TITLE: hard diophantine equation: $x^3 + y^5 = z^7$
QUESTION [11 upvotes]: 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 montana@dm.unibo.it

REPLY [2 votes]: 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.