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 satisfying the equation for .
In this post we will consider a variation of Fermat’s last theorem. We wish to show that there is no three integers satisfying the equation for any integers , and where is the imaginary number ().
Let us start by assuming that such integers exists. It implies that . And since are supposed to be integers it follows that and are integers as well. Let us define , and so that one can rewrite .
From now, all is left to do is to show that we can not find three integers satisfying . To proceed, we observer that . Therefore .
Which leads to ,
and finally .
Rewriting our last equation as , we can now use . We get
Rewriting and , all we need to show is that it exist no two integers satisfying
Let us now write under its exponential form , where is an arbitrary integer. Note that and does not modify the previous equality. Identically, we have , where again is an integer. It follows that
. Taking the we get
, which can be rewritten as
, which finally leads to
Finally we should point out that and are transcendental numbers. It implies that to any power can not be written as a rational number. In other words it is impossible to write for any and integers. This finally ends the proof. It exists no three integers satisfying
Note that it is known that for any integer is a transcendental number. See more on
- Weisstein, Eric W., “Irrational Number”, MathWorld.
- Modular functions and transcendence questions, Yu. V. Nesterenko, Sbornik: Mathematics(1996), 187(9):1319