TITLE: Applications of Liouville's theorem
QUESTION [9 upvotes]: I'm looking for "nice" applications of Liouville's theorem (every bounded entire map is constant) outside the area of complex analysis.
An example of what I'm not looking for : a non-constant entire function has dense image (this is essentially a corollary).
An example of the kind of thing I'm looking for : a complex matrix whose conjugacy class is bounded must be a homothety (if $A$ is such a matrix and $B$ is an other matrix, then $z \mapsto e^{-z B} A e^{z B}$ is entire and bounded hence constant, but its derivative at $0$ is $[A,B]$ : thus $[A,B]=0$).
In a similar vein : a subalgebra of $M_n (\mathbb{C})$ on which the spectral radius is submultiplicative is simultaneously triangularizable.

REPLY [9 votes]: There are many cool applications when combined with the uniformization theorem.  Not sure if you count them as "complex analysis" or not - you could really think of them as algebraic geometry.  For example:
The only meromorphic functions $f$ and $g$ satisfying $f^n+g^n = 1$ for $n>3$ are constant.  
Proof sketch:  If there were such functions they would define a map $F$ from the plane to the Riemann surface $x^n+y^n = 1$.  We can compute the genus of this guy using Hurwitz - it is $(n-1)(n-2)/2$.  So for $n>3$ the genus is bigger than two - the corresponding Riemann surface has negative curvature, and by uniformization it has the disk as its holomorphic universal cover. But then,  $F$ would factor through the disk, and so by Louville F had to be constant.
Note:
for n=1 the equation has lots of solutions
for n=2 sine and cosine work (for instance)
for n=3 the Riemann surface has genus 1, and so its universal cover is the plane.  You can find explicit solutions to the equation by using theta functions.
EDIT: another very important example of the same sort of reasoning is the little Picard theorem.  At a high level, the proof just says that the plane with 2 points deleted has a holomorphic universal covering by the disk, so any entire function which misses 2 points factors through the disk - Louville gives the contradiction.