How $f^{2}=g^{6}-1$ implies $f$ and $g$ are constant? From Harvard qualification exam, 1990. 
Let $f,g$ be two entire holomorphic functions satisfy the property $$f(z)^{2}=g(z)^{6}-1,\forall z\in \mathbb{C}$$ Prove that $f,g$ are constant functions. Would this be the same if $f,g$ are allowed to be meromorphic functions?
The problem comes with a hint that I should think about the algebraic curve $$y^{2}=x^{6}-1$$but I do not see how they are related. I know this curve is hyperellipitic (from Riemann-Hurwitz or simply the wiki article). But how this help?(this curve should be of genus 2). Taking a short look at the entire function article also seems to be no help. Is the author implying $f,g$ is not best to be treated by classical Riemann Surface applications (as opposed to algebraic geometry ones)? 
 A: Enlightened by various hints, here is a 'proof' which is most likely to be wrong somewhere. I have not touched entire functions and covering maps for a long time. So suggestions on improvement is welcome. 
If $f,g$ are meromorphic, then we can write them as quotients of rational functions. Thus we can extend $f,g$ to the Riemann Sphere by allowing $\infty$. Riemann-Hurwitz would imply we cannot map from low genus surface to a higher genus one: therefore there is no map from $S^{2}$ to $X$. And thus $f,g$ must be constants. 
Assume $f,g$ are holomorphic over $\mathbb{C}$ with possible non-removable singularity at $\infty$. Since $f,g$ are both open maps if they are non-constant, together $(f,g)$ should map the open set $\overline{\mathbb{C}}-\{\infty\}$ to a connected open component to $X=\{(z,w),z^{2}=w^{6}-1\}$. $X$ is a Riemann surface that can be compactified to be homeomorphic to a two hole torus. Since $X$ is connected the map must be surjective. Further, by inverse mapping theorem since $f,g$ are assumed to non-constant, $(f',g')$ are not zero except in a discrete set of points. Ignore this for now (should be tractable by using local biholomorphic transformation to a locally ramified function) we may view the map $$F=(f,g):\mathbb{C}\rightarrow X$$ as a covering map since we have a discrete inverse image at every neighborhood of $X$ according to $F$'s degree at that point. 
The universal cover of $X$ is a closed disk $D^{2}$ by the fundamental diagram. Therefore by the covering property we have a unique lift $p$ from $\mathbb{C}$ to $D^{2}$ that preserves $F$ such that $p:D^{2}\rightarrow \mathbb{C}$ satisfies $$p\circ F=q$$ where $q$ is the covering map from $D^{2}$ to $X$. Further $p$ is holomorphic.  But this is contradictory since by mean value theorem (or maximal modulo principle) a non-constant holomorphic function attains its maximal absolute at the boundary. Thus $p$ must be bounded by some constant and cannot reach the whole complex plane. This showed at least one of $f,g$ must be a constant. And by definition this showed both $f,g$ are constants. 
