Entire functions $f,g,h$ with $e^f+e^g+e^h \equiv 0$ implies $f,g,h$ only differ by constant How to show complex entire functions $f,g,h$ with $e^f+e^g+e^h \equiv 0$ implies $f,g,h$ only differ by constant?
Progress
I tried using Picard's theorem. In the equation $e^{f−h}+e^{g−h}+1\equiv 0$, is it possible for  $e^{f−h}$ and $e^{g−h}$ to cancel each other out?

After multiplying with $e^{-h}$ and renaming, we have the identity
$$e^{\tilde{f}} + e^{\tilde{g}} \equiv -1.$$
We know one value that neither $e^{\tilde{f}}$ nor $e^{\tilde{g}}$ ever attains. Therefore we know another value that these functions never attain. By Picard's (little) theorem, $e^{\tilde{f}}$ and $e^{\tilde{g}}$ are constant.
