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If I had a very shallow question, then I am sorry. $x,y,z\in\mathbb{N}^{+}$ and$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}=4$$find $x,y,z$.

I try with AM-GM, just get$$ \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\geq\frac{3}{2}$$

This means that the equation must have a real solution, but can not be sure there is an integer solution.

Let: $x=ay=abz$, then the equation becomes:$$\frac{ab}{a+b}+\frac{b}{ab+1}+\frac{1}{a^2+ab}=4$$

Which makes the problem become non-homogeneous, and seems to become more difficult. I have no more ideas. Could anyone help me? Thanks a lot.

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  • $\begingroup$ If $y+z=a$ etc. $$2(4+1+1+1)=(a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)$$ $\endgroup$ – lab bhattacharjee Mar 18 '17 at 17:14
  • $\begingroup$ Maybe a track using $\frac{x}{y+z} = \frac{x+y+z}{y+z}-1$, I get the equation $(x+y+z)^2=7(x+y)(y+z)(x+z)$. Then I suppose that $7$ divides $x+y+z$. $\endgroup$ – M. Boyet Mar 18 '17 at 17:19
  • $\begingroup$ I wrote a small brute force program and find no solutions with $x,y,z \le 1000$ $\endgroup$ – Ross Millikan Mar 18 '17 at 17:40
  • $\begingroup$ See the Online Encyclopedia of Integer Sequences oeis.org/A283564, where one comment says the first solution has 81 digits. $\endgroup$ – Michael Mar 19 '17 at 5:10
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    $\begingroup$ This should help: mathoverflow.net/questions/227713/… $\endgroup$ – Mike Miller Mar 25 '17 at 12:43
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There are indeed positive integer solutions. The smallest solution with $x, y, z$ in $\mathbb{N}_+$ is

$\small x = 154476802108746166441951315019919837485664325669565431700026634898253202035277999\\ \small y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579\\ \small z = 4373612677928697257861252602371390152816537558161613618621437993378423467772036$

(Source: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4 )

This solution was found using elliptic curves. Quite cool, eh?

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Well, why not? Multiplying through by denominators leads to cubic surface $$ x^3 + y^3 + z^3 - 3 \left(x y^2 + x^2 y + y z^2 + y^2 z + z x^2 + z^2 x \right) - 5xyz = 0 $$ which has integer points $$ (1,1,-1), $$ $$ (11,4,-1), $$ $$ (11,9,-5) $$ and perhaps no others except for permuting and changing all signs. Or multiplying by any common integer factor.

The first triple cannot be used in the original problem, the second and third work fine.

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    $\begingroup$ Can you say how did you find these numbers ? $\endgroup$ – S.H.W Mar 18 '17 at 18:11
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    $\begingroup$ OP asked for positive solutions, but it is nice to see these. I was worried my program hadn't worked, but I required positive. $\endgroup$ – Ross Millikan Mar 18 '17 at 18:16

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