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Does the equation $x^3+y^5=z^7$ have a solution $(x,y,z)$ with $x,y,z$ positive integers and $(x,y)=1$? In his book H. Cohen (Number theory,2007) said "[...] seems presently out of reach". I couldn't find any suggestion beyond Cohen's book. Thanks in advance,

Montanari Fabio department of math university of bologna italy e-mail

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There is no claim in my cv or elsewhere that me and Sander have solved the equation x^3+y^5+z^7=0. All my cv claims is that we're writing a paper on it! That's not the same thing.

All the best, Samir

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Ha! Good point:) – David Zureick-Brown Feb 22 '10 at 22:09

I'm surprised that Bjorn Poonen hasn't chimed in yet. However, you can read about the approaches to solving equations like this in Frits Beuker's lectures "The generalized Fermat Equation"

Added later: You should also look at Siksek's "Edingurgh lectures" for a good outline of how he might have approached this equation:

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From what little I can understand, the 2 papers above deal with the problem over Z. Can one generalize the distinction among hyperbolic (finite solution sets), euclidean (elliptic curves) and spherical (polynomial parametrization) cases to any number field? As an aside, one small "hyperbolic" example stands out in Z[i]: (-2+i)^3+(-2-i)^3=(1+i)^4. – Yaakov Baruch Feb 21 '10 at 12:32
As far as I know the only difference is that over a general number field the elliptic curves showing up might have positive rank, so you might get infinitely many solutions in an elliptic case. I have forgotten Darmon's argument that in the hyperbolic case the number of solutions are finite, but I do remember that he reduces to Faltings' theorem so surely the argument will work in the general number field case...(famous last words) – Kevin Buzzard Feb 21 '10 at 12:44
I haven't looked at the Darmon/Granville argument lately, but, from what I remember, they first show, in the hyperbolic case, that the primitive solutions must lie on a finite set of curves of genus 2, and then invoke Falting's theorem. However, I'm not sure if the first applies over general number fields. There might be some subtlety with units (i.e. if there is an infinite unit group). But, I would think that you might be able to figure this out from Beukers' paper above. – Victor Miller Feb 21 '10 at 14:29
One point about the equation over number fields: the most natural way of defining a solution to be primitive would be $\min(\text{ord}_{\pi}(a),\text{ord}_{\pi}(b),\text{ord}_{\pi}(c)) = 0$, for all primes $\pi$, but then you could always multiply such a solution by suitable units and get another, so you'd need a stronger definition of "primitive". – Victor Miller Feb 21 '10 at 14:48
I took the trouble to look at the Darmon/Granville paper. It says (on the version on Darmon's web page) near the bottom of p16 "It is easy to see that the proof extends to...arbitrary number fields". They take care to formulate the statement in such a way in the number field case that units don't mess them up. – Kevin Buzzard Feb 22 '10 at 11:01

Sander Dahmen and Samir Siksek are [Edit] writing a paper [end Edit] about this (according to Samir's cv, under papers in preparation), but there is no draft on his web page.

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Do they find a solution to the equation? – Petya Feb 20 '10 at 0:33
They have since posted this paper to the arxiv but it does not appear to cover the case of the OP – j.c. Sep 30 '14 at 0:13

I may be misunderstanding the question, but I do not believe that it has any integer solutions. At the very least, none are known to exist at the moment. Any solutions would be counterexamples to the Fermat-Catalan conjecture with {m,n,k} = {3,5,7} (since 1/3 + 1/5 + 1/7 = 71/105 < 1). The most I can tell you is that, for coprime {x,y,z}, there are finitely many solutions to your equation. I think your (x,y) = 1 means that they're coprime, anyway, so it follows that z must be coprime. Therefore, any solution at all would disprove the related Beal's conjecture.

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The Fermat-Catalan conjecture says that there are only finitely many solutions to this Diophantine equation, not that there are none. – Michael Lugo Feb 23 '10 at 21:17
I said it would be a counterexample, not that it would disprove. However, this particular solution, since the exponents are all higher than 2, WOULD be a counterexample to Beal's conjecture. – Gabriel Benamy Feb 23 '10 at 21:24
@Gabriel: "I said it would be a counterexample, not that it would disprove.". I think that in common mathematical parlance, if you find a counterexample to something, you've disproved it. And the issue is not really whether there are no solutions, but whether mathematics in its current state has enough techniques to prove this. To give you some background---these sorts of equations (explicit m,n,k the sum of whose reciprocals is only just less than one) are slowly being picked off by experts in the area nowadays. No-one is close to a general argument though. – Kevin Buzzard Feb 24 '10 at 7:43

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