How to find a transformation from the integer points of a curve in 3 variables to the rational points in an elliptic curve?

Question

Here in one of the steps to find positive integers $a,b,c$ so that $\frac a{b+c}+\frac b{a+c}+\frac c{a+b}=4$, Alon Amit has

$$a^3+b^3+c^3−3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)−5abc=0$$

Then he applies the transformation

$$x = \frac{-28(a+b+2c)}{6a+6b-c}$$

$$y = \frac{364(a-b)}{6a+6b-c}$$

Which implies $y^2=x^3+109x^2+224x$.

I am baffled. Of course reducing to two variables is easy: just divide through by $c^3$. But getting an elliptic curve out the other side is another matter entirely. It's more useful to look at the inverse transforms:

$$a=\frac{56-x+y}{56-14x}$$

$$b=\frac{56-x-y}{56-14x}$$

$$\begin{align*} c&=\frac{-28-6x}{28-7x}\\ &=\frac{-56-12x}{56-14x} \end{align*}$$

Even if we assume that this transformation exists and all three are the ratios of first degree polynomials in $x,y$ and the denominator is the same for all three, we get:

$$a=\frac{p_1+q_1x+r_1y}{k_1+k_2x+k_3y}$$

$$b=\frac{p_2+q_2x+r_2y}{k_1+k_2x+k_3y}$$

$$c=\frac{p_3+q_3x+r_3y}{k_1+k_2x+k_3y}$$

Twelve variables! We could further assume that $k_3=0$ but that one feels less justifiable than the others. Then

$$(p_1+q_1x+r_1y)^3+(p_2+q_2x+r_2y)^3+(p_3+q_3x+r_3y)^3-3{\huge(}(p_1+q_1x+r_1y)^2(p_2+q_2x+r_2y)+(p_1+q_1x+r_1y)(p_2+q_2x+r_2y)^2+(p_1+q_1x+r_1y)^2(p_3+q_3x+r_3y)+(p_1+q_1x+r_1y)(p_3+q_3x+r_3y)^2+(p_2+q_2x+r_2y)^2(p_3+q_3x+r_3y)+(p_2+q_2x+r_2y)(p_3+q_3x+r_3y)^2{\huge)}-5(p_1+q_1x+r_1y)(p_2+q_2x+r_2y)(p_3+q_3x+r_3y)=0$$

And then the LHS has to be of the form $-y^2+x^3+\alpha x^2+\beta x + \gamma$ so we have 10 coefficients, 3 of which can be whatever and the other 7 are required to be -1 for $y^2$, 0 for $y^3$, 1 for $x^3$ and 0 for terms in both $x$ and $y$. Which looks intractable, to say the least.

Answer 1

There are methods to find Weierstrass models of elliptic curves, which are described in a couple of references that I give below. A computer can do this, for example Magma can do this with the following code:

P:=ProjectiveSpace(Rationals(),2);

f:=a^3+b^3+c^3-3*(a^2*b+a*b^2+a^2*c+a*c^2+b^2*c+b*c^2)-5*a* b* c;

C:=Curve(P,f);

P:=C![1,-1,0];

E,map:=EllipticCurve(C,P);

E; map;

MinimalModel(E);

WeierstrassModel(E);

If you are interested in the actual method, it is based on algebraic geometry and the theory of divisors. There is a brief summary in Silverman, Tate, "Rational Points on Elliptic Curves" (Section 1.3), and a much longer and detailed description in my book "Number Theory and Geometry" in Section 15.3.

Answer 2

If you don't have Magma or the books, Google $\mathbf{Matsuura Weierstrass}$ and you get a lovely simple report by Matsuura giving explicit instructions.