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I consider the problem: ask the rational roots of $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$ I try to use the theory of ellipic curves (If we have two rational roots, we could have the third collinear one) and calculte the genus of $\Sigma_{p}:=\{[a,b,c]: p(a,b,c)=0\}$, where $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c).(*)$$

How to get the genus? I know I should use:

The Riemann-Hurwitz theorem: $$2g(\Sigma)-2=B_{p}(f)-2\deg(f),$$ where $g(\Sigma)$ is the genus of Riemann surface $\Sigma$ and $B_{p}(f)$ branch numbers of $f$ at $p$.

I try to bring $(*)$ into the form $y^2=x(x-1)(x-\lambda), \lambda \ne0,1,$ since I know the genus is one.

Maybe help? $p(a,b,c)=a^3+b^3+c^3-3a^2b-3ab^2-3a^2c-3ac^2-3cb^2-3bc^2.$

  • Do you consider $a,b,c$ as variables? Then you have 4 variables (with $y$) and only one equation, so it is a 3-fold or a surface (depending on whether you consider this as an equation in an affine or projective space), but definitely not a curve. By the way, the genus of any ELLIPTIC curve is 1 by definition, what you probably mean is a HYPERELLIPTIC curve. – Sasha May 13 '18 at 16:40
  • @Sasha oops! I am sorry for making you confused. I have edited the confused. I mean I want to calculate the genus of the surface. And I consider it in a projection space. –  May 13 '18 at 16:45
  • https://en.wikipedia.org/wiki/Genus%E2%80%93degree_formula – Elle Najt May 13 '18 at 17:15
  • @AreaMan How to prove it is irreducible? –  May 14 '18 at 01:00
  • @Tinzoe-Yui maybe show it is smooth (with Jacobian criterion) and connected? (I haven't worked this out.) Or direct computation trying to show that your polynomial can't factor? (I'd try that first..., Though you will eventually need to check it is smooth anyway.) – Elle Najt May 14 '18 at 02:55
  • @AreaMan May I consider the projective map $f: \Sigma \to a$? And then consider the branch numbers and degree of $f$. –  May 14 '18 at 03:07
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    @Tinzoe-Yui I think you need to show the curve is smooth and irreducible to use Riemann Hurwitz? But once you know that you can use degree genus. (You can prove degree genus formula with RH, if I'm remembering right.) – Elle Najt May 14 '18 at 03:22
  • Probably a duplicate, but it's fine 'cause the answer is amazing – Akiva Weinberger May 25 '18 at 06:39

3 Answers3

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There is one idea. To search for the solution of the equation.

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=q$$

If we know any solution $(a,b,c)$ of this equation. Then it is possible to find another $(a_2, b_2, c_2)$. Make such a change.

$$y=(a+b+2c)(q(a+b)-c)-(a+b)^2-(a+c)(b+c)$$

$$z=(a+2b+c)(2b-qa-(q-1)c)+(b+2a+c)(2a-qb-(q-1)c)$$

Then the following solution can be found by the formula.

$$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$

$$+(3c+(q-1)(a-b))yz^2-az^3$$

$$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$

$$+(3c+(q-1)(b-a))yz^2-bz^3$$

$$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$

$$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$

individ
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This is more of a comment as opposed to an answer

Just go here and it will take you to my answer on a related post; it will give you the solutions, viz.,

$$a= 154476802108746166441951315019919837485664325669565431700026634898253202035277999$$ $$b= 36875131794129999827197811565225474825492979968971970996283137471637224634055579$$ $$c= 4373612677928697257861252602371390152816537558161613618621437993378423467772036.$$

Knowing that these are the solutions, you can apply @individ 's answer.

Also, this post is a possible duplicate.

Mr Pie
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This is a special case of MO question "Estimating the size of solutions of a diophantine equation". Using the results there, the elliptic curve $\,E: y^2 = x^3 + 109 x^2 + 224 x\,$ corresponds to your $\,\Sigma_p.\,$ The curve has $j$-invariant $1408317602329/2153060$ and $\,\lambda = (845 + 109 \sqrt{65})/1690\,$ gives the Legendre form. The curve $\,E\,$ has a torsion point $\,(56,728)\,$ of order $6$ and a point $\,(-100,260)\,$ of infinite order.

Somos
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