Let $n$ be a strict positive integer and $x$ is a complex number. Define $f_n(x)$ as one of the solutions to
$$ f_n (x)^{2n} + f_n ' (x)^{2n} = 1$$
Where the derivative is with respect to $x$.
Why is every entire solution $f_n(x) $ periodic on the complex plane ?
For instance the solutions for $f_1(x)$ are ${-1, 1, -\sin(x),\sin(x),-\cos(x),\cos(x)}$. All of them are periodic with period $2 \pi$.
I assume implicit differentiation helps.
Actually they are not all periodic. For example, one solution (for $n=1$) is $$ f_1(x) = \cases{-1 & $x \le -\pi/2$ \cr \sin(x) & $-\pi/2 \le x < \pi/2$\cr 1 & $x \ge \pi/2$\cr} $$
EDIT: For $n > 1$, the non-constant solutions will not be entire, or even real-analytic: they won't be analytic at $x_0$ where $f(x_0) = 1$. If $f(x_0) = 1$ and the first nonzero term after the constant term in the Taylor series of $f$ at $x_0$ is $a (x-x_0)^m$, we would have $f(x)^{2n} = 1 + 2 n a (x-x_0)^m + \ldots$ while $f'(x)^{2n} = (am)^{2n} (x-x_0)^{2n(m-1)} + \ldots$, and $2n(m-1) > m$ if $m \ge 2$, so we can't get this series to work.
For example, for $n=2$ a non-constant solution with $f(0) = 1$ has the Puiseux series $$ 1 - \frac{1}{4} (3x)^{4/3} + \frac{3}{224} (3x)^{8/3}- \frac{53}{68992} (3x)^4 + \frac{1081}{115906560} (3x)^{16/3} + \ldots$$ and thus it's not analytic: $x=0$ is a branch point.