# Find all positive integral soludions of $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$ [duplicate]

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

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

I think we have to use a computer because the solutions are: $a=\color{red}{154476802108746166441951315019919837485664325669565431700026634898253202035277999}$ $b=\color{orange}{36875131794129999827197811565225474825492979968971970996283137471637224634055579}$ $c=\color{green}{4373612677928697257861252602371390152816537558161613618621437993378423467772036}$
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 $N=4$.