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I am trying to solve the following Diophantine equation with Mathematica:

$\frac{x}{y+z}+\frac{z}{x+y}+\frac{y}{x+z}=4$

It is known that there are three positive numbers that satisfy the equation above, which are on the order of $10^{81}$. Which is a big number, but not that big for Mathematica, I think.

However, neither Solve nor FindInstance can give me the solutions. The best I could do was:

    Solve[x/(y + z) + y/(x + z) + z/(x + y) == 4 && z > y > x > 1001, {x, y, z}, Integers]

which gives me a ConditionalExpression I can't work with.

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    $\begingroup$ 10^81 is still an enormous search space, and you did not even provide that bound to Solve. I would not expect a solution in a practical amount of time. $\endgroup$ – Mr.Wizard Jun 4 '17 at 13:08
  • $\begingroup$ Then if not with the above functions, how else can I solve this function with Mathematica? $\endgroup$ – Peter Dam Jun 4 '17 at 14:23
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    $\begingroup$ I think you will need to find a more intelligent (i.e. user-involved, less brute force) approach. The paper you link seems like a good place to start. Remember that Mathematica is just a tool; it cannot take the place of a mathematician. $\endgroup$ – Mr.Wizard Jun 5 '17 at 12:28
  • $\begingroup$ That seems quite reasonable. Still, Mathematica can sometimes do astounding thingy, so I thought it is worth a try. $\endgroup$ – Peter Dam Jun 6 '17 at 9:12
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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?

Use the given formula and first find the other solutions.

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