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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.

Polutropon
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  • 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. – The Count May 23 '17 at 14:01
  • Don't know if it helps, but putting everything into a common denominator results in $${x^3 + y^3 + z^3 + xyz \over (x+y)(y+z)(x+z)} = 3$$ – John Lou May 23 '17 at 14:09
  • 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.

     – Thomas Andrews May 23 '17 at 14:14

1 Answers1

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Yes, there are BIG positive solutions. For example:

x:=16666476865438449865846131095313531540647604679654766832109616387367203990642764342248100534807579493874453954854925352739900051220936419971671875594417036870073291371;

y:=184386514670723295219914666691038096275031765336404340516686430257803895506237580602582859039981257570380161221662398153794290821569045182385603418867509209632768359835;

z:=32343421153825592353880655285224263330451946573450847101645239147091638517651250940206853612606768544181415355352136077327300271806129063833025389772729796460799697289;

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

Robert Z
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