112

This problem turned out to be much more interesting than I originally
thought. Let me give my solution, which seems to be slightly different from
(but essentially the same as) the solution in the paper by Bremner and
MacLeod (see Allan MacLeod's answer).
Theorem. Let $a,b,c$ be positive integers. Then
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$ can ...

66

This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et Informaticae, pages 29-41.
It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can ...

40

Yes, there is another solution. The next one I found is a bit big, namely
$$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$
This solution also satisfies the property that
$$ \frac{a^{2}}{b+c} + \frac{b^{2}}{a+c} + \frac{c^{2}}{a+b} = \frac{31}{21} (a+b+c),
$$
which was true of $a = 13$, $b = 77$ and $c = 183$. If you clear ...

32

Yes, this is a counterexample to Theorem 1.3. But it looks like the issue is with the proof of Theorem 1.3, and is not relevant to Lang's conjecture. Namely, your example contradicts Theorem 4.2 of that paper, which does not rely on Lang's conjecture. Then the proof of Theorem 1.3 relies on Theorem 4.2, in addition to relying on Lang's conjecture. The ...

31

Not directly, as far as I know, since explicitly computing large multiples of points in $E(\mathbb Q)$ is infeasible. However, people have considered lifting points from $E(\mathbb F_p)$ to $E(\mathbb Q)$ or to the $p$-adics $E(\mathbb Q_p)$ in order to devise algorithms to solve the discrete log problem in $E(\mathbb F_p)$ (although, unsuccessfully so far). ...

answered May 9 '19 at 11:45

Joe Silverman

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30

Yes. I'll talk about why elliptic curve discrete log is harder than ordinary discrete log.
Suppose we have $g, h$ and want to find $n$ such that $g^n = h$.
The usual methods for solving the discrete logarithm problem in $\mathbb{Z} / p \mathbb{Z}$ (as opposed to on an elliptic curve) is to compute $g^k \mod p$ for lots of values $k$ and look for powers ...

answered Nov 22 '19 at 19:41

Adam P. Goucher

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28

To expand my comment, there are at least 3 subtle ways to get BSD wrong:
1) The BSD period over ${\mathbb Q}$ is the real period $\Omega_\infty$ when $E({\mathbb R})$ is connected ($\Delta(E)<0$) and $2\Omega_\infty$ when it has two connected components ($\Delta(E)>0$). The same thing happens over number fields, at every real place. So in your example ...

28

Yes. Once you know $E$ is modular, a dominant map $\varphi: X_0(N) \rightarrow E$ can be computed effectively. That's because one can effectively compute (a bound on) $\deg\varphi$.
Of course you need to compute a model for $X_0(N)$ for the question to make sense, but we know how to do that and to write $q$-expansions for the coordinates.
By integrating ...

answered Mar 2 '14 at 2:38

Noam D. Elkies

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26

In general, for any integer $N$ and any fixed elliptic curve $E$, the elliptic curves $E'$ for which $E[N]\cong E'[N]$ as Galois modules (and such that the isomorphism respects the Weil pairing) are parametrised by a twist $Y_E(N)$ of the modular curve $Y(N)$.
Tom Fisher has worked out equations of these twists for $N=7,9$ and 11. He explicitly writes down ...

25

The argument that allows you to show that an elliptic curves defined as you say is a group scheme, and even a commutative one is the construction of a functorial and natural
isomorphism $E(T) \rightarrow Pic_{E/S}^0(T)$ for every $S$-scheme $T$. This allows
to see the functor $T \mapsto E(T)$ as a funtor in group, and since this functor is representable by ...

24

I actually only wrote the part that says that this curve is a model for $X_1(11)$, not the first part, which I think was written by John Cremona.
It is standard to order elliptic curves by conductor (e.g. for statistics), and 11 is the smallest possible conductor. However, there are 3 curves with conductor 11, and no canonical way to order them as far as I ...

23

Let me give an analytic argument, to complement Noam's algebraic one. Suppose $E\to \mathbb{C}^*$ is an elliptic curve. Then pulling back $E$ along the universal covering map (also known as the exponential map) $\mathbb{C}\to \mathbb{C}^*$, one obtains an elliptic curve $\tilde E\to \mathbb{C}$. Choosing a basis for the first homology of $\tilde E$ ...

23

Well, although there is a typo (Wiles forgot to close his parenthesis, and wrote $H^2(G,\mu_{p^r}\to H^2(G,\mu_{p^s})$ in his proof), his claim is correct. Let, as ibid. $F$ be the finite extension of $\mathbb{Q}_p$ fixed by $G$, so that your arrow can be written $H^2(F,\mu_{p^r})\to H^2(F,\mu_{p^s})$. Consider the Kummer sequence (for any $n\geq 1$)
$$
1\...

answered Jan 18 '20 at 23:08

Filippo Alberto Edoardo

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22

In fact, this cannot happen: an elliptic curve over $\mathbb{Q}_p$ is supersingular if and only if its associated mod $p$ Galois representation is irreducible, but if it is irreducible as a representation of $\mathbb{F}_p[G_{\mathbb{Q}_p}]$ then it is certainly irreducible as a representation of $\mathbb{F}_p[G_{\mathbb{Q}}]$ and thus it has no $p$-isogenies....

answered Aug 28 '13 at 15:31

David Loeffler

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22

The conjecture is true. To prove it, we can restrict to $k=1$ as explained in the comments. Let $\chi$ denote a cubic Dirichlet character modulo $p$; this exists as $p\equiv 1\pmod{3}$, and it is unique up to complex conjugation. Let $\psi$ denote the (unique) quadratic Dirichlet character modulo $p$.
The number of points of the affine elliptic curve $E$ ...

22

As many people have said, the reason that people believe that factoring and discrete log in $\mathbb F_p^*$ are hard, and that discrete log in $E(\mathbb F_p)$ is even harder, is because a lot of smart people have worked on these problems, and we know the best algorithms they've come up with. ECDLP is delicate, there are certain elliptic curves on which ...

answered Nov 22 '19 at 23:16

Joe Silverman

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21

The short answer to your specific question is no, the resolution of FLT via modularity of elliptic curves does not seem to be helpful in dealing with rational points on higher dimensional varieties. The first two equations you list, $a^5+b^5=c^5+d^5$ and $a^6+b^6=c^6+d^6$, are surfaces of general type in $\mathbb{P}^3$, so conjectures of Bombieri, Lang, ...

answered Nov 30 '13 at 23:13

Joe Silverman

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21

The correct setting for this construction turns out to be projective varieties, so let me suppose we have a smooth variety $X$ inside $\mathbf{P}^N$, for some $N \ge 1$, defined by the vanishing of some homogenous polynomials $F_1, \dots, F_r$ in variables $x_0, \dots, x_N$, with the $F_i$ having coefficients in $\mathbf{Q}$. Actually, let me assume the $F_i$...

answered Jan 22 '17 at 10:18

David Loeffler

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20

Question 1: Yes, there is a nontrivial section, namely
$s: (x,y) = (-1/36, y_0)$ where $y_0$ is either solution of $y^2-y/36=-1/36^3$
(i.e. $y = 1/72 \pm 1/18^{3/2}$).
[Note that the numerator $36$ in the equation for $E_1$
is a typo should be $36x$.]
Using the theory of elliptic surfaces we can show that in fact
the group of section is infinite cyclic ...

answered Nov 17 '14 at 5:05

Noam D. Elkies

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20

No, the conjecture is still wide open for rank $r\geq 2$.
The closest thing to progress is the work of Bhargava and Shankar that quantifies the rank $0$ case and shows that BSD holds for a positive proportion of elliptic curves. You can find it here:
Manjul Bhargava & Arul Shankar, "Ternary cubic forms having bounded invariants, and the existence of a ...

20

This isn't really a full answer, but it's too long for a comment, and perhaps it's informative all the same.
Your sum $S_k[\mathcal{O}]$ can be written as the value at $s = k$ of the sum
$$\sum_{0 \ne \lambda \in \mathcal{O}} \frac{\lambda^k}{Nm(\lambda)^s} = \sum_{n \ge 1} a^{(k)}_n n^{-s},$$
where $a^{(k)}_n := \sum_{N(\lambda) = n} \lambda^k$.
Now, I ...

answered Sep 27 '20 at 13:26

David Loeffler

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19

More interesting is the question of which fields admit what sorts of elliptic curves having everywhere good reduction. For example, how about quadratic fields? Here are a couple of results.
Theorem (Rohrlich [1]) Let $K$ be an imaginary quadratic field and $j$ the invariant of some fixed isomorphism class of elliptic curves with complex multiplication by ...

answered Sep 24 '15 at 2:02

Joe Silverman

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19

The fatal error in the reasoning in the paper "L-functions and rational points on a CM elliptic curve via the classical number theory" at the top of the list of papers at
http://kazuma-morita.jimdo.com
occurs near the top of page 3 (when "proving" a statement near the bottom of page 2 that we'll see is false); it comes down to an elementary but unfixable ...

19

The structure of $E(K)$ for $K$ a complete local field, say a finite extension of $\mathbb Q_p$ or $\mathbb C_p$, is quite standard. Let $E_0(K)$ denote the set of points with good reduction. Then there are exact sequences
$$ 0\to E_0(K)\to E(K) \to \Phi \to 0 $$
and
$$ 0 \to E_1(K) \to E_0(K) \to \tilde E^{\text{ns}}_p(\mathbb F) \to 0 .$$
Here $\Phi$ is a ...

answered Sep 6 '17 at 13:21

Joe Silverman

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19

The following reference seems to answer the question of which Fermat curves admit a non-constant map to an elliptic curve:
Neal Koblitz, David Rohrlich, Simple factors in the Jacobian of a Fermat curve, Canadian Journal of Mathematics 30 No. 6 (1978) pp. 1183–1205, doi:10.4153/CJM-1978-099-6 (free pdf).
As a particular case of Theorem 2 there, if $N \...

answered Dec 20 '18 at 11:01

François Brunault

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19

Milne's and Knapp's books are excellent. As for my AEC, it covers lots of stuff that's not needed if you just want to get to Mordell-Weil. So for AEC, here's a path to MW:
If you know some algebraic geometry, skip chapters I and II, refer back as needed.
Read Ch. III through Sec. 7
Reach Ch. IV through Sec. 7
Skip chapter V and VI.
Read Ch. VII through ...

answered Jun 19 '19 at 19:09

Joe Silverman

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18

There is no such curve.
One way to see this is via the action of
${\rm Gal}\bigl(\overline{{\mathbb C}(t)}\big/{\mathbb C}(t)\bigr)$
on the group $E[p]$ of $p$-torsion points of a putative elliptic curve
$E / {\mathbb C}(t)$ that has good reduction at all $t \neq 0, \infty$.
The image of Galois would be abelian, because the coordinates of $E[p]$
would ...

answered Dec 12 '14 at 4:38

Noam D. Elkies

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18

Benedict Gross recently gave a series of lectures here at the University of Virginia on things related to the Birch and Swinnerton-Dyer Conjecture. One of the recent notable developments he mentioned is the work of Yun and Zhang. It is about the function field analogue but they obtain information about the full Taylor series of the $L$-function. The paper ...

answered May 26 '17 at 13:33

Abdelmalek Abdesselam

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18

Roughly speaking, the height of an arithmetic object (number, variety, ...) is a natural measure of its complexity, say in the sense of "how many bits does it take to describe the object." (This is not meant to be rigorous, but you seem to want to know why people use "heights".) One can then ask for heights (complexity measures) that have nice properties, ...

answered Jun 9 '19 at 23:17

Joe Silverman

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17

I felt that there couldn’t be such an isogeny, but when I saw @DavidLoeffler’s answer, I realized that I had an argument, too. The formal group of the elliptic curve would be of height two, defined over $\mathbb Z_p$, but such things don’t have $p$-isogenies: the quotient formal group is definable only over a suitably ramified extension.

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