Solution to a Diophantine equation Find all the non-trivial integer solutions to the equation
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$
 A: FWIW, using Michael Stoll' answer to Estimating the size of solutions of a diophantine equation, we use Magma to do:
F := EllipticCurve(x^3 + 109* x^2 + 224 * x);
IntegralPoints(F);
[ (-100 : -260 : 1), (-56 : 392 : 1), (-9 : 78 : 1), (-4 : -28 : 1), (0 : 0 : 1), (4 : -52 : 1), (56 : 
-728 : 1) ]
[ <(-100 : -260 : 1), 1>, <(-56 : 392 : 1), 1>, <(-9 : 78 : 1), 1>,   <(-4 : -28 : 1), 1>, <(0 : 0 : 1), 
1>, <(4 : -52 : 1), 1>, <(56 : -728 : 1), 1> ]

And for more edification:
MordellWeilShaInformation(F);
Torsion Subgroup = Z/6
   Analytic rank = 1
     ==> Rank(E) = 1
   The 2-Selmer group has rank 2
    New point of infinite order (x = -56/25)
   After 2-descent: 
        1 <= Rank(E) <= 1
        Sha(E)[2] is trivial
    (Searched up to height 100 on the 2-coverings.)
[ 1, 1 ]
[ (-100 : 260 : 1) ]
[
    <2, [ 0, 0 ]>
]

Make of this what you will.
A: 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?
A: Below mentioned equation has numerical solution: 
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$
Known solution is $(a,b,c)=(11,9,-5)$
Another solution is $(a,b,c)=(11,4,-1)$

