Suppose $n>2$. By Fermat's Last Theorem, we know that $a^{n}+b^{n}=c^{n}$ has no non-trivial solutions. Can we quantify it more?

More specifically, given $a,b,c,n\in\mathbb{N}$ with $n>2$ and $a^{n}+b^{n}\neq c^{n}$, can we prove lower bounds on $\lvert a^{n}+b^{n}-c^{n}\rvert$ in terms of $a,b,c,n$? Also assume that $a,b,c$ are distinct.

  • 2
    $\begingroup$ I think for $n=3$ the difference $a^3+b^3-c^3$ is equal to $1$ infinitely often, so no nontrivial lower bounds probably exist. $\endgroup$
    – Wojowu
    Nov 26, 2019 at 11:49
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    $\begingroup$ Not duplicate but closely related: mathoverflow.net/q/214422/30186 $\endgroup$
    – Wojowu
    Nov 26, 2019 at 11:50
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    $\begingroup$ I think the $a^n+b^n\neq c^n$ condition is unnecessary... $\endgroup$ Nov 26, 2019 at 12:06
  • $\begingroup$ @Wojowu, yeah I had looked at that related question before posting this. Here we are more interested in the asymptotics of any lower bounds we can prove. $\endgroup$ Nov 26, 2019 at 13:40
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    $\begingroup$ @GoravJindal You probably won't have much luck proving any lower bounds if we can't even (unconditionally) prove that this value tends to infinity! $\endgroup$
    – Wojowu
    Nov 26, 2019 at 13:42


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