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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.

Trevor Gunn
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Zuo
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  • If $y+z=a$ etc. $$2(4+1+1+1)=(a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)$$ – lab bhattacharjee Mar 18 '17 at 17:14
  • 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 $x+y+z$. – M. Boyet Mar 18 '17 at 17:19
  • I wrote a small brute force program and find no solutions with $x,y,z \le 1000$ – Ross Millikan Mar 18 '17 at 17:40
  • See the Online Encyclopedia of Integer Sequences https://oeis.org/A283564, where one comment says the first solution has 81 digits. – Empy2 Mar 19 '17 at 05:10
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    This should help: https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation – Mike Miller Mar 25 '17 at 12:43
  • @Mike Miller Thanks,I already known it,and you can see the Online Encyclopedia of Integer Sequences https://oeis.org/A283564 – Zuo Mar 25 '17 at 14:10

2 Answers2

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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?

Tito Piezas III
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Hunter Vallejos
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    If there is a solution. Then infinitely many of them. https://mathoverflow.net/questions/264754/solution-to-a-diophantine-equation/275193#275193 – individ Aug 26 '17 at 09:01
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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.

John Gowers
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Will Jagy
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