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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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    If $f, g$ are non-constant holomorphic functions such that $f(z)^2 = g(z)^6 - 1$, then $(f(z), g(z))$ defines a non-constant holomorphic map from... something to... something. Can you fill in the rest?2012-08-05
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    The basic way they're related is that for every $z\in\mathbb C,$ the ordered pair $(g(z),f(z))\in X,$ where $X$ is the curve defined by $y^2-x^6+1.$2012-08-05
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    This is a good hint, let me think about it.2012-08-05
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    Then $(f(z),g(z))$ defines a non-constant holomorphic map from $\mathbb{C}$ to $X=\{z^{2}=w^{6}-1\}$. This is impossible since by Riemann-Hurwitz formula we can only map higher genus curves to lower ones. I do not know if this is an appropriate proof.2012-08-05
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    Could any one please answer? I prefer a detailed proof to obscure hints.2012-08-06
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    @Makoto: you are not the one asking the question. I do not think it is in the best interest of the OP to be presented with an answer as opposed to finding one him/herself (especially if the OP is studying for quals, which I presume is the case).2012-08-06
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    @user32240: I don't think that works. The version of Riemann-Hurwitz I know only applies to _compact_ Riemann surfaces, and neither of the Riemann surfaces you're using are compact (you can compactify the codomain but you can't compactify the domain if $f$ or $g$ has an essential singularity at $\infty$).2012-08-06
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    @QiaochuYuan: that is true, I have no idea if $f$, $g$ are surjective(being open maps they should). $f,g$ can be $e^{z}$ that has essential singularities at $\infty$. My argument was too cheap and does not relate to the meromorphic case. Let me think about it.2012-08-06
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    @QiaochuYuan: I am not studying for the quals, but I need to have a sound basis before I go to graduate school. Yes, hints would be more helpful than an actual proof since a lot of time it is more important to know why this is true instead of how it is true.2012-08-06
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    This hint might be too useful, so I encoded it in ROT13: Jung vf gur havirefny pbire bs gung ulcreryyvcgvp pheir?2012-08-06
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    @QiaochuYuan I'm just curious. If the OP find the solution by your hint, the thread remains unanswered?2012-08-06
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    I still have not find the solution despite various hints, and I would like the thread to remain unanswered so I can struggle for a while. Obviously if I ask a friend for a solution I would get one, but that would be undermining my own understanding of the problem.2012-08-06
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    @Makoto: no. If the OP finds the solution, the OP can post an answer.2012-08-06
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    @user32240 You always have an option not to read the answer if any. In other words, the answer won't hurt you.2012-08-06
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    @MakotoKato: well, I am already being criticized for asking for lazy questions in the other thread, so I do not want to cultivate this laziness. Also, I think anyone (at graduate math level) interested in this problem should be able to work it out by the hints.2012-08-06
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    @user32240 I don't understand how your laziness has anything to do with the answer. As I said, even if someone posts the answer, you don't have to read it. This applies to anybody. If he thinks reading the answer is harmful, he can always ignore it. On the other hand, there may be some people who would like to read the answer.2012-08-06
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    @MakotoKato: Well, if you or others would like to provide an answer, I will not be able to prevent it as I am not the moderator at this site. But as I stated I believe hints are more helpful than solutions to me. I am posting an answer and you can criticize if it can be improved. I doubt if I can ignore other's helpful answers.2012-08-07
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    @user32240 This thread is not your personal thread even though you are the OP. Please think about others. Some people would like to know the answer. Please don't read the answer if you think it's harmful to read it.2012-08-07
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    @MakotoKato: I do not think it is harmful to learn from others, as you can see I accept answers in other posts. You can post an answer if you wish, but I hope not at this stage as I am still working on it.2012-08-07
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    @user32240 I don't see why it's so difficult to postpone to read the answer until you have your own solution.2012-08-07
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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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