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10

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 ...


5

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 ...


4

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


3

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 ...


3

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.


3

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.


3

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 ...


2

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 ...


2

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 ...


2

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.


1

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, ...


1

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 ...


1

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 $\sigma$ ...


1

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 ...


1

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 = ...


1

@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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