Find all the nontrivial integer solutions to the equation $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$$

5$\begingroup$ Possible duplicate of Estimating the size of solutions of a diophantine equation $\endgroup$ – Sam Hopkins Mar 16 '17 at 2:33

1$\begingroup$ No, I want integer solutions. At least $(a,b,c)=(11,9,5)$ is an solution. $\endgroup$ – var Mar 16 '17 at 2:36

1$\begingroup$ Perhaps the equivalent form is helpful? $a^3 + b^3 + c^3 + abc = 3(a+b)(a+c)(b+c)$ Gerhard "Don't Know What Is Next" Paseman, 2017.03.15. $\endgroup$ – Gerhard Paseman Mar 16 '17 at 3:27

2$\begingroup$ OK, so it's not an exact duplicate, but have you tried applying the answer given there to see how much of the analysis goes through? $\endgroup$ – Gerry Myerson Mar 16 '17 at 3:42

$\begingroup$ It is a related subject, but I think there are some differences. $\endgroup$ – var Mar 16 '17 at 13:05
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 2Selmer group has rank 2 New point of infinite order (x = 56/25) After 2descent: 1 <= Rank(E) <= 1 Sha(E)[2] is trivial (Searched up to height 100 on the 2coverings.)
[ 1, 1 ]
[ (100 : 260 : 1) ]
[
<2, [ 0, 0 ]>
]
Make of this what you will.

$\begingroup$ Thank you very much. Actually in the former comments, some people mentioned this paper. Thank him at the same time. $\endgroup$ – var Mar 17 '17 at 20:14
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.
$$y=(a+b+2c)(q(a+b)c)(a+b)^2(a+c)(b+c)$$
$$z=(a+2b+c)(2bqa(q1)c)+(b+2a+c)(2aqb(q1)c)$$
Then the following solution can be found by the formula.
$$a_2=((54q)c(q2)(3b+a))y^3+((54q)b+4(1q)c)zy^2+$$
$$+(3c+(q1)(ab))yz^2az^3$$
$$b_2=((54q)c(q2)(3a+b))y^3+((54q)a+4(1q)c)zy^2+$$
$$+(3c+(q1)(ba))yz^2bz^3$$
$$c_2=2(q2)cy^3+3(2q)(a+b)zy^2+$$
$$+((54q)(a+b)+2(1q)c)yz^2+(2c(q1)(a+b))z^3$$
I tried this formula to simplify, but nothing happens. Maybe someone will check in Maple?
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 $(a,b,c)=(11,4,1)$

1$\begingroup$ Can you add any explanation of how you found the additional solution? Would it be possible for you to say whether there are more solutions (and maybe how to find them?), or whether you think that there are no other solutions? $\endgroup$ – Zach Teitler Jul 6 '17 at 3:00

$\begingroup$ I think the above comment which mentions the answer of Michael Stoll is a complete answer to this question. $\endgroup$ – var Jul 7 '17 at 9:15