Complex numbers satisfying Fermat’s identity

Alexander Bogomolny (@CutTheKnotMath) has a post on his website Cut The Knot dealing with proofs of the following fact:

The proofs rely, among other things, on the fact that prime numbers $p>3$ are of the form $6m\pm 1$ , which is clear since a prime greater than 3 cannot leave a remainder of 0, 2, 3 or 4 upon division by 6.

I want to consider a slight extension of this fact – basically replacing $z=2$ by an arbitrary non-zero complex number – by considering something relevant to the first solution offered at Cut the Knot.

In that solution the facts $x+y=z, \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ are utilized.

Suppose we choose $z\neq 0$ arbitrarily as a complex number.

Can we find complex numbers $x, y$ such that $x+y=z, \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$ ?

Yes, we can, notably & uniquely:

$x = e^{-i\pi/3}z, y= e^{i\pi/3}z$

$x = e^{i\pi/3}z, y= e^{-i\pi/3}z$

So, we now have the following:

Let $z \neq 0$ be a complex number and let either

$x = e^{-i\pi/3}z, y= e^{i\pi/3}z$

$x = e^{i\pi/3}z, y= e^{-i\pi/3}z$ .

If $n=6m\pm1$ then $x^n+y^n = z^n$ .

For $n =6m+1$ this follows directly from the fact that

$( e^{i\pi/3})^{6m+1} + (e^{-i\pi/3})^{6m+1} =e^{i\pi/3} +e^{-i\pi/3} = 1$ .

A similar argument holds when $n=6m-1$ .

Since primes $p>3$ are of the form $6m\pm1$ we have the following:

Let $z \neq 0$ be a complex number and let either

$x = e^{-i\pi/3}z, y= e^{i\pi/3}z$

$x = e^{i\pi/3}z, y= e^{-i\pi/3}z$ .

If $p>3$ is prime then $x^p+y^n = z^p$ .

Another collection of positive integers that are necessarily of the form $n=6m\pm1$ are those $n$ for which $\phi(n)>2n/3$ , where $\phi(n)$ is Euler’s phi function: the number of $k<n$ that are coprime with $n$ . See the Online Encyclopedia of Integer Sequences, for example.