Fermat’s Last Theorem to the Imaginary Power

Fermat’s last theorem is one of the most popular theorem of the history of mathematics. It was first conjectured, without proof, by Pierre de Fermat, a French Mathematician, in 1637. For 356 years the proof of Fermat’s statement remained a challenge to  all mathematician. If you have never heard of this before, I strongly advice you to watch the BBC documentary. Andrew Wiles, a british mathematician, published a complete proof of Fermat’s Last Theorem in 1995.

Fermat’s Last Theorem states that there exist no three integers x, y,z satisfying the equation x^n+y^n=z^n for n\ge3.

In this post we will consider a variation of Fermat’s last theorem. We wish to show that there is no three integers x,y,z satisfying the equation x^{in}+y^{in}=z^{in} for any integers n\ge1, and where i is the imaginary number (i^2=-1).

Let us start by assuming that such integers exists. It implies that (x^n)^i+(y^n)^i=(z^n)^i. And since x, y, z are supposed to be integers it follows that x^n, y^n and z^n are integers as well. Let us define X=x^n, Y=y^n and Z=z^n so that one can rewrite X^i+Y^i=Z^i.

From now, all is left to do is to show that we can not find three integers satisfying X^i+Y^i=Z^i. To proceed, we observer that |Z^i|=1. Therefore (X^i+Y^i)(X^{-i}+Y^{-i})=1.

Which leads to X^iX^{-i}+X^iY^{-i}+Y^iX^{-i}+Y^iY^{-i}=1,

and finally (X/Y)^i+(Y/X)^i=-1.

Rewriting our last equation as X^{2i}+Y^{2i}=-(XY)^i, we can now use (X^i+Y^i)^2=X^{2i}+Y^{2i}+2(XY)^i. We get


Rewriting Z^2=n and XY=m, all we need to show is that it exist no two integers satisfying


Let us now write n^i under its exponential form n^i=\exp(i\ln(n)+2i\pi k), where k is an arbitrary integer. Note that e^{2i\pi k}=1 and does not modify the previous equality. Identically, we have m=\exp(i\ln(m)+2i\pi k'), where again k' is an integer. It follows that

\exp(i\ln(n)+2i\pi k)=\exp(i\ln(m)+2i\pi k'). Taking the log we get

\ln(n)+2\pi k=\ln(m)+2\pi k', which can be rewritten as

\ln(n/m)=2\pi(k'-k), which finally leads to

n/m=e^{2\pi (k'-k)}.

Finally we should point out that e and e^{\pi} are transcendental numbers. It implies that (e^{\pi}) to any power q can not be written as a rational number. In other words it is impossible to write e^{2\pi q}=n/m for any q, n and m integers. This finally ends the proof. It exists no three integers x,y,z satisfying

x^{in}+y^{in}=z^{in} for n\ge1.

Note that it is known that e^{\pi\sqrt{n}} for any integer n is a transcendental number. See more on

  1.  Weisstein, Eric W.“Irrational Number”MathWorld.
  2. Jump up^ Modular functions and transcendence questions, Yu. V. Nesterenko, Sbornik: Mathematics(1996), 187(9):1319

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s