Is this Diophantine problem solvable without invoking Fermat's Last Theorem?

Question

Let $a,b,c,n$ be positive integers with $a<b<c$ and $n\geq 3$ odd. Given that $a^n + b^n < 2c^n$, can one prove that $a^{n+2}+b^{n+2}\neq c^{n+2}$ without invoking Wiles' theorem ? Or is this actually equivalent to Fermat's Last Theorem ?

Answer 1

As stated in the comments, $a<b<c$ implies trivially $a^n+b^n<2c^n$, hence this condition is extraneous. The equation also immediately implies $a,b<c$ so that is not useful either. Finally the equation is symmetric in $a,b$ so we may WLOG assume $a\leq b$, so your condition just becomes $a\neq b$. In fact, $a\neq b$ is quite extraneous too, because from the Fermat equation we will obtain $\sqrt[n+2]{2}$ is rational if the equation has solutions with $a=b$. So what you're left with is, "prove that if $n$ is odd, then $a^{n+2}+b^{n+2}=c^{n+2}$ has no solutions." Now it is entirely obvious that this is just Fermat's Last Theorem in the odd case.