Solution to a Diophantine equation

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.

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?