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Let $f$ and $g$ be entire functions such that $f^n+g^n=1$, where $n\geq 3$ is an integer. Prove that $f$ and $g$ are constant.

I suppose I should somehow prove that either $f$ or $g$ is bounded so that I can apply Liouville's Theorem, but I don't see how. I tried setting the derivative of the left hand side equal to zero and work with that but that did not seem to work.

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  • $\begingroup$ Is the equation supposed to hold for all $n\ge 3$? $\endgroup$ – Mark Bennet Aug 18 '13 at 20:51
  • $\begingroup$ @MarkBennet For a single specific $n$ satisfying $n\ge 3$. $\endgroup$ – Potato Aug 18 '13 at 20:52
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    $\begingroup$ No, just for one $n\ge3$ (and note that $n=2$ wonÄt work becasue of $\sin, \cos$) $\endgroup$ – Hagen von Eitzen Aug 18 '13 at 20:53
  • $\begingroup$ See here for the case $n>3$. $\endgroup$ – Potato Aug 18 '13 at 21:23
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Here's a proof adapted from Remmert's book Classical Topics in Complex Function Theory, page 236.

Suppose $g\neq 0$. Since $f$ and $g$ cannot have common zeros, $f/g$ is a meromorphic function that takes the value $w$ at $z$ if and only if $f(z)=wg(z)$.

We can factor the given equation as

$$1=\prod_1^n (f-\zeta_ig),$$

where the $\zeta_i$ are roots of $x^n+1$. Dividing through by $g$, we see $f/g$ cannot take the (distinct) values $\zeta_i$. By Picard's theorem for meromorphic functions, a meromorphic function that omits $3$ values is constant. So $f/g$ is constant, $f=cg$ for a constant $c$, and the rest follows easily.

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  • $\begingroup$ +1, but I don't see that the assumption that $g$ never vanishes is required. $\endgroup$ – Jonathan Y. Aug 18 '13 at 21:45
  • $\begingroup$ @JonathanY. That notation means that $g$ is not the constant $0$ function. $\endgroup$ – Potato Aug 18 '13 at 22:06
  • $\begingroup$ At least, that's what I intended it to mean... $\endgroup$ – Potato Aug 18 '13 at 22:07
  • $\begingroup$ Potato, well then, that makes perfect sense ;) $\endgroup$ – Jonathan Y. Aug 18 '13 at 22:20
  • $\begingroup$ @Potato Did you mean to say "Dividing through by $g^n$"? $\endgroup$ – user87317 Aug 18 '13 at 22:29

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