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Variations on Fermat: an agony in four fits

Fermat’s Last Theorem with complex powers, wrapped in a story every mathematician can relate to


Fermat’s Last Theorem has been a source of fascination and the motivation for an enormous amount of mathematics over the last few centuries, both in attempts (eventually successful) to prove it and as the inspiration for other related questions.

This is the story of how an algebraic question inspired by Fermat’s Last Theorem morphed into an analytic question, which subsequently turned out to be expressed best as a geometric question, which could be answered using basic methods of plane geometry.

Fit the first: curiosity

A common strategy in mathematical research is to consider something that is already known, and to try to generalise or vary it hoping for something interesting to appear in the process. Our initial question was motivated by Fermat’s Last Theorem: are there solutions to


where $x, y, z$ and $n$ are positive integers and $\mathrm{i}^{2}=-1$?

An obvious simplification is to replace $x^{n}$, $y^{n}$ and $z^{n}$ by $x$, $y$ and $z$: if there are no solutions to

$x^{\,\mathrm{i}} + y^{\,\mathrm{i}} = z^{\,\mathrm{i}}$

for (non-zero) integers $x$, $y$ and $z$, then there are clearly none to the original problem, where $x$, $y$ and $z$ have to be $n$th powers.

So let’s assume that $x,y,z$ satisfy this relationship. Now, any complex number $a+\mathrm{i}b$ can be written in the form $r\,(\cos\theta+\mathrm{i}\sin\theta)$, where $r$ is called the modulus and $\theta$ the argument of the complex number (see figure below). In this case, $z^{\,\mathrm{i}} = \mathrm{e}^{i\ln z} = \cos(\ln z)+\mathrm{i}\sin(\ln z)$, so $z^{\,\mathrm{i}}$ has modulus $1$ and argument $\ln z$.

The modulus, $r$, and argument, $\theta$, of a complex number. Using properties of right angled triangles, it can be shown that $r^2=a^2+b^2$ and $\tan\theta=b / a$

So now we know that

$(x^{\,\mathrm{i}}+y^{\,\mathrm{i}})(x^{-\mathrm{i}}+y^{-\mathrm{i}}) = z^{\,\mathrm{i}} z^{-\mathrm{i}} = 1$

and expanding the brackets gives


which rearranges to


Using $(x/y)^{\,\mathrm{i}} = \cos(\ln (x/y))+\mathrm{i}\sin(\ln(x/y))$, we obtain



$\cos(\ln(x/y))= -\tfrac{1}{2}.$

Therefore an integer $k$ should exist such that $\ln(x/y)=2\pi k \pm 2\pi/3$ or, in other words,

$x/y= \mathrm{e}^{\,2\pi k \pm  2\pi/3}.$

So, remembering that $x$ and $y$ are integers, all we need to do now is to check whether $\mathrm{e}^{\,2\pi k \pm  2\pi/3}$ can be rational. It certainly seems unlikely to be rational: however, there are unlikely truths in mathematics. This seemed like a good time to climb up onto the shoulders of giants, and we did a little digging into transcendental number theory (using Transcendental Number Theory by A Baker). A number is rational if it is a zero of an expression of the form $mx+n$, where $m$ and $n$ are integers, ie it is a root of a linear polynomial with integer coefficients; it is algebraic if it is a root of any polynomial with integer coefficients; and it is transcendental if it is not algebraic. Our search showed us that $\mathrm{e}^{\,\pi}$ is called Gelfond’s constant, which is known (from the Gelfond–Schneider theorem) to be transcendental. Furthermore, any algebraic function of it is transcendental, and so we can now conclude that $\mathrm{e}^{\,2\pi k \pm 2\pi/3}$ is transcendental for all integer $k$.

It then follows that there are no integer solutions to the equation

$x^{\,\mathrm{i}} + y^{\,\mathrm{i}} = z^{\,\mathrm{i}}$

and so there are certainly none where $x$, $y$ and $z$ are all $n$th powers, ie there are no integer solutions to

$x^{\,n\mathrm{i}} + y^{\,n\mathrm{i}} = z^{\,n\mathrm{i}}$

for any positive integer values of $n$.

Fit the second: despair

We were initially happy to conclude this. We had found something interesting. Unfortunately, while looking up transcendental number theory, we found out that (by much the same argument) it was already well known from this theory that there are no solutions to

$x^{\,m+\mathrm{i} n}+y^{\,m+\mathrm{i} n}=z^{\,m+\mathrm{i} n}$

for positive rational numbers $x, y, z$ and exponents of the form $m+\mathrm{i} n$, where $m,n \in \mathbb{Z}$ with $n \neq 0$. So we had managed to prove a special case of a known result—and not for the first time.

For a while, we found this sufficiently off-putting that we stopped thinking about the problem. But finally we rallied, and did what mathematicians do. We changed the question.

Fit the third: enthusiasm

We asked a new question instead: are there positive real solutions to


and, if so, is there anything interesting to say about them?
In fact, we already saw the clue as to how to solve this: for any positive real number $x$, the number $x^{\,\mathrm{i}} = \cos(\ln x)+\mathrm{i}\sin(\ln x)$ is a complex number of unit modulus. So now, instead of considering $x, y, z$, we realise that we are looking for three complex numbers $a\; (=x^{\,\mathrm{i}})$, $b\;(=y^{\,\mathrm{i}})$, $c\;(=z^{\,\mathrm{i}})$, such that

$a+b=c \quad \text{and} \quad|a|=|b|=|c| = 1.$

Algebra becomes geometry

At first sight, this has not helped a great deal. But interpreted geometrically, the problem suddenly becomes easy. Given $a$, since $a+b=c$, we immediately know that $|a+b|=1$. $|a+b|$ is the distance between $a$ and $-b$, which is 1. Similarly, we can use $a-c=-b$ to see that $|a-c|$, the distance from $a$ to $c$, must also be $1$. So we need to find a pair of points on the unit circle, centred at the origin, both a distance $1$ from $a$, as shown in the figure to the right.

We should note that $-b$ and $c$ both being a distance $1$ from $a$ is necessary; we must also check that it is sufficient. But it is now clear from the geometry that the segment connecting $a$ to $c$ is parallel to that connecting $-b$ to $0$, so we do indeed have $a+b=c$.

We can now use this geometry to find $b$ and $c$ given $a$: for if we let
$\alpha,\beta,\gamma$ be the arguments of $a,b,c$ (or equivalently, the natural
logarithms of $x,y,z$), the relationship between $\alpha, \beta$ and $\gamma$ is easily seen. Since each of $a,b,c$ has unit modulus, the triangle with vertices at $0$, $a$ and $c$ is equilateral, and so all internal angles are $\pi /3$. It immediately follows that $c$ has argument $\gamma=\alpha+\pi/3$ and that $b$ has argument $\alpha+2\pi/3$.

From a similar picture, with the roles of $b$ and $c$ reversed, the other solution is given by $\gamma=\alpha-\pi/3$ and $\beta=\alpha-2\pi/3$.

Assembling this, we see that given an arbitrary positive real number $x=e^\alpha$, there is a pair of solutions to


given by

$y_\pm = \mathrm{e}^{\,\alpha\pm2\pi/3}=x\mathrm{e}^{\,\pm2\pi/3},\qquad z_\pm=\mathrm{e}^{\,\alpha\pm\pi/3}=x\mathrm{e}^{\,\pm\pi/3}.

Given this, we can now see that these real solutions have the following properties, which we found interesting:

  1. Any solution $x,y,z$ satisfies the relationship $xy=z^{2}$.
  2. Since $\mathrm{e}^\pi$ is transcendental, it follows that the ratio of any two of
    $x,y,z$ is transcendental, and so at most one of $x,y,z$ is algebraic.
  3. Furthermore, since there is a solution for any positive choice of $x$, and since there are only countably many algebraic numbers, for almost all solutions the three values $x,y,z$ are all transcendental.
  4. Finally, this tells us that not only are there no integer solutions, which is equivalent to there being no rational solutions, but that the real solutions are not only irrational, they are very irrational, in the sense that they are (almost all) transcendental.

Fit the fourth: the mathematical endeavour

In closing, let’s think about how this small investigation fits into what mathematicians do all day. Some people think of mathematicians as people who try to solve problems using mathematics; others think of them as people who try to prove theorems. Very crudely speaking, we could think of these activities as applied and pure mathematics respectively. But both of these are really different approaches to a bigger objective: finding out something interesting. What we went through above is a small version of this, which shows the typical features. You start off with something that you want to understand better, you find out about it—and sometimes what you find out is that somebody else has already understood it—and you refine your question until you have something that you can understand better. Then you share it with other people who find it interesting. At least, I hope we’ve done the last part!

References and further reading

  1. Cipra B 1999 Fermat’s Theorem—at last! What’s Happening in the Mathematical Sciences 3 American Mathematical Society
  2. Ribenboim P 2000 Fermat’s last theorem for amateurs, Springer
  3. Zuehlke JA 1999 Fermat’s last theorem for Gaussian integer exponents American Mathematical Monthly 106 (49)
  4. Baker A 1975 Transcendental number theory Cambridge University Press

[Pictures: Banner adapted from one of Henry Holiday’s original illustrations to The Hunting of the Snark (An Agony in 8 Fits) by Lewis Carroll; other pictures by Chalkdust]

Rob teaches mathematics at Coventry University.
Twitter  @RobJLow    + More articles by Robert

Thierry also teaches mathematics at Coventry University.
+ More articles by Thierry

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