Amateur here. I'm trying to understand how the smallest positive integer solutions (which are eighty digit numbers) for$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$$
are calculated. This equation can be expanded out to the elliptic curve$$a^{3}+b^{3}+c^{3}-3\left(a^{2}b+a^{2}c+ab^{2}+ac^{2}+b^{2}c+bc^{2}\right)-5abc=0,$$
which can be transformed to the Weierstrass form$$y^{2}=x^{3}+109x^{2}+224x.$$
This problem for any non-zero integer $N$ (not just $N=4$) is presented in the paper An unusual cubic representation problem by Bremner and Macleod (B&M) here. Pitched more at my level, there is also a nice, less technical discussion for $N=4$ given by Alon Amit in his Quora answer here.
On page 32, B&M give a generator for the above Weierstrass form as $G=\left(x,y\right)=\left(-4,28\right)$,which corresponds to $\left(a,b,c\right)=\left(11,4,-1\right)$. They don't explain how they choose this $G$. They then go to say:
“The point $9G$ is the smallest multiple of $G$ that corresponds to a positive solution (in which $a,b,c\sim10^{80}$).”
Now, I can understand most of this problem. I've derived the Weierstrass form from the above cubic. I've derived the equations linking $a,b,c$ and $x,y$ (page 30, B&M). Using the chord and tangent technique, I've calculated the first few values of $nG$ and, when that got too complicated, used SageMath to compute $9G$ and hence the desired positive $a,b,c$ solution. So far, so good.
What I can't see is why 9G is the smallest possible solution.
So my question is, how to show that there isn't another generator that gives a smaller positive solution?
In general terms, Amit answers my question when he writes:
“As we continue to add the point $P$ to itself, the denominators just keep growing. It is not quite easy to prove that, as there's always a possibility of some cancellation, but the theory of heights on an elliptic curve allows us to show that those astronomical numbers are, indeed, the simplest solutions to the question.”
I've looked up heights online, found that $h\left(nP\right)$ is roughly $n^{2}$ times a constant, but cannot relate that to why $G=\left(x,y\right)=\left(-4,28\right)$ is the smallest generator to give the desired solution.
Here is how i would try to work to get a solution, assisted by sage, then also humanly cover the details. The decision to post an answer came to honor such a good question in such a nice arithmetic framework.
We may and do equivalently assume that $c=1$ (homogenity) and search for positive rational solutions $a,b$ of the given equation under this norming condition for $c$. From it, a linear change of (projective) variables makes an elliptic curve enter the scene: $$ \begin{aligned} x &= -28(a+b+c)\ , & a &= -x+y+56z\ ,\\ y &= 364(a-b)\ , & b &= -x-y+56z\ ,\\ z &= 6a+6b-c\ , & c &= -12x-56z\ , \end{aligned} $$ and we may use the following sage code to eliminate:
The last elimination gives:
So we are eliminating $a,b,c$ (with $c=1$ or $c\ne 0$) from the given homogeneous equation (in numerator) polynomial $$ \tag{$1$} 4(a^3 + b^3 + c^3) - 3(a+b+c)(a^2+b^2+c^2) - 5abc = 0\ , $$ and obtain an equation in $x,y,z$ (equivalent to the given one). We may norm $z=1$, so the affine the equation in $x,y$ is: $$ \tag{$2$} E\ :\qquad y^2 = x^3 +109x^2+224x\ . $$ The change of variables moves the special point $(a_0:b_0:c_0)=(1:-1:0)$ that satisfies $(1)$ to the infinity point $O=(x_0:y_0:z_0)=(0:1:0)$ of the curve $E$ from $(2)$. The point $O$ is usually taken as neutral element for the group of $\Bbb Q$-rational points $E(\Bbb Q)$ of $E$, with the usual "addition of points". Working with $(1)$ is thus better done with $(2)$, in the sense, that most formulas and algorithms are described for elliptic curves in such Weierstraß form.
So far, OP details were recalled.
Let us go some steps further. Denote by $E(\Bbb Q)/2=E(\Bbb Q)/2E(\Bbb Q)$ the group of rational points modulo double( point)s. Denote by $E(\Bbb Q)[2]$ the group of two-torsion rational points.
There is a structural result on the abelian group $E(\Bbb Q)$, it is finitely generated, it has a free part isomorphic to $\Bbb Z^r$, with a finite rank $r$, and a torsion part $E(\Bbb Q)_{\text{tors}}$, so that $$ (\ E(\Bbb Q)\ , \ + \ ) \cong \Bbb Z^r\oplus E(\Bbb Q)_{\text{tors}}\ . $$ The torsion points of $E$ are computed algorithmically, we have at least six, those from the following list:
and in fact these are all of them, since $\Bbb Z/12$ would be the only other candidate (in view of the torsion conjecture, shown by Barry Mazur), but halving the $6$-torsion point $(56, 728)$ cannot be done over $\Bbb Q$.
For the free part we use a two-descent technique, which goes through the understanding of the finitely generated group $\Bbb E(\Bbb Q)/2$, and this step is more or less algorithmic (modulo problems that may or may not occur in the given special situation). So which structural data is naturally connected to this $2$-descent group $\Bbb E(\Bbb Q)/2$ ?
There exists a morphism $$ E(\Bbb Q) \text{ modulo doubles of points}\overset f\longrightarrow \Bbb Q^\times \text{ modulo squares} $$ sending a generic representative $P=(x,y)$ to $x$ (modulo squares). For the special two-torsion point $T=(0,0)$, which is three times $(56,728)$, the image $0$ is forbidden, but since $f$ is a morphism, we map $T=3(56,728)$ to $56^3$ modulo squares, which is $2\cdot 7=14$ modulo squares.
We would like to compute the image of $f$, thus having information on $E(\Bbb Q)/2E(\Bbb Q)$.
The images of these six torsion points are $14,1,1,1,14,14$ modulo squares. It is known that the image of $f$ is contained in the group of divisors of the discriminant of $E$, taken modulo squares,
so it is inside the group with $2^5$ elements generated by $S=\{-1,2,5,7,13\}$ $$ K(S,2)=\langle -1,2,5,7,13\rangle\subset \Bbb Q^\times\text{ modulo squares .} $$ This morphism $f$ is useful in many situations. Below we have a cousin of $f$ playing the main role.
Related to the $2$-descent group $E(\Bbb Q)/2$, and involving it, there are some (rather technical) steps, as a reference i will use Proposition 4.6, Remark 4.7, Example 4.8, and Proposition 4.9 (Descent via isogeny) form pages 300-305, Chapter X, in Silverman, Arithmetic of elliptic curves, GTM106, 1986. See also Theorem 1.1 page 278. It is not possible to recall here all the ingredients, but roughly we are in the following situation.
This proposition can be applied for the pair $(E,E')$ for curves joined by a two-isogeny, but then also for the dual pair $(E',E)$ joined by the dual isogeny $\hat \phi$.
Then $E(\Bbb Q)/2E(\Bbb Q)$ is obtained from the exact sequence (from Remark 4.7): $$ 0 \longrightarrow \frac{E'(\Bbb Q)[\hat\phi]}{\phi\ E(\Bbb Q)[2]} \longrightarrow \frac{E'(\Bbb Q)}{\phi\ E(\Bbb Q)} \longrightarrow \bbox[yellow]{ \frac{E(\Bbb Q)}{2E(\Bbb Q)} } \longrightarrow \frac{E(\Bbb Q)}{\hat\phi\ E'(\Bbb Q)} \longrightarrow 0\ . $$ These ideas are implemented in sage, one can use a higher verbosity for more details:
And the output has the following parts:
computation of the two-torsion, and of some simple points.
computation of $K(\Bbb Q, 2)$ for $(a,b)$, number of generators and explicit representatives:
conclusion on $E(\Bbb Q)/\phi'E'(\Bbb Q)$, and Selmer and Sha-group for $\phi'$.
computation of $K(\Bbb Q, 2)$ for $(A,B)$, number of generators and explicit representatives:
conclusion on $E'(\Bbb Q)/E(\Bbb Q)$, and Selmer and Sha-group for $\phi'$.
conclusion on $E(\Bbb Q)/2E(\Bbb Q)$, and Selmer and Sha-group for $[2]$ (multiplication by two on $E$).
finalization of the descent, rank decision and explicit generator(s) for the free part.
So far we have an explicit computation of the structure of rational points on $E$. The free part is generated by the point $G=(-4,28)$. Let $T_6$ be the torsion point of order six $T_6=(56, 728)$. Then any $\Bbb Q$-rational point on $E$ is of the shape $$ nG+jT_6\ ,\qquad n\in\Bbb Z\ ,\ j\in\{0,1,2,3,4,5\}\ . $$ We can take every point for small values of $n$ (say $|n|\le 20$) and all $j$, and compute the corresponding point $(a,b,c)$ with relatively primes $a,b,c$ and $c>0$ on the curve $(1)$. The sage code searching for a solution is simple:
Results:
$$ \tiny \begin{aligned} a &= 1054210182683112310528012408530531909717229064191793536540847847817849001214642792626066010344383473173101972948978951703027097154519698536728956323881063669558925110120619283730835864056709609662983759100063333396875182094245046315497525532634764115913236450532733839386139526489824351\\ b &= 1440354387400113353318275132419054375891245413681864837390427511212805748408072838847944629793120889446685643108530381465382074956451566809039119353657601240377236701038904980199109550001860607309184336719930229935342817546146083848277758428344831968440238907935894338978800768226766379\\ c &= 9391500403903773267688655787670246245493629218171544262747638036518222364768797479813561509116827252710188014736501391120827705790025300419608858224262849244058466770043809014864245428958116544162335497194996709759345801074510016208346248254582570123358164225821298549533282498545808644\\[3mm] \end{aligned} $$ $$ \tiny \begin{aligned} a &= 4373612677928697257861252602371390152816537558161613618621437993378423467772036\\ b &= 36875131794129999827197811565225474825492979968971970996283137471637224634055579\\ c &= 154476802108746166441951315019919837485664325669565431700026634898253202035277999\\[3mm] \end{aligned} $$ $$ \tiny \begin{aligned} a &= 16666476865438449865846131095313531540647604679654766832109616387367203990642764342248100534807579493874453954854925352739900051220936419971671875594417036870073291371\\ b &= 32343421153825592353880655285224263330451946573450847101645239147091638517651250940206853612606768544181415355352136077327300271806129063833025389772729796460799697289\\ c &= 184386514670723295219914666691038096275031765336404340516686430257803895506237580602582859039981257570380161221662398153794290821569045182385603418867509209632768359835 \end{aligned} $$ The naive height of such a solution $(a,b,c)$ is roughly a measure of the "length" of the solution. This can be formalized. There is also a naive height on the elliptic curve $E$, and from it we obtain a "true height" (which has some properties of a norm, in particular we can extract also a bilinear pairing from it). We have used all $n$ values up to $20$, and see how big the solutions become.
Playing with the code shows the next few positive solutions $(a,b,c)$, and the point when new solutions come in. It turns out that for $n$ and $-n$ we produce the same solutions, so we restrict to $n\ge 0$. We can comment in the above code the lines with the solutions, and print the first time when a new solution enters the scene:
It remains now to have some estimation (modulo constants, modulo $O(1)$) as usual between the naive heights on the curves $(1)$ and $(2)$, and between the naive and true heights on $E$ to conclude, that the height $123.0148\dots$ leads to the smallest solution. This estimation is technical, but i hope that the growing number of decimals needed to show $a,b,c$ in the numerical height statistics gives a good heuristic argument. I have to stop here, time and the hard job...