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Complex numbers satisfying Fermat’s identity

January 14, 2018 By Gary Ernest Davis

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

or

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

or

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

or

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.

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