Entire functions that satisfy certain equation in $\mathbb{C}$ I want to prove the following problem:
Let $f$ and $g$ be entire functions satisfying the equation $f^n(z) +g^n(z) = 1$ for all $z \in\mathbb{C}$, where $n \ge 2,\; n \in\mathbb{N}$.
Prove that if $g$ has no zeros, then both $f$ and $g$ are constant. 
Thanks for your help in advance.
 A: If $g(z)$ is never zero, then $f(z)$ can never be an $n$-th root of unity.
What theorem might prevent an entire function from ever being an $n$-th root of unity?
A: If $g$ is never 0, $f$ avoids $n$ values. But the (small) Picard theorem tells you that an entire function which avoids 2 values is constant
A: $g^n(z)=1-f^n(z)$. As $g$ has no zero , $g(z)\not =0$ for all $z\in \Bbb C$ implies $g^n(z)\not =0$ for all $z\in \Bbb C$. So $f^n(z) \not=1$ for all $z\in \Bbb C$. So $f$ omits the values of $n-th$ roots of unity. then by Picards Little Theorem $f$ is constant. Then $g$ is also constant.
A: Consider the follow-up where we look for entire $f$ and $g$ with $f^n+g^n=1$. I contend there are no non-constant meromorphic functions on $\Bbb C$ with $f^n+g^n=1$ for any $n\ge3$.
Consider the Riemann surface $S_n$ corresponding to the algebraic equation
$w^n+z^n=1$. This is the nonsingular curve in $\Bbb P^2(\Bbb C)$ defined
by an algebraic equation $X^n+Y^n=Z^n$. If $f$ and $g$ are meromorphic with $f^n+g^n=1$ they define a holomorphic map $F:\Bbb C\to S_n$.
Now $S_n$ has genus $\frac12(n-1)(n-2)\ge1$. By the uniformisation theorem its
universal covering can be taken to be $D$, the unit disc. As $\Bbb C$ is
simply connected, $F$ lifts to a holomorphic map to the universal covering of $S_n$, that is a holomorphic map $\Bbb C\to D$. By Liouville, this is constant. So $F$ is constant, and $f$ and $g$ are constant.
