Given an equation $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$, how to solve this problem in positive integers?

I've tried to assume $a\le b\le c$ and that $b=a+k_1, c=a+k_2$. So the equation become

$$\frac{a}{2a+k_1+k_2} + \frac{a+k_1}{2a+k_2} + \frac{a+k_2}{2a+k_1} = 4$$

or equivalently,

$$\frac{1}{2+\frac{k_1}{a}+\frac{k_2}{a}} + \frac{1+\frac{k_1}{a}}{2+\frac{k_2}{a}} + \frac{1+\frac{k_2}{a}}{2+\frac{k_1}{a}} = 4$$

Now let $x= \frac{k_1}{a}, y= \frac{k_2}{a}$, it is sufficient to find all positive rational solutions of $\frac{1}{2+x+y} + \frac{1+x}{2+y} + \frac{1+y}{2+x} = 4$.

  • $\begingroup$ What have you attempted thus far? $\endgroup$ – Mr Pie Feb 18 '18 at 10:02
  • $\begingroup$ Please give your inputs on this problem. $\endgroup$ – Agile_Eagle Feb 18 '18 at 10:23

I think we have to use a computer because the solutions are: $a=\color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$ $b=\color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$ $c=\color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$

This is not a joke. The three gargantuan numbers are the least positive integer that satisfies the equation.

Here you can find the demonstration about the number of digit that the solution should have in the general case of $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = N$$ This MathOverflow link as a full discussion about $N=4$.

  • $\begingroup$ If memory serves, there's a mathoverflow question answering, or trying to answer this question too. $\endgroup$ – Mike Miller Feb 18 '18 at 10:21
  • $\begingroup$ @MikeMiller I found it. Thank you for the suggestion. $\endgroup$ – user507623 Feb 18 '18 at 10:22
  • $\begingroup$ Could you give me a link please? @Pippo $\endgroup$ – Victor Chen Feb 18 '18 at 10:23

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