I was wondering if anyone knew of any positive integer solutions to this Diophantine equation, or had a proof there are none. Integer solutions exist with negative values, such as (11,9,-5) and (4,11,-1), but checking positive integers up through 10,000 yielded nothing and I don't see a way to show there are none.


marked as duplicate by Jose Arnaldo Bebita-Dris, Thomas Andrews, JavaMan, Jaideep Khare, Ross Millikan May 23 '17 at 14:23

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  • $\begingroup$ I'm sure you've noticed that if $(x,y,z)$ is a solution, then so is any permutation thereof. Doesn't answer your question, but might help to optimize a computer search. $\endgroup$ – The Count May 23 '17 at 14:01
  • $\begingroup$ Don't know if it helps, but putting everything into a common denominator results in $${x^3 + y^3 + z^3 + xyz \over (x+y)(y+z)(x+z)} = 3$$ $\endgroup$ – John Lou May 23 '17 at 14:09
  • $\begingroup$ If you add $3$ to both sides, you get: $$(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)=7$$ Not sure if that helps. $\endgroup$ – Thomas Andrews May 23 '17 at 14:14

Yes, there are BIG positive solutions. For example:




See Oleg567's answer here: Find integer in the form: $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

  • $\begingroup$ Sorry, misread what the answer implied. $\endgroup$ – Thomas Andrews May 23 '17 at 14:24

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