TITLE: Solution to a Diophantine equation
QUESTION [7 upvotes]: Find all the non-trivial integer solutions to the equation
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$

REPLY [2 votes]: 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.
$$a_2=((5-4q)c-(q-2)(3b+a))y^3+((5-4q)b+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(a-b))yz^2-az^3$$
$$b_2=((5-4q)c-(q-2)(3a+b))y^3+((5-4q)a+4(1-q)c)zy^2+$$
$$+(3c+(q-1)(b-a))yz^2-bz^3$$
$$c_2=2(q-2)cy^3+3(2-q)(a+b)zy^2+$$
$$+((5-4q)(a+b)+2(1-q)c)yz^2+(2c-(q-1)(a+b))z^3$$
I tried this formula to simplify, but nothing happens. Maybe someone will check in Maple?