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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 $\frac{1}{2+x+y} + \frac{1+x}{2+y} + \frac{1+y}{2+x} = 4$.

Victor Chen
  • 796

1 Answers1

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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 $N=4$.