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2

For $n=3$, taking $x_3 = 1 - x_1 - x_2$ we get $-1 + x_1 x_2 - x_1 x_2^2 - x_1^2 x_2 = 0$. This is an elliptic curve with Weierstrass form $s^3+(23/48)s-181/864+t^2$, according to Maple. I think it ought to be a finite computation (not one that I know how to do) to find generators for the rational points. EDIT: Well, apparently this is an unsolved ...

1

Geometrically, adding two points on the elliptic curve involves the line through those two points. Algebraically, we want a line such that the curve and line agree on those two points, equivalently the difference between the curve and line have those two points as roots. Thus adding a point to itself geometrically involves the tangent to the curve at that ...

0

(This is a long comment re MacLeod's answer.) We can also combine the two cases together. Assume a quartic polynomial to be made a square, $$pu^4+qu^3+ru^2+su+t=z^2\tag1$$ has a known rational point, call it $w$. We substitute $u=v+w$ and collect the new variable $v$, $$c_4v^4+c_3v^3+c_2v^2+c_1v+\color{blue}{c_0^2}=z^2\tag2$$ where the $c_i$ are ...

1

Answering my own question. Turns out the $\alpha$ referred to in the notes is exactly the endomorphism which we are trying to reduce mod $\psi_\ell$, although it isn't made very clear. Explanation: Let $\alpha = (\alpha_x(x), \alpha_y(x) y)$ be an endomorphism in $\text{End}(E)$, where $\alpha_x$ and $\alpha_y$ are rational functions. If the denominator of ...

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Fact : $\mathbb{Z}[\sqrt{-2}]$ is Unique factorization domain. lemma : in every UFD, if the product of two numbers, which are relatively prime is a cube, then each of them must be a cube. There is no any solution $$x^3 = (y+\sqrt{-2})\times(y-\sqrt{-2})$$ the greatest common divisor of these factors will divide 2-times the $\sqrt{-2}$, which is lead to ...

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The only integral solutions to your first problem are $(3, \pm 5)$. The general class of equations are known as Mordell's equation. A fairly elaborate discussion and case by case analysis is provided here.

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As requested in the comments, I will upgrade my comment to an answer. Keith Conrad's excellent article settles the question. On page $10$, it is noted that for the values $k = -5, -6, 6, 7, -24, 45$, the Mordell curve $y^{2} = x^{3}+k$ has exactly one rational point. Proof of this fact is not given, so the references may be helpful in this regard. He does ...

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Here is the general picture. Let $f \in \mathcal{M}_k(N,\chi)$ be a modular form of weight $k$, level $N$, and nebentypus $\chi$. Suppose that $f$ has the Fourier expansion $f(z) = \sum_{m = 0}^{\infty} a_f(m) e(mz).$ Then by replacing $z$ with $nz$, we see that $f(nz) = \sum_{m = 0}^{\infty} a_f(m) e(mnz) = \sum_{m = 0}^{\infty} a_{f_n}(m) e(mz)$ with \[...

3

Such a map is called an isogeny, and isogenies over finite fields preserve the number of points (in fact, two elliptic curves over a finite field are isogenous iff they have the same number of points). On the function field side, this corresponds to an embedding of the function field of the original curve into the second one. By the existence of the dual ...

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The reference of this answer is D. Cox 'Primes of the form $x^2+ny^2$'. The formulation 1 is Theorem 10.14 and Corollary 10.20, and 2 is Theorem 7.7 (ii). Then it is more natural that 1 is inferred from 2, than 2 is inferred from 1. The elements in the set of triples described in 2, can be considered as a set of equivalence classes of quadratic forms $C(D)$...

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The largest value for $f(k)$ that I know of is an example due to Noam Elkies with $k = 509142596247656696242225$, where there are (at least) $125$ pairs of solutions (so $f(k)=250$ in your notation). If ranks of elliptic curves over the rationals are absolutely bounded, then so is $f(k)$ (as long as one restricts to $6$th power free values of $k$ to avoid ...

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You could calculate the Hilbert symbols $(\frac{p,-1}{q})_2$ for varying primes $q$.

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Part 2 is easy to see as his algorithm allow us to take square root of "small numbers" modulo p(As the complexity depends is = $\sqrt{|x|} \log^{9}p$). So one can evaluate $\zeta_2,\zeta_4,\zeta_8$ as all of them involve taking square roots of numbers of small magnitude like $\sqrt{-1},\sqrt{2}$ and as Jyrki mention if $p \neq 1 \mod 16$ then one of them ...

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The cubic \begin{equation*} u^3-u^2-(n^2+n)u/3-n^3/27-w^3=0 \end{equation*} can be shown to be equivalent to the elliptic curve $$y^2=x^3+1296n^2(n^2+n+1)^2$$ by using Nagell's algorithm and a computer algebra package. It does not have to be state-of-the-art software since I used an ancient MS-DOS version of Derive. The curve has ...

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If $W$ contains more than one component, then it is the union of projective subvarieties of codimension 1, which necessarily have non-empty intersection. (This is immediate from Bezout but if you prefer a smaller hammer, also follows from the Krull height theorem.) Then, we need to check that $W$ cannot possibly be smooth at a point lying on two or more of ...

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A handful for $n$ odd 13 a: 230153 b: 12792 c: 283945 d: 284233 u: 507 v: 164 15 a: 32625 b: 3472 c: 61455 d: 61553 u: 56 v: 31 23 a: 523367 b: 57072 c: 1413145 d: 1414297 u: 984 v: 667 27 a: 206703 b: 3848 c: 231345 d: 231377 u: 156 v: 37 89 a: 11534489 b: 700920 c: 63439289 d: 63443161 u: 649 v: 540 105 a:...

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An Attempt You have $a^2+\left(n^2+1\right)b^2=d^2$. All integral solutions $(a,b,d)$ to this equation satisfies (1) $|a|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right)$ for some $r\in\mathbb{Q}$. Now, if $b^2+c^2=d^2$, then (...

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$P$ isn't an $\ell$th root of unity, but $\tau_\ell(P, P)$ is. The Tate-Lichtenbaum pairing maps into $\mathbf{F}_q^\times/ (\mathbf{F}_q^\times)^\ell$. Since $\ell \mid q - 1$, this group is isomorphic to the group $\mu_\ell$ of $\ell$th roots of unity. Since $\ell$ is prime all nontrivial elements of $\mu_\ell$ are generators, or in other words, as long as ...

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