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.

  • $\begingroup$ 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
  • 5
    $\begingroup$ 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

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.

  • $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.