^2}=\Psi^{(1)}(1,\frac{z}{\pi})$$ $$B\equiv\sum_{m=-\infty}^0\frac{1}{(z/\pi+m)^2}=\sum_{m=0}^\infty\frac{1}{(-z/\pi+m)^2}=\Psi^{(1)}(1,-\frac{z}{\pi})$$$$=-\Psi^{(1)}(1,\frac{z}{\pi})+\frac{\pi^2}{z^2}+\pi^2(1+\cot(z)^2)$$ Then ...


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Show that $f(z)=\sum_{m=-\infty}^\infty (z+\pi m)^{-2} -(\sin z)^ {-2}$ is a bounded holomorphic function in $\Bbb C$, whence it is constant. Since it is $0$ in one point (for example at $i\infty$), it is identically $0$. To bound $f$ consider vertical strips (motivated by the series and the periodicity of the sine. To show it is holomorphic note the poles ...


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I don't have a solution, but I have a simplification. I'd start with: $$\frac{d}{dx} -\cot(z) = \sin^{-2}z$$ $$\int \frac{1}{(z + \pi m)^2}dz = -\frac{1}{z + \pi m}$$ So then you would just need to prove that $\cot(z)$ is equal to $\sum_m \frac{1}{1+m\pi}$.


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Hint: Differentiate twice the natural logarithm of Euler's infinite product expression for the sine function.


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Am not sure what you mean by discreet parameters, degree and rank? That will not be enough. As you said, take $C$ to be a smooth cubic curve. Then $\mathcal{O}_C$ and a line bundle $L$ of degree zero but $L\neq \mathcal{O}_C$ have the same discreet parameters. But ...


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I hope you can find this useful. All elliptic curve $C(\mathbb {Q})$ is canonically write as $y^2=4x^3-g_2x-g_3$... (1) in which the three roots of the right side are distinct; by a birational transformation $(x,y)\to (\frac x4, \frac y4)$ you have $y^2=x^3-h_2x-h_3$... (2) where $h_2$ and $h_3$ can be supposed rational integers. Being $e_1, e_2, e_3$ ...


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The proof of that result, usually called (an explicit version of) the Weak Mordell-Weil theorem, can be found in Silverman's Arithmetic of Elliptic Curves book. The proof uses Galois cohomology and some minor arithmetic that can be followed with the knowledge of a few of the main theorems of global class field theory.



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