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I am trying to prove the following: Fix $n \geq 3 \in \mathbb{Z}$, then for any non zero integer $a,b,c$, there are only finitely many integer solutions to $ax^n+by^n=c.$.

I think the solution uses the deep theorem of Roth that asserted that for an algebraic number $\alpha$ and for any $\epsilon >0$, there exists only finitely many $p/q \in \mathbb{Q}$ such that $|\alpha - p/q|<C/q^{2+\epsilon}$. But I have trouble in establishing the inequality, someone please help.

Rue
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Yes, already the result of Thue from $1908$ (which is weaker than Roth's result) implies the following fact. If $f(x, y)$ is a form of degree $d \ge 3$, which has rational coefficients and is irreducible over the rationals, then the equation $$ f(x, y) = c, $$ where $c$ is a constant, has at most finitely many integer solutions $(x, y)$. For the proof, see the paper of Thue.

Of course, even the generalized Fermat equation $ax^p+by^q=cz^r$ with $1/p+1/q+1/r<1$ has only finitely many coprime integer solutions, as has been proved by Darmon and Granville.

Dietrich Burde
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  • Doesn't the finiteness of the number of integer solutions to the generalized Fermat equation follow from Faltings's theorem (when n>3) and Siegel's theorem (when n=3)? In fact, the projective curve defined by the generalized Fermat equation is smooth (assuming a, b and c are all non-zero) and thus has only finitely many rational points (when n>3 by Faltings) and only finitely many integral points by Siegel (when n=3). Thus, it has only finitely many integral points. Moreover, this holds over any number field. – Ariyan Javanpeykar Nov 26 '13 at 18:15
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    Well, $ax^3+by^3=cz^3$ can certainly have infinitely many coprime integral solutions... – Mike Bennett Nov 26 '13 at 18:23