Does Fermat's equation have an associated linear equation?

Question

When $xyz$ is non-divisible by an odd prime $n$, we have: $$x^{n-1}=an+1$$ $$y^{n-1}=bn+1$$ $$z^{n-1}=cn+1$$ Hence, $$x^n+y^n+z^n=x(an+1)+y(bn+1)+z(cn+1)=0$$ Can this be considered as the linear equation of a plane? If no, why not? Any hints?

Answer 1

To any counterexample $(x_0, y_0, z_0)$ to FLT for a given $n$, we can indeed associate a plane. Namely, letting $a_0n+1=x_0^{n-1}, b_0n+1=y_0^{n-1}, c_0n+1=z_0^{n-1}$, the equation $$x(a_0n+1)+y(b_0n+1)+z(c_0n+1)=0$$ defines a plane passing through $(x_0, y_0, z_0)$. However, different counterexamples will yield different planes, since $a, b, c$ depend on $x, y, z$.

Meanwhile, I don't know any obvious way to read interesting information off these planes, or to use them; but I could easily be missing something.