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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)?

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    @MakotoKato: I have no control on other's willingness to answer this post. Maybe they considered this problem too plain to write an answer, since the math involved is quite basic to an expert. The situation you raised up is hypothetical; and I do prefer hints instead of answers for that give me some elbow room to think about it independently. I do not have anything more to say. You may criticize my answer (which may be wrong) or write one on your own if you are interested.2012-08-07

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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.

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    Hopefully fixed now.2012-08-07