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Let $A, B, C \in \mathbb{N}$ be such that $\gcd(A,B,C)=1$. Is it known if the equation $A x^n + By^n = C z^n$ has any non-trivial solutions $x,y,z \in \mathbb{N}$? I know there are no such solutions if $A = B = C =1$, because this is Fermat's last theorem. I was just wondering if something similar was known when there are positive coefficients as well or not. Thank you very much!

PS By non-trivial solutions I mean the solutions that do not arise from degenerate cases.

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  • $\begingroup$ You need to exclude some other degenerate cases to make the question interesting: for example, if $x = y = z$ and $A+B = C$, there will be a solution. $\endgroup$
    – Geoff Robinson
    Commented Jan 23, 2016 at 20:56
  • $\begingroup$ ... or $z=1$ and $C=Ax^n+By^n$ ... $\endgroup$
    – Seva
    Commented Jan 23, 2016 at 20:57
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    $\begingroup$ You can choose any $x,y,z$ so that $(x,y,z) = 1$, and then pick $A,B,C$ so that $Ax^n + By^n = Cz^n$. $\endgroup$
    – Matt Young
    Commented Jan 23, 2016 at 21:04
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    $\begingroup$ The right question seems to be whether, for any given $A,B,C \in \mathbb{Z}$ (not all zero), the equation $Ax^n + By^n = Cz^n$ has only finitely many solutions $(x,y,z;n)$ with $\mathrm{gcd}(x,y,z) = 1$ and $n > 3$. This follows from the $abc$-conjecture. $\endgroup$
    – Vesselin Dimitrov
    Commented Jan 23, 2016 at 22:44
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    $\begingroup$ In the case $A = B = 1, C= 2^r$, this has been proved in work of Ribet and Darmon-Merel. $\endgroup$
    – Vesselin Dimitrov
    Commented Jan 23, 2016 at 22:53

1 Answer 1

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This is just a slight extension of Matt Young's comment and Geoff Robinson's question: given any relatively prime $x, y$ any sufficiently large $X$ can be written as a positive linear combination of $x^n$ and $y^n$ (how large is "the Frobenius problem"). In particular, for any fixed $z,$ you can find a $C>0$ such that $A x^n + By^n = Cz^n,$ for some $A, B> 0.$ Of course, since the OP does not define what "non-trivial" means, this might not be what he wants.

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