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

There is an action of $\mu_5$, the group of fifth roots of unity, on your curve,
given by $\zeta \cdot (x,y) = (\zeta x, \zeta^{-1} y)$. The quotient by this
group action is the hyperelliptic curve
$$C \colon Y^2 = X^5 + \frac{49}{4},$$
the map being given by $(X, Y) = (-xy, x^5 - \frac{7}{2})$. So it is enough
to find all the rational points on $C$. $C$ is ...

25

Well if you want to count rational points on varieties than you probably want to know what abelian varieties are, and general type varieties, and Fano varieties, and K3 surfaces, and what Azumaya algebras are, and so on to understand the main conjectures and theorems of the subject. You should probably understand how spreading out varieties into schemes over ...

24

As Terry mentions in the comments, the reason for the $\sqrt{5}$ is that the limiting case, the golden ratio, forces it. There is a very neat explanation of all of this in the classic number theory book by Hardy and Wright, pages 209 to 212. I give a brief sketch of the ideas.
Why $\phi$ is the worst case.
As Hardy and Wright put it, "from the point of ...

22

It has been conjectured by Euler that this equation has no solutions in positive integers when $n\geq 4$.
When $n=4$, this was disproved by Elkies in the paper [Elkies, On A4+B4+C4=D4] in a very strong way: he proves that the rational points of this K3 surface are dense in the real points for the euclidean topology.
When $n\geq 5$ is odd, your surface ...

answered Dec 14 '15 at 18:48

Olivier Benoist

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19

The answer to the third question is no. This is a rather counter-intuitive discovery of Micha Perles from the sixties. See this paper of Ziegler, for a simpler construction and other pertinent information. However, for polytopes in dimension $3$, the answer is yes, as mentioned in the same paper of Ziegler.

19

Yes. It is not hard to find an example: Take
$$E \colon y^2 = x^3 - 12 x - 1\,.$$
Then $E(\mathbb Q) \cong \mathbb Z$ and $P = (5, 8)$ is a generator (according
to Magma). Since $P$ is on the component of the identity, all rational points are on that component.

19

In general one can say very little. There are some positive results (as indicated in the comments) in special cases, but the below example kills any hope that one can say something in general. NB "they have the same $\ell$-adic Galois representations" is vague -- I will interpret as "they have isomorphic $\ell$-adic etale cohomology in all degrees" which is ...

answered Jan 18 '17 at 8:20

Kevin Buzzard

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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

Here are a few facts about this problem, quoting mostly from Local statistics of lattice points on the sphere by Jean Bourgain, Peter Sarnak, Zeév Rudnick:
''A celebrated result of Legendre/Gauss asserts that $n$ is a sum of three squares if and only if $n\ne 4^a(8k+7)$. Let
$$ N(n) = \bigl\{ (x,y,z)\in\mathbb Z^3 : x^2+y^2+z^2=n \bigr\}. $$
The behaviour ...

answered Jan 30 '18 at 21:50

Joe Silverman

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18

By now there is a fairly rich literature on computing the set of rational points on curves of higher genus, see for example my survey paper on "Rational points on curves".
What one can do for your concrete curve is this. First, one can simplify the curve equation (even more than Noam did); your curve is isomorphic to
$$y^2 = x^6 - 12 x^5 - 6 x^4 + 82 x^3 + ...

17

A set of points that generate $E(\mathbb{Q})$ modulo torsion is given by
(1955516573881233507049678279 : -86467145649172260650105545143411861089140 : 1),
(49225691888888099223656060329/10201 : 67749663895993353685065159554645568700902610/1030301 : 1),
(61339810590192565389735634 : -440289331793622522908840423931186017125 : 1),
(...

16

By Mordell-Weil, for any number field $K$ we have
$$C(K)=\mathbb{Z}^r \times E(K)_{\mathrm{tors}}$$
As you mention, Mazur showed all the possible options for $E(\mathbb{Q})_{\mathrm{tors}}$ in his famous 1977 paper.
The only other $K$ for which we have a torsion theorem are the quadratic fields. This is the result of a long series of papers by Kamienny, ...

16

The answer to one possible interpretation of the title question -- vary over all elliptic curves over all fields and ask which groups arise -- is given in this paper.
With regard to the structure of Mordell-Weil groups of elliptic curves over local fields, I think you will find that $\S$5.1 of this joint paper with Allan Lacy relevant. (The title is "...

answered Sep 6 '17 at 18:57

Pete L. Clark

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16

It turns out that $C(K) = C(\mathbb Q) = \{\infty_+, \infty_-, (0,1), (0,-1), (1,1), (1,-1)\}$.
To see this, consider a point $P \in C(K)$ and write $\bar{P}$ for its image under the nontrivial automorphism of $K$. Then $P + \bar{P}$ is an effective divisor of degree 2 on $C$, which is defined over $\mathbb Q$ (this is what Xarles is alluding to in his ...

15

(Collecting comments into a community wiki answer.)
There is a standard method for transforming a smooth cubic into Weierstrass form. See for example Section 1.3 or Appendix B of Silverman and Tate's book Rational Points on Elliptic Curves. It is also implemented in Sage as $\mathtt{EllipticCurve\_from\_cubic}$.
For $n=6$ your curve is isomorphic to $Y^2 + ...

14

First, may I change your notation a bit? Usually one uses $H(p/q)=\max\{|p|,|q|\}$ for the (multiplicative) height of a rational number, and $h(p/q)=\log H(p/q)$ is the logarithmic height. So I'll use that notation.
One natural way to study the distribution of the infinitely many rational points on a curve is to use a Dirichlet series. So if your curve is $...

answered Apr 25 '15 at 2:37

Joe Silverman

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14

The number of rational solutions to your equation is finite. In short: your equation defines a genus $3$ curve, as follows from a straightforward computation and an application of Riemann--Hurwitz; finally, by Faltings' theorem, the number of rational points on a curve of genus $>1$ is finite.
One shows this as follows. Your equation defines a smooth ...

14

In his 1825 paper, Lejeune Dirichlet proved that the equation $x^5 + y^5 = cz^5$ has no nontrivial solution with x and y coprime integers and z integer for a rather large class of integers c. His paper can be read on the site Gallica :
http://gallica.bnf.fr/ark:/12148/bpt6k5842885h/f2.item.zoom
but the visualization is very bad.

14

Specialize $a,b$ to functions giving the universal elliptic curve over the modular curve $X_0(N)$. These are known to have rank zero over the function field of the modular curve with coefficients over $\mathbb{C}$ even. They can have torsion but, by varying $N$, you can show that the torsion is trivial too.
T. Shioda, On elliptic modular surfaces, J. Math. ...

14

The conjecture is false, and in fact there exists a positive integer solution for
$$\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=5,$$
though I was unable to find it explicitly. I will explain how to prove it, though I remark that my method is no different that the one of Bremner and MacLeod.
Take the following SageMath code:
n = 5
F = a*(2*a+3*b)*(2*c+3*...

13

See e.g. http://www.ics.uci.edu/~eppstein/junkyard/integer-distances.html for a proof (originally due to Erdos) that there is no infinite non-collinear integer-distance set in the plane.

13

Taking the equation in Joe Silverman's comment as the defining equation and asking Magma:
> A := AffineSpace(Rationals(), 2);
> C := Curve(A, q^2*p^4 + (-4*q^3+4*q)*p^3 - 2*q^2*p^2 + (4*q^3-4*q)*p
+ 4*q^4-7*q^2+4);
Genus(C);
3
> IsHyperelliptic(C);
true Hyperelliptic Curve defined by y^2 = x^8 + 4*x^7 + 6*x^6 + 4*x^5 + x^4 +
...

13

Problems of this type are considered by Serre in the paper:
Serre - Spécialisation des éléments de $\mathrm{Br}_2(\mathbb{Q}(T_1,\ldots, T_n))$
The case relevant to you is Exemple 4. Here Serre shows that
\begin{align*}
&\#\{|A|,|B|,|C|,|D|,|E|,|F| \leq N : \\
& Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \text{ has a rational point} \} \ll \frac{N^6}{(\log N)^{1/2}}....

answered Dec 23 '15 at 19:27

Daniel Loughran

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13

The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian
of $X_0(35)$) has rank 0 (as shown for example by a 2-descent
computation in Magma); it is isomorphic to
${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$,
with generators the difference of the two points at infinity
on $X_0(35)$ and the 2-torsion point corresponding to the
...

12

So, as Igor points out, there will be a curve with $k+1$ lattice points $(k^2-t^2,2kt)$ running from $(k^2,0)$ down to $(0,2k^2).$ Call this the $k$-curve or better the $(k,m)=(k,1)$ curve as there are also $k+1$ lattice points $(m(k^2-t^2),2ktm).$ Of course it only makes sense to consider $m$ square-free.
Will correctly identifies the red curve as pretty ...

12

I assume in the question that $C = E$ is an elliptic curve.
First your claim that $E(\mathbb{R}) = U(1)$ is false; I mean $E(\mathbb{R})$ can be disconnected. The correct result is that $E(\mathbb{R})$ has at most two connected components, and the connected component of the identity is isomorphic to $U(1)$ as a topological group.
As for points over other ...

answered Sep 6 '17 at 11:04

Daniel Loughran

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12

A conjecture of Coleman asserts that only finitely many rings arise as the endomorphism ring of an abelian variety of given dimension defined over a number field of given degree. See [1] for an account of this conjecture. In your case, the relevant conjecture is denoted there by $C(1,2)$. To my knowledege, the only results on Coleman's conjecture in ...

answered Dec 13 '18 at 15:10

François Brunault

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11

We (with Keith Merrill) recently wrote a paper further quantifying the density of rational points on $S^n$, see http://arxiv.org/abs/1301.0989. Hope it helps!

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