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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 ?

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    If $a<b<c$ isn't $a^n+b^n<2c^n$ trivial? You see, $a<c$ implies that $a^n<c^n$ and $b<c$ implies that $b^n<c^n$. – Yanko Dec 10 '18 at 15:44
  • This seems trivially equivalent to Fermat. True, you exclude the case $a=b$ but $2a^n=c^n$ is clearly impossible (considering the powers of $2$ on both sides) so that's not a significant exclusion. Am I missing something? – lulu Dec 10 '18 at 15:46
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    @Yanko, ah, that's it, thanks ! –  Dec 10 '18 at 15:47

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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.

YiFan Tey
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